A.ACIDS AND BASES

At a school laboratory:

(i)An acid may be defined as a substance that turn litmus red.

(ii)A base  may be defined as a substance that turn litmus blue.

Litmus is a lichen found mainly in West Africa. It changes its colour depending on whether the solution it is in, is basic/alkaline or acidic.It is thus able to identify/show whether

 

  1. An acid is a substance that dissolves in water to form H+/H3O+ as the only positive ion/cation. This is called the Arrhenius definition of an acid. From this definition, an acid dissociate/ionize in water releasing H+ thus:

 

HCl(aq)                 ->       H+ (aq)       +        Cl(aq)

HNO3(aq)              ->       H+ (aq)       +        NO3(aq)

CH3COOH(aq)      ->         H+ (aq)      +        CH3COO(aq)

H2SO4(aq)             ->       2H+ (aq)     +        SO42-(aq)

H2CO3(aq)             ->       2H+ (aq)     +        CO32-(aq)

H3PO4(aq)             ->       3H+ (aq)     +        PO43-(aq)

 

2.A base is a substance which dissolves in water to form OH as the only negatively charged ion/anion.

This is called Arrhenius definition of a base.

From this definition, a base dissociate/ionize in water releasing OH thus:

 

KOH(aq)               ->       K+(aq)          +        OH(aq)

NaOH(aq)              ->       Na+(aq)        +        OH(aq)

NH4OH(aq)                     ->       NH4+(aq)     +        OH(aq)

Ca(OH)2(aq)          ->       Ca2+(aq)       +        2OH(aq)

Mg(OH)2(aq)         ->       Mg2+(aq)      +        2OH(aq)

 

  1. An acid is a proton donor.

A base is a proton acceptor.

This is called Bronsted-Lowry definition of acids and bases.

From this definition, an acid donates H+ .

H+ has no electrons and neutrons .It contains only a proton.

Examples

  1. From the equation:

 

          HCl(aq)       +        H2O(l)        ===    H3O+(aq)     +        Cl(aq)

 

(a)(i)For the forward reaction from left to right, H2O  gains a proton to form H3O+  and thus H2O is a proton acceptor .It is a Bronsted-Lowry base

 

(ii) For the backward reaction from right to left, H3O+   donates a proton to    form H2O and thus H3O+  is an ‘opposite’ proton donor. It is a Bronsted-Lowry conjugate acid

 

(b)(i)For the forward reaction from left to right, HCl  donates a proton to form Cl and thus HCl is a proton donor .

It is a Bronsted-Lowry acid

 

(ii) For the backward reaction from right to left, Cl  gains a proton to form HCl and thus Cl is an ‘opposite’ proton acceptor.

It is a Bronsted-Lowry conjugate base.

Every base /acid from Bronsted-Lowry definition thus must have a conjugate product/reactant.

 

  1. I From the equation:

 

          HCl(aq)       +        NH3(aq)      ===    NH4+(aq)     +        Cl(aq)

 

(a)(i)For the forward reaction from left to right, NH3  gains a proton to form NH4+ and thus NH3 is a proton acceptor .

It is a Bronsted-Lowry base

 

(ii) For the backward reaction from right to left, NH4+   donates a proton to    form NH3 and thus NH4+  is an ‘opposite’ proton donor.

It is a Bronsted-Lowry conjugate acid

 

(b)(i)For the forward reaction from left to right, HCl  donates a proton to form Cl and thus HCl is a proton donor .

It is a Bronsted-Lowry acid

 

(ii) For the backward reaction from right to left, Cl  gains a proton to form HCl and thus Cl is an ‘opposite’ proton acceptor.

It is a Bronsted-Lowry conjugate base.

 

  1. Acids and bases show acidic and alkaline properties/characteristics only in water but not in other solvents e.g.

 

(a)Hydrogen chloride gas dissolves in water to form hydrochloric acid Hydrochloric acid dissociates/ionizes in water to free  H+(aq)/H3O+(aq) ions. The free H3O+(aq) / H+(aq)  ions are responsible for:

 

(i)turning blue litmus paper/solution red.

(ii)show pH value 1/2/3/4/5/6

(iii)are good electrolytes/conductors of electricity/undergo electrolysis.

 

(iv)react with metals to produce /evolve hydrogen gas and a salt. i.e.

Ionically:

-For a monovalent metal: 2M(s) +  2H+(aq)   ->  2M+(aq)  +  H2(g)

-For a divalent metal:      M(s)   +  2H+(aq)   ->   M2+(aq)  +  H2(g)

-For a trivalent metal:    2M(s)   +  6H+(aq)   ->   2M3+(aq)  +  3H2(g)

 

Examples:

-For a monovalent metal: 2Na(s)  +   2H+(aq)  ->  2Na+(aq)    +  H2(g)

-For a divalent metal:      Ca(s)      +   2H+(aq)   ->   Ca2+(aq)   +  H2(g)

-For a trivalent metal:    2Al(s)      +   6H+(aq)   ->   2Al3+(aq)  +  3H2(g)

 

          (v)react with metal carbonates and hhydrogen carbonates  to produce /evolve carbon(IV)oxide gas ,water and a salt. i.e.

Ionically:

-For a monovalent metal: M2CO3(s)+ 2H+(aq) -> 2M+(aq) + H2O (l)+ CO2(g)

MHCO3(s)+ H+(aq) -> M+(aq) + H2O (l)+ CO2(g)

 

-For a divalent metal: MCO3(s)+ 2H+(aq) -> M2+(aq) + H2O (l)+ CO2(g)

M(HCO3) 2(aq)+2H+(aq) ->M2+(aq)+2H2O(l)+2CO2(g)

 

Examples:

-For a monovalent metal: K2CO3(s)+ 2H+(aq) -> 2K+(aq) + H2O (l)+ CO2(g)

NH4HCO3(s)+ H+(aq) -> NH4+(aq) + H2O (l)+ CO2(g)

 

-For a divalent metal: ZnCO3(s)+ 2H+(aq) -> Zn2+(aq) + H2O (l)+ CO2(g)

Mg(HCO3) 2(aq)+2H+(aq) ->Mg2+(aq)+2H2O(l)+2CO2(g)

 

(vi)neutralize metal oxides/hydroxides  to  salt and water only. i.e.

Ionically:

-For a monovalent metal: M2O(s) + 2H+(aq)  ->  2M+(aq)  +  H2O (l)

MOH(aq) +  H+(aq)  ->  M+(aq)  +  H2O (l)

 

-For a divalent metal:       MO(s)  + 2H+(aq) -> M2+(aq) + H2O (l)

M(OH) 2(s)  +  2H+(aq)  -> M2+(aq) + 2H2O(l)

 

-For a trivalent metal:       M2O3(s)  + 6H+(aq) -> 2M3+(aq) + 3H2O (l)

M(OH) 3(s)  +  3H+(aq)  -> M3+(aq) + 3H2O(l)

 

Examples:

-For a monovalent metal: K2O(s) + 2H+(aq)  ->  2K+(aq)  +  H2O (l)

NH4OH(aq) +  H+(aq)  -> NH4+(aq)  +  H2O (l)

 

-For a divalent metal:  ZnO (s) +  2H+(aq) ->  Zn2+(aq)  + H2O (l)

Pb(OH) 2(s)  + 2H+(aq) -> Pb2+(aq) + 2H2O(l)

 

(b)Hydrogen chloride gas dissolves in methylbenzene /benzene but does not dissociate /ionize into free ions.

It exists in molecular state showing none of the above properties.

 

(c)Ammonia gas dissolves in water to form aqueous ammonia which dissociate/ionize to free NH4+ (aq) and OH(aq) ions.

This dissociation/ionization makes aqueous ammonia to:

 

(i)turn litmus paper/solution blue.

 

(ii)have pH 8/9/10/11

 

(iii)be a good electrical conductor

 

(iv)react with acids to form ammonium salt and water only.

 

NH4OH(aq)  +   HCl(aq)  ->   NH4Cl(aq)   +  H2O(l)

 

(d)Ammonia gas dissolves in methylbenzene/benzene /kerosene but does not dissociate into free ions therefore existing as molecules

 

  1. Solvents are either polar or non-polar.

A polar solvent is one which dissolves ionic compounds and other polar solvents.

Water is polar solvent that dissolves ionic and polar substance by surrounding the free ions as below:

H ð+           H ð+                                                      O ð-

 

H ð+                    H ð+

H ð+              O ð-              H ð+                               H ð+

H ð+

O ð-    H+   O ð-                            O ð-                              Cl                O ð-

H ð+                         Hð+

H ð+              O ð-              H ð+                                       H+          H ð+

 

H ð+             H ð+                                                      O ð-

 

 

 

Beaker

Cl

Cl      H+                          water

H+

Cl      H+                                                    Free ions

Note:Water is polar .It is made up of :

Oxygen atom is partially negative and two hydrogen atoms which are partially  positive.

They surround the free H+ and Clions.

A non polar solvent is one which dissolved non-polar substances and covalent compounds.

If a polar ionic compound is dissolved in non-polar solvent ,it does not ionize/dissociate into free ions as below:

 

 

 

 

H-Cl

H-Cl                                                                                                                                                         methyl benzene                                                                                    H-Cl                                                                                                                                      H-Cl                                                                                                                                                                                                                                                                                                                                                                                    Covalent bond

 

  1. Some acids and bases are strong while others are weak.

(a)A strong acid/base is one which is fully/wholly/completely dissociated / ionized into many free H+ /OH ions i.e.

  1. Strong acids exists more as free H+ ions than molecules. e.g.

 

HCl(aq)                           H+(aq)          +            Cl (aq)                                             (molecules)                         (cation)                   (anion)

HNO3(aq)                        H+(aq)          +            NO3(aq)                                           (molecules)                         (cation)                   (anion)

 

H2SO4(aq)                       2H+(aq)        +            SO42-(aq)                                          (molecules)                         (cation)                   (anion)

  1. Strong bases/alkalis exists more as free OH ions than molecules. e.g.

 

KOH(aq)                         K+(aq)          +            OH (aq)                                           (molecules)                         (cation)                   (anion)

NaOH(aq)                        Na+(aq)        +            OH(aq)                                            (molecules)                         (cation)                   (anion)

 

(b) A weak base/acid is one which is partially /partly dissociated /ionized in water into free OH (aq) and H+(aq) ions.

  1. Weak acids exists more as molecules than as free H+ ions. e.g.

 

CH3COOH(aq)                          H+(aq)          +            CH3COO (aq)                             (molecules)                              (cation)                   (anion)

H3PO4(aq)                       3H+(aq)        +            PO43-(aq)                                          (molecules)                         (cation)                   (anion)

 

H2CO3(aq)                       2H+(aq)        +            CO32-(aq)                                          (molecules)                         (cation)                   (anion)

 

  1. Weak bases/alkalis exists more as molecules than free OH ions. e.g.

 

NH4OH(aq)                               NH4+(aq)    +            OH (aq)                                       (molecules)                             (cation)                   (anion)

Ca(OH)2(aq)                              Ca2+(aq)       +            2OH(aq)                                          (molecules)                         (cation)                   (anion)

 

Mg(OH)2(aq)                             Mg2+(aq)      +            2OH(aq)                                          (molecules)                         (cation)                   (anion)

  1. The concentration of an acid/base/alkali is based on the number of moles of acid/bases dissolved in a decimeter(litre)of the solution.

An acid/base/alkali with more acid/base/alkali in a decimeter(litre)  of solution is said to be concentrated while that with less is said to be dilute.

 

  1. (a) (i)strong acids have pH 1/2/3 while weak acids have high pH 4/5/6.

 

(ii)a neutral solution have pH 7

(iii)strong alkalis/bases have pH 12/13/14 while weak bases/alkalis have pH        11/10 /9 / 8.

(b) pH is a measure of H+(aq) concentration in a solution.

The higher the H+(aq)ions concentration ;

-the higher the acidity

-the lower the pH

-the lower the concentration of OH(aq)

-the lower the alkalinity

At pH 7 , a solution has equal concentration of H+(aq) and  OH(aq).

Beyond pH 7,the concentration of the OH(aq) increases as the H+(aq) ions decreases.

 

10.(a) When acids /bases dissolve in water, the ions present in the solution conduct electricity.

The more the dissociation the higher the yield of ions and the greater the electrical conductivity of the solution.

A compound that conducts electricity in an electrolyte and thus a compound showing high electrical conductivity is a strong electrolyte while a compound showing low electrical conductivity is a weak electrolyte.

 

(b) Practically, a bright light on a bulb ,a high voltage reading from a voltmeter high ammeter reading from an ammeter, a big deflection on a galvanometer is an indicator of strong electrolyte(acid/base) and the opposite for weak electrolytes(acids/base)

 

  1. Some compounds exhibit/show both properties of acids and bases/alkalis.

A substance that reacts with both acids and bases is said to be amphotellic.

The examples below show the amphotellic properties of:

 

(a)  Zinc (II)oxide(ZnO) and Zinc hydroxide(Zn(OH)2)

(i)When ½ spatula full of Zinc(II)oxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e.

 

(i) when reacting with nitric(V)acid, the oxide shows basic properties by reacting with an acid to form a simple salt and water only.

 

Basic oxide      +     Acid      ->     salt     +      water

Examples:

Chemical equation

ZnO(s)    +    2HNO3(aq)    ->    Zn(NO3) 2 (aq)   +    H2O(l)

ZnO(s)    +    2HCl(aq)       ->    ZnCl2 (aq)          +    H2O(l)

ZnO(s)    +    H2SO4(aq)     ->    ZnSO4 (aq)          +    H2O(l)

Ionic equation

ZnO(s)    +    2H+ (aq)         ->    Zn 2+ (aq)      +    H2O(l)

 

(ii) when reacting with sodium hydroxide, the oxide shows acidic properties by reacting with a base to form a complex salt.

 

Basic oxide      +     Base/alkali  + Water     ->    Complex salt

Examples:

Chemical equation

1.When Zinc oxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxozincate(II) complex salt.

ZnO(s)    +    2NaOH(aq) +    H2O(l)   ->   Na2Zn(OH) 4(aq)

 

2.When Zinc oxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxozincate(II) complex salt.

ZnO(s)    +    2KOH(aq) +    H2O(l)   ->   K2Zn(OH) 4(aq)

 

Ionic equation

ZnO(s)    +    2OH(aq)   +    H2O(l)   ->   2[Zn(OH) 4]2- (aq)

 

(ii)When Zinc(II)hydroxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e.

 

(i) when reacting with nitric(V)acid, the hydroxide shows basic properties. It reacts with an acid to form a simple salt and water only.

 

Basic hydroxide      +     Acid         ->     salt     +      water

Examples:

Chemical equation

Zn(OH) 2 (s)    +    2HNO3(aq)    ->    Zn(NO3) 2 (aq)   +    2H2O(l)

Zn(OH) 2 (s)    +    2HCl(aq)       ->    ZnCl2 (aq)          +    2H2O(l)

Zn(OH) 2 (s)    +   H2SO4(aq)     ->    ZnSO4 (aq)          +    2H2O(l)

Ionic equation

Zn(OH) 2 (s)    +    2H+ (aq)         ->    Zn 2+ (aq)      +    2H2O(l)

 

(ii) when reacting with sodium hydroxide, the hydroxide shows acidic properties by reacting with a base to form a complex salt.

 

Basic hydroxide      +     Base/alkali    ->    Complex salt

Examples:

Chemical equation

1.When Zinc hydroxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxozincate(II) complex salt.

 

Zn(OH) 2 (s)    +    2NaOH(aq)  ->   Na2Zn(OH) 4(aq)

 

2.When Zinc hydroxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxozincate(II) complex salt.

 

Zn(OH) 2 (s)   +    2KOH(aq)   ->   K2Zn(OH) 4(aq)

Ionic equation

Zn(OH) 2 (s)    +    2OH(aq)     ->   2[Zn(OH) 4]2- (aq)

 

(b)  Lead (II)oxide(PbO) and Lead(II) hydroxide (Pb(OH)2)

(i)When ½ spatula full of Lead(II)oxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e.

 

(i) when reacting with nitric(V)acid, the oxide shows basic properties by reacting with an acid to form a simple salt and water only. All other Lead salts are insoluble.

 

Chemical equation

PbO(s)    +    2HNO3(aq)    ->    Pb(NO3) 2 (aq)   +    H2O(l)

 

Ionic equation

PbO(s)    +    2H+ (aq)         ->    Pb 2+ (aq)      +    H2O(l)

 

(ii) when reacting with sodium hydroxide, the oxide shows acidic properties by reacting with a base to form a complex salt.

 

Chemical equation

1.When Lead(II) oxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoplumbate(II) complex salt.

PbO(s)    +    2NaOH(aq) +    H2O(l)   ->   Na2Pb(OH) 4(aq)

 

2.When Lead(II) oxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoplumbate(II) complex salt.

PbO(s)    +    2KOH(aq) +    H2O(l)   ->   K2Pb(OH) 4(aq)

 

Ionic equation

PbO(s)    +    2OH(aq)   +    H2O(l)   ->   2[Pb(OH) 4]2- (aq)

 

(ii)When Lead(II)hydroxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e.

 

(i) when reacting with nitric(V)acid, the hydroxide shows basic properties. It reacts with the acid to form a simple salt and water only.

 

Chemical equation

Pb(OH) 2 (s)    +    2HNO3(aq)    ->    Pb(NO3) 2 (aq)   +    2H2O(l)

 

Ionic equation

Pb(OH) 2 (s)    +    2H+ (aq)         ->    Pb 2+ (aq)      +    2H2O(l)

 

(ii) when reacting with sodium hydroxide, the hydroxide shows acidic properties. It reacts with a base to form a complex salt.

 

Chemical equation

1.When Lead(II) hydroxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoplumbate(II) complex salt.

 

Pb(OH) 2 (s)    +    2NaOH(aq)  ->   Na2Pb(OH) 4(aq)

 

2.When Lead(II) hydroxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoplumbate(II) complex salt.

 

Pb(OH) 2 (s)   +    2KOH(aq)   ->   K2Pb(OH) 4(aq)

 

Ionic equation

Pb(OH) 2 (s)    +    2OH(aq)     ->   2[Pb(OH) 4]2- (aq)

 

(c)Aluminium(III)oxide(Al2O3) and Aluminium(III)hydroxide(Al(OH)3)

(i)When ½ spatula full of Aluminium(III)oxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e.

 

(i) when reacting with nitric(V)acid, the oxide shows basic properties by reacting with an acid to form a simple salt and water only.

 

Chemical equation

Al2O3 (s)    +    6HNO3(aq)    ->    Al(NO3)3 (aq)     +    3H2O(l)

Al2O3 (s)    +    6HCl(aq)        ->    AlCl3 (aq)         +    3H2O(l)

Al2O3 (s)    +    3H2SO4(aq)     ->    Al2(SO4)3 (aq)   +    3H2O(l)

Ionic equation

Al2O3 (s)    +    3H+ (aq)         ->    Al 3+ (aq)      +    3H2O(l)

 

(ii) when reacting with sodium hydroxide, the oxide shows acidic properties by reacting with a base to form a complex salt.

 

Chemical equation

1.When Aluminium(III) oxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoaluminate(III) complex salt.

 

Al2O3 (s)    +    2NaOH(aq) +    3H2O(l)   ->   2NaAl(OH) 4(aq)

 

2.When  Aluminium(III) oxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoaluminate(II) complex salt.

 

Al2O3 (s)   +    2KOH(aq) +    3H2O(l)   ->   2NaAl(OH) 4(aq)

 

Ionic equation

Al2O3 (s)   +    2OH(aq)   +    3H2O(l)   ->   2[Al(OH) 4] (aq)

 

(ii)When Aluminium(III)hydroxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e.

 

(i) when reacting with nitric(V)acid, the hydroxide shows basic properties. It reacts with the acid to form a simple salt and water only.

 

Chemical equation

Al(OH) 3 (s)     +    3HNO3(aq)    ->    Al(NO3)3 (aq)      +    3H2O(l)

Al(OH)3 (s)      +    3HCl(aq)        ->    AlCl3 (aq)                    +    3H2O(l)

2Al(OH)3 (s)    +    3H2SO4(aq)     ->    Al2(SO4)3 (aq)    +    3H2O(l)

 

Ionic equation

Al(OH)3 (s)    +    3H+ (aq)         ->    Al 3+ (aq)      +    3H2O(l)

 

(ii) when reacting with sodium hydroxide, the hydroxide shows acidic properties. It reacts with a base to form a complex salt.

 

Chemical equation

1.When aluminium(III) hydroxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoaluminate(III) complex salt.

Al(OH) 3 (s)    +    NaOH(aq)  ->   NaAl(OH) 4(aq)

 

2.When aluminium(III) hydroxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoaluminate(III) complex salt.

Al(OH) 3 (s)   +    KOH(aq)   ->   KAl(OH) 4(aq)

 

Ionic equation

Al(OH) 3 (s)    +    OH(aq)     ->   [Al(OH) 4] (aq)

 

Summary of amphotellic oxides/hydroxides

 

Oxide Hydroxide Formula of simple salt from nitric (V)acid Formula of complex salt

from sodium hydroxide

ZnO Zn(OH)2 Zn(NO3)2 Na2Zn(OH)4

[Zn(OH)4]2-(aq)

Sodium tetrahydroxozincate(II)

 

PbO Pb(OH)2 Pb(NO3)2 Na2Pb(OH)4

[Pb(OH)4]2-(aq)

Sodium tetrahydroxoplumbate(II)

 

Al2O3 Al(OH) 3 Al(NO3)3 NaAl(OH)4

[Al(OH)4](aq)

Sodium tetrahydroxoaluminate(II)

 

 

12.(a) A salt is an ionic compound formed when the cation from a base combine with the anion derived from an acid.

A salt is therefore formed when the hydrogen ions in an acid are replaced wholly/fully or partially/partly ,directly or indirectly by a metal or ammonium radical.

 

(b) The number of ionizable/replaceable hydrogen in an acid is called basicity of an acid.

Some acids are therefore:

(i)monobasic acids generally denoted HX e.g.

HCl, HNO3,HCOOH,CH3COOH.

 

(ii)dibasic acids ; generally denoted H2X e.g.

H2SO4, H2SO3, H2CO3,HOOCOOH.

 

(iii)tribasic acids ; generally denoted H3X e.g.

H3PO4.

 

(c) Some salts are normal salts while other are acid salts.

(i)A normal salt is formed when all the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical.

 

(ii)An acid salt is formed when part/portion the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical.

 

Table showing normal and acid salts derived from common acids

 

   Acid  name Chemical formula Basicity     Normal salt      Acid salt
Hydrochloric acid HCl Monobasic Chloride(Cl) None

 

Nitric(V)acid HNO3 Monobasic Nitrate(V)(NO3) None

 

Nitric(III)acid HNO2 Monobasic Nitrate(III)(NO2) None

 

Sulphuric(VI)acid H2SO4 Dibasic Sulphate(VI) (SO42-) Hydrogen sulphate(VI)

(HSO4)

 

Sulphuric(IV)acid H2SO3 Dibasic Sulphate(IV) (SO32-) Hydrogen sulphate(IV)

(HSO3)

 

Carbonic(IV)acid H2CO3 Dibasic Carbonate(IV)(CO32-) Hydrogen carbonate(IV)

(HCO3)

 

Phosphoric(V)

acid

H3PO4 Tribasic Phosphate(V)(PO43-) Dihydrogen phosphate(V)

(H2PO42-)

 

Hydrogen diphosphate(V)

(HP2O42-)

 

The table below show shows some examples of salts.

Base/alkali Cation Acid Anion Salt Chemical name of salts
NaOH Na+ HCl Cl NaCl Sodium(I)chloride
Mg(OH)2 Mg2+ H2SO4 SO42- MgSO4

Mg(HSO4)2

Magnesium sulphate(VI)

Magnesium hydrogen sulphate(VI)

Pb(OH)2 Pb2+ HNO3 NO3 Pb(NO3)2 Lead(II)nitrate(V)
Ba(OH)2 Ba2+ HNO3 NO3 Ba(NO3)2 Barium(II)nitrate(V)
Ca(OH)2 Ba2+ H2SO4 SO42- MgSO4 Calcium sulphate(VI)
NH4OH NH4+ H3PO4 PO43- (NH4 )3PO4

(NH4 )2HPO4

NH4 H2PO4

Ammonium phosphate(V)

Diammonium phosphate(V)

Ammonium diphosphate(V)

KOH K+ H3PO4 PO43- K3PO4 Potassium phosphate(V)
Al(OH)3 Al3+ H2SO4 SO42- Al2(SO4)2 Aluminium(III)sulphate(VI)
Fe(OH)2 Fe2+ H2SO4 SO42- FeSO4 Iron(II)sulphate(VI)
Fe(OH)3 Fe3+ H2SO4 SO42- Fe2(SO4)2 Iron(III)sulphate(VI)

 

(d) Some salts undergo hygroscopy, deliquescence and efflorescence.

(i) Hygroscopic salts /compounds are those that absorb water from the atmosphere but do not form a solution.

Some salts which are hygroscopic include anhydrous copper(II)sulphate(VI), anhydrous cobalt(II)chloride, potassium nitrate(V) common table salt.

 

(ii)Deliquescent salts /compounds are those that absorb water from the atmosphere and form a solution.

Some salts which are deliquescent include: Sodium nitrate(V),Calcium chloride, Sodium hydroxide, Iron(II)chloride, Magnesium chloride.

 

(iii)Efflorescent salts/compounds are those that lose their  water of crystallization to  the atmosphere.

Some salts which effloresces include: sodium carbonate decahydrate, Iron(II)sulphate(VI)heptahydrate, sodium sulphate (VI)decahydrate.

 

(e)Some salts contain water of crystallization.They are hydrated.Others do not contain water of crystallization. They are anhydrous.

 

Table showing some hydrated salts.

Name of hydrated salt Chemical formula
Copper(II)sulphate(VI)pentahydrate CuSO4.5H2O
Aluminium(III)sulphate(VI)hexahydrate Al2 (SO4) 3.6H2O
Zinc(II)sulphate(VI)heptahydrate ZnSO4.7H2O
Iron(II)sulphate(VI)heptahydrate FeSO4.7H2O
Calcium(II)sulphate(VI)heptahydrate CaSO4.7H2O
Magnesium(II)sulphate(VI)heptahydrate MgSO4.7H2O
Sodium sulphate(VI)decahydrate Na2SO4.10H2O
Sodium carbonate(IV)decahydrate Na2CO3.10H2O
Potassium carbonate(IV)decahydrate K2CO3.10H2O
Potassium sulphate(VI)decahydrate K2SO4.10H2O

 

(f)Some salts exist as a simple salt while some as complex salts. Below are some complex salts.

Table of some complex salts

 

Name of complex salt Chemical formula Colour of the complex salt
Tetraamminecopper(II)sulphate(VI) Cu(NH3) 4 SO4 H2O Royal/deep blue solution
Tetraamminezinc(II)nitrate(V) Zn(NH3) 4 (NO3 )2 Colourless solution
Tetraamminecopper(II) nitrate(V) Cu(NH3) 4 (NO3 )2 Royal/deep blue solution
Tetraamminezinc(II)sulphate(VI) Zn(NH3) 4 SO4 Colourless solution

 

(g)Some salts exist as two salts in one. They are called double salts.

 

Table of some double salts

Name of double salts Chemical formula
Trona(sodium sesquicarbonate) Na2CO3 NaHCO3.2H2O
Ammonium iron(II)sulphate(VI) FeSO4(NH4) 2SO4.2H2O
Ammonium aluminium(III)sulphate(VI) Al2(SO4) 3(NH4) 2SO4.H2O

 

(h)Some salts dissolve in water to form a solution. They are said to be soluble. Others do not dissolve in water. They form a suspension/precipitate in water.

 

Table of solubility of salts

 

          Soluble salts            Insoluble salts
All nitrate(V)salts  
All sulphate(VI)/SO42- salts          except    Barium(II) sulphate(VI)/BaSO4

Calcium(II) sulphate(VI)/CaSO4

Lead(II) sulphate(VI)/PbSO4

All sulphate(IV)/SO32- salts          except    Barium(II) sulphate(IV)/BaSO3

Calcium(II) sulphate(IV)/CaSO3

Lead(II) sulphate(IV)/PbSO3

All chlorides/Cl–                                         except   Silver chloride/AgCl

Lead(II)chloride/PbCl2(dissolves in hot water)

All phosphate(V)/PO43-  
All sodium,potassium and ammonium salts  
All hydrogen carbonates/HCO3  
All hydrogen sulphate(VI)/ HSO4  
Sodium carbonate/Na2CO3,

potassium carbonate/ K2CO3,

ammonium carbonate (NH4) 2CO3

except    All carbonates
All alkalis(KOH,NaOH, NH4OH) except     All bases

 

13 Salts can be prepared in a school laboratory by a method that uses its solubility in water.

  • Soluble salts may be prepared by using any of the following methods:

 

(i)Direct displacement/reaction of a metal with an acid.

By reacting a metal higher in the reactivity series than hydrogen with a dilute acid,a salt is formed and hydrogen gas is evolved.

Excess of the metal must be used to ensure all the acid has reacted.

When effervescence/bubbling /fizzing has stopped ,excess metal is filtered.

The filtrate is  heated to concentrate then allowed to crystallize.

Washing with distilled water then drying between filter papers produces a sample crystal of the salt. i.e.

M(s)    +   H2X      ->   MX(aq)   +    H2(g)

Examples

Mg(s)  +  H2SO4(aq)        ->  MgSO4 (aq)      + H2(g)

Zn(s)   +  H2SO4(aq)        ->  ZnSO4 (aq)       + H2(g)

Pb(s)   +  2HNO3(aq)      -> Pb(NO3) 2(aq)  + H2(g)

Ca(s)   +  2HNO3(aq)      -> Ca(NO3) 2(aq)  + H2(g)

Mg(s)  +  2HNO3(aq)      -> Mg(NO3) 2(aq)            + H2(g)

Mg(s)  +  2HCl(aq)          -> MgCl 2(aq)        + H2(g)

Zn(s)   +  2HCl(aq)          -> ZnCl 2(aq)         + H2(g)

 

(ii)Reaction of an insoluble base with an acid

By adding an insoluble base (oxide/hydroxide )to a dilute acid until no more dissolves, in the acid,a salt and water are formed. Excess of the base is filtered off. The filtrate is heated to concentrate ,allowed to crystallize then washed with distilled water before drying between filter papers e.g.

PbO(s)   +  2HNO3(aq)    -> Pb(NO3) 2(aq)  + H2O (l)

Pb(OH)2(s)   +  2HNO3(aq)       -> Pb(NO3) 2(aq)  + 2H2O (l)

CaO (s)   +  2HNO3(aq)   -> Ca(NO3) 2(aq)  + H2O (l)

MgO (s)  +  2HNO3(aq)  -> Mg(NO3) 2(aq)            + H2O (l)

MgO (s)  +  2HCl(aq)      -> MgCl 2(aq)        + H2O (l)

ZnO (s)   +  2HCl(aq)      -> ZnCl 2(aq)         + H2O (l)

Zn(OH)2(s)   +  2HNO3(aq)       -> Zn(NO3) 2(aq)  + 2H2O (l)

CuO (s)   +  2HCl(aq)      -> CuCl 2(aq)         + H2O (l)

CuO (s)   +  H2SO4(aq)    -> CuSO4(aq)        + H2O (l)

Ag2O(s)   +  2HNO3(aq)  -> 2AgNO3(aq)     + H2O (l)

Na2O(s)   +  2HNO3(aq)  -> 2NaNO3(aq)      + H2O (l)

 

(iii)reaction of insoluble /soluble carbonate /hydrogen carbonate with an acid.

By adding an excess of a soluble /insoluble carbonate or hydrogen carbonate to adilute acid, effervescence /fizzing/bubbling out of carbon(IV)oxide gas shows the reaction is taking place. When effervescence /fizzing/bubbling out of the gas is over, excess of the insoluble carbonate is filtered off. The filtrate is heated to concentrate ,allowed to crystallize then washed with distilled water before drying between filter paper papers e.g.

PbCO3 (s)     +  2HNO3(aq)       -> Pb(NO3) 2(aq)  + H2O (l)+ CO2(g)

ZnCO3 (s)     +  2HNO3(aq)       -> Zn(NO3) 2(aq)  + H2O (l)+ CO2(g)

CaCO3 (s)     +  2HNO3(aq)       -> Ca(NO3) 2(aq)  + H2O (l)+ CO2(g)

MgCO3 (s)    + H2SO4(aq)         -> MgSO4(aq)        + H2O (l)+ CO2(g)

Cu CO3 (s)    +  H2SO4(aq)        -> CuSO4(aq)        + H2O (l) + CO2(g)

Ag2CO3 (s)   +  2HNO3(aq)      -> 2AgNO3(aq)     + H2O (l) + CO2(g)

Na2CO3 (s)    +  2HNO3(aq)      -> 2NaNO3(aq)      + H2O (l) + CO2(g)

K2CO3 (s)     +  2HCl(aq)           -> 2KCl(aq)           + H2O (l) + CO2(g)

NaHCO3 (s)    +  HNO3(aq)       -> NaNO3(aq)        + H2O (l) + CO2(g)

KHCO3 (s)     +  HCl(aq)           -> KCl(aq)                       + H2O (l) + CO2(g)

(iv)neutralization/reaction of soluble base/alkali with dilute acid

 

By adding an acid to a burette into a known volume of an alkali with 2-3 drops of an indicator, the colour of the indicator changes when the acid has completely reacted with an alkali at the end point. The procedure is then repeated without the indicator .The solution mixture is then heated to concentrate , allowed to crystallize ,washed with distilled water before drying with filter papers. e.g.

 

NaOH (aq)       +  HNO3(aq)     -> NaNO3(aq)       + H2O (l)

KOH (aq)         +  HNO3(aq)     -> KNO3(aq)         + H2O (l)

KOH (aq)         +  HCl(aq)         -> KCl(aq)            + H2O (l)

2KOH (aq)       +  H2SO4(aq)    -> K2SO4(aq)        + 2H2O (l)

2 NH4OH (aq)  +  H2SO4(aq)    -> (NH4)2SO4(aq)  + 2H2O (l)

NH4OH (aq)     +  HNO3(aq)     -> NH4NO3(aq)      +  H2O (l)

 

(iv)direct synthesis/combination.

When a metal burn in a gas jar containing a non metal , the two directly combine to form a salt. e.g.

2Na(s)         +        Cl2(g)           ->       2NaCl(s)

2K(s)           +       Cl2(g)           ->       2KCl(s)

Mg(s)           +       Cl2(g)           ->       Mg Cl2 (s)

Ca(s)            +       Cl2(g)           ->       Ca Cl2 (s)

 

Some salts once formed undergo sublimation and hydrolysis. Care should be taken to avoid water/moisture into the reaction flask during their preparation.Such salts include aluminium(III)chloride(AlCl3) and iron (III)chloride(FeCl3)

 

  1. Heated aluminium foil reacts with chlorine to form aluminium(III)chloride that sublimes away from the source of heating then deposited as solid again

2Al(s)                    +       3Cl2(g)         ->       2AlCl3 (s/g)

 

Once formed  aluminium(III)chloride hydrolyses/reacts with water vapour / moisture present to form aluminium hydroxide solution and highly acidic fumes of hydrogen chloride gas.

AlCl3(s)+     3H2 O(g)     ->       Al(OH)3 (aq) + 3HCl(g)

 

  1. Heated iron filings reacts with chlorine to form iron(III)chloride that sublimes away from the source of heating then deposited as solid again

2Fe(s)                    +       3Cl2(g)         ->       2FeCl3 (s/g)

 

Once formed , aluminium(III)chloride hydrolyses/reacts with water vapour / moisture present to form aluminium hydroxide solution and highly acidic fumes of hydrogen chloride gas.

FeCl3(s)+     3H2 O(g)     ->       Fe(OH)3 (aq) + 3HCl(g)

 

(b)Insoluble salts can be prepared by reacting two suitable soluble salts to form one soluble and one insoluble. This is called double decomposition or precipitation. The mixture is filtered and the residue is washed with distilled water then dried.

CuSO4(aq)        +   Na2CO3 (aq)         ->       CuCO3 (s)  +  Na2 SO4(aq)

BaCl2(aq)        +   K2SO4 (aq)           ->       BaSO4 (s)   +  2KCl (aq)

Pb(NO3)2(aq)   +   K2SO4 (aq)            ->       PbSO4 (s)   +  2KNO3 (aq)

2AgNO3(aq)               +  MgCl2 (aq)             ->       2AgCl(s)     +  Mg(NO3)2 (aq)

Pb(NO3)2(aq)   +   (NH4) 2SO4 (aq)    ->       PbSO4 (s)   +  2NH4NO 3(aq)

BaCl2(aq)        +   K2SO3 (aq)            ->       BaSO3 (s)   +  2KCl (aq)

 

  1. Salts may lose their water of crystallization , decompose ,melt or sublime on heating on a Bunsen burner flame.

The following shows the behavior of some salts on heating gently /or strongly in a laboratory school burner:

 

(a)effect of heat on chlorides

All chlorides have very high melting and boiling points and therefore are not affected by laboratory heating except ammonium chloride. Ammonium chloride sublimes on gentle heating. It dissociate into the constituent ammonia and hydrogen chloride gases on strong heating.

NH4Cl(s)               NH4Cl(g)                         NH3(g) + HCl(g)

(sublimation)              (dissociation)

 

(b)effect of heat on nitrate(V)

(i) Potassium nitrate(V)/KNO3 and sodium nitrate(V)/NaNO3 decompose on heating to form Potassium nitrate(III)/KNO2 and sodium nitrate(III)/NaNO2  and producing Oxygen gas in each case.

2KNO3 (s)    -> 2KNO2(s) +   O2(g)

2NaNO3 (s)  -> 2NaNO2(s) +   O2(g)

 

(ii)Heavy metal nitrates(V) salts decompose on heating to form the oxide and a mixture of brown acidic nitrogen(IV)oxide and oxygen gases. e.g.

2Ca(NO3)2 (s)          ->    2CaO(s)    +   4NO2(g)   +    O2(g)

2Mg(NO3)2(s)         ->    2MgO(s)    +   4NO2(g)   +    O2(g)

2Zn(NO3)2(s)           ->    2ZnO(s)    +   4NO2(g)   +    O2(g)

2Pb(NO3)2(s)           ->    2PbO(s)    +   4NO2(g)   +    O2(g)

2Cu(NO3)2(s)          ->    2CuO(s)    +   4NO2(g)   +    O2(g)

2Fe(NO3)2(s)           ->    2FeO(s)    +   4NO2(g)   +    O2(g)

 

(iii)Silver(I)nitrate(V) and mercury(II) nitrate(V) are lowest in the reactivity series. They decompose on heating to form the metal(silver and mercury)and the Nitrogen(IV)oxide and oxygen gas. i.e.

2AgNO3(s)   ->    2Ag (s)    +   2NO2(g)   +    O2(g)

2Hg(NO3)2 (s)         ->    2Hg (s)    +   4NO2(g)   +    O2(g)

 

(iv)Ammonium nitrate(V) and Ammonium nitrate(III) decompose on heating to Nitrogen(I)oxide(relights/rekindles glowing splint) and nitrogen gas respectively.Water is also formed.i.e.

NH4NO3(s)      ->      N2O (g)     +     H2O(l)

NH4NO2(s)      ->      N2 (g)        +     H2O(l)

 

(c) effect of heat on nitrate(V)

Only Iron(II)sulphate(VI), Iron(III)sulphate(VI) and copper(II)sulphate(VI) decompose on heating. They form the oxide, and produce highly acidic fumes of acidic sulphur(IV)oxide gas.

2FeSO4 (s)                ->      Fe2O3(s)      +     SO3(g) +     SO2(g)

Fe2(SO4) 3(s)              ->      Fe2O3(s)     +     SO3(g)

CuSO4 (s)                 ->      CuO(s)       +     SO3(g)

 

(d) effect of heat on carbonates(IV) and hydrogen carbonate(IV).

(i)Sodium carbonate(IV)and potassium carbonate(IV)do not decompose on heating.

(ii)Heavy metal nitrate(IV)salts decompose on heating to form the oxide and produce carbon(IV)oxide gas. Carbon (IV)oxide gas forms a white precipitate when bubbled in lime water. The white precipitate dissolves if the gas is in excess. e.g. CuCO3 (s)                     ->      CuO(s)       +     CO2(g)

CaCO3 (s)                       ->      CaO(s)       +     CO2(g)

PbCO3 (s)                       ->      PbO(s)       +     CO2(g)

FeCO3 (s)                       ->      FeO(s)       +     CO2(g)

ZnCO3 (s)                       ->      ZnO(s)       +     CO2(g)

 

(iii)Sodium hydrogen carbonate(IV) and Potassium hydrogen carbonate(IV)decompose on heating to give the corresponding carbonate (IV) and form water and carbon(IV)oxide gas. i.e.

2NaHCO 3(s)              ->      Na2CO3(s)     +     CO2(g)   +  H2O(l)

2KHCO 3(s)                 ->      K2CO3(s)      +     CO2(g)   +  H2O(l)

(iii) Calcium hydrogen carbonate (IV) and Magnesium hydrogen carbonate(IV) decompose on heating to give the corresponding carbonate (IV) and form water and carbon(IV)oxide gas. i. e.

Ca(HCO3) 2(aq)              ->      CaCO3(s)      +     CO2(g)   +  H2O(l)

Mg(HCO3) 2(aq)             ->      MgCO3(s)     +     CO2(g)   +  H2O(l)

 

  1. Salts contain cation(positively charged ion) and anions(negatively charged ion).When dissolved in polar solvents/water.

The cation and anion in a salt is determined/known usually by precipitation of the salt using a precipitating reagent.

The colour of the precipitate is a basis of qualitative analysis of a compound.

 

16.Qualitative analysis is the process of identifying an unknown compound /salt by identifying the unique qualities of the salt/compound.

It involves some of the following processes.

 

(a)Reaction of cation with sodium/potassium hydroxide solution.

Both sodium/potassium hydroxide solutions are precipitating reagents.

The alkalis produce unique colour of a precipitate/suspension when a few/three drops is added and then excess alkali is added to unknown salt/compound solution.

NB: Potassium hydroxide is not commonly used because it is more expensive than sodium hydroxide.

The table below shows the observations, inferences / deductions and explanations from the following test tube  experiments:

Procedure 

Put about 2cm3 of  MgCl2, CaCl2, AlCl3, NaCl, KCl, FeSO4, Fe2(SO4) 3, CuSO4, ZnSO4NH4NO3, Pb(NO3) 2, Ba(NO3) 2 each into separate test tubes. Add three drops of 2M sodium hydroxide solution then excess (2/3 the length  of a standard test tube).

 

 

 

 

 

 

 

 

Observation Inference                              Explanation

 

No white precipitate Na+ and K+ Both Na+ and K+ ions react with OHfrom 2M sodium hydroxide solution to form soluble colourless  solutions

 

Na+(aq)  + OH(aq) ->   NaOH(aq)

K+(aq)  + OH(aq) ->   KOH(aq)

No white precipitate  then pungent smell of ammonia /urine NH4+ ions NH4+ ions react with 2M sodium hydroxide solution to produce pungent smelling ammonia gas

 

NH4+ (aq)  + OH(aq)  ->  NH3 (g) +  H2O(l)

 

White precipitate insoluble in excess Ba2+ ,Ca2+, Mg2+ ions Ba2+ ,Ca2+ and  Mg2+ ions react with OHfrom 2M sodium hydroxide solution to form insoluble white precipitate of their hydroxides.

 

Ba2+(aq)  + 2OH(aq)  ->   Ba(OH) 2(s)

Ca2+(aq)  + 2OH(aq)  ->   Ca(OH) 2(s)

Mg2+(aq)  + 2OH(aq)  ->   Mg(OH) 2(s)

 

White precipitate soluble in excess Zn2+ ,Pb2+, Al3+ ions Pb2+ ,Zn2+ and  Al3+ ions react with OHfrom 2M sodium hydroxide solution to form insoluble white precipitate of their hydroxides.

 

Zn2+(aq)  + 2OH(aq)  ->   Zn(OH) 2(s)

Pb2+(aq)  + 2OH(aq)  ->   Pb(OH) 2(s)

Al3+(aq)  + 3OH(aq)  ->   Al(OH) 3(s)

 

The hydroxides formed react with more OHions to form complex salts/ions.

 

Zn(OH) 2(s) + 2OH(aq)  -> [ Zn(OH) 4]2-(aq)

Pb(OH) 2(s) + 2OH(aq)  -> [ Pb(OH) 4]2-(aq)

Al(OH) 3(s) +  OH(aq)  -> [ Al(OH) 4](aq)

 

 

 

 

   
Blue precipitate insoluble in excess Cu2+ Cu2+ ions react with OHfrom 2M sodium hydroxide solution to form insoluble blue precipitate of copper(II) hydroxide.

 

Cu2+(aq)  + 2OH(aq)  ->   Cu(OH) 2(s)

 

Green precipitate insoluble in excess

 

On adding 3cm3 of hydrogen peroxide, brown/yellow solution formed

Fe2+

 

 

 

 

Fe2+ oxidized to Fe3+

Fe2+ ions react with OHfrom 2M sodium hydroxide solution to form insoluble green precipitate of Iron(II) hydroxide.

 

Fe2+(aq)  + 2OH(aq)  ->   Fe(OH) 2(s)

Hydrogen peroxide is an oxidizing agent that oxidizes  green Fe2+ oxidized to brown Fe3+

 

Fe(OH) 2(s)  +  2H+   ->   Fe(OH) 3(aq)

Brown precipitate insoluble in excess Fe3+ Fe3+ ions react with OHfrom 2M sodium hydroxide solution to form insoluble brown precipitate of Iron(II) hydroxide.

Fe3+(aq)  + 3OH(aq)  ->   Fe(OH) 3(s)

 

 

 

(b)Reaction of cation with aqueous ammonia

 

 

Aqueous ammonia precipitating reagent that can be used to identify the cations present in a salt.

Like NaOH/KOH the OH ion in NH4OH react with the cation to form a characteristic hydroxide .

Below are the observations ,inferences and explanations of the reactions of aqueous ammonia with salts from the following test tube reactions.

 

Procedure

 

Put about 2cm3 of  MgCl2, CaCl2, AlCl3, NaCl, KCl, FeSO4, Fe2(SO4) 3, CuSO4, ZnSO4NH4NO3, Pb(NO3) 2, Ba(NO3) 2 each into separate test tubes.

Add three drops of 2M aqueous ammonia then excess (2/3 the length  of a standard test tube).

 

Observation Inference                              Explanation

 

 

 

No white precipitate Na+ and K+ NH4+,Na+ and K+ ions react with OHfrom 2M aqueous ammonia to form soluble colourless  solutions

NH4+ (aq)   +  OH(aq)  ->   NH4+OH(aq)

Na+(aq)         +  OH(aq)  ->   NaOH(aq)

K+(aq)            +  OH(aq)  ->   KOH(aq)

 

White precipitate insoluble in excess Ba2+ ,Ca2+, Mg2+ ,Pb2+, Al3+,  ions Ba2+ ,Ca2+,Mg2+ ,Pb2+ and Al3+,  ions react with OHfrom 2M aqueous ammonia  to form insoluble white precipitate of their hydroxides.

 

Pb2+ (aq)  +  2OH(aq)   ->    Pb(OH) 2(s)

Al3+ (aq)  +  3OH(aq)   ->    Al(OH) 3(s)

Ba2+ (aq)  +  2OH(aq)   ->    Ba(OH) 2(s)

Ca2+ (aq)  +  2OH(aq)   ->    Ca(OH) 2(s)

Mg2+ (aq)  +  2OH(aq)  ->   Mg(OH) 2(s)

White precipitate soluble in excess Zn2+  ions Zn2+ ions react with OHfrom 2M aqueous ammonia to form insoluble white precipitate of Zinc hydroxide.

 

Zn2+(aq)  + 2OH(aq)  ->   Zn(OH) 2(s)

The Zinc hydroxides formed react NH3(aq) to form a complex salts/ions.

Zn(OH) 2(s) + 4NH3(aq)

->[ Zn(NH3) 4]2+(aq)+ 2OH(aq)

Blue precipitate that dissolves in excess ammonia solution to form a deep/royal blue solution Cu2+ Cu2+ ions react with OHfrom 2M aqueous ammonia to form blue precipitate of copper(II) hydroxide.

 

Cu2+(aq)  + 2OH(aq)  ->   Cu(OH) 2(s)

The copper(II) hydroxide formed react NH3(aq) to form a complex salts/ions.

 

Cu(OH) 2 (s) + 4NH3(aq)

->[ Cu(NH3) 4]2+(aq)+ 2OH(aq)

 

Green precipitate insoluble in excess.

 

On adding 3cm3 of hydrogen peroxide, brown/yellow solution formed

Fe2+

 

 

 

 

Fe2+ oxidized to Fe3+

Fe2+ ions react with OHfrom 2M aqueous ammonia to form insoluble green precipitate of Iron(II) hydroxide.

Fe2+(aq)  + 2OH(aq)  ->   Fe(OH) 2(s)

 

Hydrogen peroxide is an oxidizing agent that oxidizes  green Fe2+ oxidized to brown Fe3+

Fe(OH) 2(s)  +  2H+   ->   Fe(OH) 3(aq)

Brown precipitate insoluble in excess Fe3+ Fe3+ ions react with OHfrom 2M aqueous ammonia to form insoluble brown precipitate of Iron(II) hydroxide.

Fe3+(aq)  + 3OH(aq)  ->   Fe(OH) 3(s)

 

 

Note

(i) Only Zn2+ ions/salts form a white precipitate that dissolve in excess of both 2M sodium hydroxide and 2M aqueous ammonia.

 

(ii) Pb2+ and Al3+ ions/salts form a white precipitate that dissolve in excess of 2M sodium hydroxide but not in 2M aqueous ammonia.

 

(iii) Cu2+ ions/salts form a blue precipitate that dissolve to form a deep/royal blue solution in excess of 2M aqueous ammonia but only blue insoluble precipitate in 2M sodium hydroxide

 

(c)Reaction of cation with Chloride (Cl)ions

 

All chlorides are soluble in water except Silver chloride and Lead (II)chloride (That dissolve in hot water).When a soluble chloride like NaCl, KCl, NH4Cl is added to about 2cm3 of a salt containing Ag+ or Pb2+ions a white precipitate of AgCl  or  PbCl2 is formed. The following test tube reactions illustrate the above.

 

Experiment

Put about 2cm3 of silver nitrate(V) andLead(II)nitrate(V)solution into separate test tubes. Add five drops of NaCl /KCl / NH4Cl/HCl. Heat to boil.

 

 

 

Observation Inference                Explanation

 

(i)White precipitate does not dissolve on heating Ag+ ions  Ag+ ions reacts with Clions from a  soluble chloride salt to form a white precipitate of AgCl
(ii)White precipitate dissolve on heating Pb2+ ions Pb2+ ions reacts with Clions from a soluble chloride salt to form a white precipitate of PbCl2. PbCl2 dissolves on heating.

Note

Both Pb2+ and Al3+ ions  forms an insoluble white precipitate  in excess aqueous ammonia. A white precipitate on adding Clions/salts shows Pb2+.

No white precipitate on adding Clions/salts shows Al3+.

Adding a chloride/ Clions/salts can thus be used to separate the identity of Al3+ and Pb2+.

 

(d)Reaction of cation with sulphate(VI)/SO42- and sulphate(IV)/SO32-  ions

 

  All sulphate(VI) and sulphate(IV)/SO32-  ions/salts are soluble/dissolve in water except Calcium sulphate(VI)/CaSO4, Calcium sulphate(IV)/CaSO3, Barium sulphate(VI)/BaSO4, Barium sulphate(IV)/BaSO3, Lead(II) sulphate(VI)/PbSO4 and Lead(II) sulphate(IV)/PbSO3.When a soluble sulphate(VI)/SO42- salt like Na2SO4, H2SO4, (NH4)2SO4 or Na2SO3 is added to a salt containing  Ca2+, Pb2+, Ba2+ ions, a white precipitate is formed.

The following test tube experiments illustrate the above.

Procedure

Place about 2cm3 of Ca(NO3)2, Ba(NO3)2, BaCl2 and  Pb(NO3)2, in separate boiling tubes. Add six drops of sulphuric(VI)acid /sodium sulphate(VI)/ammonium sulphate(VI)solution. Repeat with six drops of sodium sulphate(IV).

 

Observation Inference                    Explanation
White precipitate Ca2+, Ba2+, Pb2+ ions  CaSO3 and CaSO4 do not form a thick precipitate as they are  sparingly soluble.

Ca2+(aq)+ SO32-(aq)   -> CaSO3(s)

Ca2+(aq)+ SO42-(aq)   -> CaSO4(s)

 

Ba2+(aq)+ SO32-(aq)   -> BaSO3(s)

Ba2+(aq)+ SO42-(aq)   -> BaSO4(s)

 

Pb2+(aq)+ SO32-(aq)   -> PbSO3(s)

Pb2+(aq)+ SO42-(aq)   -> PbSO4(s)

 

(e)Reaction of cation with carbonate(IV)/CO32-  ions

 

All carbonate salts are insoluble except sodium/potassium carbonate(IV) and ammonium carbonate(IV).

They dissociate /ionize to release CO32-  ions. CO32-  ions produce a white precipitate when the soluble carbonate salts is  added to any metallic cation.

 

Procedure 

Place about 2cm3 of Ca(NO3)2, Ba(NO3)2, MgCl2 ,Pb(NO3)2 andZnSO4 in separate boiling tubes.

Add six drops of Potassium /sodium carbonate(IV)/ ammonium  carbonate (IV)solution.

 

Observation Inference Explanation
Green precipitate Cu2+ ,Fe2+,ions

 

CO32-(aq)

Copper(II)carbonate(IV) and Iron(II) carbonate (IV) are precipitated as insoluble green precipitates.

 

Cu2+(aq)+ CO32-(aq)   -> CuCO3(s)

Fe2+(aq)+ CO32-(aq)   -> FeCO3(s)

 

When sodium carbonate(IV)is added to CuCO3(s)  the CO32-(aq) ions  are first hydrolysed to produce CO2(g) and OH(aq)ions.

 

CO32-(aq)  +  H2O (l)  ->  CO2 (g)  + 2OH (aq)

The OH(aq) ions further react to form basic copper(II) carbonate(IV). Basic copper(II) carbonate(IV) is the only green salt of copper.

Cu2+(aq)+ CO32-(aq)+2OH (aq)

->CuCO3.Cu(OH)2 (s)

 

 

White precipitate CO32- White ppt of the carbonate(IV)salt is precipitated

Ca2+(aq)    +    CO32- (aq)    ->   CaCO3(s)

Mg2+(aq)   +    CO32- (aq)    ->   MgCO3(s)

Pb2+(aq)     +    CO32- (aq)    ->   PbCO3(s)

Zn2+(aq)     +    CO32- (aq)      ->   ZnCO3(s)

Note

(i)Iron(III)carbonate(IV) does not exist.

(ii)Copper(II)Carbonate(IV) exist only as the basic CuCO3.Cu(OH) 2

(iii)Both  BaCO3 and BaSO3 are insoluble white precipitate. If hydrochloric acid is added to the white precipitate;

  1. BaCO3 produces CO2 gas. When bubbled/directed into lime water solution,a white precipitate is formed.
  2. I. BaSO3 produces SO2 gas. When bubbled/directed into orange acidified potassium dichromate(VI) solution, it turns to green/decolorizes acidified potassium manganate(VII).

 

(f) Reaction of cation with sulphide / S2-  ions

All sulphides are insoluble black solids/precipitates except sodium sulphide/ Na2S/ potassium sulphide/K2S.When a few/3drops of the soluble sulphide is added to a metal cation/salt, a black precipitate is formed.

Procedure 

Place about 2cm3 of Cu(NO3)2, FeSO4, MgCl2,Pb(NO3)2 and ZnSO4 in separate boiling tubes.

Add six drops of Potassium /sodium sulphide solution.

 

Observation Inference                              Explanation
Black ppt S2- ions CuS, FeS,MgS,PbS, ZnS are black insoluble precipitates

Cu2+(aq)   +   S2-(aq)   ->   CuS(s)

Pb2+(aq)   +   S2-(aq)   ->   PbS(s)

Fe2+(aq)   +   S2-(aq)   ->   FeS(s)

Zn2+(aq)   +   S2-(aq)   ->   ZnS(s)

 

Sample qualitative analysis guide

You are provided with solid Y(aluminium (III)sulphate(VI)hexahydrate).Carry out the following tests and record your observations and inferences in the space provided.

1(a) Appearance

Observations                                                inference                       (1mark)

 

White crystalline solid                                   Coloured ions Cu2+ , Fe2+ ,Fe3+ absent

 

 

 

(b)Place about a half spatula full of the solid into a clean dry boiling tube. Heat gently then strongly.

Observations                                                            inference                       (1mark)

 

Colourless droplets formed on the cooler             Hydrated compound/compound

part of the test tube                                              containing water of crystallization

Solid remains a white residue

 

(c)Place all the remaining portion of the solid in a test tube .Add about 10cm3 of distilled water. Shake thoroughly. Divide the mixture into five portions.

 

Observation                                                   Inference     (1mark)

Solid dissolves to form                                  Polar soluble compound

a colourless solution                                      Cu2+ , Fe2+ ,Fe3+ absent

 

(i)To the first portion, add three drops of sodium hydroxide then add excess of the alkali.

 

Observation                                                   Inference     (1mark)

White ppt, soluble in excess                                     Zn2+ , Pb2+ , Al3+

 

(ii)To the second portion, add three drops of aqueous ammonia then add excess of the alkali.

 

Observation                                                   Inference     (1mark)

White ppt, insoluble in excess                                        Pb2+ , Al3+

 

(iii)To the third portion, add three drops of sodium sulphate(VI)solution.

 

Observation                                                   Inference     (1mark)

No white ppt                                                              Al3+

 

(iv)I.To the fourth portion, add three drops of Lead(II)nitrate(IV)solution. Preserve

 

Observation                                                   Inference     (1mark)

White ppt                                                         CO32-,  SO42-,  SO32-,  Cl,

 

II.To the portion in (iv) I above , add five drops of  dilute hydrochloric acid.

 

Observation                                                   Inference     (1mark)

White ppt persist/remains                                         SO42-,  Cl,

 

III.To the portion in (iv) II above, heat to boil.

 

Observation                                                   Inference     (1mark)

White ppt persist/remains                                         SO42-,

 

 

Note that:

(i)From test above, it can be deduced that solid Y is hydrated aluminium(III)sulphate(VI) solid

 

(ii)Any ion inferred from an observation below must be derived from previous correct observation and inferences above. e.g.

Al3+ in c(iii) must be correctly inferred in either/or  in c(ii) or c(i)above

SO42- in c(iv)III must be correctly inferred in either/or  in c(iv)II or c(iv)I above

 

(iii)Contradiction in observations and inferences should be avoided.e.g.

“White ppt soluble in excess” to infer presence of Al3+ ,Ba2+ ,Pb3+

 

(iv)Symbols of elements/ions should be correctly capitalized. e.g.

“SO4-2” is wrong, “sO42-” is wrong, “cu2+” is wrong.

 

Sample solutions of salt were labeled as I,II, III and IV.  The actual solutions, not in that order are lead nitrate, zinc sulphate potassium chloride and calcium chloride.

 

a)When aqueous sodium carbonate was added to each sample separately, a white precipitate was formed in I, III and IV only.  Identify solution II.

 

b)When excess sodium hydroxide was added to each sample separately, a white precipitate was formed in solutions III and I only.

 

                    Identify solution I

 

 

17.When solids/salts /solutes are added to a solvent ,some dissolve to form a solution.

Solute          +        Solvent        ->       Solvent

 

If a solution has a lot of solute dissolved in a solvent ,it is said to be concentrated.

 

If a solution has little solute dissolved in a solvent ,it is said to be dilute.

 

There is a limit to how much solute can dissolve in a given /specified amount of solvent/water at a given /specified temperature.

 

The maximum mass of salt/solid/solute that dissolve  in 100g of solvent/water at a specified temperature is called solubility of a salt.

 

When no more solute can dissolve in a given amount of solvent at a specified temperature, a saturated solution is formed.

 

For some salts, on heating, more of the salt/solid/solute dissolve in the saturated solution to form a super saturated solution.

 

The solubility of a salt is thus calculated from the formula

 

Solubility = Mass of solute/salt/solid   x 100

                    Mass/volume of water/solvent

 

Practice examples

(a)Calculate the solubility of potassium nitrate(V) if 5.0 g of the salt is dissolved in 50.0cm3 of water.

 

Solubility = Mass of solute/salt/solid x 100   =>( 5.0 x 100 )   = 10.0 g /100g H2O

Mass/volume of water/solvent             50.0

 

(b)Calculate the solubility of potassium chlorate(V) if 50.0 g of the salt is dissolved in 250.0cm3 of water.

 

Solubility = Mass of solute/salt/solid x 100   =>( 50.0 x 100 )   = 20.0 g /100g H2O

Mass/volume of water/solvent             250.0

 

(c)If the solubility of potassium chlorate(V) is 5g/100g H2O at 80oC,how much can dissolve in 5cm3 of water at 80oC .

 

Mass of solute/salt/solid = Solubility  x Mass/volume of water/solvent

100

=>      5  x    5  =      0.25g of KClO3 dissolve

100

 

(d)If the solubility of potassium chlorate(V) is 72g/100g H2O at 20oC,how much can saturate 25g of water at 20oC .

 

Mass of solute/salt/solid = Solubility  x Mass/volume of water/solvent

100

=>      72  x    25  =    18.0g of KClO3 dissolve/saturate

100

 

(e) 22g of potassium nitrate(V) was dissolved in 40.0g of water at 10oC. Calculate the solubility of potassium nitrate(V) at 10oC.

 

Solubility = Mass of solute/salt/solid x 100   =>( 22 x 100 )   = 55.0 g /100g H2O

Mass/volume of water/solvent             40.0.

 

(f)What volume of water should be added to 22.0g of water at 10oC if the solubility of KNO3 at 10oC is 5.0g/100g H2O?

 

 Solubility is mass/100g H2O => 22.0g + x  = 100cm3/100g H2O

X= 100 – 22  = 78 cm3 of H2O

 

  1. A graph of solubility against temperature is called solubility curve.

It shows the influence of temperature on solubility of different substances/solids/salts.

Some substances dissolve more with increase in temperature while for others dissolve less with increase in temperature

 

 

 

  Solubility

/100g water

Temperature (0C)
unsaturated  solution of KClO3

KClO3

KClO3

 

Saturated  solution of KClO3
NaCl
KClO3
KNO3
B
A

 

Note:

(i)solubility of KNO3 and KClO3 increase with increase in temperature.

 

(ii)solubility of KNO3 is always higher than that of KClO3 at any specified temperature.

 

(iii)solubility of NaCl decrease with increase in temperature.

(iv)NaCl has the highest solubility at low temperature while KClO3 has the lowest solubility at low temperature.

 

(v)At point A both NaCl and KNO3 are equally soluble.

 

(vi)At point B both NaCl and KClO3 are equally soluble.

 

(vii) An area above the solubility curve of the salt shows a saturated /supersaturated solution.

 

(viii) An area below the solubility curve of the salt shows an  unsaturated solution.

 

19.(a) For salts whose solubility increases with increase in temperature, crystals form when the salt solution at higher temperatures is cooled to a lower temperature.

 

  • For salts whose solubility decreases with increase in temperature, crystals form when the salt solution at lower temperatures is heated to a higher temperature.

 

The examples below shows determination of the mass of crystals deposited with changes in temperature.

 

1.The solubility of KClO3   at 100oC is 60g/100g water .What mass of KClO3  will be deposited at:

 

(i)75 oC if the solubility is now 39g/100g water.

At 100oC                                    =   60.0g

Less at 75oC                              = – 39.0g

Mass of  crystallized out                 21.0g

 

(i)35 oC if the solubility is now 28 g/100g water.

At 100oC                                    =   60.0g

Less at 35oC                              = – 28.0.0g

Mass of  crystallized out                 32.0g

 

  1. KNO3 has a solubility of 42 g/100g water at 20oC.The salt was heated and added 38g more of the solute which dissolved at100oC. Calculate the solubility of KNO3 at 100oC.

 

Solubility of KNO3 at 100oC  =  solubility at 20oC  +  mass of KNO3 added

=> 42g + 38g = 80g KNO3 /100g H2O

 

  1. A salt solution has a mass of 65g containing 5g of solute. The solubility of this salt is 25g per 100g water at 20oC. 60g of the salt are added to the solution at 20oC.Calculate the mass of the solute that remain undissolved.

Mass of solvent at 20oC = mass of solution – mass of solute

=>         65    –    5     =    60g

 

Solubility before adding salt = mass of solute x 100

                                                  Volume of solvent

 

=> 5 x 100                     =   8.3333g/100g water

60

 

Mass of solute to equalize with solubility =  25  –  8.3333g   = 16.6667g

 

Mass of solute undissolved    =   60.0  –  16.6667g   =  43.3333 g

 

  1. Study the table below

 

Salt

 

 

Solubility in gram at
        50oC          20oC
KNO3         90           30
KClO3         20            6

 

(i)What happens when the two salts are dissolved in water then cooled from 50oC to 20oC.

 

(90 – 30) = 60.0 g of  KNO3 crystals precipitate

(20 – 6) = 14.0 g of  KClO3 crystals precipitate

 

(ii)State the assumption made in (i) above.

Solubility of one salt has no effect on the solubility of the other.

 

  1. 5. 0 g of hydrated potassium carbonate (IV) K2CO3.xH2O on heating leave 7.93 of the hydrate.

 

(a)Calculate the mass of anhydrous salt obtained.

          Hydrated on heating leave anhydrous   =  7.93 g

 

(b)Calculate the mass of water of crystallization in the hydrated salt

 

          Mass of water of crystallization = hydrated – anhydrous

=> 10.0   –  7.93  =  2.07 g

(c)How many moles of anhydrous salt are there in 10of hydrate? (K= 39.0,C=12.0.O= 16.0)

 

Molar mass K2CO3= 138

Moles K2CO3 = mass of  K2CO3        =>       7.93    =   0.0515 moles

Molar mass K2CO3                      138

 

(d)How many moles of water are present in the hydrate for every one mole of K2CO3 ? (H=1.0.O= 16.0)

 

Molar mass H2O  = 18

Moles H2 = mass of  H2O        =>       2.07    =   0.115 moles

Molar mass H2O                 18

Mole ratio H2O : K2CO3       =          0.115 moles              2       =  2

0.0515 moles            1

 

(e)What is the formula of the hydrated salt?

K2CO3 .2 H2O

 

  1. The table below shows the solubility of Potassium nitrate(V) at different temperatures.

 

Temperature(oC) 5.0 10.0 15.0 30.0 40.0 50.0 60.0
mass KNO3/ 100g water 15.0 20.0 25.0 50.0 65.0 90.0 120.0

 

(a)Plot a graph of mass of  in 100g water(y-axis) against temperature in oC

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(b)From the graph show and determine

(i)the mass of KNO3 dissolved at:

  1. 20oC

From a correctly plotted graph = 32g

  1. 35oC

From a correctly plotted graph = 57g

                    III. 55oC

From a correctly plotted graph = 104g

          (ii)the temperature at which the following mass of KNO3 dissolved:

  1. 22g

From a correctly plotted graph =13.0oC

  1. 30g

From a correctly plotted graph =17.5oC

                    III.100g

From a correctly plotted graph =54.5oC

 

(c)Explain the shape of your graph.

Solubility of KNO3 increase with increase in temperature/More KNO3 dissolve as temperature rises.

 

(d)Show on the graph the supersaturated and unsaturated solutions.

Above the solubility curve write; “supersaturated”

          Below the solubility curve write; “unsaturated”

 

(e)From your graph, calculate the amount of crystals obtained when a saturated solution of KNO3 containing 180g of the salt is cooled from 80oC to:

 

  1. 20oC

Solubility before heating                                 = 180 g

Less   Solubility after heating(from the graph)          =  32 g

Mass of KNO3crystals                                   =  148 g

 

  1. 35oC

Solubility before heating                                 = 180 g

Less   Solubility after heating(from the graph)          =  58 g

Mass of KNO3crystals                                   =  122 g

 

          III. 55oC

Solubility before heating                                 = 180 g

Less   Solubility after heating(from the graph)          =  102 g

Mass of KNO3crystals                                   =  78 g

 

  1. The table below shows the solubility of salts A and B at various temperatures.

 

Temperature(oC) 0.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0

 

Solubility of A 28.0 31.0 34.0 37.0 40.0 43.0 45.0 48.0 51.0

 

Solubility of B 13.0 21.0 32.0 46.0 64.0 85.0 110.0 138.0 169.0

 

 

(a)On the same axis plot a graph of solubility (y-axis) against temperature for each salt.

(b)At what temperature are the two salts equally soluble.

The point of intersection of the two curves = 24oC

 

(c)What happens when a mixture of 100g of salt B with 100g if water is heated to 80oC

From the graph, the solubility of B at 80oC is 169g /100g water. All the 100g crystals of B dissolve.

 

(d)What happens when the mixture in (c) above is then cooled from 50oC to 20oC.

Method I.

                                 Total mass  before cooling at 50oC                       =  100.0 g

(From graph)  Solubility/mass  after cooling at 20oC         =    32.0 g

Mass of crystals deposited                        68.0 g

 

Method II.

Mass of soluble salt crystals at 50oC added           = 100   g

(From graph)Solubility/mass before cooling at 50oC                =  85.0 g

Mass of crystals that cannot dissolve at 50oC               15.0  g

 

(From graph) Solubility/mass  before cooling at 50oC              =  85.0 g

(From graph)  Solubility/mass  after cooling at 20oC                =   32.0 g

Mass of crystals deposited after cooling                               53.0 g

 

Total mass of crystals deposited = 15.0 + 53.0        = 68.0 g

 

(e)A mixture of 40g of A and 60g of B is added to 10g of water and heated to 70oC.The solution is then allowed to cool to 10oC.Describe clearly what happens.

 

I.For salt A

Solubility of A before heating = mass of A   x  100 

Volume of water added

=> 40    x   100      =    400g/100g Water

                                                                         10

(Theoretical)Solubility of A before heating                       =  400 g

Less   (From graph ) Solubility of A after  heating at 70oC          =    48g

Mass of crystals that can not dissolve at70oC                       = 352 g

 

(From graph ) Solubility of A after  heating at 70oC          =    48g

Less  (From graph ) Solubility of A after  cooling to 10oC          =    31g

Mass of crystals that crystallize out on cooling  to10oC      =   17 g

 

Mass of crystals that can not dissolve at70oC                       = 352 g

Add    Mass of crystals that crystallize out on cooling  to10oC      =   17 g

Total mass of A that does not dissolve/crystallize/precipitate   = 369 g

 

I.For salt B

Solubility of B before heating = mass of B   x  100 

Volume of water added

=> 60    x   100      =    600g/100g Water

                                                                         10

 

(Theoretical)Solubility of B before heating                        =   600 g

Less   (From graph ) Solubility of B after  heating at 70oC          =   138g

Mass of crystals that cannot dissolve at70oC                        = 462 g

 

(From graph ) Solubility of B after  heating at 70oC          =    138g

Less  (From graph ) Solubility of B after  cooling to 10oC          =     21g

Mass of crystals that crystallize out on cooling  to10oC      =   117 g

 

Mass of crystals that cannot dissolve at70oC                        =   462 g

Add    Mass of crystals that crystallize out on cooling  to10oC      =   117 g

Total mass of A that does not dissolve/crystallize/precipitate   =   579 g

 

(f)State the assumption made in (e)above

Solubility of one salt has no effect on the solubility of the other

 

  1. When 5.0 g of potassium chlorate (V) was put in 10cm3 of water and heated, the solid dissolves. When the solution was cooled , the temperature at which crystals reappear was noted. Another 10cm3 of water was added and the mixture heated to dissolve then cooled for the crystals to reappear .The table below shows the the results obtained

 

Total volume of water added(cm3) 10.0 20.0 30.0 40.0 50.0
Mass of KClO3 5.0 5.0 5.0 5.0 5.0
Temperature at which crystals appear 80.0 65.0 55.0 45.0 30.0
Solubility of KclO3 50.0 25.0 16.6667 12.5 10.0

 

(a)Complete the table to show the solubility of KclO3 at different temperatures.

 

(b)Plot a graph of mass of  KClO3 per 100g water against temperature at  which crystals form.

 

(c)From the graph, show and determine ;

 (i)the solubility of KClO3 at

  1. 50oC

From a well plotted graph = 14.5 g KClO3/100g water

  1. 35oC

 

From a well plotted graph = 9.0 g KclO3/100g water

(ii)the temperature at which the solubility is:

 

I.10g/100g water

From a well plotted graph = 38.0 oC

II.45g/100g water

From a well plotted graph = 77.5 oC

 

(d)Explain the shape of the graph.

Solubility of KClO3 increase with increase in temperature/more KclO3dissolve as temperature rises.

 

(e)What happens when 100g per 100g water is cooled to 35.0 oC

                                        Solubility before heating = 100.0

(From the graph)   Solubility after cooling    =    9.0

    Mass of salt precipitated/crystallization =   91.0 g

 

  1. 25.0cm3 of water dissolved various masses of ammonium chloride crystals at different temperatures as shown in the table below.

 

Mass of ammonium chloride(grams) 4.0 4.5 5.5 6.5 9.0

 

Temperature at which solid dissolved(oC) 30.0 50.0 70.0 90.0 120.0

 

Solubility of NH4Cl 16.0 18.0 22.0 26.0 36.0

 

 

(a)Complete the table

(b)Plot a solubility curve

 

(c)What happens when a saturated solution of ammonium chloride is cooled from 80oC to 40oC.

 

                    (From the graph )Solubility at 80oC  = 24.0 g

Less   (From the graph )Solubility at 40oC  = 16.8 g

Mass of crystallized/precipitated     =   7.2 g

 

  1. Solubility and solubility curves are therefore used

       (i) to know the effect of temperature on the solubility of a salt

(ii)to fractional crystallize two soluble salts by applying their differences in   solubility at different temperatures.

(iii)determine the mass of crystal that is obtained from crystallization.

 

21.Natural fractional crystallization takes place in Kenya/East Africa at:

(i)  Lake Magadi during extraction of soda ash(Sodium carbonate) from Trona(sodium sesquicarbonate)

(ii) Ngomeni near Malindi at the Indian Ocean Coastline during the extraction of common salt(sodium chloride).

 

22.Extraction of soda ash from Lake Magadi in Kenya

Rain water drains underground in the great rift valley and percolate underground where it is heated geothermically.

The hot water dissolves underground soluble sodium compounds and comes out on the  surface as  alkaline springs  which are found around the edges of Lake Magadi in Kenya.

Temperatures around the lake are very high (30-40oC) during the day.

The solubility of trona decrease with increase in temperature  therefore solid crystals of trona grows on top of the lake (upto or more than 30metres thick)

 

A bucket dredger mines the trona which is then crushed ,mixed with lake liquor and pumped to washery plant where it is further refined to a green granular product called CRS.

The CRS is then heated  to chemically decompose trona to soda ash(Sodium carbonate)

 

Chemical equation

2Na2CO3.NaHCO3.2H2O(s)  ->  3Na2CO3 (s)  +  CO2(g)  +  5H2O(l)

 

Soda ash(Sodium carbonate) is then stored .It is called Magadi Soda. Magadi Soda is used :

  • make glass
  • for making soapless detergents
  • softening hard water.

Common salt is colledcted at night because its solubility decreases with decrease in temperature. It is used as salt lick/feed for animals.

 

Summary flow diagram showing the extraction of Soda ash  from Trona

 

Sodium chloride and Trona dissolved in the sea
Natural fractional crystallization
Crystals of Trona (Day time)
Crystals of sodium chloride(At night)
Dredging /scooping/ digging
Crushing
Furnace (Heating)
                 Carbon(IV) oxide
Soda ash
Bagging

NaCl(s)

Bagging Na2CO3 (s)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

23.Extraction of common salt from Indian Ocean at Ngomeni in Kenya

 

Oceans are salty.They contain a variety of dissolved salts (about 77% being sodium chloride).

During high tide ,water is collected into shallow pods and allowed to crystallize as evaporation takes place.The pods are constructed in series to increase the rate of evaporation.

At the final pod ,the crystals are scapped together,piled in a heap and washed with brine (concentrated sodium chloride).

It contains MgCl2 and CaCl2 . MgCl2 and CaCl2are hygroscopic. They absorb water from the atmosphere and form a solution.

This makes table salt damp/wet on exposure to the atmosphere.

 

24.Some water form lather easily with soap while others do not.

Water which form lather easily with soap is said to be “soft

Water which do not form lather easily with soap is said to be “hard

Hardness of water is caused by the presence of  Ca2+ and Mg2+ ions.

Ca2+ and Mg2+ ions react with soap to form an insoluble grey /white suspension/precipitate called Scum/ curd. Ca2+ and Mg2+ ions in water come from the water sources passing through rocks containing soluble salts of Ca2+ and Mg2+ e.g. Limestone or gypsum

There are two types of water hardness:

(a)temporary hardness of water

(b)permanent  hardness of water

 

(a)temporary hardness of water

Temporary hardness of water is caused by the presence of dissolved calcium hydrogen carbonate/Ca(HCO3)2 and magnesium hydrogen carbonate/Mg(HCO3)2

When rain water dissolve carbon(IV) oxide from the air it forms waek carbonic(IV) acid i.e.

CO2(g)     +      H2O(l)    ->      H2CO3(aq)

When carbonic(IV) acid passes through limestone/dolomite rocks it reacts to form soluble salts i.e.

In limestone areas;  H2CO3(aq)    +   CaCO3(s)   ->    Ca(HCO3)2 (aq)

In dolomite areas;    H2CO3(aq)   +   MgCO3(s)   ->    Mg(HCO3)2 (aq)

 

(b)permanent hardness of water

Permanent hardness of water is caused by the presence of dissolved calcium sulphate(VI)/CaSO4 and magnesium sulphate(VI)/Mg SO4 Permanent hardness of water is caused by water dissolving  CaSO4 and MgSO4  from ground rocks.

 

Hardness of water can be removed by the following methods:

 

(a)Removing temporary hardness of water

 

(i)Boiling/heating.

 

Boiling decomposes insoluble calcium hydrogen carbonate/Ca(HCO3)2 and magnesium hydrogen carbonate/Mg(HCO3)2 to insoluble CaCO3  and MgCO3 that precipitate away. i.e

Chemical equation 

Ca(HCO3)2(aq)     ->   CaCO3 (s)   +   CO2(g)    +   H2O(l)

Mg(HCO3)2(aq)     ->   MgCO3 (s)   +   CO2(g)    +   H2O(l)

 

(ii)Adding sodium carbonate (IV) /Washing soda.

 

Since boiling is expensive on a large scale ,a calculated amount of sodium carbonate decahydrate /Na2CO3.10H2precipitates insoluble Ca2+(aq) and Mg2+(aq) ions as carbonates to remove both temporary and permanent hardness of water .This a double decomposition reaction where two soluble salts form an insoluble and soluble salt. i.e.

 

(i)with temporary hard water

Chemical equation

Na2CO3   (aq)     +  Ca(HCO3) 2 (aq)    ->   NaHCO3(aq)  + CaCO3 (s)

Na2CO3   (aq)     +  Mg(HCO3) 2 (aq)    ->   NaHCO3(aq) + MgCO3 (s)

Ionic equation

           CO32-   (aq)      +       Ca2+  (aq)     ->     CaCO3 (s)

           CO32-   (aq)      +       Mg2+  (aq)     ->     MgCO3 (s)

 

(ii)with permanent  hard water

Chemical equation

     Na2CO3   (aq)     +  MgSO4  (aq)    ->   Na2SO4 (aq) + MgCO3 (s)

               Na2CO3   (aq)     +  CaSO4  (aq)      ->   Na2SO4 (aq) + MgCO3 (s)

Ionic equation

           CO32-   (aq)      +       Ca2+  (aq)     ->     CaCO3 (s)

                    CO32-   (aq)      +       Mg2+  (aq)     ->     MgCO3 (s)

 

(iii)Adding calcium (II)hydroxide/Lime water

 

Lime water/calcium hydroxide removes only temporary hardness of water  from by precipitating insoluble calcium carbonate(IV).

Chemical equation

Ca(OH)2   (aq)     +  Ca(HCO3) 2 (aq)    ->   2H2O(l)  + 2CaCO3 (s)

 

Excess of Lime water/calcium hydroxide should not be used because it dissolves again to form soluble calcium hydrogen carbonate(IV) causing the hardness again.

         

(iv)Adding aqueous ammonia

Aqueous ammonia removes  temporary hardness of water by precipitating insoluble calcium carbonate(IV)  and magnesium carbonate(IV)

 

Chemical equation

2NH3  (aq)     +  Ca(HCO3) 2 (aq)    ->   (NH4) 2CO3(aq)  + CaCO3 (s)

          2NH(aq)     +  Mg(HCO3) 2 (aq)    ->   (NH4) 2CO3(aq)  + MgCO3 (s)

(v)Use of ion-exchange permutit

 

This method involves packing a chamber with a resin made of insoluble complex of sodium salt called sodium permutit.

The sodium permutit releases sodium ions that are exchanged with Mg2+ and Ca2+ ions in hard water making the water to be soft. i.e.

 

Na2X(aq)  +  Ca2+ (aq)   ->   Na+ (aq)  +   CaX(s)

Na2X(aq)  +  Mg2+ (aq)   ->  Na+ (aq)  +   MgX(s)

 

 

 

 

 

Hard water containing Mg2+ and Ca2+    

 

 

 

 

 

 

 

Ion exchange resin as

Sodium permutit

 

 

 

 

 

 

 

 

 

 

 

——-             Na+ ions replace Mg2+

                  and Ca2+ to make the water soft.

When all the Na+ ions in the resin is fully exchanged with Ca2+ and Ng2+ ions in the permutit column ,it is said to be exhausted.

Brine /concentrated sodium chloride solution is passed through the permutit column to regenerated /recharge the column again.

 

Hard water containing Mg2+ and Ca2+

 

 

 

 

 

 

 

Ion exchange resin as

Sodium permutit

 

 

 

 

 

 

 

 

 

 

 

——-             Na+ ions replace Mg2+

                  and Ca2+ to make the water soft.

 

(vi)Deionization /demineralization

 

This is an advanced ion exchange method of producing deionized water .Deionized water is extremely pure water made only of hydrogen and oxygen only without any dissolved substances.

Deionization involve using the resins that remove all the cations by using:

(i)A cation exchanger which  remove /absorb all the cations present in water and leave only H+ ions.

(ii)An anion exchanger which  remove /absorb all the anions present in water and leave only OH ions.

The H+(aq) and OH(aq)  neutralize each other to form pure water.

 

Chemical equation

H+(aq)   +   OH(aq)   ->     H2O(l)

When exhausted the cation exchanger is regenerated by adding H+(aq) from sulphuric(VI)acid/hydrochloric acid.

When exhausted the anion exchanger is regenerated by adding OH(aq) from sodium hydroxide.

 

 Advantages of hard water

Hard water has the following advantages:

(i)Ca2+(aq) in hard water are useful in bone and teeth formation

(ii) is good for brewing beer

(iii)contains minerals that cause it to have better /sweet taste

(iv)animals like snails and coral polyps use calcium to make their shells and coral reefs respectively.

(v)processing mineral water

 

Disadvantages of hard water

 

Hardness of water:

(i)waste a lot of soap during washing before lather is formed.

(ii)causes  stains/blemishes/marks on clothes/garments

(iii)causes fur on electric appliances like kettle ,boilers and pipes form    decomposition of carbonates on heating .This reduces their efficiency hence more/higher cost of power/electricity.

 

 

 

 

Sample revision questions

 

In an experiment, soap solution was added to three separate samples of water. The table below shows the volumes of soap solution required to form lather with 1000cm3 of each sample of water before and after boiling.

  Sample I Sample II Sample III
Volume of soap before water is boiled (cm3) 27.0 3.0 10.0
Volume of soap after water is boiled(cm3) 27.0 3.0 3.0

 

  1. a) Which water sample is likely to be soft? Explain.    (2mks)

Sample II: Uses little sample of soap .

  1. c) Name the change in the volume of soap solution used in sample III          (1mk)

On heating the sample water become soft bcause it is temporary hard.

 

2.Study the scheme below and use it to aanswer the questions that follow:

 

(a)Write the formula of:

(i)Cation in solution K

Al3+

(ii)white ppt  L

          Al(OH)3

 

(iii) colourless solution  M

          [Al(OH)4]

 

(iv) colourless solution  N

AlCl3

 

(v)white ppt  P

          Al(OH)3

 

(b)Write the ionic  equation for the reaction for the formation of:

(i)white ppt  L

Al3+(aq)      +  3OH(aq)     ->   Al(OH)3(s)

 

(v)white ppt  P

          Al3+(aq)      +  3OH(aq)     ->   Al(OH)3(s)

 

(c)What property  is illustrated  in the formation of colourless solution M and  N.

Amphotellic

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

UPGRADE

CHEMISTRY

                

FORM 4

Thermochemistry

      

 

                       Comprehensive tutorial notes

 

 

 

MUTHOMI S.G

www.kcselibrary.info

                                          0720096206

 

 

 

 

 

 

                    1.Introduction to Energy changes

Energy is the capacity to do work. There are many/various forms of energy like heat, electric, mechanical, and/ or chemical energy.There are two types of energy:

(i)Kinetic Energy(KE) ;the energy in motion.

(ii)Potential Energy(PE); the stored/internal energy.

Energy like matter , is neither created nor destroyed but can be transformed /changed from one form to the other/ is interconvertible. This is the principle of conservation of energy. e.g. Electrical energy into heat through a filament in bulb.

Chemical and physical processes take place with absorption or evolution/production of energy mainly in form of heat

The study of energy changes that accompany physical/chemical reaction/changes is called Thermochemistry. Physical/chemical reaction/changes that involve energy changes are called thermochemical reactions. The SI unit of energy is the Joule(J).Kilo Joules(kJ)and megaJoules(MJ) are also used. The Joule(J) is defined as the:

(i) quantity of energy transferred when a force of one newton acts through a distance of one metre.

(ii) quantity of energy transferred when one coulomb of electric charge is passed through a potential difference of one volt.

All thermochemical reactions  should be carried out  at standard conditions of:

(i) 298K /25oC temperature

(ii)101300Pa/101300N/m2 /760mmHg/1 atmosphere pressure.

 

2.Exothermic and endothermic processes/reactions

 

Some reactions / processes take place with evolution/production of energy. They are said to be exothermic while others take place with absorption of energy. They are said to be endothermic.

Practically exothermic reactions / processes cause a rise in temperature (by a rise in thermometer reading/mercury or alcohol level rise)

Practically endothermic reactions / processes cause a fall in temperature (by a fall in thermometer reading/mercury or alcohol level decrease)

To demonstrate/illustrate exothermic and endothermic processes/reactions

 

  1. Dissolving Potassium nitrate(V)/ammonium chloride crystals

 

Procedure:

Measure 20cm3 of water in a beaker. Determine and record its temperature T1.Put about 1.0g of Potassium nitrate(V) crystals into the beaker. Stir the mixture carefully and note the highest temperature rise /fall T2.Repeat the whole procedure by using ammonium chloride in place of Potassium nitrate (V) crystals.

Sample results

 

Temperture (oC) Using Potassium nitrate(V) crystals Using Ammonium chloride crystals
T2(Final temperature)            21.0                23.0
T1 (Initial temperature)            25.0                26.0
Change in temperature(T2 –T1)             4.0                 3.0

Note:

(i)Initial(T1) temperature of dissolution of both potassium nitrate(V) crystals and ammonium chloride crystals is higher than the final temperature(T2)

(ii) Change in temperature(T2 –T1) is not a mathematical “-4.0” or “-3.0”.

(iii)Dissolution of both potassium nitrate(V)  and ammonium chloride crystals is an endothermic process because initial(T1) temperature is higher than the final temperature(T2) thus causes a fall/drop in temperature.

 

  1. Dissolving concentrated sulphuric(VI) acid/sodium hydroxide crystals

 

Procedure:

Measure 20cm3 of water in a beaker. Determine and record its temperature T1.Carefully put about 1.0g/four pellets of sodium hydroxide crystals into the beaker. Stir the mixture carefully and note the highest temperature rise /fall T2.Repeat the whole procedure by using 2cm3 of concentrated sulphuric(VI) acid in place of sodium hydroxide crystals.

CAUTION:

(i)Sodium hydroxide crystals are caustic and cause painful blisters on contact with skin.

(ii) Concentrated sulphuric (VI) acid is corrosive and cause painful wounds on contact with skin.

 

Sample results

Temperture (oC) Using Sodium hydroxide pellets Using Concentrated sulphuric(VI) acid
T2(Final temperature)            30.0                32.0
T1 (Initial temperature)            24.0                25.0
Change in temperature(T2 –T1)             6.0                 7.0

 

Note:

(i)Initial (T1) temperature of dissolution of both concentrated sulphuric (VI) acid and sodium hydroxide pellets is lower than the final temperature (T2).

(ii)Dissolution of both Sodium hydroxide pellets and concentrated sulphuric (VI) acid is an exothermic process because final (T2) temperature is higher than the initial temperature (T1) thus causes a rise in temperature.

The above reactions show heat loss to and heat gain from the surrounding as illustrated by a rise and fall in temperature/thermometer readings.

Dissolving both potassium nitrate(V)  and ammonium chloride crystals causes heat gain from the surrounding that causes fall in thermometer  reading.

Dissolving both Sodium hydroxide pellets and concentrated sulphuric (VI) acid causes heat loss to the surrounding that causes rise in thermometer reading.

 

At the same temperature and pressure ,heat absorbed and released is called enthalpy/ heat content denoted H.

Energy change is measured from the heat content/enthalpy of the final and initial products. It is denoted ∆H(delta H).i.e.

Enthalpy/energy/ change in heat content ∆H = Hfinal – Hinitial

 

For chemical reactions:

∆H = Hproducts – Hreactants

For exothermic reactions, the heat contents of the reactants is more than/higher than the heat contents of products, therefore the ∆H is negative (-∆H)

For endothermic reactions, the heat contents of the reactants is less than/lower than the heat contents of products, therefore the ∆H is negative (+∆H)

 

Graphically, in a sketch energy level diagram:

(i)For endothermic reactions the heat content of the reactants should be relatively/slightly lower than the heat content of the products

(ii)For exothermic reactions the heat content of the reactants should be relatively/slightly higher than the heat content of the products

 

Sketch energy level diagrams for endothermic dissolution

 

 

 

Energy

(kJ)        H2                                  KNO3(aq)

 

 

+∆H = H2 – H1

 

H1   KNO3(s)

 

 

Reaction path/coordinate/progress

 

 

 

 

Energy

(kJ)        H2                                  NH4Cl  (aq)

 

 

+∆H = H2 – H1

 

H1    NH4Cl (s)

                                                                                                                                                             

 

Reaction path/coordinate/progress

 

 

Sketch energy level diagrams for exothermic dissolution

 

 

 

H2      NaOH (s)

 

Energy(kJ)

-∆H = H2 – H1

 

H1                                 NaOH (aq)

                                                                                                                                                            

Reaction path/coordinate/progress

 

 

 

H2        H2SO4 (l)

Energy

(kJ)

-∆H = H2 – H1

 

H1                                 H2SO4 (aq)

                                                                                                                                                             

Reaction path/coordinate/progress

 

3.Energy changes in physical processes

 

Melting/freezing/fusion/solidification and boiling/vaporization/evaporation are the two physical processes. Melting /freezing point of pure substances is fixed /constant. The boiling point of pure substance depend on external atmospheric pressure.

Melting/fusion is the physical change of a solid to liquid. Freezing is the physical change of a liquid to solid.

Melting/freezing/fusion/solidification are therefore two opposite but same reversible physical processes. i.e

A (s)   ========A(l)

Boiling/vaporization/evaporation is the physical change of a liquid to gas/vapour. Condensation/liquidification is the physical change of gas/vapour to liquid. Boiling/vaporization/evaporation and condensation/liquidification are therefore two opposite but same reversible physical processes. i.e

B (l)   ========B(g)

Practically

(i) Melting/liquidification/fusion involves heating a solid to weaken the strong bonds holding the solid particles together. Solids are made up of very strong bonds holding  the particles very close to each other (Kinetic Theory of matter).On heating these particles gain energy/heat from the surrounding heat source to form a liquid with weaker bonds holding the particles close together but with some degree of freedom. Melting/freezing/fusion is an endothermic (+∆H)process that require/absorb energy from the surrounding.

 

(ii)Freezing/fusion/solidification involves cooling a a liquid to reform /rejoin the very strong bonds to hold  the particles very close to each other as solid and thus lose their degree of freedom (Kinetic Theory of matter). Freezing /fusion / solidification is an exothermic (∆H)process that require particles holding the liquid together to lose energy  to the surrounding.

 

(iii)Boiling/vaporization/evaporation involves heating a liquid to completely break/free the bonds holding the liquid particles together. Gaseous particles have high degree of freedom (Kinetic Theory of matter). Boiling /vaporization / evaporation is an endothermic (+∆H) process that require/absorb energy from the surrounding.

 

(iv)Condensation/liquidification is reverse process of boiling /vaporization / evaporation.It involves gaseous particles losing energy to the surrounding to form a liquid.It is an exothermic(+∆H) process.

 

The quantity of energy required to change one mole of a solid to liquid or to form one mole of a solid from liquid at constant temperature is called molar enthalpy/latent heat of fusion. e.g.

 

H2O(s)   -> H2O(l) ∆H = +6.0kJ mole-1  (endothermic process)

H2O(l)   -> H2O(s) ∆H = -6.0kJ mole-1 (exothermic process)

The quantity of energy required to change one mole of a liquid to gas/vapour or to form one mole of a liquid from gas/vapour at constant temperature is called molar enthalpy/latent heat of vapourization. e.g.

 

H2O(l)   -> H2O(g) ∆H = +44.0kJ mole-1  (endothermic process)

H2O(g)   -> H2O(l) ∆H = -44.0kJ mole-1 (exothermic process)

 

The following experiments illustrate/demonstrate practical determination of melting and boiling

 

  1. To determine the boiling point of water

Procedure:

Measure 20cm3 of tap water into a 50cm3 glass beaker. Determine and record its temperature.Heat the water on a strong Bunsen burner flame and record its temperature after every thirty seconds for four minutes.

Sample results

Time(seconds) 0 30 60 90 120 150 180 210 240
Temperature(oC) 25.0 45.0 85.0 95.0 96.0 96.0 96.0 97.0 98.0

 

Questions

1.Plot a graph of temperature against time(y-axis)

 

Sketch graph of temperature against time

 

 

 

boiling point

96 oC

 

Temperature(0C)

 

25oC

time(seconds)

2.From the graph show and determine the boiling point of water

Note:

Water boils at 100oC at sea level/one atmosphere pressure/101300Pa but boils at below 100oC at higher altitudes. The sample results above are from Kiriari Girls High School-Embu County on the slopes of Mt Kenya in Kenya. Water here boils at 96oC.

3.Calculate the molar heat of vaporization of water.(H= 1.0,O= 16.O)

Working:

Mass of water = density x volume => (20  x  1) /1000 = 0.02kg

Quantity of heat produced

=  mass of water x specific heat capacity of water x temperature change

=>0.02kg  x  4.2  x ( 96  –  25 ) = 5.964kJ

Heat of vaporization of one mole H2O  =    Quantity of heat

Molar mass of H2O

=>5.964kJ   =      0.3313 kJ mole -1

18

To determine the melting point of candle wax

Procedure

Weigh exactly 5.0 g of candle wax into a boiling tube. Heat it on a strongly Bunsen burner flame until it completely melts. Insert a thermometer and remove the boiling tube from the flame. Stir continuously. Determine and record the temperature after every 30seconds for four minutes.

Sample results

Time(seconds) 0 30 60 90 120 150 180 210 240
Temperature(oC) 93.0 85.0 78.0 70.0 69.0 69.0 69.0 67.0 65.0

 

Questions

1.Plot a graph of temperature against time(y-axis)

Sketch graph of temperature against time

 

 

 

 

93 oC

 

Temperature(0C)                      melting point

69oC

 

time(seconds)

2.From the graph show and determine the melting point of the candle wax

4.Energy changes in chemical processes

Thermochemical reactions measured at standard conditions of 298K(25oC) and 101300Pa/101300Nm2/ 1 atmospheres/760mmHg/76cmHg produce standard enthalpies denoted ∆Hᶿ.

Thermochemical reactions are named from the type of reaction producing the energy change. Below are some thermochemical reactions:

  • Standard enthalpy/heat of reaction ∆Hᶿr
  • Standard enthalpy/heat of combustion ∆Hᶿc
  • Standard enthalpy/heat of displacement ∆Hᶿd
  • Standard enthalpy/heat of neutralization ∆Hᶿn
  • Standard enthalpy/heat of solution/dissolution ∆Hᶿs
  • Standard enthalpy/heat of formation ∆Hᶿf

 

(a)Standard enthalpy/heat of reaction ∆Hᶿr

The molar standard enthalpy/heat of reaction may be defined as the energy/heat change when one mole of products is formed at standard conditions

A chemical reaction involves the reactants forming products. For the reaction to take place the bonds holding the reactants must be broken so that new bonds of the products are formed. i.e.

AB + C-D -> A-C + B-D

Old Bonds broken   A-B and C-D on reactants

New Bonds formed A-C and B-D on products

The energy required to break one mole of a (covalent) bond is called bond dissociation energy. The SI unit of bond dissociation energy is kJmole-1

The higher the bond dissociation energy the stronger the (covalent)bond

 

Bond dissociation energies of some (covalent)bonds

Bond Bond dissociation energy

(kJmole-1)

  Bond dissociation energy         (kJmole-1)
H-H 431 I-I 151
C-C 436 C-H 413
C=C 612 O-H 463
C = C 836 C-O 358
N = N 945 H-Cl 428
N-H 391 H-Br 366
F-F 158 C-Cl 346
Cl-Cl 239 C-Br 276
Br-Br 193 C-I 338
H-I 299 O=O 497
Si-Si 226 C-F 494

The molar enthalpy of reaction can be calculated from the bond dissociation energy by:

(i)adding the total bond dissociation energy of the reactants(endothermic process/+∆H) and total bond dissociation energy of the products(exothermic process/-∆H).

(ii)subtracting total bond dissociation energy of the reactants from the total bond dissociation energy of the products(exothermic process/-∆H less/minus endothermic process/+∆H).

 

Practice examples/Calculating ∆Hr

 

1.Calculate ∆Hr from the following reaction:

  1. H2(g)     +     Cl2(g)       ->   2HCl(g)

 

Working

Old bonds broken (endothermic process/+∆H )

= (H-H + Cl-Cl)  => (+431 + (+ 239))  =  + 670kJ

New bonds broken (exothermic process/-∆H )

= (2(H-Cl )   =>  (- 428 x 2))  =  -856kJ

∆Hr =( + 670kJ  +  -856kJ)  =   186  kJ   =   -93kJ mole-1

2

The above reaction has negative -∆H enthalpy change and is therefore practically exothermic.

The thermochemical reaction is thus:

½ H2(g)        +     ½ Cl2(g)                ->   HCl(g)    ∆Hr =   -93kJ

 

  1. CH4(g)     +     Cl2(g)       ->   CH3Cl  + HCl(g)

Working

Old bonds broken (endothermic process/+∆H )

= (4(C-H) + Cl-Cl)

=> ((4 x +413) + (+ 239))  =  + 1891kJ

New bonds broken (exothermic process/-∆H )

= (3(C-H  + H-Cl  + C-Cl)

=>  (( 3 x – 413) + 428  + 346)  =  –2013 kJ

∆Hr =( + 1891kJ +  -2013 kJ)    =   -122 kJ mole-1

 

The above reaction has negative -∆H enthalpy change and is therefore practically exothermic.

The thermochemical reaction is thus:

CH4(g)                 +     Cl2(g)       ->   CH3Cl(g)   +   HCl(g)     ∆H  = -122 kJ

 

 

  1. CH2CH2(g)                   +     Cl2(g)       ->   CH3Cl CH3Cl (g)

Working

Old bonds broken (endothermic process/+∆H )

= (4(C-H) + Cl-Cl + C=C)

=> ((4 x +413) + (+ 239) +(612))  =  + 2503kJ

New bonds broken (exothermic process/-∆H )

= (4(C-H  + C-C + 2(C-Cl)  )

=>  (( 3 x – 413)  +   -436  +2 x 346  =  –2367 kJ

∆Hr =( + 2503kJ + -2367 kJ)     =   +136 kJ mole-1

 

The above reaction has negative +∆H enthalpy change and is therefore practically endothermic.

The thermochemical reaction is thus:

CH2CH2(g)       +     Cl2(g)       ->   CH3Cl CH3Cl (g)     ∆H    = +136 kJ

Note that:

(i)a reaction is exothermic if the bond dissociation energy of reactants is more than bond dissociation energy of products.

(ii)a reaction is endothermic if the bond dissociation energy of reactants is less than bond dissociation energy of products.

 

 

 

 

 

(b)Standard enthalpy/heat of combustion ∆Hᶿc

 

The molar standard enthalpy/heat of combustion(∆Hᶿc) is defined as the energy/heat change when one mole of a substance is burnt in oxygen/excess air at standard conditions.

Burning is the reaction of a substance with oxygen/air. It is an exothermic process producing a lot of energy in form of heat.

A substance that undergoes burning is called a fuel. A fuel is defined as the combustible substance which burns in air to give heat energy for domestic or industrial use. A fuel may be solid (e.g coal, wood, charcoal) liquid (e.g petrol, paraffin, ethanol, kerosene) or gas (e.g liquefied petroleum gas/LPG, Water gas-CO2/H2, biogas-methane, Natural gas-mixture of hydrocarbons)

 

To determine the molar standard enthalpy/heat of combustion(∆Hᶿc) of ethanol

 

Procedure

Put 20cm3 of distilled water into a 50cm3 beaker. Clamp the beaker. Determine the temperature of the water T1.Weigh an empty burner(empty tin with wick).

Record its mass M1.Put some ethanol into the burner. Weigh again the burner with the ethanol and record its mass M2. Ignite the burner and place it below the clamped 50cm3 beaker. Heat the water in the beaker for about one minute. Put off the burner. Record the highest temperature rise of the water, T2. Weigh the burner again and record its mass M3

 

Sample results:

 

Volume of water used    20cm3
Temperature of the water before heating T1    25.0oC
Temperature of the water after heating T2    35.0oC
Mass of empty burner M1    28.3g
Mass of empty burner + ethanol before igniting M2    29.1g
Mass of empty burner + ethanol after igniting M3    28.7g

 

Sample calculations:

1.Calculate:

(a) ∆T the change in temperature

∆T = T2 – T1   => (35.0oC – 25.0oC)  = 10.0oC

         

(b) the mass of ethanol used in burning

mass of ethanol used = M2 – M1 => 29.1g – 28.7g  =  0.4g

 

(c) the number of moles of ethanol used in burning

moles of ethanol =    mass used                =>  0.4   = 0.0087 /8.7 x 10-3 moles

molar mass of ethanol        46

 

  1. Given that the specific heat capacity of water is 4.2 kJ-1kg-1K-1,determine the heat produced during the burning.

Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T

=>  20 x 4.2 x 10 =  840 Joules =  0.84 kJ

                                                        1000

3.Calculate the molar heat of combustion of ethanol

Molar heat of combustion ∆Hc = Heat produced ∆H

Number of moles of fuel

=>      0.84 kJ                   =  96.5517 kJmole-1

0.0087 /8.7 x 10-3 moles

 

4.List two sources of error in the above experiment.

   (i)Heat loss to the surrounding lowers the practical value of the molar heat of combustion of ethanol.

A draught shield tries to minimize the loss by protecting wind from wobbling the flame.

(ii) Heat gain by reaction vessels/beaker lowers ∆T and hence ∆Hc

 

5.Calculate the heating value of  the fuel.

 Heating value = molar heat of combustion => 96.5517 kJmole-1  = 2.0989 kJg-1

                                   Molar mass of fuel                   46 g

Heating value is the enrgy produced when a unit mass/gram of  a fuel is completely burnt

 

6.Explain other factors used to determine the choice of fuel for domestic and industrial use.

          (i) availability and affordability-some fuels are more available cheaply in rural than in urban areas at a lower cost.

(ii)cost of storage and transmission-a fuel should be easy to transport and store safely. e.g LPG is very convenient to store and use. Charcoal and wood are bulky.

(iii)environmental effects –Most fuels after burning produce carbon(IV) oxide gas as a byproduct. Carbon(IV) oxide gas is green house gas that causes global warming. Some other fuel produce acidic gases like sulphur(IV) oxide ,and nitrogen(IV) oxide. These gases cause acid rain. Internal combustion engines exhaust produce lead vapour from leaded petrol and diesel. Lead is carcinogenic.

(iv)ignition point-The temperature at which a fuel must be heated before it burns in air is the ignition point. Fuels like petrol have very low ignition point, making it highly flammable. Charcoal and wood have very high ignition point.

 

7.Explain the methods used to reduce pollution from common fuels.

(i)Planting trees-Plants absorb excess carbon(IV)oxide for photosynthesis and release oxygen gas to the atmosphere.

(ii)using catalytic converters in internal combustion engines that convert harmful/toxic/poisonous gases like carbon(II)oxide and nitrogen(IV)oxide to harmless non-poisonous carbon(IV)oxide, water and nitrogen gas by using platinum-rhodium catalyst along the engine exhaust pipes.

 

Further practice calculations

 

1.Calculate the heating value of methanol CH3OH given that 0.87g of the fuel burn in air to raise the temperature of 500g of water from 20oC to 27oC.(C-12.0,H=1.0 O=16.0).

Moles of methanol used = Mass of methanol used => 0.87 g  =  0.02718 moles

Molar mass of methanol        32

Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T

=>  500 x 4.2 x 7 =  14700 Joules =  14.7 kJ

                                                                 1000

Molar heat of combustion ∆Hc = Heat produced ∆H

Number of moles of fuel

=>         14.7 kJ                   =  540.8389 kJmole-1

0.02718 moles

Heating value = molar heat of combustion => 540.8389 kJmole-1  = 16.9012 kJg-1

                                   Molar mass of fuel                   32 g

 

  1. 1.0 g of carbon burn in excess air to raise the temperature of 400g of water by 18oC.Determine the molar heat of combustion and hence the heating value of carbon(C-12.0,).

Moles of carbon used = Mass of carbon used =>      1.0 g  =  0.0833 moles

Molar mass of carbon        12

Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T

=>  400 x 4.2 x 18 =  30240 Joules =  30.24 kJ

                                                                 1000

Molar heat of combustion ∆Hc = Heat produced ∆H

Number of moles of fuel

=>         30.24 kJ                   =  363.0252 kJmole-1

0.0833 moles

Heating value = molar heat of combustion => 363.0252 kJmole-1= 30.2521 kJg-1

                                   Molar mass of fuel                  12 g

 

(c)Standard enthalpy/heat of displacement ∆Hᶿd

 

The molar standard enthalpy/heat of displacement ∆Hᶿd is defined as the energy/heat change when one mole of a substance is displaced from its solution.

A displacement reaction takes place when a more reactive element/with less electrode potential Eᶿ / negative Eᶿ /higher in the reactivity/electrochemical series remove/displace another with less reactive element/with higher electrode potential Eᶿ / positive Eᶿ /lower in the reactivity/electrochemical series from its solution.e.g.

 

(i)Zn(s) + CuSO4(aq) -> Cu(s) + ZnSO4(aq)

Ionically: Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+ (aq)

(ii)Fe(s) + CuSO4(aq) -> Cu(s) + FeSO4(aq)

Ionically: Fe(s) + Cu2+(aq) -> Cu(s) + Fe2+ (aq)

(iii)Pb(s) + CuSO4(aq) -> Cu(s) + PbSO4(s)

 

This reaction stops after some time as insoluble PbSO4(s) coat/cover unreacted lead.

 

(iv)Cl2(g) + 2NaBr(aq) -> Br2(aq) + 2NaCl(aq)

Ionically: Cl2(g)+ 2Br(aq) -> Br2(aq) + 2Cl(aq)

 

Practically, a displacement reaction takes place when a known amount /volume of a solution is added excess of a more reactive metal.

 

To determine the molar standard enthalpy/heat of displacement(∆Hᶿd) of copper

 

Procedure

Place 20cm3 of 0.2M copper(II)sulphate(VI)solution into a 50cm3 plastic beaker/calorimeter. Determine and record the temperature of the solution T1.Put all the Zinc powder provided into the plastic beaker. Stir the mixture using the thermometer. Determine and record the highest temperature change to the nearest 0.5oC- T2 . Repeat the experiment to complete table 1 below

Table 1

 

Experiment I II
Final temperature of solution(T2) 30.0oC 31.0oC
Final temperature of solution(T1) 25.0oC 24.0oC
Change in temperature(∆T) 5.0 6.0

 

Questions

1.(a) Calculate:

          (i)average ∆T

Average∆T = change in temperature in experiment I and II

=>5.0 + 6.0  = 5.5oC

2

 

(ii)the number of moles of solution used

Moles used = molarity x volume of solution   =   0.2 x 20   =  0.004 moles

1000                               1000

 

(iii)the enthalpy change ∆H for the reaction

Heat produced ∆H = mass of solution(m) x specific heat capacity (c)x ∆T

=>  20 x 4.2 x 5.5 =  462 Joules =  0.462 kJ

                                                                 1000

(iv)State two assumptions made in the above calculations.

Density of solution = density of water = 1gcm-3

Specific heat capacity of solution=Specific heat capacity of solution=4.2 kJ-1kg-1K

This is because the solution is assumed to be infinite dilute.

 

  1. Calculate the enthalpy change for one mole of displacement of Cu2+ (aq) ions.

Molar heat of displacement ∆Hd = Heat produced ∆H

Number of moles of fuel

=>         0.462 kJ          = 115.5 kJmole-1

0.004

 

3.Write an ionic equation for the reaction taking place.

Zn(s) +  Cu2+(aq)  ->   Cu(s) +  Zn2+(aq)

 

4.State the observation made during the reaction.

Blue colour of copper(II)sulphate(VI) fades/becomes less blue/colourless.

Brown solid deposits are formed at the bottom of reaction vessel/ beaker.

 

5.Illustrate the above reaction using an energy level diagram.

 

 

                              Zn(s) + Cu2+(aq)

 

 

Energy                                ∆H = -115.5 kJmole-1

                 (kJ)

                                                            Cu(s) + Zn2+(aq)

 

                                        Reaction progress/path/coordinates

 

 

  1. Iron is less reactive than Zinc. Explain the effect of using iron instead of Zinc on the standard molar heat of displacement ∆Hd of copper(II)sulphate (VI) solution.

No effect. Cu2+ (aq) are displaced from their solution.The element used to displace it does not matter.The reaction however faster if a more reactive metal is used.

 

7.(a)If the standard molar heat of displacement ∆Hd of copper(II)sulphate (VI) solution is 209kJmole-1 calculate the temperature change if 50cm3 of 0.2M solution was displaced by excess magnesium.

Moles used = molarity x volume of solution   =   0.2 x 50   =    0.01 moles

1000                               1000

 

Heat produced ∆H = Molar heat of displacement ∆Hd x Number of moles

=>209kJmole-1x 0.01 moles       =       2.09 kJ

 

T (change in temperature)  =                      Heat produced ∆H                                                                                Molar heat of displacement ∆Hd x Number of moles

=>2.09 kJ              =        9.9524Kelvin

0.01 moles

 

(b)Draw an energy level diagram to show the above energy changes

 

                              Mg(s) + Cu2+(aq)

 

 

Energy                                ∆H = -209 kJmole-1

                 (kJ)

                                                            Cu(s) + Mg2+(aq)

 

                                        Reaction progress/path/coordinates

 

  1. The enthalpy of displacement ∆Hd of copper(II)sulphate (VI) solution is 12k6kJmole-1.Calculate the molarity of the solution given that 40cm3 of this solution produces 2.204kJ of energy during a displacement reaction with excess iron filings.

Number of moles  =                      Heat produced ∆H                                                                                                          Molar heat of displacement ∆Hd

 

=>2.204 kJ              =      0.0206moles

126 moles

 

Molarity of the solution   =               moles x 1000                                                                                                                Volume of solution used

=         0.0206moles  x 1000         =      0.5167 M

40

 

  1. If the molar heat of displacement of Zinc(II)nitrate(V)by magnesium powder is 25.05kJmole-1 ,calculate the volume of solution which must be added 0.5 moles solution if there was a 3.0K rise in temperature.

 

Heat produced ∆H = Molar heat of displacement ∆Hd x Number of moles

=>25.08kJmole-1x 0.5 moles      = 1.254 kJ x 1000 =1254J

Mass of solution (m)  =         Heat produced ∆H

specific heat capacity (c)x ∆T

=>          1254J         =        99.5238 g

4.2  x  3

 

Volume =  mass x density  =  99.5238 g  x  1    =   99.5238cm3

 

Note: The solution assumes to be too dilute /infinite dilute such that the density and specific heat capacity is assumed to be that of water.

 

Graphical determination of the molar enthalpy of displacement of copper

Procedure:

 

Place 20cm3 of 0.2M copper(II)sulphate (VI) solution into a calorimeter/50cm3 of plastic beaker wrapped in cotton wool/tissue paper.

Record its temperature at time T= 0.

Stir the solution with the thermometer carefully and continue recording the temperature after every 30 seconds .

Place all the (1.5g) Zinc powder provided.

Stir the solution with the thermometer carefully and continue recording the temperature after every 30 seconds for five minutes.

Determine the highest temperature change to the nearest 0.5oC.

 

Sample results

Time oC 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 240.0 270.0
Temperature 25.0 25.0 25.0 25.0 25.0 xxx 36.0 35.5 35.0 34.5

 

Sketch graph of temperature against time

 

 

 

                              36.5

Extrapolation

                Temperature         point                 ∆T

oC

 

 

                                                                   130                           Time(seconds)

Questions

  1. Show and determine the change in temperature ∆T

From a well constructed graph ∆T= T2 –T1 at 150 second by extrapolation

∆T = 36.5 – 25.0 = 11.5oC

 

2.Calculate the number of moles of copper(II) sulphate(VI)used given the molar heat of displacement of Cu2+ (aq)ions is 125kJmole-1

 

Heat produced ∆H = mass of solution(m) x specific heat capacity (c)x ∆T

=>  20 x 4.2 x 11.5 =  966 Joules =  0.966 kJ

                                                                 1000

Number of moles  =                      Heat produced ∆H                                                                                                            Molar heat of displacement ∆Hd

 

=>.966 kJ              =        0.007728moles

125 moles                   7.728 x 10-3moles

 

  1. What was the concentration of copper(II)sulphate(VI) in moles per litre.

 

Molarity =   moles x 1000   =>  7.728 x 10-3moles x 1000  = 0.3864M

Volume used                        20

 

4.The actual concentration of copper(II)sulphate(VI) solution was 0.4M.Explain the differences between the two.

 

          Practical value is lower than theoretical. Heat/energy loss to the surrounding and that absorbed by the reaction vessel decreases  ∆T hence lowering the practical number of moles and molarity against the theoretical value

 

 

5.a)    In an experiment to determine the molar heat of reaction when magnesium displaces copper ,0.15g of magnesium powder were added to 25.0cm3 of 2.0M copper (II) chloride solution. The temperature of copper (II) chloride solution was 25oC.While that of the mixture was 43oC.

i)Other than increase in temperature, state and explain the observations which were made during the reaction.(3mks)

 

ii)Calculate the heat change during the reaction (specific heat capacity of the solution = 4.2jg-1k-1and the density of the solution = 1g/cm3(2mks)

 

iii)Determine the molar heat of displacement of copper by magnesium.(Mg=24.0).

iv)Write the ionic equation for the reaction.(1mk)

 

v)Sketch an energy level diagram for the reaction.(2mks)

 

 

 

 

 

(c)Standard enthalpy/heat of neutralization ∆Hᶿn

 

The molar standard enthalpy/heat of neutralization ∆Hᶿn is defined as the energy/heat change when one mole of a H+ (H3O+)ions react completely with one mole of OH ions to form one mole of H2O/water.

Neutralization is thus a reaction of an acid /H+ (H3O+)ions with a base/alkali/ OH ions to form salt and water only.

Strong acids/bases/alkalis are completely dissociated to many free ions(H+ /H3O+ and OH–  ions).

Weak acids/bases/alkalis are partially dissociated to few free ions(H+ (H3O+ and OH–  ions) and exist more as molecules.

Neutralization is an  exothermic(-∆H) process.The enrgy produced during neutralization depend on the amount of  free ions (H+ H3O+ and OH)ions existing in the acid/base/alkali reactant:

          (i)for weak acid-base/alkali neutralization,some of the energy is used to dissociate /ionize the molecule into free H+ H3O+ and OH ions therefore the overall energy evolved is comparatively lower/lesser/smaller than strong acid / base/ alkali neutralizations.

(ii) (i)for strong acid/base/alkali neutralization, no energy is used to dissociate /ionize since molecule is wholly/fully dissociated/ionized into free H+ H3O+ and OH ions.The overall energy evolved is comparatively higher/more than weak acid-base/ alkali neutralizations. For strong acid-base/alkali neutralization, the enthalpy of neutralization is constant at about 57.3kJmole-1 irrespective of the acid-base used. This is because ionically:

 

OH(aq)+  H+(aq)  ->   H2O(l)  for any wholly dissociated acid/base/alkali

Practically ∆Hᶿn can be determined as in the examples below:

 

To determine the molar enthalpy of neutralization ∆Hn of Hydrochloric acid

Procedure

Place 50cm3 of 2M hydrochloric acid into a calorimeter/200cm3 plastic beaker wrapped in cotton wool/tissue paper. Record its temperature T1.Using a clean measuring cylinder, measure another 50cm3 of 2M sodium hydroxide. Rinse the bulb of the thermometer in distilled water. Determine the temperature of the sodium hydroxide T2.Average T2 andT1 to get the initial temperature of the mixture T3.

Carefully add all the alkali into the calorimeter/200cm3 plastic beaker wrapped in cotton wool/tissue paper containing the acid. Stir vigorously the mixture with the thermometer.

Determine the highest temperature change to the nearest 0.5oC T4 as the final temperature of the mixture. Repeat the experiment to complete table 1.

 

Table I . Sample results

Experiment  I         II
Temperature of acid T1 (oC) 22.5 22.5
Temperature of base T2 (oC) 22.0 23.0
Final temperature of solution T4(oC) 35.5 36.0
Initial temperature of solution T3(oC) 22.25 22.75
Temperature change( T5) 13.25 13.75

(a)Calculate T6 the average temperature change                                                                        T6 =   13.25 +13.75   = 13.5 oC                                                                                                     2

(b)Why should the apparatus be very clean?

Impurities present in the apparatus reacts with acid /base lowering the overall temperature change and hence   ∆Hᶿn.

 

(c)Calculate the:

(i)number of moles of the acid used

number of moles = molarity x volume           => 2 x 50    = 0.1moles                                                                 1000                                     1000

 

(ii)enthalpy change ∆H  of neutralization.

∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T(T6)   => (50 +50)  x  4.2 x 13.5  = 5670Joules  = 5.67kJ

 

(iii) the molar heat of neutralization the acid.

∆Hn  = Enthalpy change ∆H          =>   5.67kJ   = 56.7kJ mole-1

Number of moles                     0.1moles

 

(c)Write the ionic equation for the reaction that takes place

          OH(aq)+  H+(aq)  ->   H2O(l)

 

(d)The theoretical enthalpy change is 57.4kJ. Explain the difference with the results above.

          The theoretical value is higher

Heat/energy loss to the surrounding/environment lowers ∆T/T6 and thus ∆Hn

Heat/energy is absorbed by the reaction vessel/calorimeter/plastic cup lowers ∆T and hence ∆Hn

 

(e)Compare the ∆Hn of the experiment above with similar experiment repeated with neutralization of a solution of:

 

          (i) potassium hydroxide with nitric(V) acid

The results would be the same/similar.

Both are neutralization reactions of strong acids and bases/alkalis that are fully /wholly dissociated into many free H+ / H3O+ and OH ions.

 

     (ii) ammonia with ethanoic acid

The results would be lower/∆Hn would be less.

Both are neutralization reactions of weak acids and bases/alkalis that are partially /partly dissociated into few free H+ / H3O+ and OH ions. Some energy is used to ionize the molecule.

 

(f)Draw an energy level diagram to illustrate the energy changes

 

 

H2     H+ (aq)+OH (aq)

Energy

(kJ)

∆H = -56.7kJ

 

H1                                            H2O (l)

                                                                                                                                                             

Reaction path/coordinate/progress

 

Theoretical examples

1.The molar enthalpy of neutralization was experimentary shown to be 51.5kJ per mole of 0.5M hydrochloric acid and 0.5M sodium hydroxide. If the volume of sodium hydroxide was 20cm3, what was the volume of hydrochloric acid used if the reaction produced a 5.0oC rise in temperature?

 

Working:

Moles of sodium hydroxide = molarity x volume =>  0.5 M x 20cm3  = 0.01 moles

                                                  1000                                1000

 

Enthalpy change ∆H =            ∆Hn                                    =>              51.5     =  0.515kJ

Moles sodium hydroxide                0.01 moles

 

Mass of base + acid =       Enthalpy change ∆H in Joules

Specific heat capacity x ∆T

=>               0.515kJ   x  1000     =   24.5238g

4.2  x  5

 

Mass/volume of HCl = Total volume – volume of NaOH

=>24.5238   –  20.0  =   4.5238 cm3

 

  1. ∆Hn of potassium hydroxide was practically determined to be 56.7kJmole-1.Calculate the molarity of 50.0 cm3 potassium hydroxide used to neutralize 25.0cm3 of dilute sulphuric(VI) acid raising the temperature of the solution from 10.0oC to 16.5o

 

∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T

            => (50 +25)  x  4.2 x 6.5     =       2047.5Joules

 

Moles potassium hydroxide =Enthalpy change ∆H

∆Hn

2047.5Joules      = 0.0361 moles

56700Joules

 

Molarity of KOH = moles x 1000       => 0.0361 moles  x  1000   =  0.722M

                                Volume used                         50cm3

 

3.Determine the specific heat capacity of a solution of a solution mixture of 50.0cm3 of 2M potassium hydroxide neutralizing 50.0cm3 of 2M nitric(V) acid if a 13.25oC rise in temperature is recorded.(1mole of potassium hydroxide produce 55.4kJ of energy)

 

Moles of potassium hydroxide = molarity KOH x volume

1000

=>  2 M x 50cm3  = 0.1 moles

                                                            1000

 

Enthalpy change ∆H       =      ∆Hn       x   Moles potassium hydroxide                                                              =>    55.4kJ  x   0.1 moles  = 5.54kJ  x 1000=5540Joules

 

Specific heat capacity =       Enthalpy change ∆H in Joules

Mass of base + acid x ∆T

=>                   5540                        =   4.1811J-1g-1K-1

(50+50)  x  13.25

 

Graphically ∆Hn can be determined as in the example below:

 

Procedure

Place 8 test tubes in a test tube rack .Put 5cm3 of 2M sodium hydroxide solution into each test tube.

Measure 25cm3 of 1M hydrochloric acid into 100cm3 plastic beaker.

Record its initial temperature at volume of base =0. Put one portion of the base into the beaker containing the acid.

Stir carefully with the thermometer and record the highest temperature change to the nearest 0.5oC.

Repeat the procedure above with other portions of the base to complete table 1 below

 

Table 1:Sample results.

olume of acid(cm3) 25.0 25.0 25.0 25.0 25.0 25.0 25.0 25.0 25.0
Volume of alkali(cm3) 0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0
Final temperature(oC) 22.0 24.0 26.0 28.0 28.0 27.0 26.0 25.0 24.0
Initial temperature(oC) 22.0 22.0 22.0 22.0 22.0 22.0 22.0 22.0 22.0
Change in temperature 0.0 2.0 4.0 6.0 6.0 5.0 4.0 3.0 2.0

 

(a)Complete the table to determine the change in temperature.

 

(b)Plot a graph of volume of sodium hydroxide against temperature change.

 

 

 

                  6.7=T2

Volume of sodium hydroxide(cm3)

 

   0 =T1

∆T (oC)

 

From the graph show and determine :

(i)the highest temperature change  ∆T

∆T =T2-T1  => highest temperature-T2 (from extrapolating a correctly plotted graph) less lowest temperature at volume of base=0 :T1

=>∆T = 6.7 – 0.0 =   6.70C

 

 (ii)the volume of sodium hydroxide used for complete neutralization

From a correctly plotted graph – 16.75cm3

 

(c)Calculate the number of moles of the alkali used

Moles NaOH = molarity x volume   =>2M x 16.75cm3   = 0.0335 moles

1000                                1000

 

(d)Calculate ∆H for the reaction

          ∆H = mass of solution(acid+base) x c x ∆T

=>(25.0 + 16.75)  x 4.2 x 6.7 = 1174.845 J   = 1.174845kJ

                                                                    1000

 

(e)Calculate the molar enthalpy of neutralization of the alkali.

∆Hn  =           ∆Hn              = 1.174845kJ =  35.0701kJ

                       Number of moles      0.0335

 

 

(d)Standard enthalpy/heat of solution ∆Hᶿs

 

The standard enthalpy of solution ∆Hᶿsis defined as the energy change when one mole of a substance is dissolve in excess distilled water to form an infinite dilute solution. An infinite dilute solution is one which is too dilute to be diluted further.

 

Dissolving a solid involves two processes:

(i) breaking the crystal of the solid into free ions(cations and anion).This process is the opposite of the formation of the crystal itself. The energy required to form one mole of a crystal structure from its gaseous ions is called Lattice energy/heat/enthalpy of lattice (∆Hl). Lattice energy /heat/enthalpy of lattice (Hl) is an endothermic process (+∆Hl).

 

The table below shows some ∆Hl in kJ for the process  MX(s)   -> M+ (g)  +  X(g)

  Li Na K Ca Mg
F +1022 +900 +800 +760 +631
Cl +846 +771 +690 +2237 +2493
Br +800 +733 +670 +2173 +2226

 

(ii)surrounding the free ions by polar water molecules. This process is called hydration. The energy produced when one mole of  ions are completely hydrated is called hydration energy/ heat/enthalpy of hydration(∆Hh).Hydration energy /enthalpy of hydration(∆Hh) is an exothermic process(∆Hh).

 

The table below shows some ∆Hh in kJ for some ions;

ion Li+ Na+ K+ Mg2+ Ca2+ F Cl Br
∆Hh -1091 -406 -322 -1920 -1650 -506 -364 -335

 

The sum of the lattice energy +∆Hl (endothermic) and hydration energy ∆Hh (exothermic) gives the heat of solution∆Hs

∆Hs = ∆Hl +∆Hh

Note

Since ∆Hl is an endothermic process and ∆Hh is an exothermic process then ∆Hs is:

          (i)exothermic if ∆Hl is less than ∆Hh and hence a solid dissolve easily in water.

(ii)endothermic if ∆Hl is more than ∆Hh and hence a solid does not dissolve easily in water.

 

(a)Dissolving sodium chloride crystal/s:

 

(i) NaCl –—breaking the crystal into free ions–->  Na +(g)+ Cl(g)  ∆Hl =+771 kJ

(ii) Hydrating the ions;

               Na +(g)  +  aq   ->  Na(aq) ∆Hh = – 406 kJ

              Cl(g)     + aq    ->  Cl(aq) ∆Hh = – 364 kJ

 ∆Hs =∆Hh +∆Hs -> (- 406 kJ  + – 364 kJ)  + +771 kJ  = + 1.0 kJmole-1

 

NaCl does not dissolve easily in water because overall ∆Hs is endothermic

 

Solubility of NaCl therefore increases with increase in temperature.

Increase in temperature increases the energy to break the crystal lattice of NaCl to free Na +(g)+ Cl(g)

 

(b)Dissolving magnesium chloride crystal/s//  MgCl2 (s) ->MgCl2 (aq)

 

(i) MgCl2-breaking the crystal into free ions-->Mg 2+(g)+ 2Cl(g) ∆Hl =+2493 kJ

(ii) Hydrating the ions;

Mg 2+(g)  +  aq   ->  Mg 2+(g) (aq) ∆Hh = – 1920 kJ

2Cl(g)    + aq    ->  2Cl(aq) ∆Hh = (- 364 x 2) kJ

∆Hs =∆Hh +∆Hs -> (- 1920 kJ  + (- 364  x 2 kJ))  + +2493 kJ  =  –155.0 kJmole-1

 

MgCl2 (s) dissolve easily in water because overall ∆Hs is exothermic .

Solubility of MgCl2 (s) therefore decreases with increase in temperature.

 

(c)Dissolving Calcium floride crystal/s//  CaF2 (s) -> CaF2 (aq) 

(i) CaF2 –>Ca 2+(g)+ 2F(g) ∆Hl =+760 kJ

(ii) Hydrating the ions;

Ca 2+(g)  +  aq   ->  Ca 2+(g) (aq) ∆Hh = – 1650 kJ

2F(g)    + aq    ->  2F(aq) ∆Hh = (- 506 x 2) kJ

∆Hs =∆Hh +∆Hs -> (- 1650 kJ  + (- 506  x 2 kJ))  + +760 kJ  =  –1902.0 kJmole-1

 

CaF2 (s) dissolve easily in water because overall ∆Hs is exothermic .

Solubility of CaF2 (s) therefore decreases with increase in temperature.

 

(d)Dissolving magnesium bromide crystal/s//  MgBr2 (s) ->MgBr2 (aq)

 

(i) MgCl2-breaking the crystal into free ions-->Mg 2+(g)+ 2Br(g) ∆Hl =+2226 kJ

(ii) Hydrating the ions;

Mg 2+(g)  +  aq   ->  Mg 2+(g) (aq) ∆Hh = – 1920 kJ

2Br(g)    + aq    ->  2Br(aq) ∆Hh = (- 335x 2) kJ

∆Hs =∆Hh +∆Hs -> (- 1920 kJ  + (- 335  x 2 kJ))  + +2226 kJ  =  –364.0 kJmole-1

 

MgBr2 (s) dissolve easily in water because overall ∆Hs is exothermic .

Solubility of MgBr2(s) therefore decreases with increase in temperature.

 

Practically the heat of solution can be determined from dissolving known amount /mass/volume of solute in known mass /volume of water/solvent.

From the temperature of solvent before and after dissolving the change in temperature(∆T) during dissolution is determined.

 

To determine the ∆Hs ammonium nitrate

 

Place 100cm3 of distilled water into a  plastic beaker/calorimeter. Determine its temperature and record it at time =0 in table I below.

Put all the 5.0g of ammonium nitrate (potassium nitrate/ammonium chloride can also be used)provided into the plastic beaker/calorimeter, stir using a thermometer and record the highest temperature change to the nearest 0.5oCafter every ½ minute to complete table I.

Continue stirring the mixture throughout the experiment.

Sample results: Table I

 

Time (minutes) 0.0 ½ 1 1 ½ 2 2 ½ 3 3 ½
Temperature()oC 22.0 21.0 20.0 19.0 19.0 19.5 20.0 20.5

 

(a)Plot a graph of temperature against time(x-axis)

 

 

 

 

 

 

                22.0=T1

 

    18.7. oC T1

temperature(oC)                                                                                                                                                                       ∆T

 

Time (minutes)

 

 

 

 

 

(b)From the graph show and determine the highest temperature change  ∆T

∆T =T2-T1  => lowest temperature-T2 (from extrapolating a correctly plotted graph) less highest temperature at volume of base=0 :T1

=>∆T =18.7 – 22.0 =   3.30C

 

(c)Calculate the number of moles of ammonium nitrate(V) used

Moles NH4NO3 =   mass used   =>      5.0   = 0.0625 moles

Molar mass           80

 

(d)Calculate ∆H for the reaction

          ∆H = mass of water x c x ∆T

->100 x 4.2 x 3.3 =  +1386 J   = +1.386kJ

                                                   1000

 

(e)Calculate the molar enthalpy of dissolution of ammonium nitrate(V).

∆Hs  =           ∆H               = +1.386kJ        =  + 22.176kJ mole-1

Number of moles      0.0625 moles

 

(f)What would happen if the distilled water was heated before the experiment was performed.

 

The ammonium nitrate(V)would take less time to dissolves. Increase in temperature reduces lattice energy causing endothermic dissolution to be faster

 

(g)Illustrate the process above in an energy level diagram

 

NH4+ (g)   +  NO3(g)

 

 

+∆H                                                                                                                       NH4+ (aq)+NO3(aq) Energy(kJ)                                                                                                                                                                                   +∆H                             ∆H = -22.176kJ                                                                                                                                                                                                                                                                                                              NH4NO3(s)                                                                                                                                                                                                                                                                                                                                                                                                                   Reaction path /progress/coordinate

 

(h) 100cm3 of distilled water at 25oC was added carefully 3cm3 concentrated sulphuric(VI)acid of density 1.84gcm-3.The temperature of the mixture rose from 250C to 38oC.Calculate the molar heat of solution of sulphuric(VI)acid         (S=32.0,H=1.0,0=16.0)

 

Working

Molar mass of H2SO4 = 98g

Mass of H2SO4= Density x volume  => 1.84gcm-3  x 3cm3 = 5.52 g

Mass of H2O   = Density x volume  => 1.00gcm-3  x 100cm3 = 100 g

Moles of H2SO4=             mass                 =>      5.52 g       =     0.0563 moles

Molar mass of H2SO4              98g

 

Enthalpy change ∆H= (mass of acid + water) x specific heat capacity of water x ∆T      => (100 +5.52 g) x 4.2 x 13oC  = 5761.392 J  = 5.761392 kJ

1000

 

∆Hs of H2SO4=            ∆H                   =>   5.761392 kJ  =  -102.33378kJmoles-1

Moles  of H2SO4                  0.0563 moles

 

 

 

 

 

 

 

 

 

 

 

(e)Standard enthalpy/heat of formation  ∆Hᶿf

 

The molar enthalpy of formation ∆Hᶿf is defined as the energy change when one mole of a compound is formed from its elements at 298K(25oC) and 101325Pa(one atmosphere)pressure. ∆Hᶿf is practically difficult to determine in a school laboratory.

It is determined normally determined by applying Hess law of constant heat summation.

Hess law of constant heat summation states that “the total enthalpy/heat/energy change of a reaction is the same regardless of the route taken from reactants to products at the same temperature and pressure”.

 

Hess law of constant heat summation is as a result of a series of experiments done by the German Scientist Henri Hess(1802-1850).

He found that the total energy change from the reactants to products was the same irrespective of the intermediate products between. i.e.

 

A(s)    —∆H1–>C(s)     =     A(s) —∆H2–>B(s)–∆H3–>C(s)

Applying Hess law of constant heat summation then:

 

A(s)                     ∆H2                                      B(s)

 

∆H1                            ∆H3

 

 

C(s)

 

The above is called an energy cycle diagram. It can be used to calculate any of the missing energy changes since:

(i) ∆H1   =∆H2    +  ∆H

(ii) ∆H=∆H1    +   -∆H3

(iii) ∆H=  – ∆H1    +   ∆H2

 

Examples of applying Hess law of constant heat summation

 

1.Calculate the molar enthalpy of formation of methane (CH4) given that ∆Hᶿc of carbon-graphite is -393.5kJmole-1,Hydrogen is -285.7 kJmole-1 and that of methane is 890 kJmole-1                                                                                                                                                                                

Working

Carbon-graphite ,hydrogen and oxygen can react to first form methane.

Methane will then burn in the oxygen present to form carbon(IV)oxide and water. Carbon-graphite can burn in the oxygen to form carbon(IV)oxide.

Hydrogen can burn in the oxygen to form water.

C(s)+ 2H2 (g)+2O2 (g) —∆H1–> CH4(g) +2O2(g) —∆H2–> CO2(g)+2H2O(l)

C(s)+ 2H2 (g)+2O2 (g) —∆H3–> CO2(g)+2H2O(l)

 

Energy cycle diagram

C(s)  +   2H2 (g) + 2O2(g)                    ∆H1=∆Hᶿc =-890.4kJ          CH4(g)+2O2(g)

 

 

 

 

∆H3=∆Hᶿc =-393.5kJ    ∆H3=∆Hᶿc =-285.7kJ x 2     ∆H2= ∆Hᶿf= x

 

 

 

CO2(g)        +     2H2O(l)

 

Substituting:

∆H3    = ∆H1   +  ∆H2

-393.5 + (-285.7 x 2)  = -890.4kJ + x

x = -74.5 kJ

Heat of formation ∆Hᶿf  CH4 = -74.5 kJmole-1

 

  1. Calculate the molar enthalpy of formation of ethyne (C2H2) given : ∆Hᶿc of carbon-graphite = -394kJmole-1,Hydrogen = -286 kJmole-1 , (C2H2) = -1300 kJmole-1

 

Working

Carbon-graphite ,hydrogen and oxygen can react to first form ethyne.

Ethyne will then burn in the oxygen present to form carbon(IV)oxide and water. Carbon-graphite can burn in the oxygen to form carbon(IV)oxide.

 

Hydrogen can burn in the oxygen to form water.

2C(s)+ H2 (g)+2 ½ O2 (g) —∆H1–> C2 H2 (g) +2 ½ O2(g) —∆H2–> CO2(g)+H2O(l)

2C(s)+ H2 (g)+ 2 ½ O2 (g) —∆H3–> 2CO2(g)+H2O(l)

 

Energy cycle diagram

 

2C(s)   + H2 (g) +2½O2(g)            ∆H1=∆Hᶿf =x                       C2 H2+2½O2(g)

 

 

 

 

∆H3=∆Hᶿc =-394kJx 2    ∆H3=∆Hᶿc =-286kJ      ∆H2= ∆Hᶿc= -1300kJ

 

 

 

2CO2(g)        +      H2O(l)

 

 

Substituting:

∆H3    = ∆H1   +  ∆H2

( -394 x 2) + -286  = -1300kJ + x

x = +244 kJ

Heat of formation ∆Hᶿf  CH4 = +244 kJmole-1

 

  1. Calculate the molar enthalpy of formation of carbon(II)oxide (CO) given : ∆Hᶿc of carbon-graphite = -393.5kJmole-1, ∆Hᶿc of carbon(II)oxide (CO)= -283 kJmole-1

 

Working

Carbon-graphite reacts with oxygen first to form carbon (II)oxide (CO).

Carbon(II)oxide (CO) then burn in the excess oxygen to form carbon(IV)oxide. Carbon-graphite can burn in excess oxygen to form carbon (IV) oxide.

 

C(s)+   ½O2 (g) —∆H1–> CO (g) + ½ O2(g) —∆H2–> CO2(g)

C(s)+  O2 (g) —∆H3–> CO2(g)

 

Energy cycle diagram

 

C(s)   + ½O2(g)                   ∆H1=∆Hᶿf =x                                     CO+½O2(g)

 

 

 

 

∆H3=∆Hᶿc =-393.5kJ                                ∆H2= ∆Hᶿc= -283kJ

 

 

CO2(g)

 

Substituting:

 

∆H3    = ∆H1   +  ∆H2

-393.5kJ  = -283kJ + x

x = -110 kJ

Heat of formation ∆Hᶿf  CO = -110 kJmole-1

 

4.Study the information below:

          H2(g) + ½ O2(g)    -> H2O(l)   ∆H1=-286 kJmole-1 

C(s) +  O2(g)    -> CO2(g)   ∆H2=-393 kJmole-1     

               2C(s) + H2(g) + ½ O2(g)    ->C2H5OH(l)   ∆H3=-277 kJmole-1  

Use the information to calculate the molar enthalpy of combustion ∆H4 of ethanol

Energy cycle diagram

 

2C(s)  + 3H2 (g) +3½O2(g)       ∆H3=∆Hᶿf =-227kJ            C2 H5OH +3O2(g)

 

 

 

 

∆H2=∆Hᶿc =-394kJx 2    ∆H1=∆Hᶿc =-286kJx 3      ∆H4= ∆Hᶿc= x

 

 

2CO2(g)        +      3H2O(l)

 

Substituting:

∆H1    +   ∆H2   =  ∆H+  ∆H4

( -394 x 2)  +  -286 x 3  =  -277   +  x

∆H4 = -1369 kJ

Heat of combustion ∆Hᶿc  C2H5OH = -1369 kJmole-1

 

5.Given the following information below:

          CuSO4(s) + (aq)    -> CuSO4(aq)   ∆H=-66.1 kJmole-1   

CuSO4(s) + (aq)   +  5H2O(l)-> CuSO4 .5H2O (aq)   ∆H=-77.4 kJmole-1         

 Calculate  ∆H  for the reaction;

CuSO4(aq) +   5H2O-> CuSO4 .5H2O (aq)   ∆H=-77.4 kJmole-1

Working

CuSO4(s) + (aq)   +  5H2O(l)->  CuSO4(aq)+ 5H2O(l)-> CuSO4 .5H2O (aq)

CuSO4(s) + (aq)   +  5H2O(l)-> CuSO4 .5H2O (aq)

 

Energy cycle diagram

 

CuSO4(s) + (aq)   +  5H2O(l)           ∆H1=+66.1kJ                         CuSO4(aq)+ 5H2O(l)

 

 

 

 

∆H3= =-77.4kJ                               ∆H2= x

 

 

CuSO4 .5H2O (aq)

 

Substituting:

 

                ∆H3      =  ∆H+  ∆H1

( -77.4kJ =  x  +  +66.1kJ

∆H4 = -10.9 kJ

Heat of dissolution of CuSO4 = -10.9kJmole-1

 

Practically, Hess’ law can be applied practically as in the following examples

 

a)Practical example 1

Determination of the enthalpy of formation of CuSO4.5H2O

Experiment I

Weigh accurately 12.5 g of copper(II)sulphate(VI)pentahydrate. Measure 100cm3 of distilled water into a beaker. Determine its temperature T1 .Put all the crystals of the copper(II)sulphate(VI)pentahydrate carefully into the beaker. Stir using a thermometer and determine the highest temperature change T2 Repeat the procedure again to complete table 1.

 

Table 1:Sample results

 

Experiment I II
Highest /lowest temperature T2 27.0 29.0
Initial temperature T1 24.0 25.0
Change in temperature ∆T 3.0 4.0

 

Experiment II

Weigh accurately 8.0g of anhydrous copper(II)sulphate(VI). Measure 100cm3 of distilled water into a beaker. Determine its temperature T1 .Put all the crystals of the copper(II)sulphate(VI)pentahydrate carefully into the beaker. Stir using a thermometer and determine the highest temperature change T2 Repeat the procedure again to complete  table II.

Table II :Sample results

 

Experiment I II
Highest /lowest temperature T2 26.0 27.0
Initial temperature T1 25.0 25.0
Change in temperature ∆T 1.0 2.0

 

Questions

(a)Calculate the average ∆T in

          (i)Table I

∆T= T2 -T1 => 3.0 +4.0   = 3.5 oC

                                                  2

(ii)Table II

∆T= T2 -T1 => 1.0 +2.0   = 1.5 oC

                                                  2

 

(b)Calculate the number of moles of solid used in:

          (i)Experiment I

Moles of CuSO4.5H2O =        Mass         =>   12.5  = 0.05 moles

                                                            Molar mass          250

(ii)Experiment II

Moles of CuSO4 =        Mass                   =>       8.0 = 0.05 moles

                                                   Molar mass                     160

 

(c)Calculate the enthalpy change for the reaction in:

          (i)Experiment I

Enthalpy change of CuSO4.5H2O= mass of Water(m) x c x ∆T

=>100cm3  x  4.2  x  3.5 oC  = -1.47kJ

1000

(ii)Experiment II

Enthalpy change of CuSO4 = mass of water(m) x c x ∆T

=>100cm3  x  4.2  x  1.5 oC  = -0.63kJ

1000

 

(c)Calculate the molar enthalpy of solution CuSO4 .5H2O (s) form the results in   (i)experiment I.

∆Hs   =   CuSO4.5H2O=                ∆H                      =>    -1.47kJ       = 29.4kJ

Number of Moles                   0.05 moles

 

(ii)experiment II.

∆Hs   =   CuSO4=                   ∆H                   =>      -0.63kJ              =   12.6kJ

Number of Moles           0.05 moles

 

 

(d) Using an energy level diagram, calculate the molar enthalpy change for the reaction:

CuSO4 .5H2O (s) -> CuSO4(s)  +  5H2O(l)

 

Energy cycle diagram

 

CuSO4(s) + (aq)   +  5H2O(l)           ∆H1=x                  CuSO4. 5H2O (s)+ (aq)

 

 

 

 

∆H3= =-29.4kJ                               ∆H2= -12.6kJ

 

 

CuSO4 .5H2O (aq)

 

∆H= ∆H+∆H2

=>-29.4kJ = -12.6kJ + x

=>-29.4kJ  –  (+12.6kJ)  =  x

x = 16.8kJ

 

 

b)Practical example II

Determination of enthalpy of solution of ammonium chloride

 

Theoretical information.

Ammonium chloride dissolves in water to form ammonium chloride solution. Aqueous ammonia can react with excess dilute hydrochloric acid to form ammonium chloride solution. The heat change taking place can be calculated from the heat of reactions:

(i) NH3(aq)   +  HCl(aq) -> NH4Cl(s)

(ii) NH4Cl(s)   +  (aq) -> NH4Cl(aq)

(iii) NH3(aq)   +  HCl(aq) -> NH4Cl(aq)

 

Experiment procedure I

Measure 50cm3 of water into a 100cm3 beaker. Record its temperature T1 as initial temperature to the nearest 0.5oC in table I. Add exactly 5.0g of ammonium chloride crystals weighed carefully into the water. Stir and record the highest temperature change  T2 as the final temperature change. Repeat the above procedure to complete table I.

 

Sample results TableI

 

Experiment I II
final temperature(oC) 19.0 20.0
initial temperature(oC) 22.0 22.0
 temperature change  ∆T(oC) 3.0 2.0

 

Experiment procedure II

Measure 25cm3 of 2M aqueous ammonia into a 100cm3 beaker. Record its temperature T1 as initial temperature to the nearest 0.5oC in table II. Measure 25cm3 of 2M hydrochloric acid solution. Add the acid into the beaker containing aqueous ammonia. Stir and record the highest temperature change T2 as the final temperature change. Repeat the above procedure to complete table II.

Sample results:Table II

 

Experiment I II
final temperature(oC) 29.0 29.0
initial temperature(oC) 22.0 22.0
 temperature change  ∆T(oC) 7.0 7.0

 

Sample Calculations:

(a)Calculate the average ∆T in

          (i)Table I

∆T= T2 -T1 => –3.0 +-2.0   = 2.5 oC

                                                  2

(ii)Table II

∆T= T2 -T1 => 7.0 +7.0   = 7.0 oC

                                                  2

 

 

(b)Calculate the enthalpy change for the reaction in:

          (i)Experiment I

Enthalpy change ∆H = mass of Water(m) x c x ∆T

=>50cm3  x  4.2  x  2.5 oC        = +0.525kJ

1000

(ii)Experiment II

Enthalpy change of CuSO4 = mass of water(m) x c x ∆T

=>25+25cm3  x  4.2  x  7 oC  = +1.47kJ

1000

 

(c)Write the equation for the reaction taking place in:

          (i)Experiment I

NH4Cl(s)   +  (aq) -> NH4Cl(aq)

 

(ii)Experiment I

NH3(aq)    +  HCl(aq)     ->  NH4Cl(aq)

 

(d)Calculate the enthalpy change ∆H for the reaction:

NH3(g)   +  HCl(g) -> NH4Cl(s) given that:

 

(i) NH3(g)   +  (aq) -> NH3(aq)           ∆H= -40.3kJ

(ii) (aq)   +  HCl(g) -> HCl(aq)            ∆H= -16.45kJ

 

(e)Applying Hess’ Law of constant heat summation:

Energy level diagram

 

N2(g)           +    1½ H2(g)         + ½ Cl2            ∆Hf                  NH4Cl(s)  + aq

 

 

+0.525kJ=∆H4

 

 

 

(aq)                 (aq)

– 40.3kJ=∆H1      -16.43kJ=∆H2

 

 

NH3 (aq)   +      HCl(aq)        -1.47kJ=∆H3              NH4Cl(s)

 

 

∆H1       +       ∆H2          +     ∆H3        =    ∆H4           +          ∆Hf

– 40.3kJ  +   -16.43kJ   +   -1.47kJ    =   +0.525kJ    +          ∆Hf

=>∆Hf  = -58.865kJ.

 

Practice theoretical examples:

 

  1. Using an energy level diagram calculate the ∆Hs of ammonium chloride crystals given that.

 ∆Hf  of NH3 (aq) = -80.54kJ mole-1

∆Hf  of HCl (aq) = -164.46kJ mole-1

∆Hf   of NH4Cl (aq) = -261.7483kJ mole-1

∆Hs   of NH4Cl (aq) = -16.8517kJ mole-1

 

N2(g)  +    1½ H2(g)    + ½ Cl2         ∆Hf=-261.7483kJ                 NH4Cl(s)  + aq

 

 

   

 

x=∆Hs

 

(aq)                   (aq)

– 80.54kJ=∆H1      -164.46kJ=∆H2

 

 

NH3 (aq)   +      HCl(aq)          16.8517kJ=∆H3                NH4Cl(s)

 

∆H1       +       ∆H2          +       ∆H3         =              ∆H4                +            ∆Hf

– 80.54kJ  +   -164.46kJ + -16.8517kJ    =   -261.7483kJ   +  ∆Hf

=>∆Hf  = -33.6kJmole-1.

 

Study the energy cycle diagram below and use it to:

 

(a)Identify the energy changes ∆H1 ∆H2 ∆H3 ∆H4 ∆H5 ∆H6

 

 

 

∆H1  – enthalpy/heat of formation of  sodium chloride  (∆Hf)

∆H2   -enthalpy/heat of atomization of  sodium   (∆Hat)

∆H3    -enthalpy/heat of  ionization/ionization energy of  sodium  (∆H i)

∆H4 -enthalpy/heat of atomization of  chlorine (∆Hat)

∆H5 -enthalpy/heat of electron affinity of  chlorine (∆He)

∆H6 enthalpy/heat of lattice/Lattice energy of  sodium chloride(∆H l)

 

(b) Calculate ∆H1  given that ∆H2 =+108kJ , ∆H3=+500kJ,  ∆H4 =+121kJ ,∆H5 =-364kJ and  ∆H6 =-766kJ

 

Working:

∆H1 =∆H2 +∆H3 +∆H4 +∆H5 +∆H6

Substituting:

∆H1= +108kJ  + +500kJ + +121kJ +-364kJ + -766kJ

∆H1= -401kJmole-1

 

(c) Given the that:

 (i) Ionization energy of  sodium = + 500kJmole-1                             

(ii)∆Hat of  sodium = + 110kJmole-1                                          

(iii) Electron affinity of  chlorine = – 363kJmole-1

 (iv)∆Hat of  chlorine = + 120kJmole-1 

 (v) ∆Hf  of  sodium chloride = -411kJ , calculate the lattice energy of  sodium chloride using an energy cycle diagram.

 

 

Working:

Applying Hess law then:

∆Hf =∆Ha  +∆Hi +∆Ha +∆He +∆Hl

Substituting:

-411= +108kJ  + +500kJ + +121kJ +-364kJ + x

-411 + -108kJ  + -500kJ + -121kJ + +364kJ   = x

x= -776kJmole-1

 

 

 

UPGRADE

CHEMISTRY

                

FORM 4

 Rates of reaction                        Equilibria

      

         Comprehensive tutorial notes

 

 

MUTHOMI S.G

www.kcselibrary.info

                                          0720096206

 

 

 

 

 

A.THE RATE OF CHEMICAL REACTION         

                 (CHEMICAL KINETICS)

 

1.Introduction

The rate of a chemical reaction is the time taken for a given mass/amount of products to be formed. The rate of a chemical reaction is also the time taken for a given mass/amount of reactant to be consumed /used up.

 

Some reactions are too slow to be determined. e.g rusting ,decomposition of hydrogen peroxide and weathering.

 

Some reactions are too fast and instantaneous e.g. neutralization of acid and bases/alkalis in aqueous solution and double decomposition/precipitation.

 

Other  reactions are explosive and very risky to carry out safely e.g. reaction of potassium with water and sodium with dilute acids.

 

The study of the rate of chemical reaction is useful in knowing the factors that influence the reaction so that efficiency and profitability is maximized in industries.

 

Theories of rates of reaction.

 

The rate of a chemical reaction is defined as the rate of change of concentration/amount of reactants in unit time. It is also the rate of formation of given concentration of products in unit time.  i.e.

 

Rate of reaction  = Change in concentration/amount of reactants

Time taken for the change to occur

 

Rate of reaction  = Change in concentration/amount of  products formed

Time taken for the products to form

 

For the above, therefore the rate of a chemical reaction is rate of decreasing reactants to form an increasing product.

The SI unit of time is second(s) but minutes and hours are also used.

 

(a)The collision theory

 

The collision theory is an application of the Kinetic Theory of matter which assumes matter is made up of small/tiny/minute particles like ions atoms and molecules.

The collision theory proposes that

(i)for a reaction to occur, reacting particles must collide.

(ii)not all collisions between reacting particles are successful in a reaction. Collisions that initiate a chemical reaction are called successful / fruitful/ effective collisions

 

(iii)the speed at which particles collide is called collision frequency.

The higher the collision frequency the higher the chances of successful / fruitful/ effective collisions to form products.

 

(iv)the higher the chances of successful collisions, the faster the reaction.

 

(v)the average distance between solid particles from one another is too big for them to meet and collide successfully.

 

(vi)dissolving substances in a solvent ,make the solvent a medium for the reaction to take place.

The solute particle distance is reduced as the particle ions are free to move in the solvent medium.

 

(vii)successful collisions take place if the particles colliding have the required energy  and right orientation which increases their vibration and intensity of successful / fruitful/ effective collisions to form products.

 

(b)The Activation Energy(Ea) theory

 

The Enthalpy of activation(Ha) /Activation Energy(Ea) is the minimum amount of energy which the reactants must overcome before they react. Activation Energy(Ea) is usually required /needed in bond breaking of the reacting particles.

Bond breaking is an endothermic process that require an energy input.

The higher the bond energy the slower the reaction to start of.

Activation energy does not influence whether a reaction is exothermic or endothermic.

 

The energy level diagrams below shows the activation energy for exothermic and endothermic processes/reactions.

 

Energy level diagram showing  the activation energy for exothermic processes /reactions.

Activated complex

   A       B

 

 

A-B    A-B
B
A
A-A    B-B
Ea
Reaction path/coordinate/path
Energy

kJ

 

 

 

Energy level diagram showing  the activation energy for endothermic processes /reactions.

 

Activated complex

   A       B

 

 

Hr
A-B    A-B
A-A    B-B
Ea
B
A
Reaction path/coordinate/path
Energy

kJ

 

The activated complex is a mixture of many intermediate possible products which  may not exist under normal physical conditions ,but can theoretically exist.

Exothermic reaction proceeds without further heating /external energy because it generates its own energy/heat to overcome activation energy.

Endothermic reaction cannot proceed without further heating /external energy because it does not generates its own energy/heat to overcome activation energy. It generally therefore requires continuous supply of more energy/heat to sustain it to completion.

 

  1. Measuring the rate of a chemical reaction.

 

The rate of a chemical reaction can be measure as:

(i)Volume of a gas in unit time;

– if reaction is producing a gas as one of the products.

– if reaction is using a gas as one reactants

(ii)Change in mass of reactants/products for solid products/reactants in unit time.

(iii)formation of a given mass of precipitate in unit time

(iv)a certain mass of reactants to completely form products/diminish.

 

Reactants may be homogenous or heterogenous.

-Homogenous reactions involve reactants in the same phase/state e.g. solid-solid,gas-gas,liquid-liquid.

         -Heterogenous reactions involve reactants in the different phase/state e.g. solid-liquid,gas-liquid,solid-gas.

 

  1. Factors influencing/altering/affecting/determining rate of reaction

 

The following factors alter/influence/affect/determine the rate of a chemical reaction:

(a)Concentration

(b)Pressure

(c) Temperature

(d)Surface area

(e)Catalyst

 

  1. Influence of concentration on rate of reaction

 

The higher the concentration, the higher the rate of a chemical reaction. An increase in concentration of the reactants reduces the distance between the reacting particles increasing their collision frequency to form products.

Practically an increase in concentration reduces the time taken for the reaction to take place.

 

Practical determination of effect of concentration on reaction rate

 

Method 1(a)

Reaction of sodium thisulphate with dilute hydrochloric acid

Procedure:

Measure 20cm3 of 0.05M sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it. Measure 20cm3 of 0.1M hydrochloric acid solution using a 50cm3 measuring cylinder. Put the acid into the beaker containing sodium thisulphate. Immediately start off the stop watch/clock. Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above. Repeat the procedure by measuring different volumes of the acid and adding the volumes of the distilled water to complete table 1.                                                                                            Sample results:Table 1.                                                                                                                                                                                                                                                                                                 

Volume of acid(cm3) Volume of water(cm3) Volume of sodium thiosulphate(cm3) Time taken for mark ‘X’ to be invisible/obscured(seconds) Reciprocal of time

     1  

     t

20.0 0.0 20.0 20.0 5.0 x 10-2
18.0 2.0 20.0 23.0 4.35 x 10-2
16.0 4.0 20.0 27.0 3.7 x 10-2
14.0 6.0 20.0 32.0 3.13 x 10-2
12.0 8.0 20.0 42.0 2.38 x 10-2
10.0 10.0 20.0 56.0 1.78 x 10-2

 

For most examining bodies/councils/boards the above results score for:

(a) complete table as evidence for all the practical work done and completed.

 

(b) (i)Consistent use of a decimal point  on time as evidence of understanding/knowledge of  the degree of accuracy of stop watches/clock.

 

(ii)Consistent use of  a minimum of four decimal points  on inverse/reciprocal of time as evidence of understanding/knowledge of  the degree of accuracy of scientific calculator.

 

(c) accuracy against  a school value based on candidate’s  teachers-results submitted.

 

(d) correct trend (time increase as more water is added/acid  is diluted) in conformity with expected theoretical results.

 

Sample questions    

                                                                                     

  1. On separate graph papers plot a graph of:

(i)volume of acid used(x-axis) against time. Label this graph I

(ii) volume of acid used(x-axis) against 1/t. Label this graph II

 

  1. Explain the shape of graph I

Diluting/adding water is causes a decrease in concentration.

Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products.

 

 

 

 

 

 

Sketch sample  Graph I

Volume of acid(cm3)
    Time

(seconds)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

                                        Sketch sample  Graph II

 

 

 

Volume of acid(cm3)

 

1/t

Sec-1  x 10-2

Volume of acid (cm3)

 

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                    3.From graph II ,determine the time taken for the cross to be obscured/invisible when the volume of the acid is:

 

(i) 13cm3

From a correctly plotted graph

1/t  at 13cm3  on the graph => 2.75 x 10-2

t  =  1 /  2.75 x 10-2   =   36.3636 seconds

 

(ii) 15cm3

From a correctly plotted graph

1/t  at 15cm3  on the graph => 3.35 x 10-2

t  =  1 /  3.35 x 10-2   =   29.8507 seconds

 

(iii) 15cm3

From a correctly plotted graph

1/t  at 17cm3  on the graph => 4.0 x 10-2

t  =  1 /  4.0 x 10-2   =   25.0 seconds

 

(iv) 19cm3

From a correctly plotted graph

1/t  at 19cm3  on the graph => 4.65 x 10-2

t  =  1 /  4.65 x 10-2   =   21.5054 seconds

 

4.From graph II ,determine  the volume of the acid used if the time taken for the cross to be obscured/invisible is:

 

(i)25 seconds

1/t  =>   1/25  = 4.0 x 10-2

Reading from a correctly plotted graph;

4.0 x 10-2 correspond to 17.0 cm3

 

(ii)30 seconds

1/t  =>   1/30  = 3.33 x 10-2

Reading from a correctly plotted graph;

3.33 x 10-2 correspond to 14.7 cm3

 

(iii)40 seconds

 

1/t  =>   1/40  = 2.5 x 10-2

Reading from a correctly plotted graph;

2.5 x 10-2 correspond to 12.3 cm3

 

  1. Write the equation for the reaction taking place

Na2S2O3 (aq) + 2HCl(aq) -> 2NaCl (aq)+ SO2 (g) + S(s) + H2O(l)

 

Ionically:

S2O32- (aq) + 2H+ (aq) ->  SO2 (g) + S(s) + H2O(l)

 

5.Name the yellow precipitate

         Colloidal sulphur

 

 

Method 1(b)

Reaction of sodium thisulphate with dilute hydrochloric acid

 

You are provided with

         2.0M Hydrochloric acid

         0.4M sodium thiosulphate solution

 

Procedure:

Measure 10cm3 of  sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it.

Add 5.0cm3 of hydrochloric acid solution using a 10cm3 measuring cylinder into the beaker containing sodium thisulphate.

Immediately start off the stop watch/clock. Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above.

Repeat the procedure by measuring different volumes of the thiosulphate and adding the volumes of the distilled water to complete table 1.

Sample results:Table 1.                                                                                                                                                                                                                                                                                                     

 

 

Volume of acid(cm3) Volume of water

(cm3)

Volume of sodium thiosulphate

(cm3)

Concentation of sodium thisulphate in molesdm-3 Time(T) taken for mark ‘X’ to be invisible/ obscured(seconds) T-1
5.0 0.0 25.0 0.4 20.0 5.0 x 10-2
5.0 5.0 20.0 0.32 23.0 4.35 x 10-2
5.0 10.0 15.0 0.24 27.0 3.7 x 10-2
5.0 15.0 10.0 0.16 32.0 3.13 x 10-2

Note concentration of diluted solution is got:

C1V1=C2V2                  => 0.4 x 25 = C2x 25  =0.4M

C1V1=C2V2                  => 0.4 x 20 = C2x 25  =0.32M

C1V1=C2V2                  => 0.4 x 15 = C2x 25  =0.24M

C1V1=C2V2                  => 0.4 x 10 = C2x 25  =0.16M

Sample questions    

                                                                                     

  1. On separate graph papers plot a graph of:

(i)Concentration of sodium thiosulphate against time. Label this graph I

(ii)Concentration of sodium thiosulphate against against T-1.Label this graph II

 

  1. Explain the shape of graph I

Diluting/adding water causes a decrease in concentration.

Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products.

From graph II

Determine the time taken if

(i)12cm3 of sodium thisulphate is diluted with 13cm3 of water.

At  12cm3 concentration of sodium thisulphate

= C1V1=C2V2                  => 0.4 x 1 2 = C2x 25  =0.192M

From correct graph at concentration 0.192M => 2.4 x10-2

I/t =  2.4 x10-2   t = 41.6667seconds

(ii)22cm3 of sodium thisulphate is diluted with 3cm3 of water.

At  22cm3 concentration of sodium thisulphate

= C1V1=C2V2                  => 0.4 x 22 = C2x 25  =0.352M

From correct graph at concentration 0.352M => 3.6 x10-2

I/t =  3.6 x10-2   t = 27.7778seconds

 

Determine the volume of water and sodium thiosulphate if  T-1 is 3.0 x10-1

From correct graph at T-1 = 3.0 x10-1   =>  concentration   =  0.65 M

= C1V1=C2V2                  => 0.4 x 25 =  0.65 M  x  V2   =   15.3846cm3

Volume of water   =  25 – 15.3846cm3 =  9.6154cm3

Determine the concentration of hydrochloric acid if 12cm3 of sodium thiosulphate and 13cm3 of water was used.

At  12cm3 concentration of sodium thisulphate

= C1V1=C2V2                  => 0.4 x 1 2 = C2x 25  =0.192M

Mole ratio Na2S2 O3 :HCl =1:2

 

Moles of Na2S2 O3 =        0.192M x 12          =>   2.304 x 10-3  moles

                                                      1000

Mole ratio  HCl =2.304 x 10-1  moles  = 1.152 x 10-3  moles

2

Molarity o f HCl =  1.152 x 10-3  moles x 1000   = 0.2304M

5.0

 

 

 

Method 2

Reaction of Magnesium with dilute hydrochloric acid

Procedure

 

Scub 10centimeter length of magnesium ribbon with sand paper/steel wool. Measure 40cm3 of 0.5M dilute hydrochloric acid into a  flask .Fill a graduated gas jar with water and invert it into a trough. Stopper the flask and set up the apparatus to collect the gas produced as in the set up below:

 

Hydrogen gas
Graduated gas jar
Magnesium ribbon
Hydrochloric acid

 

Carefully remove the stopper, carefully put the magnesium ribbon into the flask . cork tightly. Add the acid into the flask. Connect the delivery tube into the gas jar. Immediately start off the stop watch and determine the volume of the gas produced after every 30 seconds to complete table II below.

 

Sample results: Table II

 

Time(seconds) 0 30 60 90 120 150 180 210 240
Volume of gas produced(cm3) 0.0 20.0 40.0 60.0 80.0 90.0 95.0 96.0 96.0

 

Sample practice questions

 

1.Plot a graph of volume of gas produced (y-axis) against time

 

 

                                                                                                                                                                                                                                                                                                                                                              2.Explain the shape of the graph.

The rate of reaction is faster when the concentration of the acid is high .

As time goes on, the concentration of the acid decreases and therefore less gas is produced.

When all the acid has reacted, no more gas is produced after 210 seconds and the graph flattens.

 

3.Calculate the rate of reaction at 120 seconds

 

From a tangent at 120 seconds rate of reaction = Change in volume of gas

Change in time

=> From the tangent at 120seconds V2 – V1  =   96-84    =  12    =   0.2cm3sec-1

T2 – T1       150-90        60

 

  1. Write an ionic equation for the reaction taking place.

         Mg2+(s)  +  2H+(aq)  ->  Mg2+(aq)  + H2 (g)

 

  1. On the same axis sketch then explain the curve that would be obtained if:

         (i) 0.1 M hydrochloric acid is used –Label this curve I

         (ii)1.0 M hydrochloric acid is used –Label this curve II

 

Observation:

Curve I is to the right

Curve II is to the left

Explanation

 

A decrease in concentration shift the rate of reaction graph to the right as more time is taken for completion of the reaction.

An increase in concentration shift the rate of reaction graph to the left as less time is taken for completion of the reaction.

Both graphs flatten after some time indicating the completion of the reaction.

 

b)Influence of pressure on rate of reaction

 

Pressure affects only gaseous reactants.

An increase in pressure reduces the volume(Boyles law) in which the particles are contained.

Decrease in volume of the container bring the reacting particles closer to each other which increases their chances of effective/successful/fruitful collision to form products.

An increase in pressure therefore increases the rate of reaction by reducing the time for reacting particles of gases to react.

At industrial level, the following are some reactions that are affected by pressure:

(a)Haber process for manufacture of ammonia

N2(g) + 3H2(g) -> 2NH3(g)

 

(b)Contact process for manufacture of sulphuric(VI)acid

2SO2(g) + O2(g) -> 2SO3(g)

 

(c)Ostwalds process for the manufacture of nitric(V)acid

4NH3(g) + 5O2(g) -> 4NO (g) + 6H2O (l)

 

The influence of pressure on reaction rate is not felt in solids and liquids.

This is because the solid and liquid particles have fixed positions in their strong bonds and therefore no degree of freedom (Kinetic Theory of matter)

 

c)Influence of temperature on rate of reaction

 

An increase in temperature increases the kinetic energy of the reacting particles by increasing their collision frequency.

Increase in temperature increases the particles which can overcome the activation energy (Ea).

A 10oC rise in temperature doubles the rate of reaction by reducing the time taken for the reaction to complete by a half.

 

Practical determination of effect of Temperature on reaction rate

Method 1

 

Reaction of sodium thisulphate with dilute hydrochloric acid

 

Procedure:

Measure 20cm3 of 0.05M sodium thisulphate into a 50cm3 glass beaker.

Place the beaker on a white piece of filter paper with ink mark ‘X’ on it.

Determine and record its temperature as room temperature in table 2 below.

Measure 20cm3 of 0.1M hydrochloric acid solution using a 50cm3 measuring cylinder.

Put the acid into the beaker containing sodium thisulphate.

Immediately start off the stop watch/clock.

Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above.

Measure another 20cm3 separate portion of the thisulphate into a beaker, heat the solution to 30oC.

Add the acid into the beaker and repeat the procedure above. Complete table 2 below using different temperatures of the thiosulphate.

 

Sample results:Table 2.

 

Temperature of Na2S2O3 Room temperature 30 40 50 60
Time taken for mark X to be obscured /invisible (seconds) 50.0 40.0 20.0 15.0 10.0
Reciprocal of time(1/t) 0.02 0.025 0.05 0.0667 0.1

 

Sample practice questions

  1. Plot a graph of temperature(x-axis) against 1/t                                                                       

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                 

 

 

2(a)From your graph determine the temperature at which:

 

(i)1/t is ;

  1. 0.03

Reading directly from a correctly plotted graph = 32.25 oC

  1. 0.07

Reading directly from a correctly plotted graph = 48.0 oC

 

(ii) t is;

  1. 30 seconds

                  30 seconds => 1/t =1/30  =0.033

Reading directly from a correctly plotted graph 0.033 => 33.5 oC

 

  1. 45 seconds

                   45 seconds => 1/t =1/45  =0.022

Reading directly from a correctly plotted graph 0.022 => 29.0 oC

 

III. 25 seconds

                   25 seconds => 1/t =1/25  =0.04

Reading directly from a correctly plotted graph 0.04 => 36.0 oC

 

 (b) From your graph determine the time taken for the cross to become invisible at:

(i) 57.5 oC

Reading directly from a correctly plotted graph at 57.5 oC= 0.094

=>1/t = 0.094

t= 1/0.094  => 10.6383 seconds

 

(ii) 45 oC

Reading directly from a correctly plotted graph at 45 oC = 0.062

=>1/t = 0.062

t= 1/0.094  => 16.1290 seconds

 

(iii) 35 oC

Reading directly from a correctly plotted graph at 35 oC = 0.047

=>1/t = 0.047

t= 1/0.047  => 21.2766 seconds

 

Method 2

Reaction of Magnesium with dilute hydrochloric acid

Procedure

 

Scub 5centimeter length of magnesium ribbon with sand paper/steel wool.

Cut the piece into five equal one centimeter smaller pieces.

Measure 20cm3 of 1.0M dilute hydrochloric acid into a glass beaker .

Put one piece of the magnesium ribbon into the acid, swirl.

Immediately start off the stop watch/clock.

Determine the time taken for the effervescence/fizzing/bubbling to stop when viewed from above.

Record the time in table 2 at room temperature.

Measure  another  20cm3 portions of 1.0M dilute hydrochloric acid into a clean beaker.

Heat separately one portion to 30oC, 40oC , 50oC and 60oC  and adding 1cm length of the ribbon and determine the time taken for effervescence /fizzing /bubbling to stop when viewed from above .

Record each time to complete table 2 below using different temperatures of the acid.

 

Sample results:Table 1.

 

Temperature of acid(oC) Room temperature 30 40 50 60
Time taken effervescence to stop (seconds) 80.0 50.0 21.0 13.5 10.0
Reciprocal of time(1/t) 0.0125 0.02 0.0476 0.0741 0.1

                                                       Sample practice questions

  1. Plot a graph of temperature(x-axis) against 1/t

 

 

 

Temperature(oC)
1/t

 

 

 

 

2.(a)Calculate the number of moles of magnesium used given that 1cm of magnesium has a mass of 1g.(Mg= 24.0)

 

Moles = Mass of magnesium   =>   1.0   =  4.167  x 10 -2 moles

Molar mass of Mg         24

 

(b)Calculate the number of moles of hydrochloric acid used

                 Moles of acid = molarity x volume of acid

1000

=> 1.0 x 20     = 2.0 x 10 -2 moles

1000

 

(c)Calculate the mass  of magnesium that remain unreacted

 

                 Mole ratio Mg: HCl = 1:2

Moles Mg = ½ moles HCl

=> ½  x 2.0 x 10 -2 moles = 1.0 x 10 -2 moles

                 Mass of reacted Mg  = moles x molar mass

=> 1.0 x 10 -2 moles x 24 = 0.24 g

                 Mass of unreacted Mg = Original total mass – Mass of reacted Mg

=> 1.0 g – 0.24 = 0.76 g

 

(b)Calculate the total volume of hydrogen gas produced during the above reactions.

 

Mole ratio Mg : H2 = 1:1

Moles of Mg that reacted per experiment = moles H2 =1.0 x 10 -2 moles

Volume of Hydrogen at s.t.p produced per experiment = moles x 24 dm3

=> 1.0 x 10 -2 moles x 24 dm3 = 0.24dm3

Volume of Hydrogen at s.t.p produced  in 5 experiments =0.24 dm3 x 5

= 1.2 dm3

 

3.(a)At what temperature was the time taken for magnesium to react equal to:

         (i)70seconds

         70 seconds => 1/t =1/70  =0.01429

Reading directly from a correctly plotted graph 0.01429 => 28.0 oC

 

(ii)40seconds

         40 seconds => 1/t =1/40  =0.025

Reading directly from a correctly plotted graph 0.025 => 32.0 oC

 

    (b)What is  the time taken for magnesium to react if the reaction was done at:

         (i) 55.0 oC

Reading directly from a correctly plotted graph at 55.0 oC=> 1/t = 8.0 x 10-2

=> t = 1/8.0 x 10-2  =  12.5 seconds

 

(ii) 47.0 oC

Reading directly from a correctly plotted graph at 47.0 oC=> 1/t = 6.0 x 10-2

=> t = 1/6.0 x 10-2  =  16.6667 seconds

 

         (iii) 33.0 oC

Reading directly from a correctly plotted graph at 33.0 oC=> 1/t = 2.7 x 10-2

=> t = 1/2.7 x 10-2  =  37.037 seconds

 

  1. Explain the shape of the graph.

 

Increase in temperature increases the rate of reaction as particles gain kinetic energy increasing their frequency and intensity of collision to form products.

 

d)Influence of surface area on rate of reaction

 

Surface area is the area of contact. An increase in surface area is a decrease in particle size. Practically an increase in surface area involves chopping /cutting solid lumps into smaller pieces/chips then crushing the chips into powder. Chips thus have a higher surface area than solid lumps but powder has a highest surface area.

An increase in surface area of solids increases the area of contact with a liquid solution increasing the chances of successful/effective/fruitful collision to form products. The influence of surface area on rate of reaction is mainly in heterogeneous reactions.

 

Reaction of chalk/calcium carbonate on dilute hydrochloric acid  

Procedure

Measure 20cm3 of 1.0 M hydrochloric acid into three separate conical flasks labeled C1 C2 and C3 .

Using a watch glass weigh three separate 2.5g  a piece of white chalk. Place the conical flask C1 on an electronic balance.

Reset the balance scale to 0.0.

Put one weighed sample of the chalk into the acid in the conical flask. Determine the scale reading and record it at time =0.0.

Simultaneously start of the stop watch.

Determine and record the scale reading after every 30 seconds to complete Table I .

Repeat all the above procedure separately with C2 and C3  to complete Table II and Table III  by cutting the chalk into small pieces/chips  for C2 and crushing the chalk to powder  for C3

Sample results:Table 1.

 

Time(seconds) 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 240.0
Mass of CaCO3 2.5 2.0 1.8 1.4 1.2 1.0 0.8 0.5 0.5
Loss in mass 0.0 0.5 0.7 1.1 1.3 1.5 1.7 2.0 2.0

Sample results:Table 1I.

 

Time(seconds) 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 240.0
Mass of CaCO3 2.5 1.9 1.5 1.3 1.0 0.8 0.5 0.5 0.5
Loss in mass 0.0 0.6 1.0 1.2 1.5 1.7 2.0 2.0 2.0

 

Sample results:Table III.

 

Time(seconds) 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 240.0
Mass of CaCO3 2.5 1.8 1.4 1.0 0.8 0.5 0.5 0.5 0.5
Loss in mass 0.0 0.7 1.1 1.5 1.7 2.0 2.0 2.0 2.0

Sample questions:

 

1.Calculate the loss in mass made at the end of each time from the original to complete table I,II and III

 

2.On the same axes plot a graph of total loss in mass against time (x-axes) and label them curve I, II, and III from Table I, II, and III.

 

3.Explain why there is a loss in mass in all experiments.

Calcium carbonate react with the acid to form carbon(IV)oxide gas that escape to the atmosphere.

 

4.Write an ionic equation for the reaction that take place

CaCO3(s) + 2H+(aq) -> Ca2+(aq) + H2O(l) + CO2(g)

 

5.Sulphuric(VI)acid cannot be used in the above reaction. On the same axes sketch the curve which would be obtained if the reaction was attempted by reacting a piece of a lump of chalk with 0.5M sulphuric(VI)acid. Label it curve IV. Explain the shape of curve IV.

 

Calcium carbonate would react with dilute  0.5M sulphuric(VI)acid to form insoluble calcium sulphate(VI) that coat /cover unreacted Calcium carbonate  stopping the reaction from reaching completion.

 

6.Calculate the volume of carbon(IV)oxide evolved(molar gas volume at room temperature = 24 dm3, C= 12.0, O= 16.O Ca=40.0)

Method I

 Mole ratio  CaCO3(s) : CO2(g) = 1:1

Moles CaCO3(s) used     = Mass CaCO3(s)                =  0.025 moles

Molar mass CaCO3(s)

 

Moles CO2(g) = 0.025 moles

 

Volume of  CO2(g)    = moles x molar gas volume

=>0.025 moles   x  24 dm3      = 0.600 dm3/600cm3

 

Method II

 Molar mass of CaCO3(s) = 100g produce 24 dm3 of CO2(g)

Mass of CaCO3(s) =2.5 g produce  2.5 x 24        =  0.600dm3

100

7.From curve I ,determine the rate of reaction (loss in mass per second)at time 180 seconds on the curve.

 

From  tangent at 180 seconds on curve I

Rate     =    M2-M1    =>  2.08 – 1.375   =  0.625   = 0.006944g sec-1

T2– T1                222-132             90

 

8.What is the effect of particle size on the rate of reaction?

A larger surface area is a reduction in particle size which increases the area of contact between reacting particles increasing their collision frequency.

 

Theoretical examples

 

  1. Excess marble chips were put in a beaker containing 100cm3 of 0.2M hydrochloric acid. The beaker was then placed on a balance and total loss in mass recorded after every two minutes as in the table below.

 

Time(minutes) 0.0 2.0 4.0 6.0 8.0 10.0 12.0
Loss in mass(g) 0.0 1.80 2.45 2.95 3.20 3.25 3.25

 

(a)Why was there a loss in mass?

Carbon (IV) oxide gas was produced that escape to the surrounding

 

(b)Calculate the average rate of loss in mass between:

         (i) 0 to 2 minutes

         Average rate =M2-M1    =>  1.80 – 0.0       =  1.8   = 9.00g min-1

T2– T1                2.0 – 0.0            2

 

         (i) 6 to 8 minutes

         Average rate =M2-M1    =>  3.20 – 2.95       =  0.25   = 0.125g min-1

T2– T1                8.0 – 6.0              2

 

(iii) Explain the difference between the average rates of reaction in (i) and(ii) above.

Between 0 and 2 minutes , the concentration of marble chips and hydrochloric acid is high therefore there is a higher collision frequency between the reacting particles leading to high successful rate of formation of products.

Between 6 and 8 minutes , the concentration of marble chips and hydrochloric acid is low therefore there is low collision frequency between the reacting particles leading to less successful rate of formation of products.

(c)Write the equation for the reaction that takes place.

CaCO3(s) + 2HCl (aq) -> CaCO3 (aq) + H2O(l) + CO2(g)

 

(d)State and explain three ways in which the rate of reaction could be increased.

         (i)Heating the  acid- increasing the temperature of the reacting particles increases their kinetic energy and thus collision frequency.

(ii)Increasing the concentration of the acid-increasing in concentration reduces the distances between the reacting particles increasing their chances of effective/fruitful/successful collision to form products faster.

(iii)Crushing the marble chips to powder-this reduces the particle size/increase surface area increasing the area of contact between reacting particles.

 

(e)If the solution in the beaker was evaporated to dryness then left overnight in the open, explain what would happen.

It becomes wet because calcium (II) chloride absorbs water from the atmosphere and form solution/is deliquescent.

 

(f)When sodium sulphate (VI) was added to a portion of the contents in the beaker after the reaction , a white precipitate was formed .

         (i)Name the white precipitate.

                 Calcium(II)sulphate(VI)

 

(ii)Write an ionic equation for the formation of the white precipitate

                 Ca2+(aq)  + SO42-(aq)->CaSO4(s)

 

(iii)State one use of the white precipitate

                 -Making plaster for building

-Manufacture of plaster of Paris

-Making sulphuric(VI)acid

 

(g)(i) Plot a graph of total loss in mass(y-axes) against time

(ii)From the graph, determine the rate of reaction at time 2 minutes.

From a tangent/slope  at 2 minutes;

Rate  of reaction = Average rate =M2-M1 => 2.25 – 1.30  = 0.95 = 0.3958g min-1

T2– T1        3.20 – 0.8             2.4

(iii)Sketch on the same axes the graph that would be obtained if 0.02M hydrochloric acid was used. Label it curve II

 

  1. e) Influence of catalyst on rate of reaction

 

Catalyst is a substance that alter the rate /speed of a chemical reaction but remain chemically unchanged at the end of a reaction. Biological catalysts are called enzymes. A catalyst does not alter the amount of products formed but itself may be altered physically e.g. from solid to powder to fine powder. Like biological enzymes, a catalyst only catalyse specific type of reactions

Most industrial catalysts are transition metals or their compounds. Catalyst works by lowering the Enthalpy of activation(∆Ha)/activation energy (Ea)  of the reactants .The catalyst lowers the Enthalpy of activation(∆Ha)/activation energy (Ea) by:

(i) forming short lived intermediate compounds called activated complex that break up to form the final product/s

(ii) being absorbed by the reactants thus providing the surface area on which reaction occurs.

A catalyst  has no effect on the enthalpy of reaction ∆Hr but only lowers the  Enthalpy of activation(∆Ha)/activation energy (Ea)It thus do not affect/influence whether the reaction is exothermic or endothermic as shown in the energy level diagrams below.

Energy level diagram showing  the activation energy for exothermic processes /reactions.

Activated complex

   A       B

 

 

Ea uncatalysed
A-B    A-B
B
A
A-A    B-B
Reaction path/coordinate/path
Energy

kJ

Ea Catalysed

 

 

 

 

 

 

Energy level diagram showing  the activation energy for endothermic processes /reactions.

Activated complex

   A       B

                                          

Hr
A-B    A-B
A-A    B-B
Ea
B
A
Reaction path/coordinate/path
Energy

kJ

 

 

 

 

 

 

 

The following are some catalysed reaction processes.

 

(a)The contact process

Vanadium(V) Oxide(V2O5)  or platinum(Pt) catalyses the oxidation of sulphur(IV)oxide during the manufacture of sulphuric(VI) acid from contact process.

SO2(g)      +      O2(g)  —-V2O5–>   SO3(g)

 

To reduce industrial cost of manufacture of sulphuric (VI) acid from contact process Vanadium(V) Oxide(V2O5)  is used because it is cheaper though it is easily poisoned by impurities.

 

(b)Ostwalds process

Platinum promoted with Rhodium catalyses the oxidation of ammonia to nitrogen(II)oxide and water during the manufacture of nitric(V)acid

4NH3(g)    +      5O2(g)  —-Pt/Rh–>   4NO (g) + 6H2O(l)

 

(c)Haber process

Platinum or iron catalyses the combination of nitrogen and hydrogen to form ammonia gas

N2(g)  + 3H2(g)  —Pt or Fe—> 2NH3(g)

 

(d)Hydrogenation/Hardening of oil to fat

 

Nickel (Ni) catalyses the hydrogenation of unsaturated compound containing              – C=C-  or   –C=C- to saturated compounds without double or triple bond

This process is used is used in hardening oil to fat.

(e)Decomposition of hydrogen peroxide

Manganese(IV)oxide speeds up the rate of decomposition of hydrogen peroxide to water and oxygen gas.

This process/reaction is used in the school laboratory preparation of Oxygen.

2H2O2 (g)   —-MnO2–>   O2(g)  + 2H2O(l)

 

(f)Reaction of metals with dilute sulphuric(VI)acid

Copper(II)sulphate(VI) speeds up the rate of production of hydrogen gas from the reaction of Zinc and dilute sulphuric(VI)acid.

This process/reaction is used in the school laboratory preparation of Hydrogen.

 

H2 SO4 (aq) +  Zn(s)  —-CuSO4–>   ZnSO4 (aq)  +  H2(g)

 

(g) Substitution reactions

When placed in bright sunlight or U.V /ultraviolet light , a mixture of a halogen and an alkane undergo substitution reactions explosively to form halogenoalkanes. When paced in diffused sunlight the reaction is very slow.

 

e.g.  CH4(g)    +    Cl2(g)  —u.v. light–>   CH3Cl(g)   +   HCl(g)

 

(h)Photosynthesis

Plants convert carbon(IV)oxide gas from the atmosphere and water from the soil to form glucose and oxygen as a byproduct using sunlight / ultraviolet  light.

 

6CO2(g)    +    6H2O(l)  —u.v. light–>   C6H12O6(g)   +   O2(g)

 

(i)Photography

Photographic film contains  silver bromide emulsion which decomposes to silver and bromine on exposure to sunlight.

2AgBr(s)  —u.v/sun light–> 2Ag(s)  + Br2(l)

 

When developed, the silver deposits give the picture of the object whose photograph was taken depending on intensity of light. A picture photographed in diffused light is therefore blurred.

 

 

 

Practical determination of effect of catalyst on decomposition of hydrogen peroxide

 

Measure 5cm3 of 20 volume hydrogen peroxide and then dilute to make 40cm3 in a measuring cylinder by adding distilled water.

Divide it into two equal portions.

(i)Transfer one 20cm3volume hydrogen peroxide into a conical/round bottomed/flat bottomed flask. Cork and swirl for 2 minutes. Remove the cork. Test the gas produced using a glowing splint. Clean the conical/round bottomed/flat bottomed flask.

 

(ii)Put 2.0g of Manganese (IV) oxide into the clean conical/round bottomed/flat bottomed flask. Stopper the flask.

Transfer the second portion of the 20cm3volume hydrogen peroxide into a conical/round bottomed/flat bottomed flask through the dropping/thistle funnel. Connect the delivery tube to a calibrated/graduated gas jar as in the set up below.

Start off the stop watch and determine the volume of gas in the calibrated/graduated gas jar after every 30 seconds to complete Table 1.

 

(iii)Weigh a filter paper .Use the filter paper to filter the contents of the conical conical/round bottomed/flat bottomed flask. Put the residue on a sand bath to dry. Weigh the dry filter paper again .Determine the new mass Manganese (IV) oxide.

 

 

Time(seconds) 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 240.0 270.0
Volume of gas (cm3) 0.0 20.0 40.0 60.0 80.0 90.0 95.0 96.0 96.0 96.0

 

 

 

 

Mass of MnO2 before reaction(g) Mass of MnO2 after reaction(g)
                    2.0                        2.0

 

Plot a graph of volume of gas produced against time(x-axes)

Catalysed reaction
Uncatalysed reaction

 

 

 

 

 

  1. b) On the same axes, plot a graph of the uncatalysed reaction.

(c) Explain the changes in mass of manganese(IV)oxide before and after the reaction.

The mass of MnO2 before and after the reaction is the same but a more fine powder after the experiment. A catalyst therefore remains unchanged chemically but may physically change.

 

 

B.EQUILIBRIA (CHEMICAL CYBERNETICS)

 

Equilibrium is a state of balance.

Chemical equilibrium is state of balance between the reactants and products.

As reactants form products, some products form back the reactants.

Reactions in which the reactants form products to completion are said to be reversible i.e.

A      +      B     ->     C     +      D

Reactions in which the reactants form products and the products can reform the reactants are said to be reversible.

A      +      B               C     +      D

Reversible reactions may be:

(a)Reversible physical changes

(b)Reversible chemical changes

(c)Dynamic equilibrium

 

(a)Reversible physical changes

 

Reversible physical change is one which involves:

(i) change of state/phase from solid, liquid, gas or  aqueous solutions. States of matter are interconvertible and a reaction involving a change from one state/phase can be reversed back to the original.

(ii) colour changes. Some substances/compounds change their colours without change in chemical substance.

 

Examples of reversible physical changes

 

(i) colour change on heating and cooling:

 

  1. Zinc(II)Oxide changes from white when cool/cold to yellow when hot/heated and back.

ZnO(s)                               ZnO(s)

(white when cold)            (yellow when hot)

 

  1. Lead(II)Oxide changes from yellow when cold/cool to brown when hot/heated and back.

PbO(s)                                PbO(s)

(brown when hot)             (yellow when cold)

 

(ii)Sublimation

 

  1. Iodine sublimes from a grey crystalline solid on heating to purple vapour. Purple vapour undergoes deposition back to the grey crystalline solid.

I2(s)                                    I2(g)

(grey crystalline solid                 (purple vapour

undergo sublimation)                 undergo deposition)

 

  1. Carbon (IV)oxide gas undergoes deposition from a colourless gas to a white solid at very high pressures in a cylinder. It sublimes back to the colourless gas if pressure is reduced

CO2(s)                         CO2(g)

(white powdery solid         (colourless/odourless gas

undergo sublimation)                 undergo deposition)

 

(iii)Melting/ freezing and boiling/condensation

Ice on heating undergo melting to form a liquid/water. Liquid/water on further heating boil/vaporizes to form gas/water vapour. Gas/water vapour on cooling, condenses/liquidifies to water/liquid. On further cooling, liquid water freezes to ice/solid.

H2O(s)
H2O(l)
H2O(s)

Melting                      boiling

Freezing                    condensing

 

(iv)Dissolving/ crystallization/distillation

Solid crystals of soluble substances (solutes) dissolve in water /solvents to form a uniform mixture of the solute and solvent/solution. On crystallization /distillation /evaporation the solvent evaporate leaving a solute back. e.g.

NaCl(s)     +  aq                          NaCl(aq)

 

(b)Reversible chemical changes

 

These are reactions that involve a chemical change of the reactants which can be reversed back by recombining the new substance formed/products.

Examples of Reversible chemical changes

 

(i)Heating Hydrated salts/adding water to anhydrous salts.

 

When hydrated salts are heated they lose some/all their water of crystallization  and become anhydrous.Heating an unknown substance /compound that forms a colourless liquid droplets on the cooler parts of a dry test/boiling tube is in fact a confirmation inference that the substance/compound  being heated is hydrated.

When anhydrous salts are added (back) some water they form hydrated compound/salts.

 

Heating Copper(II)sulphate(VI)pentahydrate and cobalt(II)chloride hexahydrate

(i)Heat about 5.0g of Copper(II)sulphate(VI) pentahydrate in a clean dry test tube until there is no further colour change on a small Bunsen flame. Observe any changes on the side of the test/boiling tube. Allow the boiling tube to cool.Add about 10 drops of distilled water. Observe any changes.

(ii)Dip a filter paper in a solution of cobalt(II)chloride hexahydrate. Pass one end the filter paper to a small Bunsen flame repeatedly. Observe any changes on the filter paper. Dip the paper in a beaker containing distilled water. Observe any changes.

 

Sample observations

Hydrated compound Observation before heating Observation after heating Observation on adding water
Copper(II)sulphate

(VI) pentahydrate

Blue crystalline solid (i)colour changes from blue to white.

(ii)colourless liquid forms on the cooler parts of boiling / test tube

(i)colour changes from white to blue

(ii)boiling tube becomes warm /hot.

Cobalt(II)chloride hexahydrate Pink crystalline solid/solution (i)colour changes from pink to blue.

(ii) colourless liquid forms on the cooler parts of boiling / test tube (if crystal are used)

 

(i)colour changes from blue to pink

(ii)boiling tube becomes warm/hot.

When blue Copper(II)sulphate (VI) pentahydrate is heated, it loses the five molecules of water of crystallization to form white anhydrous Copper(II)sulphate (VI).Water of crystallization  form and condenses as colourless droplets on the cooler parts of a dry boiling/test tube.

This is a chemical change that produces a new substance. On adding drops of water to an anhydrous white copper(II)sulphate(VI) the hydrated compound is formed back. The change from hydrated to anhydrous and back is therefore reversible chemical change.Both anhydrous white copper(II)sulphate(VI) and blue cobalt(II)chloride hexahydrate are therefore used to test for the presence of water when they turn to blue and pink respectively.

CuSO4(s)       +      5H2 O(l)                     CuSO4.5H2 O(s/aq)

      (white/anhydrous)                                                   (blue/hydrated)

 

CoCl2(s)       +      6H2 O(l)                               CoCl2.6H2 O(s/aq)

      (blue/anhydrous)                                                      (pink/hydrated)

 

(ii)Chemical sublimation

Some compounds sublime from solid to gas by dissociating into new different compounds. e.g.

 

Heating ammonium chloride

(i)Dip a glass rod containing concentrated hydrochloric acid. Bring it near the mouth of a bottle containing concentrated ammonia solution. Explain the observations made.

When a glass rod containing hydrogen chloride gas is placed near ammonia gas, they react to form ammonium chloride solid that appear as white fumes.

This experiment is used interchangeably to test for the presence of hydrogen chloride gas (and hence Cl ions) and ammonia gas (and hence NH4+ ions)

 

(ii)Put 2.0 g of ammonium chloride in a long dry boiling tube. Place wet / moist /damp blue and red litmus papers separately on the sides of the mouth of the boiling tube. Heat the boiling tube gently then strongly. Explain the observations made.

When ammonium chloride is heated it dissociates into ammonia and hydrogen chloride gases. Since ammonia is less dense, it diffuses faster to turn both litmus papers blue before hydrogen chloride turn red because it is denser. The heating and cooling of ammonium chloride is therefore a reversible chemical change.

NH3(g)           +        HCl(g)                       NH4Cl(s)

(Turns moist              (Turns moist          (forms white fumes)

litmus paper blue)        litmus paper red)

 

 

 

 

 

 

(c)Dynamic equilibria

 

For reversible reactions in a closed system:

(i) at the beginning;

-the reactants are decreasing in concentration with time

-the products are increasing in concentration with time

(ii) after some time a point is reached when as the reactants are forming products the products are forming reactants. This is called equilibrium.

Reactants concentration decreases to form products

Sketch showing the changes in concentration of reactants and products in a closed system

 

Equilibrium  established /rate of formation of products equal to rate of formation of reactants.

 

 

 

 

Reaction progress/path/coordinate
Concentration

Mole dm-3

Products concentration increases from time=0.0

For a system in equilibrium:

(i) a reaction from left to right (reactants to products) is called forward reaction.

(ii) a reaction from right to left (products to reactants) is called backward reaction.

(iii)a reaction in which the rate of forward reaction is equal to the rate of backward reaction is called a dynamic equilibrium.

 

A dynamic equilibrium is therefore a balance of the rate of formation of products and reactants. This balance continues until the reactants or products are disturbed/changed/ altered.

The influence of different factors on a dynamic equilibrium was first investigated from 1850-1936 by the French Chemist Louis Henry Le Chatellier. His findings were called Le Chatelliers Principle which states that:

 

if a stress/change is applied to a system in dynamic equilibrium, the system readjust/shift/move/behave so as to remove/ reduce/ counteract/ oppose the stress/change

 

Le Chatelliers Principle is applied in determining the effect/influence of several factors on systems in dynamic equilibrium. The following are the main factors that influence /alter/ affect systems in dynamic equilibrium:

(a)Concentration

(b)Pressure

(c)Temperature

(d)Catalyst

 

(a)Influence of concentration on dynamic equilibrium

 

An increase/decrease in concentration of reactants/products at equilibrium is a stress. From Le Chatelliers principle the system redjust so as to remove/add the excessreduced concentration.

 

Examples of influence of concentration on dynamic equilibrium

(i)Chromate(VI)/CrO42- ions in solution are yellow. Dichromate(VI)/Cr2O72- ions in solution are orange. The two solutions exist in equilibrium as in the equation:

 

2H+ (aq)  +  2CrO42- (aq)              Cr2O72-  (aq) + H2O(l)

(Yellow)                    (Orange)

 

  1. I. If an acid is/H+ (aq) is added to the equilibrium mixture a stress is created on the reactant side where there is already H+ The equilibrium shift forward to the right to remove/reduce the excess H+ ions added. Solution mixture becomes More Cr2O72- ions formed in the solution mixture make it to be more orange in colour.

 

  1. II. If a base/OH (aq) is added to the equilibrium mixture a stress is created on the reactant side on the H+ H+ ions react with OH (aq) to form water.

H+ (aq) +OH (aq) -> H2O(l)

 

The equilibrium shift backward to the left to add/replace the H+ ions that have reacted with the OH (aq) ions . More of the CrO42- ions formed in the solution mixture makes it to be more yellow in colour.

 

2OH (aq)  +  2Cr2O72- (aq)                                      CrO42-  (aq)    +   H2O(l)

(Orange)                                (Yellow)

 

  1. I. If an acid/ H+ (aq) is added to the equilibrium mixture a stress is created on the reactant side on the OH (aq). H+ ions react with OH (aq) to form water.

H+ (aq) +OH (aq) -> H2O(l)

 

The equilibrium shift backward to the left to add/replace the 2OH (aq) that have reacted with the H+ (aq) ions . More Cr2O72- (aq)ions formed in the solution mixture makes it to be more Orange in colour.

 

  1. II. If a base /OH (aq) is added to the equilibrium mixture a stress is created on the reactant side where there is already OH (aq) ions. The equilibrium shift forward to the right to remove/reduce the excess OH (aq) ions added. More of the Cr2O72- ions are formed in the solution mixture making it to be more orange in colour.

 

(i)Practical determination of the influence of alkali/acid on Cr2O72-  / CrO42-  equilibrium mixture

 

Measure about 2 cm3 of Potassium dichromate (VI) solution into a test tube.

Note that the solution mixture is orange.

Add three drops of 2M sulphuric(VI) acid. Shake the mixture carefully.

Note that the solution mixture is remains orange.

Add about six drops of 2M sodium hydroxide solution. Shake carefully.

Note that the solution mixture is turns yellow.

Explanation

The above observations can be explained from the fact that both the dichromate(VI)and chromate(VI) exist in equilibrium. Dichromate(VI) ions are stable in acidic solutions while chromate(VI)ions are stable in basic solutions. An equilibrium exist thus:

Cr2O72-
CrO42-

            OH-                             

                        H+

When an acid is added, the equilibrium shift forward to the right and the mixture become more orange as more Cr2O72- ions exist.

When a base is added, the equilibrium shift backward to the left and the mixture become more yellow as more CrO42- ions exist.

 

(ii)Practical determination of the influence of alkali/acid on bromine water in an equilibrium mixture

 

Measure 2cm3 of bromine water into a boiling tube. Note its colour.

Bromine water is yellow

Add three drops of 2M sulphuric(VI)acid. Note  any colour change

Colour becomes more yellow

Add seven drops of 2M sodium hydroxide solution. Note any colour change.

Solution mixture becomes colourless/Bromine water is decolourized.

Explanation

When added distilled water,an equilibrium exist between bromine liquid (Br2(aq)) and the bromide ion(Br), hydrobromite ion(OBr) and hydrogen ion(H+) as in the equation:

 

H2O(l)  + Br2(aq)                          OBr (aq) +  H+ (aq)  +  Br (aq)

If an acid (H+)ions is added to the equilibrium mixture, it increases the concentration of the ions on the product side which shift backwards to the left to remove the excess H+ ions on the product side making the colour of the solution mixture more yellow.

If a base/alkali OH is added to the equilibrium mixture, it reacts with H+ ions on the product side to form water.

H+ (aq)+ OH(aq) -> H2O(l)

This decreases the concentration of the H+ ions on the product side which shift the equilibrium forward to the right to replace H+ ions making the solution mixture colourless/less yellow (Bromine water is decolorized)

 

(iii)Practical determination of the influence of alkali/acid on common acid-base indicators.

Place 2cm3 of phenolphthalein ,methyl orange and litmus solutions each in three separate test tubes.

To each test tube add two drops of water. Record your observations in Table 1 below.

To the same test tubes, add three drops of 2M sulphuric(VI)acid. Record your observations in Table 1 below.

To the same test tubes, add seven drops of 2M sodium hydroxide solution. Record your observations in Table 1 below.

To the same test tubes, repeat adding four drops of 2M sulphuric(VI)acid.    Table 1

Indicator Colour of indicator in
Water Acid(2M sulphuric (VI) acid) Base(2M sodium hydroxide)
Phenolphthalein Colourless Colourless Pink
Methyl orange Yellow Red Orange
Litmus solution Colourless Red Blue

 

Explanation

An indicator is a substance which shows whether another substance is an acid , base or neutral.

Most indicators can be regarded as very weak acids that are partially dissociated into ions.An equilibrium exist between the undissociated molecules and the dissociated anions. Both the molecules and anions are coloured. i.e.

 

HIn(aq)                                       H+ (aq)  +  In (aq)

(undissociated indicator                                       (dissociated indicator

molecule(coloured))                                             molecule(coloured))

When an acid H+ is added to an indicator, the H+ ions increase and equilibrium shift backward to remove excess H+ ions and therefore the colour of the undissociated (HIn) molecule shows/appears.

When a base/alkali OH is added to the indicator, the OH reacts with H+ ions from the dissociated indicator to form water.

H+ (aq)      +      OH(aq)    -> H2O(l)

(from indicator)    (from alkali/base)

The equilibrium shift forward to the right to replace the H+ ion and therefore the colour of dissociated (In) molecule shows/appears.

The following examples illustrate the above.

(i)Phenolphthalein indicator exist as:

HPh                   H+ (aq)      +      Ph(aq)

(colourless molecule)                                  (Pink anion)

On adding an acid ,equilibrium shift backward to the left to remove excess H+ ions and the solution mixture is therefore colourless.

When a base/alkali OH is added to the indicator, the OH reacts with H+ ions from the dissociated indicator to form water.

H+ (aq)      +      OH(aq)    -> H2O(l)

(from indicator)    (from alkali/base)

The equilibrium shift forward to the right to replace the removed/reduced H+ ions. The pink colour of dissociated (Ph) molecule shows/appears.

   (ii)Methyl Orange indicator exists as:

HMe                  H+ (aq)      +      Me(aq)

(Red molecule)                           (Yellow/Orange anion)

On adding an acid ,equilibrium shift backward to the left to remove excess H+ ions and the solution mixture is therefore red.

When a base/alkali OH is added to the indicator, the OH reacts with H+ ions from the dissociated indicator to form water.

H+ (aq)      +      OH(aq)    -> H2O(l)

(from indicator)    (from alkali/base)

The equilibrium shift forward to the right to replace the removed/reduced H+ ions. The Orange colour of dissociated (Me) molecule shows/appears.

 

(b)Influence of Pressure on dynamic equilibrium

Pressure affects gaseous reactants/products. Increase in pressure shift/favours  the equilibrium towards the side with less volume/molecules. Decrease in pressure shift the equilibrium towards the side with more volume/molecules. More yield of products is obtained if high pressures produce less molecules / volume of products are formed.

If the products and reactants have equal volume/molecules then pressure has no effect on the position of equilibrium

The following examples show the influence of pressure on dynamic equilibrium:

 

(i)Nitrogen(IV)oxide /Dinitrogen tetroxide mixture

 

Nitrogen(IV)oxide and dinitrogen tetraoxide can exist in dynamic equilibrium in a closed test tube. Nitrogen(IV)oxide is a brown gas. Dinitrogen tetraoxide is a yellow gas.

Chemical equation :                    2NO2(g)    =====          N2 O4 (g)

Gay Lussacs law               2Volume                      1Volume

Avogadros law                           2molecule                     1molecule

 

2 volumes/molecules of Nitrogen(IV)oxide form 1 volumes/molecules of dinitrogen tetraoxide

Increase in pressure shift the equilibrium forward to the left where there is less volume/molecules.The equilibrium mixture become more yellow.

Decrease in pressure shift the equilibrium backward to the right where there is more volume/molecules. The equilibrium mixture become more brown.

 

(ii)Iodine vapour-Hydrogen gas/Hydrogen Iodide mixture.

Pure hydrogen gas reacts with Iodine vapour to form Hydrogen Iodide gas.

Chemical equation :            I2(g)         +      H2(g)      =====          2HI (g)

Gay Lussacs law      1Volume           1Volume                             2Volume

Avogadros law                  1molecule                  1molecule                  2molecule

 

(1+1) 2 volumes/molecules of Iodine and Hydrogen gasform 2 volumes/molecules of Hydrogen Iodide gas.

Change in pressure thus has no effect on position of equilibrium.

 

(iii)Haber process.

Increase in pressure  of the Nitrogen/Hydrogen mixture favours the formation of more molecules of Ammonia gas in Haber process.

The yield of ammonia is thus favoured by high pressures

 

Chemical equation :   N2(g)      +      3H2 (g)      ->      2NH3 (g)

Gay Lussacs law      1Volume       3Volume           2Volume

Avogadros law                  1molecule       3molecule         2molecule

 

(1 + 3) 4 volumes/molecules of Nitrogen and Hydrogen react to form 2 volumes/molecules of ammonia.

Increase in pressure shift the equilibrium forward to the left where there is less volume/molecules.

The yield of ammonia increase.

Decrease in pressure shift the equilibrium backward to the right where there is more volume/molecules.

The yield of ammonia decrease.

 

(iv)Contact  process.

Increase in pressure  of the Sulphur(IV)oxide/Oxygen mixture favours the formation of more molecules of Sulphur(VI)oxide gas in Contact process. The yield of Sulphur(VI)oxide gas is thus favoured by high pressures.

 

Chemical equation :   2SO2(g)      +       O2 (g)      ->      2SO3 (g)

Gay Lussacs law      2Volume       1Volume           2Volume

Avogadros law                  2molecule       1molecule         2molecule

 

(2 + 1) 3 volumes/molecules of Sulphur(IV)oxide/Oxygen mixture react to form 2 volumes/molecules of Sulphur(VI)oxide gas.

Increase in pressure shift the equilibrium forward to the left where there is less volume/molecules. The yield of Sulphur(VI)oxide gas increase.

Decrease in pressure shift the equilibrium backward to the right where there is more volume/molecules. The yield of Sulphur(VI)oxide gas decrease.

 

(v)Ostwalds process.

Increase in pressure of the Ammonia/Oxygen mixture favours the formation of more molecules of Nitrogen(II)oxide gas and water vapour in Ostwalds process. The yield of Nitrogen(II)oxide gas and water vapour is thus favoured by low pressures.

 

Chemical equation :   4NH3(g)      +  5O2 (g)      ->  4NO(g)     +    6H2O (g)

Gay Lussacs law      4Volume       5Volume        4Volume        6Volume

Avogadros law                  4molecule      5molecule      4molecule       6Molecule

 

(4 + 5)  9 volumes/molecules of Ammonia/Oxygen mixture react to form 10 volumes/molecules of Nitrogen(II)oxide gas and water vapour.

Increase in pressure shift the equilibrium  backward to the left where there is less volume/molecules. The yield of Nitrogen(II)oxide gas and water vapour decrease.

Decrease in pressure shift the equilibrium  forward to the right where there is more volume/molecules. The yield of Nitrogen(II)oxide gas and water vapour increase.

Note

If the water vapour is condensed on cooling, then:

Chemical equation :   4NH3(g)      +  5O2 (g)      ->  4NO(g)     +    6H2O (l)

Gay Lussacs law      4Volume       5Volume        4Volume         0Volume

Avogadros law                  4molecule      5molecule      4molecule       0Molecule

 

(4 + 5)  9 volumes/molecules of Ammonia/Oxygen mixture react to form 4 volumes/molecules of Nitrogen(II)oxide gas and no vapour.

 

Increase in pressure shift the equilibrium  forward to the right where there is less volume/molecules. The yield of Nitrogen(II)oxide gas increase.

 

Decrease in pressure shift the equilibrium  backward to the left where there is more volume/molecules. The yield of Nitrogen(II)oxide gas decrease.

 

(c)Influence of Temperature on dynamic equilibrium

 

A decrease in temperature favours the reaction that liberate/generate more heat thus  exothermic reaction(-ΔH).

An increase in temperature favours the reaction that do not liberate /generate more heat thus endothermic reaction(+ΔH).

Endothermic reaction are thus favoured by high temperature/heating

Exothermic reaction are favoured by low temperature/cooling.

If a reaction/equilibrium mixture is neither exothermic or endothermic, then a change in temperature/cooling/heating has no effect on the equilibrium position.

 

(i)Nitrogen(IV)oxide /Dinitrogen tetroxide mixture

 

Nitrogen(IV)oxide and dinitrogen tetraoxide can exist in dynamic equilibrium in a closed test tube. Nitrogen(IV)oxide is a brown gas. Dinitrogen tetraoxide is a yellow gas.

Chemical equation :                    2NO2(g)    =====          N2 O4 (g)

On heating /increasing temperature, the mixture becomes more brown. On cooling the mixture become more yellow.

This show that

(i)the forward reaction to the right is exothermic(-ΔH).

On heating an exothermic process the equilibrium shifts to the side that generate /liberate less heat.

(ii)the backward reaction to the right is endothermic(+ΔH).

On cooling an endothermic process the equilibrium shifts to the side that do not  generate /liberate heat.

 

(c)Influence of Catalyst on dynamic equilibrium

 

A catalyst has no effect on the position of equilibrium. It only speeds up the rate of attainment. e.g.

Esterification of alkanols and alkanoic acids naturally take place in fruits.In the laboratory concentrated sulphuric(VI)acid catalyse the reaction.The equilibrium mixture forms the ester faster but the yield does not increase.

CH3CH2OH(l)+CH3COOH(l) ==Conc.H2SO4== CH3COOCH2CH3(aq) + H2O(l)

 

 

(d)Influence of  rate of reaction  and dynamic equilibrium (Optimum conditions) on industrial processes

 

 

Industrial processes are commercial profit oriented. All industrial processes take place in closed systems and thus in dynamic equilibrium.

For manufacturers, obtaining the highest yield at minimum cost and shortest time is paramount.

The conditions required to obtain the highest yield of products within the shortest time at minimum cost are called optimum conditions

Optimum condition thus require understanding the effect of various factors on:

 

(i)rate of reaction(Chemical kinetics)

(ii)dynamic equilibrium(Chemical cybernetics)

 

1.Optimum condition in Haber process

Chemical equation

 

          N2 (g)  +   3H2 (g)   ===Fe/Pt===  2NH3 (g)     ΔH = -92kJ

 

Equilibrium/Reaction rate considerations

(i)Removing ammonia gas once formed shift the equilibrium forward to the right to replace the ammonia. More/higher yield of ammonia is attained.

(ii)Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules . More/higher yield of ammonia is attained. Very high pressures raises the cost of production because they are expensive to produce and maintain. An optimum pressure of about 500atmospheres is normally used.

 

(iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -92kJ) . Ammonia formed decomposes back to Nitrogen and Hydrogen to remove excess heat therefore a less yield of ammonia is attained. Very low temperature decrease the collision frequency of Nitrogen and Hydrogen and thus the rate of reaction too slow and uneconomical.

An optimum temperature of about 450oC is normally used.

 

(iv)Iron and platinum can be used as catalyst. Platinum is a better catalyst but more expensive and easily poisoned by impurities than Iron. Iron is promoted /impregnated with AluminiumOxide(Al2O3) to increase its surface area/area of contact with reactants and thus efficiency.The catalyst does not increase the yield of ammonia but it speed up its rate of formation.

 

2.Optimum condition in Contact process

 

Chemical equation

 

          2SO2 (g)  +   O2 (g)   ===V2O5/Pt===  2SO3 (g)     ΔH = -197kJ

 

Equilibrium/Reaction rate considerations

(i)Removing sulphur(VI)oxide gas once formed shift the equilibrium forward to the right to replace the sulphur(VI)oxide. More/higher yield of sulphur(VI) oxide  is attained.

(ii)Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules . More/higher yield of sulphur(VI)oxide is attained. Very high pressures raises the cost of production because they are expensive to produce and maintain. An optimum pressure of about 1-2 atmospheres is normally used to attain about 96% yield  of SO3.

 

(iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -197kJ) . Sulphur(VI)oxide formed decomposes back to Sulphur(IV)oxide and Oxygen to remove excess heat therefore a less yield of Sulphur(VI)oxide is attained. Very low temperature decrease the collision frequency of Sulphur(IV)oxide and Oxygen and thus the rate of reaction too slow and uneconomical.

An optimum temperature of about 450oC is normally used.

 

(iv)Vanadium(V)Oxide and platinum can be used as catalyst. Platinum is a better catalyst and less easily poisoned by impurities but more expensive. Vanadium(V)Oxide is very cheap even if it is easily poisoned by impurities. The catalyst does not increase the yield of Sulphur (VI)Oxide but it speed up its rate of formation.

 

3.Optimum condition in Ostwalds process

 

Chemical equation

 

          4NH3 (g)  +  5O2 (g)   ===Pt/Rh===  4NO (g) + 6H2O (g) ΔH = -950kJ

 

Equilibrium/Reaction rate considerations

(i)Removing Nitrogen(II)oxide gas once formed shift the equilibrium forward to the right to replace the Nitrogen(II)oxide. More/higher yield of Nitrogen(II) oxide  is attained.

(ii)Increase in pressure shift the equilibrium backward to the left where there is less volume/molecules . Less/lower yield of Nitrogen(II)oxide is attained. Very low pressures increases the distance between reacting NH3and O2 molecules.

An optimum pressure of about 9 atmospheres is normally used.

 

(iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -950kJ) . Nitrogen(II)oxide and water vapour formed decomposes back to Ammonia and Oxygen to remove excess heat therefore a less yield of Nitrogen(II)oxide is attained. Very low temperature decrease the collision frequency of Ammonia and Oxygen and thus the rate of reaction too slow and uneconomical.

An optimum temperature of about 900oC is normally used.

 

(iv)Platinum can be used as catalyst. Platinum is very expensive.It is:

-promoted with Rhodium to increase the surface area/area of contact.

-added/coated on the surface of asbestos to form platinized –asbestos to  reduce the amount/quantity used.

The catalyst does not increase the yield of Nitrogen (II)Oxide but it speed up its rate of formation.

 

 

C.SAMPLE REVISION QUESTIONS

 

1.State two distinctive features  of a dynamic equilibrium.

 

(i)the rate of forward reaction is equal to the rate of forward reaction

(ii)at equilibrium the concentrations of reactants and products do not change.

 

  1. Explain the effect of increase in pressure on the following:

 

(i) N2(g)    +      O2(g)  =====  2NO(g)

 

Gay Lussacs law         1Volume        1Volume              2 Volume

Avogadros law            1 molecule      1 molecule           2 molecule

2 volume on reactant side produce 2 volume on product side.

Increase in pressure thus have no effect on position of equilibrium.

 

(ii) 2H2(g)            +   CO(g)    =====   CH3OH (g)

 

Gay Lussacs law         2Volume         1Volume              1 Volume

Avogadros law            2 molecule      1 molecule           1 molecule

3 volume on reactant side produce 1 volume on product side.

Increase in pressure shift the equilibrium forward  to the left. More yield of CH3OH is formed.

 

  1. Explain the effect of increasing temperature on the following:

   2SO2(g)    +    O2 (g)    =====   2SO3 (g) ΔH = -189kJ                       Forward reaction is exothermic. Increase in temperature shift the equilibrium backward to reduce the excess heat.

5.120g of brass an alloy of copper and Zinc was put it a flask containing dilute hydrochloric acid. The flask was placed on an electric balance. The readings on the balance were recorded as in the table below

 

Time(Seconds) Mass of flask(grams) Loss in mass(grams)
0 600  
20 599.50  
40 599.12  
60 598.84  
80 598.66  
100 598.54  
120 598.50  
140 598.50  
160 598.50  

 

(a)Complete the table by calculating the loss in mass

 

(b)What does the “600” gram reading on the balance represent

The initial mass of brass and the acid before any reaction take place.

 

(c)Plot a graph of Time (x-axes) against loss in mass.

 

(d)Explain the shape of your graph

The reaction produce hydrogen gas as one of the products that escape to the atmosphere. This decreases the mass of flask.After 120 seconds,the react is complete. No more hydrogen is evolved.The mass of flask remain constant.

 

(d)At what time was the loss in mass equal to:

(i)1.20g

Reading from a correctly plotted graph =

 

(ii)1.30g

Reading from a correctly plotted graph =

 

(iii)1.40g

Reading from a correctly plotted graph =

(e)What was the loss in mass at:

(i)50oC

Reading from a correctly plotted graph =

 

(ii) 70oC

Reading from a correctly plotted graph =

 

(iii) 90oC g

Reading from a correctly plotted graph =

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

       

 

 

UPGRADE

CHEMISTRY

                

                        FORM 4

Electrochemistry

      

 

                 Comprehensive tutorial notes

 

 

 MUTHOMI S,G

www.kcselibrary.info

                                     0720096206

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ELECTROCHEMISTRY

Electrochemistry can be defined as the study of the effects of electricity on a substance/ compound and how chemical reactions produce electricity. Electrochemistry therefore deals mainly with:

  1. Reduction and oxidation
  2. Electrochemical (voltaic) cell
  • Electrolysis (electrolytic) cell

 

(i)REDUCTION AND OXIDATION (REDOX)

 

  1. In teams of oxygen transfer:
  2. i) Reduction is removal of oxygen.
  3. ii) Oxidation is addition of oxygen.

iii) Redox is simultaneous addition and removal of oxygen.

  1. iv) Reducing agent is the species that undergoes oxidation, therefore gains
  2. v) Oxidizing agent is the species that undergoes reduction, therefore looses/donates

e.g.          When hydrogen is passed through heated copper (II) oxide, it is oxidised to copper metal as in the equation below:

CuO (s)           +           H2 (g)                     ->       Cu (s)        +      H2O (l)

(Oxidising agent)     (Reducing agent)

 

  1. In terms of hydrogen transfer:
  2. i) Oxidation is the removal of hydrogen.
  3. ii) Reduction is the addition of hydrogen.

iii) Redox is simultaneous addition and removal of hydrogen.

  1. iv) Reducing agent is the species that undergoes oxidation, therefore looses/ donates
  2. v) Oxidizing agent is the species that undergoes reduction, therefore gains

e.g.    When hydrogen sulphide gas is bubbled into a gas jar containing chlorine gas it is oxidized (loose the hydrogen) to sulphur (yellow solid). The chlorine is reduced (gain hydrogen) to hydrogen chlorine gas.

Cl2 (g)           +              H2S (g)     ->               S(S)            +           2HCl (g)

(Oxidizing agent)     (Reducing agent)

 

  1. In terms of electron transfer:
  2. i) Oxidation is donation/ loss/ removal of electrons.
  3. ii) Reduction is gain/ accept/ addition of electrons.

iii) Redox is simultaneous gain/ accept/ addition and donation/ loss/ removal of electrons.

  1. iv) Reducing agent is the species that undergoes oxidation, therefore looses/ donates
  2. v) Oxidizing agent is the species that undergoes reduction, therefore gains/ accepts

 

Example

  1. Displacement of metals from their solutions:

 

Place 5cm3 each of Iron (II) sulphate (VI) solution into three different test tubes. Add about 1g of copper tunings / powder into one test tube then zinc and magnesium powders separately into the other test tubes. Shake thoroughly for 2 minutes each. Record any colour changes in the table below.

 

Metal added to Iron (II) sulphate (VI) solution Colour changes
Copper Solution remains green
Zinc Green colour fades
Magnesium Green colour fades

 

Explanation

 

-When a more reactive metal is added to a solution of less reactive metal, it displaces it from its solution.

-When a less reactive metal is added to a solution of a more reactive metal, it does not displace it from its solution.

-Copper is less reactive than iron therefore cannot displace iron its solution.

-Zinc is more reactive than iron therefore can displace iron from its solution.

-Magnesium is more reactive than iron therefore can displace iron from its solution.

 

In terms of electron transfer:

– the more reactive metal undergoes oxidation (reducing agent) by donating/loosing electrons to form ions

-the less reactive metal undergoes reduction (oxidizing agent) by its ions in solution gaining /accepting/acquiring the electrons to form the metal.

-displacement of metals involves therefore electron transfer from a more reactive metal to ions of another less reactive metal.

Examples

 

  1. Zn(s) ->      Zn2+(aq) +    2e        (oxidation/donation of electrons)

Fe2+(aq)  +   2e      ->       Fe(s)     (reduction/gain of electrons)

Fe2+(aq)  +    Zn(s) ->   Zn2+(aq) +  Fe(s)  (redox/both donation and gain of electrons)

 

  1. Mg(s) -> Mg2+(aq) +   2e        (oxidation/donation of electrons)

Fe2+(aq)  +   2e      ->       Fe(s)     (reduction/gain of electrons)

Fe2+(aq)  +    Mg(s) ->   Mg2+(aq) +  Fe(s)  (redox/both donation and gain of electrons)

 

  1. Zn(s) ->      Zn2+(aq) +    2e        (oxidation/donation of electrons)

Cu2+(aq)  +  2e      ->       Cu(s)     (reduction/gain of electrons)

Cu2+(aq)  +    Zn(s) ->   Zn2+(aq) +  Cu(s)  (redox/both donation and gain of electrons)

 

  1. Fe(s) ->      Fe2+(aq) +    2e        (oxidation/donation of electrons)

2Ag+(aq)  + 2e      ->       2Ag(s)     (reduction/gain of electrons)

2Ag+(aq)  +    Fe(s) ->   Fe2+(aq) +  2Ag(s)  (redox/both donation and  gain of electrons)

 

  1. Zn(s)       ->      Zn2+(aq) +    2e        (oxidation/donation of electrons)

Cl2(g)  +      2e      ->       2Cl(aq)     (reduction/gain of electrons)

Cl2(g)   +    Zn(s) ->   Zn2+(aq) +  2Cl(aq)  (redox/both donation and  gain of electrons)

 

  1. 2Mg(s)    ->      2Mg2+(aq) + 4e        (oxidation/donation of electrons)

    O2(g)  +   4e      ->       2O2-(aq)     (reduction/gain of electrons)

O2(g)   +   2Mg(s) ->   2Mg2+(aq)2O2-(aq)   (redox/both donation and  gain of electrons)

 

Note

(i)The number of electrons donated/lost MUST be equal to the number of electrons gained/acquired.

(i)During displacement reaction, the colour of ions /salts fades but does not if displacement does not take place. e.g

 

a)Green colour of Fe2+(aq) fades if Fe2+(aq) ions are displaced from their solution. Green colour of Fe2+(aq) appear if Fe/iron displaces another salt/ions  from their solution.

 

b)Blue colour of Cu2+(aq) fades if  Cu2+(aq) ions are displaced from their solution and brown copper deposits appear. Blue colour of Cu2+(aq) appear if Cu/copper displaces another salt/ions  from their solution.

 

c)Brown colour of Fe3+(aq) fades if Fe3+(aq) ions are displaced from their solution. Brown colour of Fe3+(aq) appear if Fe/iron displaces another salt/ions  from their solution to form Fe3+(aq).

 

(iii)Displacement reactions also produce energy/heat. The closer/nearer the metals in the reactivity/electrochemical series the less energy/heat of displacement.

 

(iv)The higher the metal in the reactivity series therefore the easier to loose/donate electrons and thus the stronger the reducing agent.

 

  1. (a)In terms of oxidation number:
  2. i) Oxidation is increase in oxidation numbers.
  3. ii) Reduction is decrease in oxidation numbers.

iii) Redox is simultaneous increase in oxidation numbers of one species/substance and a decrease in oxidation numbers of another species/substance.

  1. iv) Reducing agent is the species that undergoes oxidation, therefore increases its oxidation number.
  2. v) Oxidizing agent is the species that undergoes reduction, therefore increases its oxidation number.

 

(b)The idea/concept of oxidation numbers uses/applies the following simple guideline rules:

 

Guidelines /rules applied in assigning oxidation number

 

1.Oxidation number of combined Oxygen is always -2 except in peroxides (Na2O2/H2O2) where its Oxidation number is -1

 

2.Oxidation number of combined Hydrogen is always +1except in Hydrides (NaH/KH) where its Oxidation number is -1

 

3.All atoms and molecules of elements have oxidation number 0 (zero)

 

Atom Oxidation number Molecule Oxidation number
Na 0 Cl2 0
O 0 O2 0
H 0 H2 0
Al 0 N2 0
Ne 0 O3 0
K 0 P3 0
Cu 0 S8 0

 

4.All combined metals and non-metals have oxidation numbers equal to their valency /oxidation state e.g.

 

Metal/non-metal ion Valency Oxidation state Oxidation number
 Fe2+ 2 -2 -2
 Fe3+ 3 -3 -3
 Cu2+ 2 -2 -2
 Cu+ 1 +1 +1
 Cl 1 -1 -1
 O2- 2 -2 -2
Na+ 1 +1 +1
Al3+ 3 +3 +3
 P3- 3 -3 -3
 Pb2+ 2 +2 +2

 

5.Sum of oxidation numbers of atoms of elements making a compound is equal  zero(0) e.g.

Using this rule ,an unknown oxidation number of an atom in a compound can be determined as below:

  1. a) CuSO4 has-

-one atom of Cu with oxidation number  +2( refer to Rule 4)

-one atom of  S with oxidation number  +6 ( refer to Rule 4)

-six atoms of O each with oxidation number  -2( refer to Rule 4)

Sum of oxidation numbers of atoms in CuSO4 = (+2 + +6 + (-2 x 6)) = 0

 

  1. b) H2SO4 has-

-two atom of H each with oxidation number  +1( refer to Rule 2)

-one atom of  S with oxidation number  +6 ( refer to Rule 4)

-four atoms of O each with oxidation number  -2( refer to Rule 4)

Sum of oxidation numbers of atoms in H2SO4 = (+2 + +6 + (-2 x 4)) = 0

  1. c) KMnO4 has-

-one atom of K with oxidation number  +1( refer to Rule 4)

-one atom of  Mn with oxidation number  +7 ( refer to Rule 4)

-four atoms of O each with oxidation number  -2( refer to Rule 4)

Sum of oxidation numbers of atoms in KMnO4 = (+1 + +7 + (-2 x 4)) = 0

 

Determine the oxidation number of:

I.Nitrogen in;

-NO   => x + -2 = 0 thus  x = 0 – (-2) = + 2

The chemical name of this compound is thus Nitrogen(II)oxide

-NO2   => x + (-2 x2)= 0 thus  x = 0 – (-4) = + 4

The chemical name of this compound is thus Nitrogen(IV)oxide

-N2O  => 2x + -2 = 0 thus  2x = 0 – (-2) = +2/2= +1

The chemical name of this compound is thus Nitrogen(I)oxide

 

  1. Sulphur in;

-SO2   => x + (-2 x2)= 0 thus  x = 0 – (-4) = + 4

The chemical name of this compound is thus Sulphur(IV)oxide

-SO3   => x + (-2 x3)= 0 thus  x = 0 – (-6) = + 6

The chemical name of this compound is thus Sulphur(VI)oxide

-H2SO4  = ((+1 x 2) + x + (-2 x 4)) thus x= 0-( +2 +-8) =+6

The chemical name of this compound is thus Sulphuric(VI)acid

-H2SO3  = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4

The chemical name of this compound is thus Sulphuric(IV)acid

 

III. Carbon in;

-CO2   => x + (-2 x2)= 0 thus  x = 0 – (-4) = + 4

The chemical name of this compound is thus carbon(IV)oxide

-CO   => x + -2 = 0 thus  x = 0 – -2 = + 2

The chemical name of this compound is thus carbon(II)oxide

-H2CO3  = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4

The chemical name of this compound is thus Carbonic(IV)acid

IV.Manganese in;

-MnO2   => x + (-2 x2)= 0 thus  x = 0 – (-4) = + 4

The chemical name of this compound is thus Manganese(IV)oxide

-KMnO4  = ((+1 + x + (-2 x 4)) thus x= 0-( +1 +-8) =+7

The chemical name of this compound is thus Potassium manganate(VII)

V.Chromium in;

– Cr2O3   => 2x + (-2 x 3)= 0 thus  2x = 0 – (-6) = +6 / 2= +3

The chemical name of this compound is thus Chromium(III)oxide

-K2Cr2O7   => (+1 x 2) + 2x + (-2 x7)= 0

thus 2x = 0 – +2 +-14 = +12 / 2= +6

The chemical name of this compound is thus Potassium dichromate(VI)

-K2CrO4   => (+1 x 2) + x + (-2 x4)= 0

thus 2x = 0 – +2 +-8 = +12 / 2= +6

The chemical name of this compound is thus Potassium chromate(VI)

 

6.The sum of the oxidation numbers of atoms of elements making a charged radical/complex ion is equal to its charge.

Using this rule ,the oxidation number of unknown atom of an element in a charged radical/complex ion can be determined as in the examples below;

  1. a) SO42- has-

-one atom of S with oxidation number  +6( refer to Rule 4)

-four atoms of O each with oxidation number  -2( refer to Rule 1)

Sum of oxidation numbers of atoms in SO42- = ( +6 + (-2 x 4)) = -2

The chemical name of this radical is thus sulphate(VI) ion

  1. b) NO3 has-

-one atom of N with oxidation number  +4( refer to Rule 4)

-three atoms of O each with oxidation number  -2( refer to Rule 1)

Sum of oxidation numbers of atoms in NO3 = ( +4 + (-2 x 3)) = -1

The chemical name of this radical is thus nitrate(IV) ion.

 

Determine the oxidation number of:

I.Nitrogen in;

-NO2   => x + (-2 x2)= -1 thus  x = -1 – (-4) = + 3

The chemical name of this compound/ion/radical is thus Nitrate(III)ion

  1. Sulphur in;

-SO32-  => x + (-2 x3)= -2 thus  x = -2 – (-6) = + 4

The chemical name of this compound/ion/radical is thus Sulphate(IV)ion

III. Carbon in;

-CO32-  = x + (-2 x 3) = -2 thus  x = -2 – (-6) = + 4

The chemical name of this compound/ion/radical is thus Carbonate(IV)ion

IV.Manganese in;

-MnO4 =  x + (-2 x 4)= -1 thus x= -1-(-2 +-8) =+7

The chemical name of this compound/ion/radical is thus manganate(VII) ion

V.Chromium in

-Cr2O72-   => 2x + (-2 x7)= -2

thus 2x = -2 – +2 +-14 = +12 / 2= +6

The chemical name of this compound/ion//radical is thus dichromate(VI) ion

-CrO42-   => x + (-2 x4)= -2

thus x =  -2 + (-2  x 4) = +6

The chemical name of this compound/ion//radical is thus chromate(VI) ion

 

(c)Using the concept/idea of oxidation numbers as increase and decrease in oxidation numbers , the oxidizing and reducing species/agents can be determined  as in the following examples;

 

(i)                                    Cu2+   (aq)   +      Zn(s)   ->       Zn2+   (aq)   +     Cu(s)

Oxidation numbers ->      +2                          0                  +2                          0

Oxidizing species/agents =>Cu2+ ;its oxidation number decrease from+2 to 0 in Cu(s)

Reducing species/agents => Zn2+ ;its oxidation number increase from 0 to +2 in Zn(s)

 

(ii)                                   2Br  (aq)   +      Cl2(g)  ->       2Cl  (aq)    +     Br2 (l)

Oxidation numbers ->      -1                           0                  -1                           0

Oxidizing agent =>Cl2(g) ;its oxidation number decrease from 0 to-1 in 2Cl  (aq)

Reducing agents => Zn2+ ;its oxidation number increase from -1 to 0 in Zn(s)

 

(iii)                                  Br2 (l)          +      Zn(s)   ->       Zn2+   (aq)   +     2Br(aq)

Oxidation numbers ->      0                           0                  +2                          -1

Oxidizing agent => Br2 (l) ;its oxidation number decrease from 0 to-1 in 2Br(aq)

Reducing agents => Zn(s) ;its oxidation number increase from 0 to +2 in Zn2+

 

(iv)                                  2HCl (aq)   +      Mg(s)   ->      MgCl2   (aq)      +   H2 (g)

Oxidation numbers ->    2 (+1  -1)                0                  +2  2(-1)                   0

Oxidizing agent => H+  in HCl;its oxidation number decrease from +1to 0 in H2 (g)

Reducing agents => Mg(s) ;its oxidation number increase from 0 to +2 in Mg2+

 

(v)                              2H2O (l)   +      2Na(s)   ->   2NaOH   (aq)      +   H2 (g)

Oxidation numbers -> +1  -2                     0            +1 -2  +1                   0

Oxidizing agent => H+  in H2O;its oxidation number decrease from +1to 0 in H2 (g)

Reducing agents => Na(s) ;its oxidation number increase from 0 to +1 in Na+

 

(vi)  5Fe2+ (aq)   +   8H+ (aq)  +  MnO4   ->  5Fe3+ (aq)   +   Mn2+ (aq)  +  4H2O (l)                                    +2                +1                  +7   -2             +3                   +2                +1   -2

Oxidizing agent => Mn in MnO4;its oxidation number decrease from +7to+2 in Mn2+

Reducing agents => Fe2+ ;its oxidation number increase from +2  to +3 in Fe3+

 

(vii) 6Fe2+ (aq) + 14H+ (aq) + Cr2O72-(aq) -> 6Fe3+ (aq)  +  Cr3+ (aq)  +  7H2O (l)                                    +2                +1              +6  -2                  +3                +3                 +1   -2

Oxidizing agent:

 Cr in Cr2O72- ;its oxidation number decrease from +6 to+3 in Cr3+

Reducing agents => Fe2+ ;its oxidation number increase from +2  to +3 in Fe3+

(viii) 2Fe2+ (aq)  +   2H+ (aq)   +   H2O2(aq)   ->   2Fe3+ (aq)  +    2H2O (l)                                           +2                  +1                 +1  -1                  +3                    +1   -2

Oxidizing agent:

 O in H2O2;its oxidation number decrease from -1 to -2 in H2O

Reducing agents => Fe2+ ;its oxidation number increase from +2  to +3 in Fe3+

 

(ix) Cr2O72-(aq)  +   6H+ (aq)   +  5H2O2(aq)   ->  2Cr3+ (aq)  +   2H2O (l) + 5O2(g)                                 

+6   -2                +1                 +1  -1                  +3                    +1   -2        0

Oxidizing agents:

 O in H2O2;its oxidation number decrease from -1 to -2 in H2O

Cr in Cr2O72- its oxidation number decrease from +6 to +3 in Cr3+

Reducing agents

O in H2O2;its oxidation number increase from -1 to O in O2(g)

O in Cr2O72- its oxidation number increase from -2 to O in O2(g)

 

(x) 2MnO4(aq)  +   6H+ (aq)   +  5H2O2(aq)   ->  2Mn2+ (aq)  +   8H2O (l) + 5O2(g)                                 

+7   -2                +1                 +1  -1                  +2                    +1   -2        0

Oxidizing agents:

 O in H2O2;its oxidation number decrease from -1 to -2 in H2O

Mn in MnO4 its oxidation number decrease from +7 to +2 in Mn2+

Reducing agents

O in H2O2;its oxidation number increase from -1 to O in O2(g)

O in MnO4 its oxidation number increase from -2 to O in O2(g)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(ii)ELECTROCHEMICAL (VOLTAIC) CELL

 

  1. When a metal rod/plate is put in a solution of its own salt, some of the metal ionizes and dissolve into the solution i.e.

M(s)      ->    M+(aq)    +   e   ( monovalent metal)

M(s)      ->    M2+(aq)   +  2e  ( divalent metal)

M(s)      ->    M3+(aq)   +  3e  ( Trivalent metal)

The ions move into the solution leaving electrons on the surface of the metal rod/plate.

 

2.The metal rod becomes therefore negatively charged while its own solution positively charged. As the positive charges of the solution increase, some of them recombine with the electrons to form back the metal atoms

M+(aq)         +   e     ->   M(s)    ( monovalent metal)

M2+(aq)        +   2e   ->   M(s)    (divalent metal)

M3+(aq)        +   3e   ->   M(s)    (Trivalent metal)

  1. When a metal rod/plate is put in a solution of its own salt, it constitutes/forms a half-cell. The tendency of metals to ionize differ from one metal to the other. The difference can be measured by connecting two half cells to form an electrochemical/voltaic cell as in the below procedure:

 

To set up an electrochemical /voltaic cell

To compare the relative tendency of metals to ionize

 

Place 50cm3 of 1M Zinc(II) sulphate(VI) in 100cm3 beaker. Put a clean zinc rod/plate into the solution. Place 50cm3 of 1M Copper(II) sulphate(VI) in another 100cm3 beaker. Put a clean copper rod/plate of equal area (length x width) with Zinc into the solution. Connect/join the two metals(to a voltmeter) using connecting wires. Dip a folded filter paper into a solution of Potassium nitrate(V) or sodium(I) chloride(I) until it soaks. Use the folded soaked filter paper to connect/join the two solutions in the two beakers. The whole set up should be as below

 

V

 

Repeat the above procedure by replacing:

(i)Zinc half cell with Magnesium rod/plate/ribbon dipped in 50cm3 of IM magnesium (II) sulphate(VI) solution

(ii)Zinc half cell with Silver rod/plate/coin dipped in 50cm3 of IM silver(I) nitrate(V) solution

(iii)Copper half cell with Iron rod/plate/spoon dipped in 50cm3 of IM Iron (II) sulphate(VI) solution

Record the observations in the table below

Changes on the  1st  metal rod (A) Changes on the  2nd   metal rod (B) Changes on the  1st  solution  (A(aq)) Changes on the  2nd solution  (B(aq)) Voltage/voltmeter reading(Volts)
Using Zn/Cu half cell

-The rod decrease in size /mass /dissolves/ erodes

 

 

-copper rod /plate increase in size /mass/ deposited

 

 

Zinc(II)sulphate

(VI)colour remain

colourless

 

 

Blue Copper (II)sulphate

(VI)colour fades. Brown solid/residue/ deposit

 

 

 

0.8

(Theoretical value=1.10V)

Using Mg/Cu half cell

-The rod decrease in size /mass /dissolves/ erodes

 

 

-copper rod /plate increase in size /mass/ deposited

 

 

Magnesium(II) sulphate(VI) colour remain

colourless

 

 

Blue Copper (II)sulphate

(VI)colour fades Brown solid/residue/ deposit

 

 

 

1.5

(Theoretical value=2.04V)

Using Ag/Cu half cell

-The rod increase in size /mass /deposited

 

 

-silver coin/ rod /plate increase in size /mass/ deposited

 

 

Blue Copper (II)sulphate

(VI)colour remains

 

 

 

Silver(I)nitrate

(V)colour remain

colourless

 

 

 

0.20

(Theoretical value=0.46V)

Using Fe/Cu half cell

-The rod decrease in size /mass /dissolves/ erodes

 

 

-copper rod /plate increase in size /mass/ deposited

 

 

Iron(II)sulphate

(VI)colour becomes more

green

 

 

Blue Copper (II)sulphate

(VI)colour fades.Brown solid/residue/ deposit

 

 

 

0.60

(Theoretical value=0.78V)

From the above observations ,it can be deduced that:

(i)in the Zn/Cu half-cell the;

-Zinc rod/plate ionizes /dissolves faster than the copper rod/plate to form Zn2+

          Ionic equation        Zn(s)           ->        Zn2+(aq)     +        2e

-blue copper ions in the Copper (II)sulphate solution gains the donated electrons to form brown copper metal/atoms

Ionic equation        Cu2+(aq)     +        2e     ->       Cu(s)

This reaction shows /imply the Zinc rod has a higher tendency to ionize than copper.The Zinc rod has a higher net accumulation of electrons and is more negative compared to the copper rod which has lower accumulation of electrons. The copper rod is therefore relatively more positive with respect to Zinc rod.

 

When the two half cells are connected , electrons therefore flow from the negative Zinc rod through the external wire to be gained by copper ions. This means a net accumulation/increase of Zn2+ positive ions on the negative half cell and a net decrease in Cu2+ positive ions on the positive half cell.

The purpose of the salt bridge therefore is:

(i)complete the circuit

(ii)maintain balance of charges /ions on both half cells.

For the negative half cell the NO3 /Cl  from salt bridge decrease/neutralise  the increased positive(Zn2+) ion.

For the positive half cell the Na+ / K+  from salt bridge increase  the decreased positive(Cu2+) ion.

 

The voltmeter should theoretically register/read a 1.10Volts as a measure of the electromotive force (e.m.f) of the cell .Practically the voltage reading is lowered because the connecting wires have some resistance to be overcomed.

A combination of two half cells that can generate an electric current from a redox reaction is called a voltaic/electrochemical cell.

 

By convention a voltaic/electrochemical cell is represented;

M(s)       /  M2+(aq)   //   N2+ (aq)   /   N(s)

(metal rod of M)(solution ofM)(solution ofN)(metal rod ofN)

 

Note;

a)(i)Metal M must be the one higher in the reactivity series.

(ii)It forms the negative terminal of the cell.

(iii)It must diagrammatically be drawn first  on the left hand side when illustrating the voltaic/electrochemical cell.

 

b)(i)Metal N must be the one lower in the reactivity series.

(ii)It forms the positive terminal of the cell.

(iii)It must diagrammatically be drawn second/after/ right hand side when illustrating the voltaic/electrochemical cell.

 

Illustration of  the voltaic/electrochemical cell.

(i)Zn/Cu cell

 

  1. Zinc rod ionizes /dissolves to form Zn2+ ions at the negative terminal

Zn(s)           ->        Zn2+(aq)     +        2e

 

  1. Copper ions in solution gain the donated electrons to form copper atoms/metal

Cu2+(aq)     +        2e     ->       Cu(s)

 

3.Overall redox equation

Cu2+(aq)     +    Zn(s)  ->           Zn2+(aq)     +        Cu(s)

 

4.cell representation.

 Zn(s) / 1M, Zn2+(aq) // 1M,Cu2+(aq) / Cu(s)  E0  = +1.10 V

5.cell  diagram

   Voltmeter rrVVVVVVVAAAV VVVVVVVV

 

(ii)Mg/Cu cell

 

  1. Magnesium rod ionizes /dissolves to form Mg2+ ions at the negative terminal

Mg(s)                    ->        Mg2+(aq)    +        2e

 

  1. Copper ions in solution gain the donated electrons to form copper atoms/metal

Cu2+(aq)     +        2e     ->       Cu(s)

 

3.Overall redox equation

Cu2+(aq)     +    Mg(s)      ->     Mg2+(aq)    +        Cu(s)

 

4.cell representation.

 Mg(s) / 1M, Mg2+(aq) // 1M,Cu2+(aq) / Cu(s)  E0  = +2.04 V

 

5.cell  diagram.

(iii)Fe/Cu cell

 

  1. Magnesium rod ionizes /dissolves to form Mg2+ ions at the negative terminal

Fe(s)            ->        Fe2+(aq)     +        2e

 

  1. Copper ions in solution gain the donated electrons to form copper atoms/metal

Cu2+(aq)     +        2e     ->       Cu(s)

 

3.Overall redox equation

Cu2+(aq)     +    Fe(s)      ->       Fe2+(aq)     +        Cu(s)

 

4.cell representation.

 Fe(s) / 1M, Fe2+(aq) // 1M,Cu2+(aq) / Cu(s)  E0  = +0.78 V

 

5.cell  diagram.

 
V
Fe++
Fe

 

(iv)Ag/Cu cell

 

  1. Copper rod ionizes /dissolves to form Cu2+ ions at the negative terminal

Cu(s)           ->        Cu2+(aq)    +        2e

 

  1. Silver ions in solution gain the donated electrons to form silver atoms/metal

2Ag+(aq)     +        2e     ->       2Ag(s)

 

3.Overall redox equation

2Ag+(aq)     +    Cu(s)      ->      Cu2+(aq)    +        2Ag(s)

 

4.cell representation.

 Cu(s) / 1M, Cu2+(aq) // 1M,2Ag+(aq) / 2Ag(s)  E0  = +0.46 V

 

5.cell  diagram.

Cu(s)  +  2Ag+ (aq)  ­­­­­­­­­­­­            Cu2+(aq) + 2Ag(s)
Ag+
Cu++

 

V
Ag
Cu

 

Standard electrode potential  (Eᶿ)

 

The standard electrode potential  (Eᶿ) is  obtained if the hydrogen half cell is used as reference. The standard electrode potential  (Eᶿ) consist of inert platinum electrode immersed/dipped in 1M solution of (sulphuric(VI) acid) H+ ions. Hydrogen gas is bubbled on the platinum electrodes at:

(i)a temperature of 25oC

(ii)atmospheric pressure of 101300Pa/101300Nm-2/1atm/760mmHg/76cmHg

(iii)a concentration of 1M(1moledm-3) of sulphuric(VI) acid/ H+ ions and 1M(1moledm-3) of the other half cell.    

        

Hydrogen is adsorbed onto the surface of the platinum. An equilibrium/balance exist between the adsorbed layer of molecular hydrogen and H+ ions in solution to form a half cell.

½ H2 (g)        ====         H+ (aq)     +    e

 

The half cell representation is:

Pt,½ H2 (g) / H+ (aq), 1M

 

The standard electrode potential (Eᶿ) is thus defined as the potential difference for a cell comprising of a particular element in contact with1M solution of its own ions and the standard hydrogen electrode.

 

If the other electrode has a higher/greater tendency to lose electrons than the hydrogen electrode, the electrode is therefore negative with respect to hydrogen electrode and its electrode potential has negative (Eᶿ) values.

If the other electrode has a lower/lesser tendency to lose electrons than the hydrogen electrode, the electrode is therefore positive with respect to hydrogen electrode and its electrode potential has positive (Eᶿ) values.

 

Table showing the standard electrode potential (Eᶿ) of some reactions

Reaction (Eᶿ) values in volts
F2 (g)+ 2e -> 2F(aq) +2.87
H2 O2 (aq)+ H+ (aq)   +2e -> H2 O (l) +1.77
Mn O4 (aq)+ 4H+ (aq)   +3e -> MnO2 (s) +H2 O (l) +1.70
2HClO (aq)+ 2H+ (aq)   +2e -> Cl2 (aq) +2H2 O (l) +1.59
Mn O4 (aq)+ 4H+ (aq)   +5e -> Mn2+ (aq) +H2 O (l) +1.51
Cl2 (g)+ 2e -> 2Cl(aq) +1.36
Mn O2 (s)+ 4H+ (aq)   +2e -> Mn2+ (aq) +2H2 O (l) +1.23
Br2 (aq)+ 2e -> 2Br(aq) +1.09
NO3 (aq)+ 2H+ (aq)   +  e -> NO2 (g) + H2 O (l) +0.80
Ag+ (aq)  +  e    ->  Ag(s) +0.80
Fe3+ (aq)  +  e    ->  Fe2+ (aq) +0.77
2H+ (aq)+ O2 (g)  -> H2 O2 (aq) +0.68
I2 (aq)+ 2e -> 2I(aq) +0.54
Cu2+ (aq)  +  2e    ->  Cu(s) +0.34
2H+ (aq)  +  2e    ->  H2(g) +0.00
Pb2+ (aq)  +  2e    ->  Pb(s) -0.13
Fe2+ (aq)  +  2e    ->  Fe(s) -0.44
Zn2+ (aq)  +  2e    ->  Zn(s) -0.77
Al3+ (aq)  +  3e    ->  Al(s) -1.66
Mg2+ (aq)  +  2e    ->  Mg(s) -2.37
Na+ (aq)  +  e    ->  Na(s) -2.71
K+ (aq)  +  e    ->  K(s) -2.92

 

Note:

(i)Eᶿ values generally show the possibility/feasibility of a reduction process/oxidizing strength.

(ii)The element/species in the half cell with the highest negative Eᶿ value easily gain / acquire electrons.

It is thus the strongest oxidizing agent and its reduction process is highly possible/feasible. The element/species in the half cell with the lowest positive Eᶿ value easily donate / lose electrons.

It is thus the strongest reducing agent and its reduction process is the least possible/feasible.

(iii)The overall redox reaction is possible/feasible is it has a positive (+) Eᶿ.

If the overall redox reaction is not possible/ not feasible/ forced, it has a  negative (-) Eᶿ

Sample standard electrochemical cell

 

 

 

 

Calculation examples on Eᶿ 

Calculate the Eᶿ value of a cell made of:

a)Zn and Cu

From the table above:

Cu2+ (aq) +  2e ->  Cu(s)  Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram)

Zn2+ (aq) +  2e ->Zn(s)    Eᶿ = -0.77V(lower Eᶿ/ Left Hand Side diagram)

Zn(s) ->Zn2+ (aq) +  2e    Eᶿ = +0.77(reverse lower Eᶿ to derive cell reaction / representation)

 

Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS – Eᶿ LHS/ Eᶿoxidized- Eᶿ reduced

Substituting:

Overall Eᶿ = +0.34 – (- 0.77) = +1.10V

 

Overall redox equation:

Cu2+ (aq) + Zn(s)   ->   Zn2+ (aq)   +  Cu(s)    Eᶿ  = +1.10V

 

Overall conventional cell representation:

Zn(s) / Zn2+ (aq) 1M,  // 1M,Cu2+ (aq)  / Cu(s)    Eᶿ  = +1.10V

 

Zn2+
1M Zn2+ (aq)
1M Cu2+ (aq)
Voltmeter(1.10V)

Overall conventional cell diagram:

 

Zinc and copper reaction has a positive(+) overall Eᶿ therefore is possible/feasible and thus Zinc can displace/reduce Copper solution.

 

b)Mg and Cu

From the table above:

Cu2+ (aq) +  2e ->  Cu(s)  Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram)

Mg2+ (aq) +  2e ->Mg(s)    Eᶿ = -2.37V(lower Eᶿ/ Left Hand Side diagram)

Mg(s) ->Mg2+ (aq) +  2e    Eᶿ = +2.37(reverse lower Eᶿ to derive cell reaction / representation)

Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS – Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced

Substituting:

Overall Eᶿ = +0.34 – (- 2.37) = +2.71V

 

Overall redox equation:

Cu2+ (aq) + Mg(s)   ->   Mg2+ (aq)   +  Cu(s)    Eᶿ  = +2.71V

 

Overall conventional cell representation:

Mg(s) / Mg2+ (aq) 1M,  // 1M,Cu2+ (aq)  / Cu(s)    Eᶿ  = +2.71V

 

c)Ag and Pb

From the table above:

2Ag+ (aq) +  2e ->  2Ag(s)  Eᶿ = +0.80V(higher Eᶿ /Right Hand Side diagram)

Pb2+ (aq) +  2e ->Pb(s)    Eᶿ = -0.13V(lower Eᶿ/ Left Hand Side diagram)

Pb(s) ->Pb2+ (aq) +  2e    Eᶿ = +0.13(reverse lower Eᶿ to derive cell reaction / representation)

 

Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS – Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced

Substituting:

Overall Eᶿ = +0.80 – (- 0.13) = +0.93V

 

Overall redox equation:

2Ag+ (aq) + Pb(s)   ->   Pb2+ (aq)   +  2Ag(s)    Eᶿ  = +0.93V

 

Overall conventional cell representation:

Pb(s) / Pb2+ (aq) 1M,  // 1M,2Ag+ (aq)  / Ag(s)    Eᶿ  =  +0.93V

 

d)Chlorine and Bromine

From the table above:

2e  +  Cl2(g) ->2Cl(aq)   Eᶿ = +1.36V(higher Eᶿ /Right Hand Side diagram)

2e  +  Br2(aq) ->2Br(aq)   Eᶿ = +0.13V(lower Eᶿ/ Left Hand Side diagram)

2Br(aq) -> Br2(aq) +  2e    Eᶿ = -0.13(reverse lower Eᶿ to derive cell reaction / representation)

 

Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS – Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced

Substituting:

Overall Eᶿ =  – 0.13 – (- 1.36) = +1.23V

 

Overall redox equation:

2Br(aq) + Cl2(g)   ->   2Cl(aq)   +  Br2(aq)    Eᶿ  = +1.23V

 

Overall conventional cell representation:

Cl2(g) / 2Cl(aq) 1M,  // 1M, 2Br(aq)  / Br2(aq)    Eᶿ  =  +1.23V

 

 

Chlorine displaces bromine  from bromine water. When chlorine gas is thus bubbled in bromine water, the pale green colour fades as displacement takes place and a brown solution containing dissolved bromine liquid is formed. This reaction is feasible /possible because the overall redox reaction has a positive Eᶿ value.

 

 

 

e)Strongest oxidizing agent and the strongest reducing agent.

From the table above:

2e  +  F2(g) ->2F(aq)   Eᶿ = +2.87V(highest Eᶿ /strongest oxidizing agent)

2e  +  2K+ (aq) ->2K (aq)   Eᶿ = -2.92V(lowest Eᶿ/ strongest reducing agent)

2K (aq) -> 2K+ (aq) +  2e    Eᶿ = +2.92V (reverse lower Eᶿ to derive cell reaction / representation)

 

Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS – Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced

Substituting:

Overall Eᶿ =  +2.87 – (-2.92) = +5.79V

 

Overall redox equation:

F2(g) + 2K(s)   ->   2F(aq)   +  2K+ (aq)    Eᶿ  = +5.79V

 

Overall conventional cell representation:

 

2K(s) / 2K+ (aq),1M,  // 1M, 2F(aq)  / F2(g)    Eᶿ  = +5.79V

The redox reactions in an electrochemical/voltaic is commercially applied to make the:

(a)Dry /primary/Laclanche cell.

 

(b)Wet /secondary /accumulators.

 

(a)Dry/primary/Laclanche cell

 

Examine a used dry cell.

Note the positive and the negative terminal of the cell. Carefully using a knife cut a cross section from one terminal to the other.

The dry cell consist of a Zinc can containing a graphite rod at the centre surrounded by a paste of;

-Ammonium chloride

-Zinc chloride

-powdered manganese (IV) oxide mixed with Carbon.

Zinc acts/serve as the negative terminal where it ionizes/dissociates:

 

Zn(s)   ->     Zn2+(aq)  +   2e

Ammonium ions in ammonium chloride serve as the positive terminal where it is converted to ammonia gas and hydrogen gas.

2NH4+(aq)  +   2e  -> 2NH3(g) + H2(g)

Ammonia forms a complex salt / compound /(Zn(NH3) 4)2+ (aq) / tetramminezinc(II) complex with the Zinc chloride in the paste.

 

Manganese (IV) oxide oxidizes the hydrogen produced at the electrodes to water preventing any bubbles from coating the carbon terminal which would reduce the efficiency of the cell.

 

Ammonium chloride is used as paste because the solid does not conduct electricity because the ions are fused/not mobile.

Since the reactants are used up, the dry /primary /Laclanche cell cannot provide continous supply of electricity.The process of restoring the reactants is called recharging.

 

b)Wet/Secondary/Accumulators

  1. Wet/Secondary/Accumulators are rechargeable unlike dry /primary /Laclanche cells.Wet/Secondary/Accumulators are made up of:

(i)Lead plate that forms the negative terminal

(ii)Lead(IV) oxide that forms the positive terminal

2.The two electrodes are dipped in concentrated sulphuric(VI) acid of a relative density 1.2/1.3

3.At the negative terminal,lead ionizes /dissolves;

Pb(s)     ->  Pb2+   +  2e 

 

4.At the positive terminal,

(i) Lead(IV) oxide reacts with the hydrogen ions in sulphuric(VI)acid to form Pb2+ (aq) ions;

PbO2(s)    + 4H+(aq) + 2e ->  Pb2+ (aq)  + H2O(l)

         

(ii) Pb2+ (aq) ions formed instantly react with sulphate (VI) ions/ SO42- (aq) from sulphuric (VI)acid to form insoluble Lead(II) sulphate (VI).

Pb2+ (aq)   +  SO42- (aq) ->  PbSO4(s)

 

5.The overall cell reaction is called discharging

PbO2(s)  +Pb(s) + 4H+(aq) + 2SO42- (aq)-> 2PbSO4(s) + 2H2O(l) Eᶿ = +2.0V

6.The insoluble Lead(II) sulphate (VI) formed should not be left for long since fine Lead(II) sulphate (VI) will change to a course non-reversible and inactive form making the cell less efficient.

As the battery discharges ,lead and lead(IV)oxide are depleted/finished/reduced and the concentration of sulphuric(VI)acid decreases.

 

  1. During recharging, the electrode reaction is reversed as below:

2PbSO4(s) + 2H2O(l) ->PbO2(s)  +Pb(s) + 4H+(aq) + 2SO42- (aq)

 

  1. A car battery has six Lead-acid cells making a total of 12 volts.

 

 

 

 

 

 

 

 

 

 

 

 

 

        (iii)ELECTROLYSIS (ELECTROLYTIC CELL)

 

1.Electrolysis is defined simply as the decomposition of a compound by an electric current/electricity.

A compound that is decomposed by an electric current is called an electrolyte. Some electrolytes are weak while others are strong.

 

2.Strong electrolytes are those that are fully ionized/dissociated into (many) ions.    Common strong electrolytes include:

(i)all mineral acids

(ii)all strong alkalis/sodium hydroxide/potassium hydroxide.

(iii)all soluble salts

 

3.Weak electrolytes are those that are partially/partly ionized/dissociated into (few) ions.

Common weak electrolytes include:

(i)all organic acids

(ii)all bases except sodium hydroxide/potassium hydroxide.

(iii)Water

 

  1. 4. A compound that is not decomposed by an electric current is called non-electrolyte. Non-electrolytes are those compounds /substances that exist as molecules and thus cannot ionize/dissociate into(any) ions .

Common non-electrolytes include:

(i) most organic solvents (e.g. petrol/paraffin/benzene/methylbenzene/ethanol)

(ii)all hydrocarbons(alkanes /alkenes/alkynes)

(iii)Chemicals of life(e.g. proteins, carbohydrates, lipids, starch, sugar)

 

  1. 5. An electrolytes in solid state have fused /joined ions and therefore do not conduct electricity but the ions (cations and anions) are free and mobile in molten and aqueous (solution, dissolved in water) state.

 

6.During electrolysis, the free ions are attracted to the electrodes. An electrode is a rod through which current enter and leave the electrolyte during electrolysis. An electrode that does not influence/alter the products of electrolysis is called an inert electrode.

 Common inert electrodes include:

(i)Platinum

(ii)Carbon graphite

Platinum is not usually used in a school laboratory because it is very expensive. Carbon graphite is easily/readily and cheaply available (from used dry cells).

 

7.The positive electrode is called Anode.The anode is the electrode through which current enter the electrolyte/electrons leave the electrolyte

8.The negative electrode is called Cathode. The cathode is the electrode through which current leave the electrolyte / electrons enter the electrolyte

 

  1. 9. During the electrolysis, free anions are attracted to the anode where they lose /donate electrons to form neutral atoms/molecules. i.e.

 

M(l)  ->       M+(l)  +  e  (for cations from molten electrolytes)

M(s)  ->       M+(aq)  +  e  (for cations from electrolytes in aqueous state / solution / dissolved in water)

 

The neutral atoms /molecules form the products of electrolysis at the anode. This is called discharge at anode

 

  1. During electrolysis, free cations are attracted to the cathode where they gain /accept/acquire electrons to form neutral atoms/molecules.

X+ (aq)  +  2e -> X(s) (for cations from electrolytes in aqueous state / solution / dissolved in water)

2X+ (l)  +  2e -> X (l)  (for cations from molten electrolytes)

 

The neutral atoms /molecules form the products of electrolysis at the cathode. This is called discharge at cathode.

 

  1. The below set up shows an electrolytic cell.

 

 

 

 

 

 

  1. For a compound /salt containing only two ion/binary salt the products of electrolysis in an electrolytic cell can be determined as in the below examples:

 

 

a)To determine the products of electrolysis of molten Lead(II)chloride

 

(i)Decomposition of electrolyte into free ions;

PbCl2 (l)        ->     Pb 2+(l)  +    2Cl(l)

(Compound decomposed into free cation and anion in liquid state)

 

(ii)At the cathode/negative electrode(-);

Pb 2+(l)    +      2e     ->    Pb (l)

(Cation / Pb 2+ gains / accepts / acquires electrons to form free atom)

 

(iii)At the anode/positive electrode(+);

2Cl(l)      ->    Cl2 (g)  +    2e

(Anion / Cldonate/lose electrons to form free atom then a gas molecule)

 

(iv)Products of electrolysis therefore are;

I.At the cathode grey beads /solid lead metal.

II.At the anode pale green chlorine gas.

 

 

b)To determine the products of electrolysis of molten Zinc bromide

 

(i)Decomposition of electrolyte into free ions;

ZnBr2 (l)        ->    Zn 2+(l)  +    2Br(l)

(Compound decomposed into free cation and anion in liquid state)

 

(ii)At the cathode/negative electrode(-);

Zn 2+(l)    +      2e     ->    Zn(l)

(Cation / Zn2+ gains / accepts / acquires electrons to form free atom)

 

(iii)At the anode/positive electrode(+);

2Br(l)      ->    Br2 (g)  +    2e

(Anion / Brdonate/lose electrons to form free atom then a liquid molecule which change to gas on heating)

 

(iv)Products of electrolysis therefore are;

I.At the cathode grey beads /solid Zinc metal.

II.At the anode red bromine liquid / red/brown bromine gas.

 

c)To determine the products of electrolysis of molten sodium chloride

 

(i)Decomposition of electrolyte into free ions;

NaCl (l)        ->      Na +(l)  +    Cl(l)

(Compound decomposed into free cation and anion in liquid state)

 

(ii)At the cathode/negative electrode(-);

2Na+(l)    +      2e     ->    Na (l)

(Cation / Na+ gains / accepts / acquires electrons to form free atom)

 

(iii)At the anode/positive electrode(+);

2Cl(l)      ->    Cl2 (g)  +    2e

(Anion / Cldonate/lose electrons to form free atom then a gas molecule)

 

(iv)Products of electrolysis therefore are;

I.At the cathode grey beads /solid sodium metal.

II.At the anode pale green chlorine gas.

 

      d)To determine the products of electrolysis of molten Aluminium (III)oxide

 

(i)Decomposition of electrolyte into free ions;

Al2O3 (l)        ->     2Al 3+(l)  +    3O2-(l)

(Compound decomposed into free cation and anion in liquid state)

 

(ii)At the cathode/negative electrode(-);

4Al 3+ (l)    +      12e     ->    4Al (l)

(Cation / Al 3+ gains / accepts / acquires electrons to form free atom)

 

(iii)At the anode/positive electrode(+);

6O2-(l)      ->    3O2 (g)  +    12e

(Anion /6O2- donate/lose 12 electrons to form free atom then three gas molecule)

 

(iv)Products of electrolysis therefore are;

I.At the cathode grey beads /solid aluminium metal.

II.At the anode colourless  gas that relights/rekindles glowing splint.

 

  1. 13. For a compound /salt mixture containing many ions in an electrolytic cell, the discharge of ions in the cell depend on the following factors:

 

  1. Position of cations and anions in the electrochemical series

 

  1. Most electropositive cations require more energy to reduce (gain electrons) and thus not readily discharged. The higher elements /metals in the electrochemical series the less easily/readily it is discharged at the cathode in the electrolytic cell.

Table I showing the relative ease of discharge of cations in an electrolytic cell

 

K+(aq)          +  e    -> K(s)        (least readily/easily discharged)

Na+(aq)        +  e   -> Na(s)

Ca2+(aq)       +  2e -> Ca(s)

Mg2+(aq)      +  2e -> Mg(s)

Al3+(aq)       +  3e -> Al(s)

Zn2+(aq)       +  2e -> Zn(s)

Fe2+(aq)       +  2e -> Fe(s)

Pb2+(aq)       +  2e -> Pb(s)

2H+(aq)        +  2e -> H2(g)        (hydrogen is usually “metallic”)

Cu2+(aq)       +  2e -> Cu(s)

Hg2+(aq)      +  2e -> Hg(s)

Ag+(aq)        +  e -> Ag(s)         (most readily/easily discharged)

 

2.The OH ion is the most readily/easily discharged anion . All the other  anionic radicals(SO42- ,SO32- ,CO32- ,HSO4 ,HCO3,NO3,PO43-)are not/never discharged. The ease of discharge of halogen ions increase down the group.

 

Table II showing the relative ease of discharge of anions in an electrolytic cell

 

4OH (aq)     ->    2H2O(l)   + O2 (g) + 4e (most readily/easily discharged)

2 I(aq)         ->    I2(aq)        +  2e

2 Br(aq)      ->    Br2(aq)     +  2e

2 Cl(aq)      ->    Cl2(aq)     +  2e

2 F(aq)        ->    F2(aq)       +  2e

 

SO42- ,SO32- ,CO32- ,HSO4 ,HCO3,NO3,PO43- not/never/rarely discharged.

 

3.(a)When two or more cations are attracted to the cathode, the ion lower in the electrochemical series is discharged instead of that which is higher as per the table I above. This is called selective/preferential discharge at cathode.

 

(b)When two or more anions are attracted to the anode, the ion higher in the electrochemical series is discharged instead of that which is lower as per the table I above. This is called selective/preferential discharge at anode.

 

4.The following experiments shows the influence /effect of selective/preferential discharge on the products of electrolysis:

 

(i)Electrolysis of acidified water/dilute sulphuric(VI) acid

 

Fill the Hoffmann voltameter with dilute sulphuric(VI) acid. Connect the Hoffmann voltameter to a d.c. electric supply. Note the observations at each electrode.

 

Electrolytic cell set up during electrolysis of acidified water/dilute sulphuric(VI) acid

 

 

 

Answer the following questions:

  1. Write the equation for the decomposition of the electrolytes during the electrolytic process.

H2O(l)         -> OH (aq)   + H+(aq)

H2 SO4(aq)   -> SO42-(aq)  + 2H+(aq)

 

  1. Name the ions in acidified water that are attracted/move to:

 Cathode- H+(aq) from either sulphuric(VI) acid (H2 SO4) or water (H2O)       

AnodeSO42-(aq) from sulphuric (VI) acid (H2 SO4)  and OH (aq) from water (H2O)       

 

III. Write the equation for the reaction during the electrolytic process at the:

Cathode      4H+(aq)        +   4e   ->  2H2(g)

Anode         4OH (aq)   ->    2H2O(l)   + O2 (g)  +  4e

(4OHions selectively discharged instead of SO42- ions at the anode)

 

  1. Name the products of electrolysis of acidified water.

Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound

AnodeOxygen gas (colourless gas that relights /rekindles glowing splint)

 

  1. Explain the difference in volume of products at the cathode and anode.

The four(4) electrons donated/lost by OH ions to form 1 molecule/1volume/1mole of oxygen (O2)gas at the anode are gained/acquired/accepted by the four H+(aq) ions to form 2 molecule/2volume/2mole of Hydrogen (H2)gas at the cathode.

The volume of Oxygen gas at the anode is thus a half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is thus a twice the volume of Oxygen produced at the anode.

 

  1. Why is electrolysis of dilute sulphuric(VI) acid called “electrolysis of (acidified) water”?

The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. This implies/means that water in the electrolyte is being decomposed into hydrogen and Oxygen gases. The electrolysis of dilute sulphuric acid is therefore called “electrolysis of acidified water.”

  1. Explain the changes in concentration of the electrolyte during electrolysis of acidified water”

The concentration of dilute sulphuric (VI) acid increases. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape. The concentration /mole of acid present in a given volume of solution thus continue increasing/rising.

 

(ii)Electrolysis of  Magnesium sulphate(VI) solution

 

Fill the Hoffmann voltameter with dilute sulphuric(VI) acid. Connect the Hoffmann voltameter to a d.c. electric supply. Note the observations at each electrode.

 

 

Answer the following questions:

  1. Write the equation for the decomposition of the electrolytes during the electrolytic process.

H2O(l)         -> OH (aq)   + H+(aq)

Mg SO4(aq)   -> SO42-(aq)  + Mg2+(aq)

 

  1. Name the ions in Magnesium sulphate(VI) solution that are attracted/move to:

 Cathode- Mg2+(aq) from Magnesium sulphate(VI) solution (Mg SO4) and H+(aq) from water (H2O)       

AnodeSO42-(aq) from Magnesium sulphate(VI) solution (Mg SO4) and OH (aq) from water (H2O)       

 

III. Write the equation for the reaction during the electrolytic process at the:

Cathode      4H+(aq)        +   4e   ->  2H2(g)

H+ ions selectively discharged instead of Mg2+ ions at the cathode)

 

Anode         4OH (aq)   ->    2H2O(l)   + O2 (g)  +  4e

(4OHions selectively discharged instead of SO42- ions at the anode)

 

  1. Name the products of electrolysis of Magnesium sulphate(VI) solution

 

Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound

AnodeOxygen gas (colourless gas that relights /rekindles glowing splint)

 

  1. Explain the difference in volume of products at the cathode and anode.

 

The four(4) electrons donated/lost by OH ions to form 1 molecule/1volume/1mole of oxygen (O2)gas at the anode are gained/acquired/accepted by the four H+(aq) ions to form 2 molecule/2volume/2mole of Hydrogen (H2)gas at the cathode.

The volume of Oxygen gas at the anode is thus a half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is thus a twice the volume of Oxygen produced at the anode.

 

  1. Explain the changes in concentration of the electrolyte during electrolysis of Magnesium sulphate(VI) solution

 

The concentration of dilute Magnesium sulphate(VI) solution increases.

The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water.

Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products.

The concentration /mole of acid present in a given volume of Magnesium sulphate(VI) solution thus continue increasing/rising.

 

 

 

 

The set – up below was used during the electrolysis of aqueous magnesium sulphate using inert electrodes.

 

 

 

 

 

 

 

 

 

 

Name a suitable pair of electrodes for this experiment

 

Identify the ions and cations in the solution

 

On the diagram label the cathode

 

Write ionic equations for the reactions that took place at the anode.

 

Explain the change that occurred to the concentration of magnesium sulphate solution during the experience.

 

During the electrolysis a current of 2 amperes was passed through the solution for 4 hours. Calculate the volume of the gas produced at the anode.(1 faraday 96500 coulombs and volume of a gas at room temperature is 24000cm3)

 

One of the uses of electrolysis is electroplating

What is meant by electroplating?

Give tow reasons why electroplating is necessary.

 

  1. Concentration of the electrolytes

 

1.High concentrations of cations and/or anions at the electrodes block the ion/s that is likely to be discharged at the electrode. This is called over voltage. A concentrated solution therefore produces different products of electrolysis from a dilute one.

 

  1. The following experiments show the influence/effect of concentration of electrolyte on the products of electrolysis.

 

 (i)Electrolysis of  dilute and concentrated(brine)sodium chloride solution

 

  1. I. Dissolve about 0.5 g of pure sodium chloride crystals in 100cm3 of water. Place the solution in an electrolytic cell. Note the observations at each electrode for 10 minutes. Transfer the set up into a fume chamber/open and continue to make observations for a further 10 minute.

 

Answer the following questions:

  1. Write the equation for the decomposition of the electrolytes during the electrolytic process.

H2O(l)         -> OH (aq)   + H+(aq)

NaCl(aq)   -> Cl(aq)  + Na+(aq)

 

  1. Name the ions in sodium chloride solution that are attracted/move to:

 Cathode- Na+(aq) from Sodium chloride solution (NaCl) and H+(aq) from water (H2O)       

AnodeCl(aq) from sodiumchloride solution (NaCl) and OH (aq) from water (H2O)       

 

III. Write the equation for the reaction during the electrolytic process at the:

Cathode      4H+(aq)        +   4e   ->  2H2(g)

H+ ions selectively discharged instead of Na+ ions at the cathode)

 

Anode         4OH (aq)   ->    2H2O(l)   + O2 (g)  +  4e

(4OHions selectively discharged instead of Cl ions at the anode)

 

  1. Name the products of electrolysis of dilute sodium chloride solution

Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound

AnodeOxygen gas (colourless gas that relights /rekindles glowing splint)

 

  1. Explain the difference in volume of products at the cathode and anode.

Four(4) electrons donated/lost by OH ions to form 1 molecule/1volume/1mole of oxygen (O2)gas at the anode are gained/acquired/accepted by four H+(aq) ions to form 2 molecule/2volume/2mole of Hydrogen (H2)gas at the cathode.

The volume of Oxygen gas at the anode is half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is twice the volume of Oxygen produced at the anode.

 

  1. Explain the changes in concentration of the electrolyte during electrolysis of sodium chloride solution

The concentration of dilute sodium chloride solution increases.

The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products. The concentration /moles of salt present in a given volume of sodium chloride solution continue increasing/rising.

 

  1. II. Dissolve about 20 g of pure sodium chloride crystals in 100cm3 of water. Place the solution in an electrolytic cell. Note the observations continuously at each electrode for 30 minutes in a fume chamber/open.

 

Answer the following questions:

  1. Write the equation for the decomposition of the electrolytes during the electrolytic process.

H2O(l)         -> OH (aq)   + H+(aq)

NaCl(aq)      -> Cl(aq)      + Na+(aq)

 

  1. Name the ions in sodium chloride solution that are attracted/move to:

 Cathode- Na+(aq) from Sodium chloride solution (NaCl) and H+(aq) from water (H2O)       

AnodeCl(aq) from sodium chloride solution (NaCl) and OH (aq) from water (H2O)       

 

III. Write the equation for the reaction during the electrolytic process at the:

Cathode      2H+(aq)        +   2e   ->  H2(g)

H+ ions selectively discharged instead of Na+ ions at the cathode)

 

Anode         2Cl (aq)   ->  Cl2(g)  +  4e

(Clions with a higher concentration block the discharge of OH ions at the anode)

 

  1. Name the products of electrolysis of concentrated sodium chloride solution/brine

Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound

AnodeChlorine gas(pale green gas that bleaches damp/moist/wet litmus papers)

 

  1. Explain the difference in volume of products at the cathode and anode.

Two (2) electrons donated/lost by Cl ions to form 1 molecule/1volume/1mole of Chlorine (Cl2)gas at the anode are gained/acquired/accepted by two H+(aq) ions to form 1 molecule/1volume/1mole of Hydrogen (H2)gas at the cathode.

The volume of Chlorine gas at the anode is equal to the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is equal to the volume of Chlorine produced at the anode.

 

  1. Explain the changes in concentration of the electrolyte during electrolysis of concentrated sodium chloride solution/brine

The concentration of concentrated sodium chloride solution/brine increases.

The ratio of Cl2 (g): H2 (g) is 1:1 as they are combined in water.

Water in the electrolyte is decomposed into only Hydrogen gas that escapes as products at cathode.

 

The concentration /moles of OH (aq) and Na+ ion (as NaOH) present in a given volume of electrolyte continue increasing/rising.

This makes the electrolyte strongly alkaline with high pH.

As the electrolysis of brine continues the concentration of Cl ions decrease and oxygen gas start being liberated at anode.

The electrolyte pH is thus lowered and the concentration of brine starts again increasing.

 

(ii)Electrolysis of dilute and concentrated Hydrochloric acid solution

 

  1. I. Prepare about 50cm3 of 0.05 M of dilute Hydrochloric acid in 100cm3 solution. Place the solution in an electrolytic cell. Note the observations at each electrode for 10 minutes.

 

Answer the following questions:

  1. Write the equation for the decomposition of the electrolytes during the electrolytic process.

H2O(l)         -> OH (aq)   + H+(aq)

HCl(aq)       -> Cl(aq)      + H+(aq)

 

  1. Name the ions in dilute Hydrochloric acid solution that are attracted/move to:

 Cathode- H+(aq) from dilute Hydrochloric acid (HCl) and H+(aq) from water (H2O)       

AnodeCl(aq) from dilute Hydrochloric acid (HCl) and OH (aq) from water (H2O)       

 

III. Write the equation for the reaction during the electrolytic process at the:

Cathode      4H+(aq)        +   4e   ->  2H2(g)

H+ ions selectively discharged instead of Na+ ions at the cathode)

 

Anode         4OH (aq)   ->     H2O(l)  +O2+  4e

(4OHions selectively discharged instead of Cl ions at the anode)

 

  1. Name the products of electrolysis of dilute Hydrochloric acid

Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound

AnodeOxygen gas (colourless gas that relights /rekindles glowing splint)

  1. Explain the difference in volume of products at the cathode and anode.

Four(4) electrons donated/lost by OH ions to form 1 molecule/1volume/1mole of oxygen (O2)gas at the anode are gained/acquired/accepted by four H+(aq) ions to form 2 molecule/2volume/2mole of Hydrogen (H2)gas at the cathode.

The volume of Oxygen gas at the anode is half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is twice the volume of Oxygen produced at the anode.

 

  1. Explain the changes in concentration of the electrolyte during electrolysis of dilute Hydrochloric acid

The concentration of dilute Hydrochloric acid increases.

The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products. The concentration /moles of HCl present in a given volume of dilute Hydrochloric acid continue increasing/rising.

 

  1. II. Prepare about 50cm3 of 2M of Hydrochloric acid in 100cm3 solution. Place the solution in an electrolytic cell. Note the observations at each electrode for 30 minutes

CautionThis experiment should be done in the open/fume chamber.

 

Answer the following questions:

  1. Write the equation for the decomposition of the electrolytes during the electrolytic process.

H2O(l)         -> OH (aq)   + H+(aq)

HCl(aq)       -> Cl(aq)      + H+(aq)

 

  1. Name the ions in 2M Hydrochloric acid solution that are attracted/move to:

 Cathode- H+(aq) from dilute Hydrochloric acid (HCl) and H+(aq) from water (H2O)       

AnodeCl(aq) from dilute Hydrochloric acid (HCl) and OH (aq) from water (H2O)       

 

III. Write the equation for the reaction during the electrolytic process at the:

Cathode      4H+(aq)        +   4e   ->  2H2(g)

H+ ions selectively discharged instead of Na+ ions at the cathode)

 

Anode         2Cl (aq)   ->     Cl2+  2e

(OHions concentration is low.Cl ions concentration is higher at the anode thus cause over voltage/block discharge of OHions)

 

  1. Name the products of electrolysis of 2M Hydrochloric acid

Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound

 

AnodeChlorine gas (Pale green gas that bleaches blue/red moist/wet/damp litmus papers)

 

  1. Explain the difference in volume of products at the cathode and anode.

Two(2) electrons donated/lost by Cl ions to form 1 molecule/1volume/1mole of Chlorine (Cl2)gas at the anode are gained/acquired/accepted by two H+(aq) ions to form 1 molecule/1volume/1mole of Hydrogen (H2)gas at the cathode.

 

The volume of Chlorine gas at the anode is equal to the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is twice the volume of Chlorine produced at the anode.

 

  1. Explain the changes in concentration of the electrolyte during electrolysis of 2M Hydrochloric acid

The concentration of Hydrochloric acid decreases.

The ratio of H2 (g): Cl2 (g) is 1:1 as they are combined in Hydrochloric acid.

Water in the electrolyte is decomposed only into Hydrogen gas that escapes as products at the cathode.

There is a net accumulation of excess OH (aq) ions in solution.

This makes the electrolyte strongly alkaline with high pH.

 

 

 

  1. Nature of electrodes used in the electrolytic cell

 

Inert electrodes (carbon-graphite and platinum) do not alter the expected products of electrolysis in an electrolytic cell. If another/different electrode is used in the electrolytic cell it alters/influences/changes the expected products of electrolysis.

The examples below illustrate the influence of the nature of electrode on the products of electrolysis:

 

(i)Electrolysis of copper(II) sulphate(VI) solution

  1. Using carbon-graphite electrodes

Weigh Carbon -graphite electrodes. Record the masses of the electrodes in table I below. Place the electrodes in 1M copper(II) sulphate(VI) solution in a beaker. Set up an electrolytic cell.

Close the switch and pass current for about 20 minutes. Observe each electrode and any changes in electrolyte. Remove the electrodes from the electrolyte. Wash with acetone/propanone and allow them to dry. Reweigh each electrode.

Sample results

 

Mass of cathode before electrolysis 23.4 g Mass of anode before

electrolysis

 22.4 g
Mass of cathode after electrolysis 25.4 g Mass of anode after

electrolysis

 22.4 g
Brown solid deposit at the cathode after electrolysis     – Bubbles of colourless gas that relight splint     –
Blue colour of electrolyte fades/become less blue     – Blue colour of electrolyte fades /become less blue     –

 

 

Answer the following questions:

  1. Write the equation for the decomposition of the electrolytes during the electrolytic process.

H2O(l)         ->   OH (aq)     + H+(aq)

CuSO4(aq)             ->    SO42-(aq)    +  Cu2+(aq)

 

  1. Name the ions in 1M copper(II) sulphate(VI) solution that are attracted/move to:

 Cathode- Cu2+ (aq) from copper(II) sulphate(VI) solution and H+(aq) from water (H2O)       

AnodeSO42-(aq) from copper(II) sulphate(VI) solution and OH (aq) from water (H2O)       

 

III. Write the equation for the reaction during the electrolytic process at the:

Cathode      2Cu2+ (aq)    +   4e   ->  2Cu(g)

Cu2+ ions are lower than H+ ions in the electrochemical series therefore selectively discharged at the cathode.)

 

Anode         4OH (aq)   ->    H2O(l) + O2+  4e

(OHions ions are higher than SO42- ions in the electrochemical series therefore selectively discharged at the cathode.))

 

  1. Name the products of electrolysis of 1M copper(II) sulphate(VI) solution

Cathode-2 moles  of copper metal as brown solid coat

 

AnodeOxygen gas (Colourless gas that relights /rekindles glowing splint)

 

  1. Explain the changes that take place at the cathode and anode.

Four(4) electrons donated/lost by OH ions to form 1 molecule/1volume/1mole of Oxygen (O2)gas at the anode are gained/acquired/accepted by two Cu2+(aq) ions to form 2 moles of  brown copper solid that deposit itself at  the cathode.

 

The moles of oxygen gas at the anode is equal to the moles of copper produced at the cathode

  1. Explain the changes in electrolyte during electrolysis of 1M copper (II) sulphate(VI) solution.

 

(i)The pH of copper(II) sulphate(VI) solution lowers/decreases. The salt becomes more acidic. Water in the electrolyte is decomposed only into Oxygen gas (from the OH  ions) that escapes as products at the anode. There is a net accumulation of excess H+ (aq) ions in solution. This makes the electrolyte strongly acidic with low pH.

 

(ii) Cu2+ (aq) ions are responsible for the blue colour of the electrolyte/ copper(II) sulphate (VI) solution. As electrolysis continues, blue Cu2+ (aq) ions gain electrons to form brown Copper. The blue colour of electrolyte therefore fades/become less blue.

(iii)Copper is deposited at the cathode. This increases the mass of the cathode.OH  ions that produce Oxygen gas at anode come from water. Oxygen escapes out/away without increasing the mass of anode.

 

  1. Using copper electrodes

Weigh clean copper plates electrodes. Record the masses  of the electrodes in table I below. Place the electrodes in 1M copper(II) sulphate(VI) solution in a beaker. Set up an electrolytic cell.

Close the switch and pass current for about 20 minutes. Observe each electrode and any changes in electrolyte. Remove the electrodes from the electrolyte. Wash with acetone/propanone and allow them to dry. Reweigh each electrode.

 

Sample results

 

Mass of cathode before electrolysis 23.4 g Mass of anode before

electrolysis

 22.4 g
Mass of cathode after electrolysis 25.4 g Mass of anode after

electrolysis

 20.4 g
Brown solid deposit at the cathode after electrolysis     – Anode decrease insize/erodes/wear off     –
Blue colour of electrolyte remain blue     – Blue colour of electrolyte remain blue     –

 

Answer the following questions:

 

  1. Write the equation for the decomposition of the electrolytes during the electrolytic process.

H2O(l)         ->   OH (aq)     + H+(aq)

CuSO4(aq)             ->    SO42-(aq)    +  Cu2+(aq)

 

  1. Name the ions in 1M copper(II) sulphate(VI) solution that are attracted/move to:

 Cathode- Cu2+ (aq) from copper(II) sulphate(VI) solution and H+(aq) from water (H2O)       

AnodeSO42-(aq) from copper(II) sulphate(VI) solution and OH (aq) from water (H2O)       

 

III. Write the equation for the reaction during the electrolytic process at the:

Cathode      Cu2+ (aq)      +   2e   ->  Cu(s)

Cu2+ ions are lower than H+ ions in the electrochemical series therefore selectively discharged at the cathode.)

 

Anode         Cu (s)   ->    Cu2+(aq)        +  2e

(Both OHions  and  SO42- ions  move to the anode but none is discharged. The copper anode itself ionizes/dissolves/dissociate because less energy is used to remove an electron/ionize /dissociate copper atoms than OHions.

 

  1. Name the products of electrolysis of 1M copper(II) sulphate(VI) solution using copper electrodes.

Cathode-1 moles  of copper metal as brown solid coat (Cathode increase/deposits)

 

Anode-Anode erodes/decrease in size

 

  1. Explain the changes that take place during the electrolytic process

(i)Cathode

-Cu2+ ions are lower than H+ ions  in the electrochemical series therefore selectively discharged at the cathode. Cu2+ ions have greater tendency to accept/gain/acquire electrons to form brown copper atoms/solid that deposit itself and increase the mass/size of  the cathode.The copper deposited at the cathode is pure

-H+ ions accumulate around the cathode. Electrolyte thus becomes strongly acidic around the cathode.

-Cu2+ ions in solution are responsible for the blue colour of electrolyte. Blue colour of electrolyte fade around the cathode.

 

(ii)Anode

Copper atom at the anode easily ionizes to release electrons. The anode therefore keeps decreasing in mass/eroding. The amount of copper that dissolve/erode is equal to the mass of copper deposited. This is called electrode ionization.

Electrode ionization is where the anode erodes/decrease and the cathode deposits/increase during electrolysis. The overall concentration of the electrolyte remains constant

 

14.In industries electrolysis has the following uses/applications:

 

(a)Extraction of reactive metals from their ores.

Potassium, sodium ,magnesium, and aluminium  are extracted from their ores using electrolytic methods.

 

(b)Purifying copper after exraction from copper pyrites ores.

Copper obtained from copper pyrites ores is not pure. After extraction, the copper is refined by electrolysing copper(II)sulphate(VI) solution using the impure copper as anode and a thin strip of pure copper as cathode. Electrode ionization take place there:

(i)At the cathode; Cu2+ (aq)      +   2e   ->  Cu(s) (Pure copper deposits on the strip

(ii)At the anode;  Cu(s) ->Cu2+ (aq)   +   2e   (impure copper erodes/dissolves)

 

(c)Electroplating

The label EPNS(Electro Plated Nickel Silver) on some steel/metallic utensils mean they are plated/coated with silver and/or Nickel to improve their appearance(add their aesthetic value)and prevent/slow corrosion(rusting of iron). Electroplating is the process of coating a metal with another metal using an electric current. During electroplating,the cathode is made of the metal to be coated/impure.

Example:

During the electroplating of a spoon with silver

(i)the spoon/impure is placed as the cathode(negative terminal of battery)

(ii)the pure silver is placed as the anode(positive terminal of battery)

(iii)the pure silver erodes/ionizes/dissociates to release electrons:

Ag(s) ->Ag+ (aq)   +   e   (impure silver erodes/dissolves)

(iv) silver (Ag+)ions from electrolyte gain electrons to form pure silver  deposits / coat /cover the spoon/impure

Ag+ (aq)      +   e   ->Ag(s)    (pure silver deposits /coat/cover on spoon)

 

 

 

 

 

15.The quantitative amount of products of electrolysis can be determined by applying Faradays 1st law of electrolysis.

Faradays 1st law of electrolysis states that “the mass/amount of substance liberated/produced/used during electrolysis is directly proportional to the quantity of of electricity passed/used.”

 

(a)The SI unit of quantity of electricity is the coulomb(C). The coulomb may be defined as the quantity of electricity passed/used when a current of one ampere flow for one second.i.e;

 

1Coulomb  = 1 Ampere x 1Second

The Ampere is the SI unit of current(I)

The Second is the SI unit of time(t) therefore;

 

Quantity of electricity(in Coulombs) = Current(I)  x  time(t)

Practice examples

  1. A current of 2 amperes was passed through an electrolytic cell for 20 minutes. Calculate the quantity of electric charge produced.

Working:

Quantity of electricity(in Coulombs)            = Current(I)  x  time(t)

Substituting /converting time to second         = 2  x (20 x 60)

= 2400 C

 

  1. A current of 2 amperes was passed through an electrolytic.96500 coulombs of charge were produced. Calculate the time taken.

Working:

Time(t) in seconds          =       Quantity of electricity(in Coulombs)

Current(I) in amperes

Substituting                     =        96500

2

=        48250 seconds

 

  1. 96500 coulombs of charge were produced after 10 minutes in an electrolytic cell . Calculate the amount of current used.

Working:

Current(I) in amperes                          =       Quantity of electricity(in Coulombs)                                                                         Time(t) in seconds

Substituting/converting time to second=                  96500

10 x 60

=        160.8333 Amperes

                              

(b)The quantity of electricity required for one mole of electrons at the anode/cathode is called the Faraday constant(F). It is about 96500 Coulombs.i.e

 

The number of Faradays used /required is equal to the number of electrons used at cathode/anode during the electrolytic process. e.g.

Cu2+ require to gain 2 moles of electrons=2 Faradays =2 x 96500 coulombs of electricity at the cathode.

Al3+ require to gain 3 moles of electrons=3 Faradays =3 x 96500 coulombs of electricity at the cathode

Na+ require to gain 1 moles of electrons=1 Faradays =1 x 96500 coulombs of electricity at the cathode

2H+ require to gain 2 moles of electrons=2 Faradays =2 x 96500 coulombs of electricity at the cathode to form 1molecule of hydrogen gas

 

2O2 require to lose/donate 4 moles of electrons=4 Faradays =4 x 96500 coulombs of electricity at the anode to form 1molecule of Oxygen O2 gas.

4OH require to lose/donate 4 moles of electrons=4 Faradays =4 x 96500 coulombs of electricity at the anode to form 1molecule of Oxygen gas and 2 molecules of water.

 

(c)The mass/amount of products at the cathode/anode is related to the molar mass of the substance and/or the volume of gases at standard/room temperature and pressure as in the below examples:

 

Practice examples

 

1.Calculate the mass of copper deposited at the cathode when a steady current of 4.0 amperes is passed through copper(II)sulphate(VI) for 30 minutes in an electrolytic cell. (Cu=63.5, 1F = 96500C)

 

Working:

Quantity of electricity(in Coulombs)            = Current(I)  x  time(t)

Substituting /converting time to second         = 4  x (30 x 60)

= 7200 C

 

Equation at the cathode:   Cu2+ (aq) +  2e   ->   Cu(s)

 

2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper  thus;

2  x   96500C    ->   63.5 g

72000C             ->    7200  x  63.5      =  2.3689 g of copper

                                                       2  x 96500

2.a)If 3.2 g of Lead were deposited when a current of 2.5 amperes was passed through an electrolytic cell of molten Lead(II)bromide for 20 minutes, determine the Faraday constant.(Pb = 207)

Working:

Quantity of electricity (in Coulombs) = Current(I)  x  time(t)

Substituting /converting time to second         = 2.5  x (20 x 60)

= 3000 C

 

If 3.2g of Lead                 ->    3000C

Then 207 g of Lead          ->      207 x 3000      =    194062.5 C

3.2

 

Equation at the cathode:   Pb2+ (l) +  2e   ->   Pb(l)

 

From the equation: 2 moles of electrons = 2 Faradays =  194062.5 C

1mole of electrons   =  1 Faraday  => 194062.5   =  97031.25 C

2

 

b)What is the volume of bromine vapour produced at the anode at room temperature(1mole of gas at room temperature and pressure = 24000cm3)

 

Method 1

Equation at the anode:        Br (l) -> Br2(g)  +  2e

From the equation: 2 moles of electrons = 2 Faradays =  194062.5 C  -> 24000cm3

3000 C -> 3000 x 24000

                                                                                                              194062.5

=371.0145cm3

Method 2

Equation at the anode:        Br (l) -> Br2(g)  +  2e

Mole ratio of products at Cathode: anode = 1:1

Moles of Lead at cathode = 3.2  = 0.0155moles = moles of Bromine

207

1 moles of bromine vapour  -> 24000cm3

0.0155moles  of Bromine -> 0.0155 x 24000  = 372 cm3

 

Method 3

Equation at the anode:        Br (l) -> Br2(g)  +  2e

Ratio of Faradays used to form products at Cathode: anode = 2:2

=>  2 x 97031.25 C produce 24000cm3 of bromine vapour

Then: 3000 C  ->  3000 x 24000cm3     = 371.0145cm3

2 x 97031.25

 

3.What mass of copper remain from 2.0 at the anode if a solution of copper(II)sulphate(VI) is electrolysed using a current of 1 ampere flowing through an electrolytic cell for 20 minutes.(Cu= 63.5, 1Faraday = 96487 coulombs)

Working:

Quantity of electricity (in Coulombs) = Current(I)  x  time(t)

Substituting /converting time to second         = 1  x (20 x 60)

= 1200 C

Equation at the cathode:      Cu2+ (aq)   +  2e  -> Cu(s)

 

2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of copper  thus;

2  x   96500C    ->   63.5 g

1200C               ->    1200  x  63.5  =  0.3948g of copper deposited

                                                       2  x 96500

Mass of copper remaining = Original mass – mass dissolved/eroded

=>  2.0 -0.3948     =      1.6052 g of copper remain

 

  1. Calculate the current passed if a mass of 0.234 g of copper is deposited in 4 minutes during electrolysis of a solution of copper (II)sulphate(VI).

(Cu= 63.5  ,1F = 96500C)

Working:

 

Equation at the cathode:      Cu(s)   ->   Cu2+ (aq)   +  2e

2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper  thus;

63.5 g     ->    2  x   96500C

0.234 g  ->  0.234   x 2 x 96500      = 711.2126 C

63.5

Current(I) in amperes                         =       Quantity of electricity(in Coulombs)                                                                         Time(t) in seconds

Substituting/converting time to second=                  711.2126 C

4x 60

=        2.9634 Amperes

 

  1. (a)What quantity of electricity will deposit a mass of 2.43 g of Zinc during electrolysis of a solution of Zinc (II)sulphate(VI).

(Zn= 65  ,1F = 96500C)

Working:

Equation at the cathode:      Zn2+ (aq)   +  2e  -> Zn(s)

 

2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of  Zinc  thus;

                         65 g         ->    2  x  96500

2.43 g     ->       2.43 x 2  x  96500    =  7215.2308 C

                                                              65

(b)Calculate the time (in minutes) it would take during electrolysis of the  solution of Zinc (II)sulphate(VI) above if a current of 4.0 Amperes is used.

 

Time(t) in seconds =       Quantity of electricity(in Coulombs)

Current(I) in amperes

Substituting      =  7215.2308     =       1803.8077 seconds   =   30.0635 minutes

4                                60

 

6.When a current of 1.5 amperes was  passed through a cell containing M3+  ions of metal M for 15 minutes, the mass at cathode increased by 0.26 g.(Faraday constant = 96500C

  1. a) Calculate the quantity of electricity used.

Quantity of electricity (in Coulombs) = Current(I)  x  time(t)

Substituting /converting time to second         = 1.5  x (15 x 60)

= 1350 C

  1. Determine the relative atomic mass of metal M

Equation at the cathode:      M3+ (aq)   +  3e  -> M(s)

1350 C of electricity    ->  0.26 g  of metal M

3 mole of electrons = 3 Faradays = 3 x 96500 C produce a mass =molar mass of  M  thus;

RAM of M    = 0.26 g  x  3  x   96500   =   55.7556(No units)

1350

7.An element “P” has a relative atomic mass 88.When a current of 0.5 amperes was passed through fused chloride of  “P” for 32 minutes and 10seconds ,0.44 g of “P” was deposited at the cathode. Determine the charge on an ion of “P”(Faraday constant = 96500C)

Working:

Quantity of electricity (in Coulombs) = Current(I)  x  time(t)

Substituting /converting time to second         = 0.5  x ((32 x 60) + 10)

= 965C

 0.44 g of metal  “P” are deposited by 965C

88g of of metal  “P” are deposited by:  88  x 965= 193000 C

0.44

96500 C  = 1 mole of electrons = 1 Faradays = single charge

193000 C         ->    193000 = 2  moles/Faradays/charges => symbol of ion = P2+

96500

 

  1. During purification of copper by electrolysis 1.48 g of copper was deposited when a current was passed through aqueous copper (II)sulphate(VI) for 2 ½ hours. Calculate the amount of current that was passed. (Cu= 63.5 ,1F = 96500C)

Working:

 

Equation at the cathode:        Cu2+ (aq)   +  2e-> Cu(s)

2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper  thus;

63.5 g     ->    2  x   96500C

1.48 g  ->  1.48   x 2 x 96500      = 4255.1181 C

63.5

Current(I) in amperes                         =       Quantity of electricity(in Coulombs)                                                                         Time(t) in seconds

Substituting/converting time to second=                  4255.1181C

(( 2  x 60) + 30) x60

=        0.4728 Amperes

  1. Practically Faraday 1st law of electrolysis can be verified as below.

Verifying Faraday 1st law of electrolysis

Procedure.

Weigh clean copper plates electrodes. Record the masses of the electrodes in table I below. Place the electrodes in 1M copper(II) sulphate(VI) solution in a beaker. Set up an electrolytic cell.

Close the switch and pass a steady current of 2 amperes by adjusting the rheostat for exactly 20 minutes.Remove the electrodes from the electrolyte. Wash with acetone/ propanone and allow them to dry. Reweigh each electrode.

 

Sample results

 

Mass of cathode before electrolysis 7.00 g Mass of anode before

electrolysis

 7.75 g
Mass of cathode after electrolysis 8.25 g Mass of anode after

electrolysis

 6.50 g
Change in mass at cathode after electrolysis  1.25 g Change in mass at anode after electrolysis    1.25 g

 

Answer the following questions:

 

  1. Write the equation for the decomposition of the electrolytes during the electrolytic process.

H2O(l)         ->   OH (aq)     + H+(aq)

CuSO4(aq)             ->    SO42-(aq)    +  Cu2+(aq)

 

  1. Name the ions in 1M copper(II) sulphate(VI) solution that are attracted/move to:

 Cathode- Cu2+ (aq) from copper(II) sulphate(VI) solution and H+(aq) from water (H2O)       

AnodeSO42-(aq) from copper(II) sulphate(VI) solution and OH (aq) from water (H2O)       

 

III. Write the equation for the reaction during the electrolytic process at the:

Cathode      Cu2+ (aq)      +   2e   ->  Cu(s)

Cu2+ ions are lower than H+ ions in the electrochemical series therefore selectively discharged at the cathode.)

 

Anode         Cu (s)   ->    Cu2+(aq)        +  2e

(Both OHions  and  SO42- ions  move to the anode but none is discharged. The copper anode itself ionizes/dissolves/dissociate as less energy is used to remove an electron/ionize /dissociate copper atoms than OHions.

 

  1. Name the products of electrolysis of 1M copper(II) sulphate(VI) solution using copper electrodes.

Cathode-1.25 g  of copper metal as brown solid coat/deposits

 

Anode-1.25 g  of copper metal erodes/decrease in size

 

  1. (i)How many moles of electrons are used to deposit/erode one mole of copper metal at the cathode/anode?

From the equation at anode/cathode= 2 moles

 

(ii)How many Faradays are used to deposit/erode one mole of copper metal at the cathode/anode?

From the equation at anode/cathode : 2 moles  = 2 Faradays

 

(iii)Calculate the quantity of electric charge used

 

Working:

Quantity of electricity (in Coulombs) = Current(I)  x  time(t)

Substituting /converting time to second         = 2  x  20 x 60

=  2400C

 

  1. (i) Calculate the quantity of electricity required to deposit/erode one mole of copper at the cathode/anode(Cu=63.5)

Since    1.25 g of copper              ->   2400C

Then    63.5 g (1mole of copper) ->  63.5  x  2400121920 C

1.25

(ii)Determine the Faraday constant from the results in V(i) above

From the equation at;

Cathode       Cu2+ (aq)      +   2e   ->  Cu(s)

Anode          Cu (s)   ->    Cu2+(aq)        +  2e

2 moles = 2 Faradays  -> 121920 C

1 moles = 1 Faradays  -> 121920  = 60960 C

2

(iii) The faraday constant obtained above is far lower than theoretical.Explain

                    -high resistance of the wires used.

-temperatures at 25oC were not kept constant

-plates/electrodes used were not made of pure copper

-plates/electrodes used were not thoroughly clean copper

 

Further practice

1.An element P has a relative atomic mass of 88. When a current of 0.5 amperes was passed through the fused chloride of P for 32 minutes and 10 seconds, 0.44g of P were deposited at the cathode. Determine the charge on an ion of P. (1 faraday = 96500 Coulombs).

 

2.During electrolysis of aqueous copper (II) sulphate, 144750 coulombs of electricity were used. Calculate the mass of copper metal that was obtained

(Cu = 64      ;1 Faraday = 96500 coulombs)   ( 3 mks)

 

3.A nitrate of a metal M was electrolysed .1.18 g of metal was deposited when a current of 4 ampheres flow for 16 minutes.Determine the formula of the sulphate(VI)salt of the metal.

(Faraday constant = 96500 , RAM of X = 59.0)

Working

Q  = It    =>( 4 x 16 x 60) = 3840 C

1.18 g of X => 3840 C

59.0 g =>     59.0 x 3840  = 192000 C

1.18

96500 C = 1Faraday

192000 C=  192000 C x1    =    2F thus charge of M = M2+

96500 C

Valency of  M  is 2 thus formula of sulphate(VI)salt  MSO4

 

  1. Below is the results obtained when a current of 2.0ampheres is passed through copper(II)sulphate(VI)solution for 15 minutes during electrolysis using copper electrode.

               Initial mass of cathode   = 1.0 g

               Final mass of cathode    = 1.6 g

Change in mass of cathode  =  0.60 g

(i)Determine  the change in mass at the anode. Explain your answer.

Mass decrease  = 0.6g.

Electrode ionization take place where the cathode increase in mass form the erosion of the anode

 

(ii)Calculate  the quantity of electricity required to deposit one mole of copper.(Cu =63.5)

Q =It => 2 x 15 x 60 = 1800 coulombs

Method 1

0.60 g  of copper              ->1800 coulombs

63.5 g                   ->          63.5 x 1800      =   190500 Coulombs

                                                                                               0.60

Method 2

Moles of Copper =                 Mass               =>     0.60  =   9.4488 x10 -3 moles

                                                                             Molar mass             63.5

 

9.4488 x10 -3 moles     -> 1800 coulombs

1 Mole                         ->                1 x 1800 coulombs     =   190500.381 coulombs

9.4488 x10 -3 moles

 

(iii)Determine the oxidation number of copper produced at the cathode and hence the formula of its nitrate (V)salt (1 Faraday = 96500 Coulombs)

 

        96500 Coulombs                      ->      1 Faraday

190500.381 coulombs               ->       190500.381 coulombs   x 1

96500 Coulombs

 

= 1.9741 Faradays    => 2F(whole number)

Charge of copper = 2+   =   Oxidation number

=>  Valency of copper  = 2 hence chemical formula of nitrate (V)salt =  Cu (NO3)2

 

 

 

 

 

 

 

 

 

 

 

                            UPGRADE

CHEMISTRY

                

                        FORM 4    

Chemistry of METALS

 

 

 

 

 

 

                 Comprehensive tutorial notes

 

 MUTHOMI S.G

www.kcselibrary.info

                                     0720096206

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

a)Introduction to metals

 

The rationale of studying metals cannot be emphasized.Since ages, the world over, metals like gold and silver have been used for commercial purposes.

The periodicity of alkali and alkaline earth metals was discussed in year 2 of secondary school education. This topic generally deals with:

(a)Natural occurrence of the chief ores of the most useful metals for industrial /commercial purposes.

(b)Extraction of these metals from their ores for industrial/ commercial purposes.

(c)industrial/ commercial uses of these metals.

(d)main physical and chemical properties /characteristic of the metals.

The metals given detailed emphasis here are; Sodium, Aluminium, Iron, Zinc, Lead and Copper.

Electrolysis of  the ore is used for reactive metals; Potassium, Sodium, Magnesium, Calcium, Aluminium
If deep on the earth’s crust deep mining is used
Position on the earth’s crust

The main criteria used in extraction of metals is based on its position in the electrochemical/reactivity series and its occurrence on the earth’s crust.

 

 

 

 

 

 

If near the surface ,open cast mining / quarrying is used

 

 

The oxide is reduced using carbon/ carbon(II) oxide in a furnace if it is made of Zinc ,Tin, Lead ,Copper and Iron
If the ore is low grade  oil, water, and air is  blown  forming a froth(froth flotation) to concentrate
The ore first roasted if it is a carbonate or sulphide  of Zinc, Iron, Tin, Lead, and Copper to form the oxide

 

 

1.SODIUM

  1. Natural occurrence

Sodium naturally occurs as:

(i)Brine-a concentrated solution of sodium chloride(NaCl(aq)) in salty seas and oceans.

(ii)Rock salt-solid sodium chloride(NaCl(s)

(iii)Trona-sodium sesquicarbonate(NaHCO3.Na2CO3.2H2O) especially in lake Magadi in Kenya.

(iv)Chile saltpeter-sodium nitrate(NaNO3)

 

b)(i)

 Extraction of Sodium from brine/Manufacture of Sodium hydroxide/The flowing mercury cathode cell/ TheCaster-Keller process

 

I.Raw materials

 

(i) Brine-concentrated solution of sodium chloride (NaCl (aq))  from salty seas and oceans.

(ii)Mercury

(iii)Water from river/lakes

 

  1. Chemical processes

 

Salty lakes, seas and oceans contain large amount of dissolved sodium chloride (NaCl (aq)) solution.

This solution is concentrated to form brine which is fed into an electrolytic chamber made of suspended Carbon graphite/titanium as the anode and a continuous flow of Mercury as the cathode.Note

Mercury is the only naturally occurring known liquid metal at room

temperature and pressure

 

Questions

 

  1. Write the equation for the decomposition of the electrolyte during the electrolytic process.

H2O(l)                       H+(aq)      +      OH(aq)

                 NaCl(aq)                   Na+(aq)    +      Cl(aq)

 

  1. Name the ions present in brine that moves to the:

(i)Mercury cathode; H+(aq) , Na+(aq)

(ii)Titanium/graphite; OH(aq), Cl(aq)

 

III. Write the equation for the reaction that take place during the electrolytic process at the;

Cathode;   2Na+(aq)  +      2e                    2Na(s)

Anode;     2Cl(aq)                   Cl2(g)    +  2e

Note

(i)Concentration of 2Cl(aq) ions is higher than OHions causing overvoltage thus blocking OH ions from being discharged at the anode.

 

(ii)Concentration of Na+(aq) ions is higher than H+ ions causing overvoltage thus blocking H+ ions from being discharged at the cathode.

 

  1. Name the products of electrolysis in the flowing mercury-cathode cell.

 

(i)Mercury cathode; Sodium metal as grey soft metal/solid

 

(ii)Titanium/graphite; Chlorine gas as a pale green gas that turns moist blue/red litmus papers red then bleaches both. Chlorine gas is a very useful by-product in;

(i)making (PVC)polyvinylchloride(polychloroethene) pipes.

(ii)chlorination/sterilization of water to kill germs.

(iii)bleaching agent

(iv)manufacture of hydrochloric acid.

 

Sodium produced at the cathode immediately reacts with the mercury at the cathode forming sodium amalgam(NaHg) liquid that flow out of the chamber.

Na(s)   +   Hg(l)                        Na Hg (l)

 

Sodium amalgam is added distilled water and reacts to form sodium hydroxide solution, free mercury and Hydrogen gas.

2Na Hg (l)  + 2H2O(l)                2NaOH (aq)   +  2Hg(l)  +  H2(g)

 

Hydrogen gas is a very useful by-product in;

(i)making ammonia gas in the Haber process

                               (ii)manufacture of hydrochloric acid

                               (iii)in weather balloons to forecast weather

(iv)as rocket fuel

As the electrolysis of brine continues, the concentration of Cl-ions decreases and oxygen gas start being liberated. Continuous feeding of the electrolyte is therefore very necessary.

 

III.Uses of sodium hydroxide

The sodium hydroxide produced is very pure and is used mainly in:
(i)Making soapy and soapless detergents.

(ii)making cellulose acetate/rayon

 

  1. Diagram showing the Manufacture of Sodium hydroxide from the flowing Mercury-cathode cell.

 

 

  1. Environmental effects of Manufacture of Sodium hydroxide from the flowing Mercury-cathode cell.

 

1.Most of the Mercury used at the cathode is recycled ;

(i)to reduce the cost because mercury is expensive

(ii)to reduce pollution because mercury kills marine life.

(iii)because it causes chromosomal/genetic mutation to human beings.

 

2.Chlorine produced at the anode;

(i)has a pungent irritating smell that causes headache to human beings.

(ii)bleaches any wet substance.

(iii)dissolves water to form both hydrochloric acid and chloric(I)acid

Both cause marine pollution and stomach upsets.

 

b)(ii)

Extraction of sodium from rock salt/The Downs cell/process

 

  1. Raw materials

(i)Rock salt/solid sodium chloride

(ii)calcium(II)chloride

 

  1. Chemical processes.

Rock salt/ solid sodium chloride is heated to molten state in a chamber lined with fire bricks on the outside.

Sodium chloride has a melting point of about 800oC. A little calcium (II) chloride is added to lower the melting point of the electrolyte to about 600oC.

The molten electrolyte is the electrolyzed in a carbon graphite anode suspended at the centre and surrounded by steel cathode.

 

Questions

  1. Write the equation for the decomposition of the electrolyte during the electrolytic process.

NaCl(l)                      Na+(l)       +      Cl(l)

Note: In absence of water, the ions are in liquid state.

  1. Name the ions present in molten rock salt that move to the;

(i)Steel cathode -Na+(l)

         (ii)Carbon graphite anode- Cl(l)

III. Write the equation for the reaction that take place during the electrolytic process at the;

(i)Steel cathode

2Na+(l)     +      2e               2Na(l)

         (ii)Carbon graphite anode

2Cl(l)                    Cl2(g)      +    2e

  1. Name the products of electrolysis in the Downs cell at;

(i)Cathode:

Grey solid Sodium metal is less dense than the molten electrolyte and therefore float on top of the cathode to be periodically tapped off.

 

(ii)Anode:

Pale green chlorine gas that turns moist/damp/wet blue/red litmus papers red then bleaches/decolorizes both. Chlorine gas is again a very useful by-product in;

(i)making (PVC)polyvinylchloride(polychloroethene) pipes.

(ii)chlorination/sterilization of water to kill germs.

(iii)bleaching agent

(iv)manufacture of hydrochloric acid.

A steel diaphragm/gauze is suspended between the electrodes to prevent recombination of sodium at the cathode and chlorine gas at the anode back to sodium chloride.

 

III. Diagram showing the Downs cell/process for extraction of sodium

 

 

 

 

  1. Uses of sodium.

1.Sodium vapour is used as sodium lamps to give a yellow light in street lighting.

2.Sodium is used in making very useful sodium compounds like;

(i)Sodium hydroxide(NaOH)

(ii)Sodium cyanide(NaCN)

(iii)Sodium peroxide(Na2O2)

(iv)Sodamide(NaNH2)

3.An alloy of Potassium and Sodium is used as coolant in nuclear reactors.

 

  1. Environmental effects of Downs cell.

 

1.Chlorine produced at the anode;

(i)has a pungent irritating smell that causes headache to human beings.

(ii)bleaches any wet substance.

(iii)dissolves water to form both hydrochloric acid and chloric(I)acid

Both cause marine pollution and stomach upsets.

2.Sodium metal rapidly react with traces of water to form alkaline Sodium hydroxide(NaOH(aq))solution. This raises the pH of rivers/lakes killing aquatic lifein case of leakages.

 

  1. Test for presence of Na.

If a compound has Na+ ions in solid/molten/aqueous state then it changes a non-luminous clear/colourless flame to a yellow coloration but does not burn

Experiment

 

Scoop a portion of sodium chloride crystals/solution in a clean metallic spatula. Introduce it to a clear /colourless Bunsen flame.

 

 

Observation Inference
Yellow coloration Na+

 

Practice

(i)Calculate the time taken in hours for 230kg of sodium to be produced in the Downs cell when a current of 120kA is used.

(ii)Determine the volume of chlorine released to the atmosphere. (Na=23.0),Faraday constant=96500C.I mole of a gas =24dm3 at r.t.p)

Working:

 

Equation at the cathode:

2Na+ (l)   +  2e  -> 2Na(l)

2 mole of electrons = 2 Faradays = 2 x 96500 C  deposits a mass = molar mass of  Na = 23.0g  thus;

         23.0 g        ->    2  x  96500 C

(230 x 1000)g   ->      230 x 1000 x 2  x  96500

                                                           23

= 1,930,000,000 / 1.93 x 10 9C

 

Time(t) in seconds    =  Quantity of electricity                                                       Current(I) in amperes

Substituting

1,930,000,000 / 1.93 x 10 9C

120 x 1000A

= 16,083,3333seconds / 268.0556 minutes

= 4.4676hours

                                                          

Volume of Chlorine

Method 1

 

Equation at the anode:

2 Cl (l) -> Cl2(g)  +  2e

From the equation:

2 moles of electrons = 2 Faradays    =2 x 96500C

2 x 96500C  -> 24dm3

1,930,000,000 / 1.93 x 10 9C->1,930,000,000 / 1.93×10 9C x 24

2 x 96500C

Volume of Chlorine = 240,000dm3 /2.4 x 105dm3

 

Method 2

Equation at the anode:          Cl (l) -> Cl2(g)  +  2e

Mole ratio of products at Cathode: anode = 1:1

Moles of  sodium  at cathode =(230 x 1000 )g= 10,000moles

23

10,000moles of  Na    =       10,000moles moles of Chlorine

1 moles of Chlorine gas    =   24000cm3

10,000moles of Chlorine-   >  10000 x 24

                                  =240,000dm3 / 2.4x 105dm3

 

Method 3

Equation at the anode: Cl (l) -> Cl2(g)  +  2e

Ratio of Faradays of products at Cathode: anode = 2:2

=>  2 x 96500C produce 24000cm3 of chlorine gas  Then: 1,930,000,000 / 1.93 x 10 9C ->

1,930,000,000 / 1.93 x 10 9C x24  = 240,000dm3

2 x 96500

 

(iij)The sodium metal produced was reacted with water to form 25000dm3  solution in a Caster-Keller tank.

(a)Calculate the concentration of the resulting solution in moles per litre.

(b)The volume of gaseous products formed at s.t.p(1 mole of gas =22.4 dm3 at s.t.p)

 

Chemical equation at Caster-Keller tank

2Na(s)   +  2H2O(l) -> 2NaOH(aq)   +  H2 (g)

 

Mole ratio Na:NaOH =  2 : 2 => 1:1

Moles Na =10000moles=10000moles of NaOH

25000dm3 ->10000moles of NaOH

1dm3        -> 10000  x 1    =  0.4M0.4 moles/dm3

25000

 

Mole ratio Na: H2 (g) =  2 : 1

Moles Na = 10000moles  = 5000moles of H2 (g)

Volume of H2 (g) = moles x molar gas volume at s.t.p
=> 5000moles x 22.4 dm3

                                  =120,000dm3

 

(iv)The solution formed was further diluted with water for a titration experiment. 25.0 cm3  of the diluted solution required 20.0cm3 of 0.2M sulphuric(VI)acid for complete neutralization. Calculate the  volume of water added to the diluted solution before titration.

 

Chemical equation

2NaOH(aq)   +  H2SO4(aq)  -> Na2SO4(aq)  +  H2O(l)

Moles ratio NaOH : H2SO4   =   2 : 1

Moles ratio  H2SO4 = molarity x volume =>   0.2M  x  20

1000                               1000

=4.0 x 10-3 moles

 

Moles NaOH = 2 x 4.0 x 10-3 moles= 8.0 x 10-3 moles

Molarity of NaOH= Moles x 1000=> 8.0 x 10-3 moles x 1000

volume                          25

=0.16 molesdm-3 /M

 

Volume used during dilution

  C1V1 = C2V2    =>    0.4M   x  V1  =  0.16 M   x  25

                     = 0.16 M x 25    = 10cm3

0.4

 

(a) Below is  a simplified diagram of the Downs  Cell  used for the  manufacture  of sodium. Study it and answer the questions that follow

 

 

(i)What material is the anode made of? Give a reason (2 mks)

Carbon graphite/Titanium

This because they are cheap and inert/do not influence/affect the products of electrolysis

 (ii) What precaution is taken to prevent chlorine and sodium from re- combination? ( 1 mks)

Using a steel gauze/diaphragm separating the cathode from anode

 (iii) Write an ionic equation for the reaction in which chlorine gas is formed    ( 1mk)

 

2Cl(l)       ->     Cl2(g)     +       2e

 

 (b) In the Downs process, (used for manufacture of sodium), a certain salt is added to lower the  melting point of sodium chloride from about 8000C to about 6000C.

         (i) Name the salt that is added                  (1mk)

Calcium chloride

         (ii) State why it is necessary to lower the temperature(1mk)

To reduce the cost of production

 

(c) Explain why aqueous sodium chloride is not suitable as  an electrolyte for the manufacture  of  sodium in the Downs process( 2mk)

The sodium produced react explosively/vigorously with water in the aqueous sodium chloride

 (d) Sodium metal reacts with air to form two oxide. Give the formulae of two oxides     ( 1mk)

Na2O   Sodium oxide(in limited air)

Na2O2   Sodium peroxide(in excess air)

 

2.ALUMINIUM

a)Natural occurrence

Aluminium is the most common naturally occurring metal. It makes 7% of the earths crust as:

(i)Bauxite ore- Hydrated aluminium oxide(Al2O3.2H2O)

(ii)Mica ore-Potassium aluminium silicate(K2Al2Si6O16)

(iii)China clay ore- aluminium silicate (Al2Si6O16)

(iv)Corrundum-Anhydrous aluminium oxide(Al2O3)

 

         b)Extraction of aluminium from Bauxite/Halls cell/process)

The main ore from which aluminium is extracted is Bauxite ore- hydrated aluminium oxide(Al2O3.2H2O).

The ore is mined by open-caste mining method/quarrying where it is scooped together with silica/sand/silicon(IV)oxide (SiO2) and soil/ iron(III)oxide (Fe2O3) as impurities.

The mixture is first dissolved in hot concentrated sodium/potassium hydroxide solution.

The alkalis dissolve both bauxite and silicon(IV)oxide.

This is because bauxite is amphotellic while silicon(IV)oxide is acidic.

Iron(III)oxide (Fe2O3) is filtered of /removed as a residue.

Carbon(IV)oxide is bubbled into the filtrate to precipitate aluminium (III) hydroxide (Al(OH)3) as residue.

 

The aluminium (III) hydroxide (Al(OH)3) residue is filtered off. Silicon (IV)oxide remain in the solution as filtrate. Aluminium (III) hydroxide (Al(OH)3) residue is then heated to form pure aluminium (III)oxide(Al2O3)

 

2Al(OH)3 (s)                               Al2O(s)    +    3H2O(l)

 

Pure aluminium (III)oxide (Al2O3) has a very high melting point of 2015oC.

Alot of energy is required to melt the oxide.

It is therefore dissolved first in molten cryolite /sodium hexafluoroaluminate (III)/Na3AlF6 to lower the melting point to about 800oC.

 

The molten electrolyte is put in the Hall cell made up of a steel tank lined with carbon graphite and an anode suspended into the electrolyte.

During the electrolysis:

 

(i)At the cathode;

4Al3+(l)  +  12e                   4Al(l)

 

(ii) At the anode;

6O2-(l)                        3O2(g)      +   12e

 

Aluminium is denser than the electrolyte therefore sink to the bottom of the Hall cell.

At this temperature ,the Oxygen evolved/produced at the anode reacts with carbon anode to form carbon(IV)oxide gas that escape to the atmosphere.

 

C(s)   + O2(g)                              CO2(g)

The anode thus should be continuously replaced from time to time.

 

Flow chart summary of extraction of aluminium from Bauxite

 

 

 

 

Hot concentrated sodium hydroxide

 

 

 

 

 

Bauxite(Al2O3.2H2O) ore with impurities Fe2O3 and SiO2

 

Powdered mixture
Crush (increase surface area)
Iron(III)oxide- Fe2Oas residue
Sodium aluminate (NaAl(OH)4)

and sodium silicate (Na2SiO3) as filtrate

Carbon(IV)oxide
Aluminium hydroxide (Al(OH)3) as residue
Sodium silicate  (Na2SiO3)
Aluminium (III) Oxide
Roast at 1000oC
Cryolite
Electrolysis
Oxygen  gas at anode
Pure aluminium sinks in Hall cell

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. c) Diagram showing the Hall cell / process for extraction of Bauxite

 

 

 

d)Uses of aluminium

(i) In making aeroplane parts, buses, tankers, furniture because aluminium is very light.

(ii)Making duralumin-an alloy which is harder and has a higher tensile strength

(iii)Making utensils,sauce pans,spoons because it is light and good conductor of electricity.

(iv)Making overhead electric cables because it is light,ductile and good conductor of electricity.

(iv)Used in the thermite process for production of Manganese, Chromium amd Titanium.

 

  1. e) Environmental effects of extracting aluminium from Bauxite.

Carbon(IV)oxide gas that escape to the atmosphere is a green house gas that causes global warming.

Bauxite is extracted by open caste mining that causes soil/environmental degradation.

 

  1. f) Test for presence of Al3+

If an ore is suspected to contain Al3+ it is;

(i)added hot concentrated sulphuric(VI)/Nitric(V)acid to free the ions present.

(ii)the free ions are then added a precipitating reagent like 2M sodium hydroxide /2M aqueous ammonia.

 

Observation Inference
White precipitate in excess 2M NaOH(aq) Pb2+ , Al3+,  Zn2+
White precipitate in excess 2M NH3(aq) Pb2+ , Al3+
No black precipitate on adding Na2S(aq) Al3+
No white precipitate on adding either NaCl(aq),HCl(aq),H2SO4(aq),Na2SO4(aq) Al3+

 

 Practice

 

1.An unknown rock X was discovered in Ukraine. Test with dilute sulphuric (VI)acid shows rapid effervescence with production of a colourless gas A that forms a white precipitate with lime water and colourless solution B. On adding 3cm3 of 2M sodium hydroxide, a white precipitate C is formed that dissolves to form a colourless solution D on adding more sodium hydroxide. On adding 2M aqueous ammonia, a white precipitate E is formed which persist in excess aqueous ammonia.On which on adding 5cm3 of 1M Lead(II)nitrate(V) to F  a white precipitate G is formed which remains on heating.

Identify:

A

Hydrogen/H2

B

Aluminium sulphate(VI)/Al2(SO4) 3

C

Aluminium hydroxide/ Al(OH4) 3

D

Tetrahydroxoaluminate(III)/ [Al(OH4) 3]

E

Aluminium hydroxide/ Al(OH) 3

F

Aluminium chloride/ AlCl3

 

2.Aluminium is obtained from the ore with the formula Al2O3. 2H2O. The ore is first heated and refined to obtain pure aluminium oxide (Al2O3).  The oxide is then electrolysed to get Aluminium and oxygen gas using carbon anodes and carbon as cathode. Give the common name of the ore from where aluminium is extracted from ½  mark

 

What would be the importance of heating the ore first before refining it?1 mark

To remove the water of crystallization

The refined ore has to be dissolved in cryolite first before electrolysis. Why is this necessary? 1½  mark

To  lower the melting point of aluminium oxide from  about 2015oC to 900oC so as to lower /reduce cost of production

 

Why are the carbon anodes replaced every now and then in the cell for electrolysing aluminium oxide?                                            1 mark

Oxygen produced at anode react with carbon to form carbon(IV)oxide gas that escape

State two uses of aluminium

In making aeroplane parts, buses, tankers, utensils,     sauce pans,spoons

          Making overhead electric cables

          Making duralumin

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. IRON

a)Natural occurrence

 

Iron is the second most common naturally occurring metal. It makes 4% of the earths crust as:

(i)Haematite(Fe2O3)

(ii)Magnetite(Fe3O4)

(iii)Siderite(FeCO3)

b)The blast furnace for extraction of iron from Haematite and Magnetite

 

a)Raw materials:

         (i)Haematite(Fe2O3)

(ii)Magnetite(Fe3O4)

(iii)Siderite(FeCO3)

         (iv)Coke/charcoal/ carbon

(v)Limestone

b)Chemical processes:

 

Iron is usually extracted from Haematite (Fe2O3), Magnetite(Fe3O4) Siderite (FeCO3).These ores contain silicon(IV)oxide(SiO2) and aluminium(III)oxide (Al2O3) as impurities.

When extracted from siderite, the ore must first be roasted in air to decompose the iron(II)Carbonate to Iron(II)oxide with production of carbon(IV)oxide gas:

 

FeCO3(s)                        FeO(s)         +      CO2(g)

 

Iron(II)oxide is then rapidly oxidized by air to iron(III)oxide(Haematite).

 

4FeO(s)    +       O2(g)                          2Fe2O3(s)

 

Haematite (Fe2O3), Magnetite(Fe3O4), coke and limestone are all then fed from top into a tall (about 30metres in height) tapered steel chamber lined with refractory bricks called a blast furnace.

The furnace is covered with inverted double cap to prevent/reduce amount of  any gases escaping .

Near the base/bottom, blast of hot air at about 1000K (827oC) is driven/forced into the furnace through small holes called Tuyeres.

As the air enters ,it reacts with coke/charcoal/carbon to form carbon(IV)oxide gas. This reaction is highly exothermic.

 

C(s)+ O2(g)                                                          CO2 (g)  ∆H = -394kJ

This raises the temperature at the bottom of the furnace to about 2000K(1650oC).As Carbon(IV)oxide gas rises up the furnace it reacts with more coke to form carbon(II)oxide gas.This reaction is endothermic.

CO2 (g) +  C(s)                                                  2CO (g)  ∆H = +173kJ

 

Carbon(II)oxide gas is a strong reducing agent that reduces the ores at the upper parts of the furnace where temperatures are about 750K(500oC) i.e.

For Haematite;

Fe2O3 (s)    +     3CO(g)                      2Fe(s) + CO2(g)

For Magnetite;

Fe3O4 (s)    +     4CO(g)                      3Fe(s) + 4CO2(g)

 

Iron is denser than iron ore. As it  falls to the hotter base of the furnace it melts and can easily be tapped off.

Limestone fed into the furnace decomposes to quicklime/calcium oxide and produce more carbon(IV)oxide gas.

 

CaCO3(s)                   CaO(s) + CO2(g)

 

Quicklime/calcium oxide reacts with the impurities silicon(IV)oxide(SiO2) and aluminium(III)oxide(Al2O3)in the ore to form calcium silicate and calcium aluminate.

 

CaO(s) + SiO2(s)                    CaSiO3 (l)

CaO(s) + Al2O3(s)                  Ca Al2O4 (l)

 

Calcium silicate and calcium aluminate mixture is called slag.Slag is denser than iron ore but less dense than iron therefore float on the pure iron. It is tapped at different levels to be tapped off for use in:

(i)tarmacing roads

(ii) cement manufacture

(iii)as building construction material

 

(c)Uses of Iron

Iron obtained from the blast furnace is hard and brittle. It is called Pig iron. It is remelted, added scrap steel then cooled. This iron is called cast iron.

Iron is mainly used to make:

(i)gates ,pipes, engine blocks, rails, charcoal iron boxes,lamp posts because it is cheap.

(ii)nails, cutlery, scissors, sinks, vats, spanners,steel rods, and railway points from steel.

Steel is an alloy of iron with carbon, and/or Vanadium, Manganese, Tungsten, Nickel ,Chromium. It does not rust/corrode like iron.

 

 

 

 

 

 

  1. e) Environmental effects of extracting Iron from Blast furnace

 

(i)Carbon(IV)oxide(CO2) gas is a green house gas that causes/increases global warming if allowed to escape/leak from the furnace.

(ii)Carbon(II)oxide(CO)gas is a highly poisonous/toxic odourless gas that can kill on leakage.

It is preferentially absorbed by the haemoglobin in mammals instead of Oxygen to form a stable compound that reduce free hemoglobin in the blood.

(iii) Haematite (Fe2O3),  Magnetite(Fe3O4) and Siderite (FeCO3) are extracted through quarrying /open cast mining that cause soil / environmental degradation .

 

  1. f) Test for the presence of Iron

Iron naturally exist in its compound as Fe2+ /Fe3+

If an ore is suspected to contain Fe2+ /Fe3+ it is;

(i)added hot concentrated sulphuric(VI)/Nitric(V)acid to free the ions present.

(ii)the free ions are then added a precipitating reagent like 2M sodium hydroxide /2M aqueous ammonia which forms;

  1. I) an insoluble green precipitate in excess of 2M sodium hydroxide /2M aqueous ammonia if Fe2+ ions are present.
  2. I) an insoluble brown precipitate in excess of 2M sodium hydroxide /2M aqueous ammonia if Fe2+ ions are present.

 

 

Observation Inference
green precipitate in excess 2M NaOH(aq) Fe2+
green precipitate in excess 2M NH3(aq) Fe2+
brown precipitate in excess 2M NaOH(aq) Fe3+
brown precipitate in excess 2M NH3(aq) Fe3+

 

Practice questions                                                               

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                      

 

 

 

 

 

 

 

 

4.COPPER

a)Natural occurrence

 

Copper is found as uncombined element/metal on the earths crust in Zambia, Tanzania, USA and Canada .The chief ores of copper are:

(i)Copper pyrites(CuFeS2)

(ii)Malachite(CuCO3.Cu(OH)2)

(iii)Cuprite(Cu2O)

b)Extraction of copper from copper pyrites.

Copper pyrites are first crushed into fine powder. The powdered ore is the added water and oil. The purpose of water is to dissolve hydrophilic substances/particle. The purpose of oil is to make cover copper ore particle so as to make it hydrophobic

Air is blown through the mixture. Air creates bubbles that stick around hydrophobic copper ore. The air bubbles raise through buoyancy small hydrophobic copper ore particles to the surface. A concentrated ore floats at the top as froth. This is called froth flotation. The concentrated ore is then skimmed off.The ore is then roasted in air to form copper(I)sulphide ,sulphur(IV)oxide and iron (II) oxide.

2CuFeS2(s) + 4O2(g)                   Cu2S(s) + 3SO2(g)  + 2FeO(s)

Limestone (CaCO3) and silicon(IV)oxide (SiO2) are added and the mixture heated in absence of air.Silicon(IV)oxide (SiO2) reacts with iron (II) oxide to form Iron silicate which constitutes the slag and is removed.

FeO(s)    +    SiO2(s)                   FeSiO3(s)

The slag separates off from the copper(I)sulphide. Copper(I)sulphide is then heated in a regulated supply of air where some of it is converted to copper (I) oxide.

2Cu2S (s)  +  3O2(g)                     2Cu2S(s)  +   2SO2(g)

The mixture then undergo self reduction in which copper(I)oxide is reduced by copper(I)sulphide to copper metal.

Cu2S (s)  +  2Cu2O (s)                  6Cu (s)  +   SO2(g)

The copper obtained has Iron, sulphur and traces of silver and gold as impurities.It is therefore about 97.5% pure. It is refined by electrolysis/electrolytic method.

During the electrolysis of refining copper, the impure copper is made the anode and a small pure strip is made the cathode.

Electrode ionization takes place where:

At the anode;

Cu(s)                          Cu2+ (aq)   +    2e

Note: Impure copper anode dissolves/erodes into solution and decreases in size.

At the Cathode;

Cu2+ (aq)   +    2e                   Cu(s)

Note: The copper ions in the electrolyte(CuSO4) are reduced and deposited as copper metal at the cathode. The copper obtained is 99.98% pure.

Valuable traces of silver and gold collect at the bottom of the electrolytic cell as sludge. It is used to finance the extraction of copper pyrites.

 

 

(c)Flow chart summary of extraction of copper from Copper pyrites

 

 

Oil
Water

 

Anode; Impure Copper eroded.

 

Cathode; Pure Copper deposited.

 

Copper pyrites(CuFeS2) ore with impurities Fe2O3 and SiO2

 

Froth flotation
Crush (increase surface area)
Concentration chamber
1st roasting chamber
Silicon(IV)

oxide

Smelting furnace
2nd roasting furnace
Calcium aluminate (CaAl2O4)slag
Limestone
Sulphur(IV)Oxide
Iron Silicate (FeSiO3)Slag

 

 

Excess  air
Limited air
Sulphur(IV)Oxide

 

Self reduction
Impure copper
Rocky impurities
Cu2S
Cu2S, Cu2O

 

 

Electrolysis using Copper electrodes

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Electrolytic purification of impure copper

 

 

 

 

 

 

 

 

        

 

 

 

  1. d) Uses of copper

Copper is mainly used in:

(i)making low voltage electric cables,contact switches, cockets and plugs because it is a good conductor of electricity.

(ii)Making solder  because it is a good thermal conductor.

(iii)Making useful alloys e.g.

-Brass is an alloy of copper and Zinc(Cu/Zn)

-Bronze is an alloy of copper and Tin(Cu/Sn)

-German silver is an alloy of copper ,Zinc and Nickel(Cu/Zn/Ni)

(iv)Making coins and ornaments.

 

  1. e) Environmental effects of extracting copper from Copper pyrites

(i)Sulphur(IV)oxide is a gas that has a pungent poisonous smell that causes head ache to human in high concentration.

(ii)Sulphur(IV)oxide  gas if allowed to escape dissolves in water /rivers/rain to form weak sulphuric(IV)acid lowering the pH of the water leading to marine pollution, accelerated corrosion/rusting of metals/roofs and breathing problems to human beings.

(iii)Copper is extracted by open caste mining leading to land /environmental /soil degradation.

  1. f) Test for the presence of copper in an ore

Copper naturally exist in its compound as Cu2+ /Cu+

Copper (I) / Cu+ is readily oxidized to copper(II)/ Cu2+

If an ore is suspected to contain Cu2+ /Cu+ it is;

(i)added hot concentrated sulphuric(VI)/Nitric(V)acid to free the ions present.

(ii)the free ions are then added a precipitating reagent; 2M sodium hydroxide /2M aqueous ammonia which forms;

  1. I) an insoluble blue precipitate in excess of 2M sodium hydroxide if Cu2+ ions are present.
  2. I) an insoluble blue precipitate in 2M aqueous ammonia that dissolve to royal/deep blue solution in excess if Cu2+ ions are present.

 

 

Observation Inference
blue precipitate in excess 2M NaOH(aq) Cu2+
blue precipitate,dissolve to royal/deep blue solution in excess 2M NH3(aq) Cu2+

 

g)Sample questions

Copper is extracted from copper pyrites as in the flow chart outlined below. Study it and answer the questions that follow

 

 

 

5.ZINC and LEAD

 

a)Natural occurrence

 

Zinc occurs mainly as:

(i)Calamine-Zinc carbonate(ZnCO3)

(ii)Zinc blende-Zinc sulphide(ZnS)

Lead occurs mainly as Galena-Lead(II)Sulphide mixed with Zinc blende:

 

b)Extraction of Zinc/Lead  from Calamine ,Zinc blende and Galena.

 

During extraction of Zinc , the ore is first roasted in air:

For Calamine Zinc carbonate decompose to Zinc oxide and carbon(IV) oxide gas.

ZnCO3(s)                   ZnO(s)      +      CO2(g)

Zinc blende does not decompose but reacts with air to form Zinc oxide and sulphur(IV) oxide gas.

Galena as a useful impurity also reacts with air to form Lead(II) oxide and sulphur(IV) oxide gas.

 

2ZnS(s)  + 3O2(g)                  2ZnO(s)    +      2SO2(g)

(Zinc blende)

 

2PbS(s)  + 3O2(g)                   2PbO(s)    +      2SO2(g)

(Galena)

 

The oxides are mixed with coke and limestone/Iron(II)oxide/ Aluminium (III) oxide and heated in a blast furnace.

At the furnace temperatures limestone decomposes to quicklime/CaO and produce Carbon(IV)oxide gas.

 

CaCO3(s)                   CaO(s) + CO2 (g)

 

Carbon(IV)oxide gas reacts with more coke to form the Carbon(II)oxide gas.

C(s)          +  CO2 (g)                          2CO (g)

 

Both Carbon(II)oxide and carbon/coke/carbon are reducing agents.

The oxides are reduced to the metals by either coke or carbon (II)oxide.

 

ZnO(s)      +      C(s)                  Zn(g)  + CO (g)

PbO(s)     +      C(s)                  Pb(l)  + CO (g)

 

PbO(s)      +      CO(s)               Pb(l)  + CO2 (g)

PbO(s)      +      CO(s)               Pb(g)  + CO2 (g)

 

At the furnace temperature:

(i)Zinc is a gas/vapour and is collected at the top of the furnace. It is condensed in a spray of molten lead to prevent reoxidation to Zinc oxide. On further cooling , Zinc collects on the surface from where it can be tapped off

(ii)Lead is a liquid and is ale to trickle to the bottom of the furnace from where it is tapped off.

Quicklime/CaO, Iron(II)Oxide, Aluminium(III)oxide are used to remove silica/silicon(IV)oxide as silicates which float above Lead preventing its reoxidation back to Lead(II)Oxide.

CaO(s) + SiO2(s)                CaSiO3(s/l)

(Slag-Calcium silicate)

FeO(s) + SiO2(s)                FeSiO3(s/l)

(Slag-Iron silicate)

Al2O3(s) + SiO2(s)              Al2SiO4(s/l)

(Slag-Aluminium silicate)

 

c)Flow chart on extraction of Zinc  from Calamine ,Zinc blende.

SO2 from Zinc blende

 

 

CO2 from calamine

 

Water
Oil

 

Zinc ore (calamine /Zinc blende
Powdered ore
Froth flotation
Roasting chamber
Reduction chamber
Iron/aluminium/ Limestone
Coke
Slag (Iron silicate/ aluminium silicate/calcium silicate)

 

Condenser
Filtration
Granulating tank
Lead liquid
Water
Granulated Zinc

 

 

 

 

 

 

 

 

 

  1. d) Flow chart on extraction of Lead from Galena

 

oil
Water
Froth flotation
LEAD VAPOUR
Zinc residue
Filtration
Condenser
Slag(Iron silicate)
SO2(g)
coke
Iron/Limestone
Reduction chamber
Roasting chamber
Powdered ore
Lead ore/Galena

 

 

  1. e) Uses of Lead

 

Lead is used in:

(i)making gun-burettes.

(ii)making protective clothes against nuclear (alpha rays/particle)     radiation in a nuclear reactor.

(iii)Mixed with tin(Sn) to make solder alloy

 

  1. f) Uses of Zinc

  Zinc is used in:

(i)Galvanization-when iron sheet is dipped in molten Zinc ,a thin layer of Zinc is formed on the surface.Since Zinc is more reactive than iron ,it reacts with elements of air(CO2/ O2 / H2O) to form basic Zinc carbonate(ZnCO3.Zn(OH)2).This sacrificial method protects iron from corrosion/rusting.

(ii)As negative terminal and casing in dry/Laclanche cells.

(iii)Making brass alloy with copper(Cu/Zn)

  1. g) Environmental effects of extracting Zinc and Lead.

(i) Lead and Lead salts are carcinogenic/causes cancer

(ii)Carbon(IV)oxide is a green house gas that causes/accelerate global warming.

(iii)Carbon(II)oxide is a colourless odourless poisonous /toxic gas that combines with haemoglobin in the blood to form stable carboxyhaemoglobin reducing free haemoglobin leading to death.

(iv) Sulphur(IV)oxide is a gas that has a pungent poisonous smell that causes headache to human if in high concentration.

(v)Any leakages in Sulphur(IV)oxide gas escapes to the water bodies to  form weak sulphuric(VI)acid lowering the pH of the water. This causes marine pollution /death of aquatic life, accelerated rusting/corrosion of metals/roofs and breathing problems to human beings.

 

  1. h) Test for presence of Zinc/ Lead.

 

If an ore is suspected to contain Zinc/Lead it is:

I.added hot concentrated Nitric(V)acid to free the ions present.

Note:

Concentrated Sulphuric(VI)acid forms insoluble PbSO4 thus cannot be used to free the ions in Lead salts.

II.the free ions are then added a precipitating reagent mostly 2M sodium hydroxide or 2M aqueous ammonia with the formation of;

(i)a soluble precipitate in excess of 2M sodium hydroxide if Zn2+, Pb2+, Al3+ions are present.

(ii)a white precipitate that dissolves to form a colorless solution in excess 2M aqueous ammonia if Zn2+ions are present.

(iii)an insoluble white precipitate in excess 2M aqueous ammonia if Pb2+,  Al3+ions are present.

(iv) Pb2+ ions form a white precipitate when any soluble SO42-, SO32-, CO32-, Cl, is added while Al3+ ions do not form a white precipitate

(v) Pb2+ ions form a yellow precipitate when any soluble I(e.g. Potassium/sodium Iodide) is added while Al3+ ions do not form a yellow precipitate.

(vi) Pb2+ ions form a black precipitate when any soluble S(e.g. Potassium/sodium sulphide) is added while Al3+ ions do not form a black precipitate.i.e;

Observation Inference
White precipitate in excess 2M NaOH (aq) Zn2+, Pb2+, Al3+ ions
White precipitate that dissolves to form a colourless solution in excess 2M NH3(aq) Zn2+ ions
White precipitate in excess 2M NH3(aq) Pb2+, Al3+ ions
White precipitate on adding about 4 drops of either  Na2CO3(aq), Na2SO3(aq), Na2SO4(aq), H2SO4(aq), HCl(aq), NaCl(aq) Pb2+ions
Yellow precipitate on adding about 4 drops of of KI(aq).NaI (aq) Pb2+ ions
Black precipitate on adding aout 4 drops of Na2S(aq)/K2S(aq) Pb2+ ions

 

6.GENERAL SUMMARY OF METALS

 

  1. a) Summary methods of extracting metal from their ore

 

Add oil, water, and blow air to form froth to concentrate the ore if it is a low grade
If near the surface use open cast mining / quarrying
Position on the earth’s crust

The main criteria used in extraction of metals is based on its position in the electrochemical/reactivity series and its occurrence on the earth’s crust.

 

 

 

 

 

 

If deep on the earth’s crust use deep mining

 

 

Roast  the ore first if it is a carbonate / sulphide  of Zinc, Iron, Tin, Lead, and copper to form the oxide
Reduce the oxide using carbon in a furnace if it is made of Zinc ,Tin, Lead ,Copper and Iron
Electrolyse the ore if it is made of reactive metals; Potassium, Sodium, Magnesium, Calcium, Aluminium

 

 

 

  1. b) Summary of extraction of common metal.
Metal Chief ore/s Chemical formula of ore Method of extraction Main equation during extraction
Sodium Rock salt NaCl(s) Downs process

Through electrolysis of molten NaCl (CaCl2 lower m.pt from 800oC-> 600oC)

Cathode:

2Na+(l) + 2e -> 2Na(l)

Anode:

2Cl(l) -> Cl2(g) + 2e

Sodium/

sodium hydroxide

Brine NaCl(aq) Flowing mercury cathode cell

Through electrolysis of concentrated NaCl(aq)

Cathode:

2Na+(aq)+2e ->2Na(aq)

Anode:

2Cl(aq) -> Cl2(g) + 2e

Aluminium Bauxite Al2O3.2H2O Halls process

Through electrolysis of molten Al2O3. (Cryolite lower m.pt from 2015oC -> 800oC)

 

Cathode:

4Al3+(l) + 12e -> 4Al(l)

Anode:

6O2-(l) -> 3O2(g) + 12e

Iron Haematite

Magnetite

Fe2O3

Fe3O4

Blast furnace

Reduction of the ore by carbon(II)oxide

 

Fe2O3(s)+ 3CO(g)

2Fe(l) +3CO2(g)

 

Fe3O4(s)+ 4CO(g)

3Fe(l) +4CO2(g)

 

 Copper Copper pyrites CuFeS2

 

Roasting the ore in air to get Cu2S.

Heating Cu2S ore in regulated supply of air.

Reduction of Cu2O  by Cu2S

 

2CuFeS2 (s)+ 4O2(g) ->

Cu2S(s)+3SO2(g) +2FeO(s)

 

2Cu2S (s)+ 3O2(g) ->

2Cu2O(s)+2SO2(g)

 

Cu2S (s)+ 2Cu2O(s) ->   6Cu(s)+ SO2(g)

 

Zinc Calamine ZnCO3 Roasting the ore in air to get ZnO

Blast furnace /reduction of the oxide by Carbon(II)Oxide/Carbon

ZnCO3(s)-> ZnO(s) + CO2(g)

 

2ZnS(s) +3O2(g) -> 2ZnO(s) + 2SO2(g)

 

ZnO(s) + CO(g)->

Zn(s) + CO2(g)

 

Lead Galena PbS Blast furnace-Reduction of the oxide by carbon(II)oxide /Carbon PbO(s) + CO(g)->

Pb(s) + CO2(g)

 

 

  1. c) Common alloys of metal.
Alloy name Constituents of the alloy Uses of the alloy
Brass Copper and Zinc Making scews and bulb caps
Bronze Copper and Tin Making clock springs,electrical contacts and copper coins
Soldier Lead and Tin Soldering, joining electrical contacts because of its low melting points and high thermal conductivity
Duralumin Aluminium, Copper and Magnesium Making aircraft , utensils ,windows frames because of its light weight and corrosion resistant.
Steel Iron, Carbon ,Manganese and other metals Railway lines , car bodies girders  and utensils.
Nichrome Nichrome and Chromium Provide resistance in electric heaters and ovens
German silver Copper,Zinc and Nickel Making coins

 

 

  1. d) Physical properties of metal.

Metals form giant metallic structure joined by metallic bond from electrostatic attraction between the metallic cation and free delocalized electrons.

This makes metals to have the following physical properties:

 

(i)High melting and boiling points

The giant metallic structure has a very close packed metallic lattice joined by strong electrostatic attraction between the metallic cation and free delocalized electrons.The more delocalized electrons the higher the melting/boiling points e.g.

Aluminium has a melting point of about 2015oC while that of sodium is about 98oC.This is mainly because aluminium has more/three delocalized electrons than sodium/has one.

Aluminium has a boiling point of about 2470oC while that of sodium is about 890oC.This is mainly because aluminium has more/three delocalized electrons than sodium/has one.

 

 (ii)High thermal and electrical conductivity

All metals are good thermal and electrical conductors as liquid or solids. The more delocalized electrons the higher the thermal and electrical conductivity. e.g.

Aluminium has an electrical conductivity of about 3.82 x 19-9 ohms per metre. Sodium has an electrical conductivity of about 2.18 x 19-9 ohms per metre.

 

(iii)Shiny/Lustrous

The free delocalized electrons on the surface of the metal absorb, vibrate and then scatter/re-emit/lose light energy. All metals are therefore usually shades of grey in colour except copper which is shiny brown.e.g.

Zinc is bluish grey while iron is silvery grey.

 

(iv)High tensile strength

The free delocalized electrons on the surface of the metal atoms binds the surface immediately when the metal is coiled/folded preventing  it from breaking /being brittle.

 

(v)Malleable.

Metals can be made into thin sheet. The metallic crystal lattice on being beaten/pressed/hammered on two sides extend its length and width/bredth and is then immediately bound by the delocalized electrons preventing it from breaking/being brittle.

(vi)Ductile.

Metals can be made into thin wires. The metallic crystal lattice on being beaten/pressed/hammered on all sides extend its length is then immediately bound by the delocalized electrons preventing it from breaking/being brittle.

 

Revision  questions

1.Given some soil , dilute sulphuric(VI)acid,mortar,pestle,filter paper,filter funnel and 2M aqueous ammonia,describe with explanation,how you would show that the soil contain Zinc.

Place the soil sample in the pestle. Crush using the mortar to reduce the particle size/increase its surface area.

Add dilute sulphuric(VI)acid to free the ions in soil sample.

Filter to separate insoluble residue from soluble filtrate

To filtrate,add three drops of aqueous ammonia as precipitating reagent. A white precipitate of  Zn(OH)2, Pb(OH)2 or Al(OH)3 is formed

Add  excess aqueous ammonia to the white precipitate. If it dissolves the Zn2+  ions are present. Zn(OH)2 react with excess ammonia to form soluble [Zn(OH)4]2+ complex.

 

2.In the extraction of aluminium,the oxide is dissolved in cryolite.

(i)What is the chemical name of cryolite?

Sodium hexafloroaluminate/Na3AlF6

 

(ii)What is the purpose of cryolite?

To lower the melting point of the electrolyte/Aluminium oxide from about 2015oC to 900oC

 

(iii)Name the substance used for similar purpose in the Down cell

Calcium chloride/CaCl2

(iv)An alloy of sodium and potassium is used as coolant in nuclear reactors.Explain.

Nuclear reactors generate a lot of heat energy. sodium and potassium alloy reduce/lower the high temperature in the reactors.

(v)Aluminium metal is used to make cooking utensils in preference to other metals.Explain.

Aluminium

(i) is a very good conductor of electricity because it has three delocalized electrons in its metallic structure

(ii)is cheap,malleable,ductile and has high tensile strength

(iii)on exposure to fire/heat form an impervious layer that prevent it from rapid corrosion.

3.Study the scheme below and use it to answer the questions that follow.

 

(a)Identify:

(i)solid residue L

Iron(III)Oxide/Fe2O3

 

(ii)Solid N

Aluminium hydroxide /Al(OH)3

 

(iii)Filtrate M

Sodium tetrahydroxoaluminate/ NaAl(OH)4 and sodium silicate/ NaSiO3

 

(iv)Solid P

Aluminium oxide/ Al2O3

 

(v)Gas Q

Oxygen/O2

 

(vi)Process K1

Filtration

 

(vii)Process K2

Electrolysis

 

(b)Write the equation for the reaction taking place in the formation of solid P from solid N

2Al(OH)3     ->       Al2O3 (s)      +        3H2O(l)

 

(c)Name a substance  added to solid N before process Process K2 take place.

Cryolite/Sodium tetrahydroxoaluminate/ NaAl(OH)4

(d)State the effect of evolution of gas Q on

(i)process K2

Oxygen produced at the anode reacts with the carbon anode to form carbon(IV) oxide which escape. The electrolytic process needs continuous replacement of the carbon anode.

(ii)the environment

Oxygen produced at the anode reacts with the carbon anode to form carbon(IV) oxide which escape to the atmosphere.CO2 is a green house gas that cause global warming.

(e)An aluminium manufacturing factory runs for  24 hours. If the total mass of aluminium produced is  27000kg,

(i)Calculate the current used. (Faraday constant=96500Coulombs, Al=27.0).

(ii)assuming  all the gas produced  react with 200kg of anode ,calculate the loss in mass  of the electrode.(Molar gas volume at room temperature = 24dm3,C=12.0)

 

Working

Equation at Cathode                                                                                                                                                                                  Al3+(l)          +        3e      ->       Al(l)

27g  Al   ->  3 Faradays   =   3   x   96500C

(27000kg x 1000) g  -> (27000kg x 1000) g x 3   x   96500C                                                                                                                                                               27g

=289500000000 Coulombs

 

Current  =    Quantity of electricity    =>289500000000 Coulombs                                                                                           Time in seconds                                24 x 60 x 60

3350690Ampheres

Working

Equation at Anode                                                                                                                                                                                        2O2-(l)       +        4e      ->       O2(g)

4 Faradays                         ->   4   x   96500C24dm3 O2(g)   –

289500000000 Coulombs   ->  289500000000 Coulombs x 24dm3                                                                                                   4   x   96500C

18,000,000dm3

Chemical equation at anode

          O2(g)           +        C (s) ->       CO2(g)

Method 1

 

24dm3 of O2(g)               ->       12.0g Carbon

18,000,000dm3 of O2(g) ->       18,000,000dm3 x 12 =   9000000g    =  9000kg                                                                                  24dm3                              1000g

Loss in mass of the carbon graphite anode =  9000kg                                              

NB:Mass of the carbon graphite anode  remaining =27000kg –  9000kg =18000kg

 

The flow chart below shows the extraction of iron metal.Use it to answer the questions that follow.

 

 

(a)Identify:

(i)gas P

Carbon(IV)oxide/CO2

 

(ii)Solid Q

Carbon/coke/charcoal

 

(iii)Solid R

Carbon/coke/charcoal

 

(iv)Solid V

Limestone/calcium carbonate/CaCO3

 

(v)Solid S

Iron/Fe

(b)Write the chemical equation for the reaction for the formation of:

          (i)Solid S

Fe2O3(s)   +   3CO(g)   ->   2Fe(s)   +   3CO2(g)

 

(ii)Carbon(II)oxide

C(s)   +   CO2 (g)   ->   2CO (g)

 

(iii)Slag

SiO2(s)      +   CaO(s)   ->   CaSiO3(s)

Al2O3 (s)   +   CaO(s)   ->   Ca Al2O4(s)

 

(iv)Gas P

C(s)         +    O2 (g)   ->   CO2 (g)

(c)State two uses of:

(i)Solid S

Iron is used in making:

(i)gates ,pipes, engine blocks, rails, charcoal iron boxes, lamp posts because it is cheap.

(ii)nails, cutlery, scissors, sinks, vats, spanners, steel rods, and railway points from steel.

Steel is an alloy of iron with carbon, and/or Vanadium, Manganese, Tungsten, Nickel ,Chromium.

It does not rust/corrode like iron.

 

(ii)Slag

           (i) tarmacing roads

(ii) cement manufacture

(iii) as building construction material

 

3.You are provided with sulphuric(VI)acid ,2M  aqueous ammonia and two ores suspected to contain copper and iron. Describe with explanation how you would  differentiate the two ores.

Crush the two ores separately in using a mortar and pestle to reduce the particle size/increase the surface area.

Add sulphuric(VI)acid  to separate portion of the ore. Filter.

To  a portion of the filtrate,add three drops of 2M  aqueous ammonia then axcess

Results

A green precipitate insoluble in excess  2M  aqueous ammonia confirms the ore contain Fe2+ ion.

A brown precipitate insoluble in excess  2M  aqueous ammonia confirms the ore contain Fe3+ ion.

A blue precipitate  that dissolve in excess  2M  aqueous ammonia to form a deep/royal blue solution confirms the ore contain Cu2+ ion.

 

  1. Use the flow chart below showing the extraction of Zinc metal to answer the questions that follow

 

(a)Name:

 (i)two ores from which Zinc can be extracted

                    Calamine(ZnCO3)

Zinc blende(ZnS)

(ii)two possible identity of gas P

                    Sulphur(IV)oxide(SO2) from roasting Zinc blende

Carbon(IV)oxide(CO2) from  decomposition of Calamine.

 

(b)Write a possible chemical equation taking place in the roasting chamber.

          2ZnS(s)       +        3O2 (g) ->    2ZnO(s)       +   2SO2(g)

ZnCO3(s)     ->      ZnO(s)        +   CO2(g)

 

(c)Explain the effect of the by-product of the roating on the environment.

Sulphur (IV)oxide from roasting Zinc blende is an acidic gas that causes “acid rain” on dissolving in rain water.

Carbon(IV)oxide(CO2) from  decomposition of Calamine is a green house gas that causes global warming.

 

(d)(i)Name a suitable reducing agent used in the furnace during extraction of  Zinc.

                    Carbon(II)oxide

 

(ii)Write a chemical equation  for the reduction process

ZnO(s)    +        CO(g) ->      Zn(s)           +   CO2(g)

 

(e)(i)Before electrolysis, the products from roasting is added dilute sulphuric (VI)acid. Write the equation for the reaction with dilute sulphuric(VI)acid.

ZnO(s) +  H2SO4 (aq)  ->  Zn SO4(aq) +  H2(g)

 

    (ii)During the electrolysis for extraction of Zinc,state the

  1. Anode used

Aluminium sheet

  1. Cathode used

Lead plate coated with silver

     (ii)Write the equation for  the electrolysis for extraction of Zinc at the:

I.Cathode;

Zn2+(aq)       +        2e   ->   Zn(s)

 

II.Anode;

4OH(aq)     ->  2H2O(l)   +  O2(s) +   4e

 

(f)(i)What is galvanization

Dipping Iron in molten Zinc to form a thin layer of Zinc to prevent iron  from rusting.

 

(ii)Galvanized iron sheet rust after some time. Explain

The thin layer of  Zinc protect Iron from rusting through sacrificial protection. When all the Zinc has reacted with elements of air, Iron start rusting.

 

(g)State two uses of Zinc other than galvanization.

Making brass(Zinc/copper alloy)

Making  german silver(Zinc/copper/nickel alloy)

As casing for dry cells/battery

 

(h)Calculate the mass of Zinc that is produced from the reduction chamber if 6400kg of Calamine ore is fed into the roaster. Assume the process is 80% efficient in each stage(Zn=64.0,C=12.0,O=16.0)

 

Molar mass ZnCO3(s)      =124g

Molar mass Zn                 = 64g

Molar mass ZnO              = 80g

 

Chemical equation

ZnCO3(s)     ->      ZnO(s)        +   CO2(g)

Method 1

124g  ZnCO3                      =>   80g  ZnO

(6400kg x1000)g ZnCO3      =>  (6400  x1000) x 80 512,000,000 g of ZnO

                                                            124

100%           =>    512,000,000 g of ZnO

80%             =>    80    x  512,000,000 g       =  409600000g of ZnO

100

Chemical equation

ZnO(s)         +        CO(g) ->      Zn(s)           +   CO2(g)

 

80g ZnO(s)                      =>      64g Zn(s)

409600000g of ZnO        =>      409600000g    x   64 =   327680000 g Zn

80

100%           =>    327680000 g Zn

80%             =>    80    x 327680000 g Zn    =   262144000g of Zn

100

Mass of Zinc produced  =  262144000g  of  Zn

 

5.An ore is suspected to  bauxite. Describe the process that can be used to confirm the presence of  aluminium in the  ore.

 

Crush the ore to fine powder to increase surface area/reduce particle size.

Add hot concentrated sulphuric(VI)/nitric(V) acid to free the ions.

Filter. Retain the filtrate

Add excess aqueous ammonia to a sample of filtrate.

A white precipitate confirms presence of either Al3+ or Pb2+.

Add sodium sulphate,dilute sulphuric(VI)to another portion of filtrate.

No white precipitate confirms presence of Al3+

    Or Add potassium iodide to another portion of filtrate.

No yellow precipitate confirms presence of Al3+

 

  6.The flow chart below illustrate the industrial extraction of Lead metal

 

 

(a)(i)Name the chief ore that is commonly used in this process

Galena(PbS)

 

(ii)Explain what take place in the roasting furnace

 

 

 

 

 

                            UPGRADE

CHEMISTRY

                

                        FORM 4

 Radioactivity

      

 

                 Comprehensive tutorial notes

 

 MUTHOMI S.G

www.kcselibrary.info

                                     0720096206

 

 

 

 

 

 

Contents

 A  INTRODUCTION/CAUSES OF RADIOCTIVITY

           Alpha (α) particle

          Beta (β) particle

          Gamma(y) particle

B .NUCLEAR FISSION AND NUCLEAR FUSSION

  1. HALF-LIFE PERIOD AND DECAY CURVES

D .CHEMICAL vs NUCLEAR REACTIONS

E .APPLICATION OF RADIOACTIVITY AND RADIO ISOTOPES.

  1. DANGERS OF RADIOACTIVITY AND RADIO ISOTOPES.
  2. COMPREHENSIVE REVISION QUESTIONS

 

A: INTRODUCTION / CAUSES OF RADIOCTIVITY

 

Radioactivity is the spontaneous disintegration/decay of an unstable nuclide.

A nuclide is an atom with defined mass number (number of protons and neutrons), atomic number and definite energy.

Radioactivity takes place in the nucleus of an atom unlike chemical reactions that take place in the energy levels involving electrons.

A nuclide is said to be stable if its neutron: proton ratio is equal to one (n/p = 1)

All nuclide therefore try to attain n/p = 1 by undergoing radioactivity.

Examples

(i)Oxygen nuclide with 168 O has 8 neutrons and 8 protons in the nucleus therefore an n/p = 1 thus stable and do not decay/disintegrate.

(ii)Chlorine nuclide with 3517 Cl has 18 neutrons and 17 protons in the nucleus therefore an n/p = 1.0588  thus unstable and decays/disintegrates to try to attain n/p = 1.

(ii)Uranium nuclide with 23792 U has 206 neutrons and 92 protons in the nucleus therefore an n/p = 2.2391 thus more unstable than  23592 U and thus more readily decays / disintegrates to try to attain n/p = 1.

(iii) Chlorine nuclide with 3717 Cl has 20 neutrons and 17 protons in the nucleus therefore an n/p = 1.1765 thus more unstable than  3517 Cl and thus more readily decays / disintegrates to try to attain n/p = 1.

(iv)Uranium nuclide with 23592 U has 143 neutrons and 92 protons in the nucleus therefore an n/p = 1.5543 thus more stable than  237 92U  but also readily decays / disintegrates to try to attain n/p = 1.

All unstable nuclides naturally try to attain nuclear stability with the production of:

 

(i)alpha(α) particle decay

 

The alpha (α) particle has the following main characteristic:

i)is positively charged(like protons)

  1. ii) has mass number 4 and atomic number 2 therefore equal to a charged Helium atom ( 42He2+)

iii) have very low penetrating power and thus can be stopped /blocked/shielded by a thin sheet of paper.

  1. iv) have high ionizing power thus cause a lot of damage to living cells.
  2. v) a nuclide undergoing α-decay has its mass number reduced by 4 and its atomic number reduced by 2

 

Examples of alpha decay

210 84 Pb                 ->       x 82 Pb          +        42He 2+

210 84 Pb                 ->       206 82 Pb       +        42He 2+

 

226 88 Ra                 ->       222 y Rn        +        42He 2+

226 88 Ra                 ->       222 86 Rn      +        42He 2+

 

x y U                      ->       23490 Th        +        42He 2+

238 92 U                   ->       23490 Th        +        42He 2+

 

x y U                       ->       23088 Ra        +        2 42He 2+

238  92 U                  ->       23088 Ra        +        2 42He 2+

 

 210 84 U                  ->       xy W             +        10 α

210 84 U                   ->       17064 W         +        10 α

 

  210  92U                 ->       xy W             +        6 α

210  92U                   ->       18680W          +        6 α

 

 

(ii)Beta (β) particle decay

 

The Beta (β) particle has the following main characteristic:

i)is negatively charged(like electrons)

ii)has  no mass number  and atomic number  negative one(-1)  therefore equal to a fast moving electron (0 -1e)

iii) have medium  penetrating power and thus can be stopped /blocked/shielded by a thin sheet of  aluminium foil.

  1. iv) have medium ionizing power thus cause less  damage to living cells than the α particle.
  2. v) a nuclide undergoing β -decay has its mass number remain the same and its atomic number increase by 1

 

Examples of beta (β) decay

1.23 x Na                ->        2312Mg                  +        0 -1e

 23 11 Na                ->        2312Mg                  +        0 -1e

 

  1. 234 x Th           ->        y91 Pa                              +        0 -1e

       234 90 Th                     ->        y91 Pa                    +        0 -1e

 

  1. 20770Y           ->        x y Pb                    +        30 -1e

      20770Y                 ->        207 73Pb                 +        30 -1e

 

  1. x y C ->          147N                    +        0 -1e

       14 6 C                ->          147N                    +        0 -1e

 

  1. 1 x n ->         y1H                      +        0 -1e

   1 0 n                               ->          11H                     +        0 -1e

 

  1. 42He            ->       411H                      +         x 0 -1e

 42He                  ->       411H                      +          2 0 -1e

 

  1. 22888Ra         ->       22890Th                   +         x β

       22888Ra              ->        22892Th                  +         4 β

 

  1. 23290Th            ->      21282Pb                   +        2 β     +         x α

                  23290Th         ->      21282Pb                  +        2 β     +         5 α

 

  1. 23892U                ->      22688 Ra                  +        x β      +       3 α

 23892U                ->        22688 Ra                 +        2 β     +       3 α

 

  1. 21884Po          ->      20682Pb                   +        x β     +       3 α

 21884Po             ->       20682Pb                   +        4β        +      3 α

 

(iii)Gamma (y) particle decay

The gamma (y) particle has the following main characteristic:

i)is neither negatively charged(like electrons/beta) nor positively charged(like protons/alpha) therefore neutral.

ii)has  no mass number  and atomic number therefore equal to   electromagnetic waves.

iii) have very high  penetrating power and thus can be stopped /blocked/shielded by a thick block of lead..

  1. iv) have very low ionizing power thus cause less damage to living cells  unless on prolonged exposure..
  2. v) a nuclide undergoing y -decay has its mass number and its atomic number remain the same.

 

Examples of gamma (y) decay

 

  • 3717Cl ->             3717Cl                    +         y
  • 146C      ->             146C             +         y

 

The sketch diagram below shows the penetrating power of the radiations from a radioactive nuclide.

 

 

 

radioactive nuclide      sheet of paper       aluminium foil       thick block of lead

(radiation source)          (block α-rays)         (block β-rays)        block y-rays)

 

 

 

                                                                                                                                                                                                                                                    α-rays                      β-rays                    y-rays

         

The sketch diagram below illustrates the effect of electric /magnetic field on the three radiations from a radioactive nuclide

 

 

 

 

Radioactive disintegration/decay naturally produces the stable 20682Pb nuclide /isotope of lead.Below is the 238 92 U natural decay series. Identify the particle emitted in each case

 

Write the nuclear equation for the disintegration from :

(i)238 92 U   to       23490 T

          238 92 U            ->          23490 T       +              4 2 He 2+

           238 92 U           ->          23490 T       +              α

 

(ii)238 92 U   to   222 84 Rn

 238 92 U        ->          22284 Rn    +             4 4 2 He 2+

  238 92 U       ->          22284 Rn    +              4α

 

 230 90 Th undergoes alpha decay to 222 86 Rn. Find the number of α particles emitted. Write the nuclear equation for the disintegration.

Working      

 230 90 Th      ->        222 86 Rn      +              x 4 2 He

Method 1

 

Using mass numbers

          230              =     222   +    4 x          =>   4 x  = 230  – 222  =  8

                    x = 8 / 4  = 2 α

Using atomic numbers

          90       =   86   +     2 x          =>    2 x  = 90  –  86  =  4

                     x = 4 / 2  = 2 α

Nuclear equation

230 90 Th       ->        222 86 Rn     +              2 4 2 He

 

 214 82 Pb undergoes beta decay to 214 84 Rn. Find the number of β particles emitted. Write the nuclear equation for the disintegration.

Working      

 214 82 Pb       ->        214 84 Rn      +              x 0 -1 e

 

Using atomic numbers only

           82      =    84   –      x          =>     -x  =  82  –  84  =  -2

                     x     = 2 β

Nuclear equation

 214 82 Pb       ->        214 84 Rn      +              2 0 -1 e

 

238 92 U undergoes beta and alpha decay to 206 82 Pb. Find the number of β  and α particles emitted. Write the nuclear equation for the disintegration.

Working      

          238 92 U            ->        206 82 Pb      +     x 0 -1 e +                  y 4 2 He

 

Using Mass numbers only

           238              =    206   +    4y    =>     4y  = 238  –   206  =  32

                y     =     32      =     8 α

                              4

Using atomic numbers only and substituting the 8 α(above)

 238 92 U        ->        206 82 Pb       +            8 4 2 He   +     x 0 -1 e

92      =            82               +        16           +     – x

=>   92 –  (82  +  16)   = – x

                               x     = 6  β

Nuclear equation

          238 92 U            ->        206 82 Pb      +     6 0 -1 e +                  8 4 2 He

 

 298 92 U undergoes alpha and beta decay to 214 83 Bi. Find the number of  α and β particles emitted. Write the nuclear equation for the disintegration.

Working      

         298 92 U                ->     210 83 Bi    +                  x 4 2 He      +     y 0 -1 e

Using Mass numbers only

  298    =   214   +    4x    => 4x  = 298  –   214  =  84

     y     =      84      =     21 α

                     4

Using atomic numbers only and substituting the 21 α (above)

 238 92 U   -> 214 83Bi          +     21 4 2 He   +     y 0 -1 e

92   =    83         +                  42       +     – y

=>   92 –  (83  +  42)   = – x

                    x     =      33  β

Nuclear equation

               298 92 U  ->   210 83 Bi    +   21 4 2 He    +   33 0 -1 e

 

B:NUCLEAR FISSION AND NUCLEAR FUSION

Radioactive disintegration/decay can be initiated in an industrial laboratory through two chemical methods:

  1. a) nuclear fission
  2. b) nuclear

a)Nuclear fission

Nuclear fission is the process which a fast moving neutron bombards /hits /knocks a heavy unstable nuclide releasing lighter nuclide, three daughter neutrons and a large quantity of energy.  

Nuclear fission is the basic chemistry behind nuclear bombs made in the nuclear reactors.

The three daughter neutrons becomes again fast moving neutron bombarding / hitting /knocking  a heavy unstable nuclide releasing lighter nuclides, three more daughter neutrons each and a larger quantity of energy setting of a chain reaction

 

Examples of nuclear equations showing nuclear fission

10 n  +  235 b U  -> 9038 Sr  + c 54Xe + 310 n  +  a

 

10 n   +        2713 Al            ->        2813 Al        +         y  +   a

 

10 n   +      28a Al     ->          b11 Na        +    42  He

 

 a0 n   +       147 N     ->          14b C  +       11 H

 

10 n   + 11 H                  ->        21 H                    +         a

 

10 n     +  235 92 U  ->   95 42 Mo +  139 57 La      + 210 n  + 7 a

 

  1. b) Nuclear fusion

Nuclear fusion is the process which smaller nuclides join together to form larger / heavier nuclides and releasing a large quantity of energy.

Very high temperatures and pressure is required to overcome the repulsion between the atoms.

Nuclear fusion is the basic chemistry behind solar/sun radiation.

Two daughter atoms/nuclides of Hydrogen fuse/join to form Helium atom/nuclide on the surface of the sun releasing large quantity of energy in form of heat and light.

21H      +    21H     ->     abHe       +      10 n

        

  21H       +    a     ->     32He

 

   21H       +       21H    ->    a       +    11 H

 

11H       ->     42He       +         a

 

147H     +   a    ->  178O      +           11 H

 

 

C: HALF LIFE PERIOD (t1/2)

 

The half-life period is the time taken for a radioactive nuclide to spontaneously decay/ disintegrate to half its original mass/ amount.

It is usually denoted t 1/2.

The rate of radioactive nuclide disintegration/decay is constant for each nuclide.

 

 

The table below shows the half-life period of some elements.

Element/Nuclide Half-life period(t 1/2 )
238 92 U 4.5 x 109 years
14 6 C 5600 years
229 88 Ra 1620 years
35 15 P 14 days
210 84 Po 0.0002 seconds

 

The less the half life the more unstable the nuclide /element.

The half-life period is determined by using a Geiger-Muller counter (GM tube)

.A GM tube is connected to ratemeter that records the count-rates per unit time.

This is the rate of decay/ disintegration of the nuclide.

If the count-rates per unit time fall by half, then the time taken for this fall is the half-life period.

Examples

a)A radioactive substance gave a count of 240 counts per minute but after 6 hours the count rate were 30 counts per minute. Calculate the half-life period of the substance.

If  t 1/2  = x

then  240 –x–>120 –x–>60 –x—>30

From 240 to 30 =3x =6 hours

=>x  = t 1/2 = ( 6 / 3 )

= 2 hours

  1. b) The count rate of a nuclide fell from 200 counts per second to 12.5 counts per second in 120 minutes.

Calculate the half-life period of the nuclide.

If  t 1/2  =x

then

200 –x–>100 –x–>50 –x—>25 –x—>12.5

From 200 to 12.5 =4x =120 minutes

=>x  = t 1/2 = ( 120 / 4 )

= 30  minutes

  1. c) After 6 hours the count rate of a nuclide fell from 240 counts per second to 15 counts per second on the GM tube. Calculate the half-life period of the nuclide.

If  t 1/2  = x

then  240 –x–>120 –x–>60 –x—>30 –x—>15

From 240 to 15 =4x =6 hours

=>x  = t 1/2 = ( 6  / 4 )=  1.5  hours

 

  1. d) Calculate the mass of nitrogen-13 that remain from 2 grams after 6 half-lifes if the half-life period of nitrogen-13 is 10 minutes.

If  t 1/2  = x then:

2 —x–>1 –2x–>0.5 –3x—>0.25 –4x–>0.125–5x—>0.0625–6x—>0.03125

After the 6th half life 0.03125 g of nitrogen-13 remain.

 

  1. e) What fraction of a gas remains after 1hour if its half-life period is 20 minutes?

If  t 1/2  = x then:

then  60 /20  = 3x

1   –x–> 1/2 –2x–> 1/4  –3x—> 1/8

After the 3rd half-life  1/8   of the gas remain

 

  1. f) 348 grams of a nuclide A was reduced to 43.5 grams after 270days.Determine the half-life period of the nuclide.

If  t 1/2  = x then:

348 –x–>174 –2x–>87 –3x—>43.5

From 348 to 43.5=3x =270days

=>x  = t 1/2 = ( 270 / 3 )

=  90  days

 

  1. g) How old is an Egyptian Pharaoh in a tomb with 2grams of 14C if the normal 14C in a present tomb is 16grams.The half-life period of 14C is 5600years.

                    If  t 1/2  = x  = 5600 years then:

16 –x–>8 –2x–>4 –3x—>2

3x = ( 3  x  5600 )

= 16800years

 

  1. h) 100 grams of a radioactive isotope was reduced 12.5 grams after 81days.Determine the half-life period of the isotope.

                    If  t 1/2  = x then:

100 –x–>50 –2x–>25 –3x—>12.5

From 100 to 12.5=3x =81days

=>x  = t 1/2

= ( 81  / 3 )

=  27  days

 

 

A graph of activity against time is called decay curve.

A decay curve can be used to determine the half-life period of an isotope since activity decrease at equal time interval to half the original

 

 

(i)From the graph show and determine the half-life period of the isotope.

 

From the graph t 1/changes in activity from:

( 100 – 50 )      =>   ( 20 – 0 )  =  20 minutes

( 50 – 25 )      =>   ( 40 – 20 )  =  20 minutes

Thus      t ½  20 minutes

 

(ii)Why does the graph tend to ‘O’?

Smaller particle/s will disintegrate /decay to half its original.

             There can never be ‘O’/zero particles

 

 

 

D:  CHEMICAL vs NUCLEAR REACTIONS

 

Nuclear and chemical reaction has the following similarities:

(i)-both involve the subatomic particles; electrons, protons and neutrons in an  atom

(ii)-both involve the subatomic particles trying to make the atom more stable.

(iii)-Some for of energy transfer/release/absorb from/to the environment take  place.

 

Nuclear and chemical reaction has the following differences:

(i) Nuclear reactions mainly involve protons and neutrons in the nucleus of an atom.

Chemical   reactions mainly involve outer electrons in the energy levels an atom.

(ii) Nuclear reactions  form a new element.

Chemical   reactions do not form new elements

(iii) Nuclear reactions mainly involve evolution/production of large quantity of heat/energy.

Chemical   reactions produce or absorb small quantity of heat/energy.

(iv)Nuclear reactions are accompanied by a loss in mass/mass defect.Do not obey the law of conservation of matter.

Chemical   reactions are not accompanied by a loss in mass/ mass defect hence obey the law of conservation of matter.

(v)The rate of decay/ disintegration of the nuclide is independent of physical conditions (temperature/pressure /purityp/article size)

The rate of a chemical reaction is dependent on physical conditions (temperature/pressure/purity/particle size/ surface area)

 

E: APPLICATION AND USES OF RADIOCTIVITY.

The following are some of the fields that apply and use radioisotopes;

a)Medicine: –Treatment of cancer to kill malignant tumors through    radiotherapy.

Sterilizing hospital /surgical instruments /equipments by   exposing them to gamma radiation.

  1. b) Agriculture:

 If a plant or animal is fed with radioisotope, the metabolic processes of the plant/animal is better understood by tracing the route of the radioisotope.

  1. c) Food preservation:

 X-rays are used to kill bacteria in tinned food to last for a long time.

  1. d) Chemistry:

 To study mechanisms of a chemical reaction, one  reactant is replaced in its structure by a radioisotope e.g.

During esterification the ‘O’ joining the ester was discovered comes from the alkanol and not alkanoic acid.

During photosynthesis the ‘O’ released was discovered comes from   water.

 

  1. e) Dating rocks/fossils:

The quantity of 14C in living things    (plants/animals) is constant.

When they die the fixed mass of 14C is trapped in the cells and continues to decay/disintegrate.

The half-life period of 14C  is 5600 years .

Comparing the mass of 14C in living and dead cells, the age of the dead can be determined.

 

F: DANGERS OF RADIOCTIVITY.

All rays emitted by radioactive isotopes have ionizing effect of changing the genetic make up of living cells.

Exposure to theses radiations causes chromosomal and /or genetic mutation in living cells.

Living things should therefore not be exposed for a long time to radioactive substances.

One of the main uses of radioactive isotopes is in generation of large cheap electricity in nuclear reactors.

Those who work in these reactors must wear protective devises made of thick glass or lead sheet.

Accidental leakages of radiations usually occur

In 1986 the Nuclear reactor at Chernobyl in Russia had a major explosion that emitted poisonous nuclear material that caused immediate environmental disaster

In 2011, an earthquake in Japan caused a nuclear reactor to leak and release poisonous radioactive waste into the Indian Ocean.

The immediate and long term effects of exposure to these poisonous radioactive waste on human being is of major concern to all environmentalists.

 

 

 

 

 

 

 

G: SAMPLE REVISION QUESTIONS

The figure below shows the behaviour of emissions by a radioactive isotope x. Use it to answer the question follow

 

(a) Explain why isotope X emits radiations.             (1mk)

-is unstable //has n/p ratio greater/less than one

 

(b) Name the radiation labeled T                             (1mk)

alpha particle

 

(c) Arrange the radiations labeled P and T in the increasing order of ability to be deflected by an electric filed.           (1mk)

T -> P

 

  1. a) Calculate the mass and atomic numbers of element B formed after 21280 X has emitted three beta particles, one gamma ray and two alpha particles.

Mass number

           = 212 – (0  beta+ o gamma + (2 x 4 ) alpha = 204

Atomic number

          = 80 –  (-1 x3) beta + 0 gamma + (2 x 2 )) alpha =79

 

b)Write a balanced nuclear equations for the decay of  21280 X to B using the information in (a) above.

            21280 X      ->     20479B   +    242He     +    3  0-1 e    +   y

 

Identify the type of radiation emitted from the following nuclear equations.

(i)       146 C              ->     147N      + ………

          β  –  Beta

  • (ii) 11 H +   10 n       ->      21H         + ……

                    y -gamma

 

(iii)     23592 U                      ->       9542Mo      +  13957La  +  10 n  +……

                    7 β – seven beta particles

 

  • 23892 U                 ->     23490Th             + … …

                    α-alpha

 

  • (v) 146 C  +       11 H         ->           157N           +     ……

                    y-gamma

 

X grams of a radioactive isotope takes 100 days to disintegrate to 20 grams. If the half-life period isotope is 25 days, calculate the initial mass X of the radio isotope.

 

Number of half-lifes    =   (  100   /   25 )   =    4

   20g —–> 40g —-> 80g—–> 160g —–> 320g

                    Original mass X = 320g

 

Radium has a half-life of 1620 years.

(i)What is half-life?

The half-life period is the time taken for a radioactive nuclide to spontaneously decay/ disintegrate to half its original mass/ amount

 

b)If one milligram of radium contains 2.68 x 10 18 atoms ,how many atoms disintegrate during 3240 years.

 

Number of half-lifes    =   (  3240   /   1620  )   =    2

                    1 mg —1620—> 0.5mg —1620—-> 0.25mg

                      If 1mg       ->       2.68 x 1018  atoms

 

Then   0.25 mg ->  ( 0.25  x   2.68  x  1018   ) =   6.7  x  1017

 Number of atoms remaining = 6.7  x  1017

 

  Number of atoms disintegrated  =

                               (2.68 x 1018     –    6.7  x  1017   )

                                                                  = 2.01  x  1018

 

The graph below shows the mass of a radioactive isotope plotted against time

 

 

Using the graph, determine the half – life of the isotope

From graph 10 g to 5 g takes 8 days

From graph 5 g to 2.5 g takes 16 – 8 = 8 days

Calculate the mass of the isotope dacayed after 32 days

Number of half lifes= 32/8 = 4

Original mass = 10g

10g—1st  –>5g—2nd–>2.5—3rd –>1.25—4th –>0.625 g

Mass remaining = 0.625 g

Mass decayed after 32 days =  10g – 0.625 g = 9.375g

 

A radioactive isotope X2 decays by emitting two alpha (a) particles and one beta (β) to form 214 83Bi

 

(a)Write the nuclear equation for the radioactive decay

 21286 X      -> 214 83Bi   +    242He     +      0-1 e

 

(b)What is the atomic number of X2?

86

(c) After 112 days, 1/16 of the mass of X2 remained. Determine the half life of X2

 

  1—x-> 1 /2 –x-> 1 /4 –x-> 1 /8–x-> 1 /16

    Number of t 1 /2 in 112 days   =  4

          t 1 /2       =           112      =      28 days

                                          4

1.Study the nuclear reaction given below and answer the questions that follow.

 

126 C   –step 1–>127 N –step 2–> 1211Na

 

(a)126 C and 146 C are isotopes. What does the term isotope mean?

Atoms of the same element with different mass number /number of neutrons.

 

(b)Write an equation for the nuclear reaction in step II

 127 N ->          1211Na              +       0 -1e

(c)Give one use of   146 C

          Dating rocks/fossils:

          Study of metabolic pathways/mechanisms on plants/animals

 

Study  the graph of a radioactive decay series for isotope H below.

 

 

  • Name the type of radiation emitted when isotope

(i) H changes to isotope J.

                    AlphaMass number decrease by 4 from 214 to 210(y-axis)

                              atomic number decrease by 2 from 83 to 81(x-axis)

(ii) J changes to isotope K

BetaMass number  remains 210(y-axis)

                             atomic number increase by  1 from 81 to 82(x-axis).

(b) Write an equation for the nuclear reaction that occur when  isotope

(i)J changes to isotope L

21081 J          ->          21084L               +      3 0 -1e

 

(i)H changes to isotope  M

21483 H         ->          20682M              +      3 0 -1e +        2 4 2He

 

Identify a pair of isotope of an element in the decay series

K and M

Have same atomic number 82 but different mass number K-210 and M-206

 

a)A radioactive substance emits three different particles.

Identify the particle:

(i)with the highest mass.

                              Alpha/ α

(ii) almost equal to an electron

                              Beta/ β

1.a)State two differences between chemical and nuclear reactions(2mks)

(i) Nuclear reactions mainly involve protons and neutrons in the nucleus of an atom.Chemical  reactions mainly involve outer electrons in the energy levels an atom.

(ii) Nuclear reactions  form a new element. Chemical   reactions do not form new elements

(iii) Nuclear reactions mainly involve evolution/production of large quantity of heat/energy.Chemical   reactions produce or absorb smaller quantity of heat/energy.

(iv)Nuclear reactions are accompanied by a loss in mass /mass defect.

Chemical   reactions are not accompanied by a loss in mass.

(v)Rate of decay/ disintegration of nuclide is independent of physical conditionsThe rate of a chemical reaction is dependent on physical conditions  of temperature/pressure/purity/particle size/ surface area

b)Below is a radioactive decay series starting from 21483 Bi and ending at 20682 Pb. Study it and answer the question that follows.

 

Identify the particles emitted in steps I and III   (2mks)

I    –  α-particle

III –  β-ray

ii)Write the nuclear equation for the reaction which takes place in (a) step  I

                              21483Bi           ->               21081Bi        + 4 2 He

(b) step  1 to 3

                              21483Bi           ->               21081Bi        + 4 2 He +  0 -1 e

(c) step  3 to 5

                              21082Pb           ->     20682Pb        + 4 2 He +  2  0 -1 e

(c) step  1 to 5

                              21483Bi ->     20682Pb        + 2 4 2 He +  3  0 -1 e

 

The table below give the percentages of a radioactive isotope of Bismuth that remains after decaying at different times.

 

Time (min) 0 6 12 22 38 62 100
Percentage of Bismuth 100 81 65 46 29 12 3

i)On the  grid below , plot a graph of the percentage of Bismuth remaining(Vertical axis) against time.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ii)Using the graph, determine the:

  1. Half – life of the Bismuth isotope
  2. Original mass of the Bismuth isotope given that the mass that remained after 70 minutes was 0.16g (2mks)

 

  1. d) Give one use of radioactive isotopes in medicine (1mk)

 

14.a)Distinguish between nuclear fission and nuclear fusion.                (2mks)

 

Describe how solid wastes containing radioactive substances should be disposed of.                               (1mk)

 

b)(i)Find the values of Z1 and Z2 in the nuclear equation below

                              Z1           1             94             140               1

U   +          n  ->     Sr   +   Xe  + 2   n

92           0             38                Z2               0   

iii)What type of nuclear reaction is represented  in b (i) above?

 

A radioactive cobalt   6128Co  undergoes decay by emitting a beta particle and forming Nickel atom,

 

Write a balanced decay equation for the above change                                                                                        1 mark

If a sample of the cobalt has an activity of 1000 counts per minute, determine the time it would take for its activity to decrease to 62.50 if  the half-life of the element is 30years                                                                       2 marks

 

Define the term half-life.

 

The diagram below shows the rays emitted by a radioactive sample

  1. Identify the rays S,R and  Q

S- Beta ( β )particle/ray

R- Alpha (α )particle/ray

          Q- Gamma (y )particle/ray

 

  1. b) State what would happen if an aluminium plate is placed in the path of ray R,S and Q:

          R-is blocked/stopped/do not pass through

Q-is not blocked/pass through

S-is blocked/stopped/do not pass through

 

(c)The diagram bellow is the radioactive decay series of nuclide A which is 24194Pu.Use it to answer the questions that follow. The letters are not the actual symbols of the elements.

 

 

 

(a)Which letter represent the : Explain.

(i)shortest lived nuclide

L-has the shortest half life

          (ii)longest lived nuclide

          P-Is stable

(iii) nuclide with highest n/p ratio

          L-has the shortest half life thus most unstable thus      easily/quickly decay/disintegrate

(iv) nuclide with lowest n/p ratio

          P-is stable thus  do not decay/disintegrate

 

(b)How long would it take for the following:

(i)Nuclide A to change to B

          10 years (half life of A)

(ii) Nuclide D to change to H

               27days +162000years+70000years+16days

                              232000 years and 43 days

 

(iii) Nuclide A to change to P

               27days +162000years+70000years+16days

                              232000 years and 43 days

Study

 

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