FORM 4 CHEMISTRY NOTES HANDBOOK -

ACIDS, BASES

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

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) + SO₄^2-(aq)

H₂CO₃(aq) -> 2H⁺ (aq) + CO₃^2-(aq)

H₃PO₄(aq) -> 3H⁺ (aq) + PO₄^3-(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)

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

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.

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.

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

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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+ H^ð+

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 Cl^–ions.

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

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.

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) +SO₄^2-(aq) (molecules) (cation) (anion)

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.

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) +PO₄^3-(aq) (molecules) (cation) (anion)

H₂CO₃(aq) 2H⁺(aq) +CO₃^2-(aq) (molecules) (cation) (anion)

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)

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.

(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)

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) -> Zn²⁺ (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) -> Zn²⁺ (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) -> Pb²⁺ (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) -> Pb²⁺ (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) -> Al³⁺ (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) -> Al³⁺ (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.

(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) (SO₄^2-)	Hydrogen sulphate(VI) (HSO₄^–)
Sulphuric(IV)acid	H₂SO₃	Dibasic	Sulphate(IV) (SO₃^2-)	Hydrogen sulphate(IV) (HSO₃^–)
Carbonic(IV)acid	H₂CO₃	Dibasic	Carbonate(IV)(CO₃^2-)	Hydrogen carbonate(IV) (HCO₃^–)
Phosphoric(V) acid	H₃PO₄	Tribasic	Phosphate(V)(PO₄^3-)	Dihydrogen phosphate(V) (H₂PO₄^2-) Hydrogen diphosphate(V) (HP₂O₄^2-)

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	SO₄^2-	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	SO₄^2-	MgSO₄	Calcium sulphate(VI)
NH₄OH	NH₄⁺	H₃PO₄	PO₄^3-	(NH₄ )₃PO₄ (NH₄ )₂HPO₄ NH₄ H₂PO₄	Ammonium phosphate(V) Diammonium phosphate(V) Ammonium diphosphate(V)
KOH	K⁺	H₃PO₄	PO₄^3-	K₃PO₄	Potassium phosphate(V)
Al(OH)₃	Al³⁺	H₂SO₄	SO₄^2-	Al₂(SO₄)₂	Aluminium(III)sulphate(VI)
Fe(OH)₂	Fe²⁺	H₂SO₄	SO₄^2-	FeSO₄	Iron(II)sulphate(VI)
Fe(OH)₃	Fe³⁺	H₂SO₄	SO₄^2-	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)/SO₄^2- salts except	Barium(II) sulphate(VI)/BaSO₄ Calcium(II) sulphate(VI)/CaSO₄ Lead(II) sulphate(VI)/PbSO₄
All sulphate(IV)/SO₃^2- 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)/PO₄^3-
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) -> MgCl₂(aq) + H₂(g)

Zn(s) + 2HCl(aq) -> ZnCl₂(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) -> MgCl₂(aq) + H₂O (l)

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

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

CuO (s) + 2HCl(aq) -> CuCl₂(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)

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)

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) + 2NH₄NO₃(aq)

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

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.

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

2KHCO₃(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)

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 OH^–from 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 OH^–from 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 OH^–from 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 OH^–ions 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 OH^–from 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 OH^–from 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 OH^–from 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 OH^–from 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 OH^–from 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 OH^–from 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 OH^–from 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 OH^–from 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 OH^–from 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 Cl^–ions from a soluble chloride salt to form a white precipitate of AgCl

(ii)White precipitate dissolve on heating

Pb²⁺ ions

Pb²⁺ ions reacts with Cl^–ions from a soluble chloride salt to form a white precipitate of PbCl₂. PbCl₂dissolves on heating.

Note

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

No white precipitate on adding Cl^–ions/salts shows Al³⁺.

Adding a chloride/ Cl^–ions/salts can thus be used to separate the identity of Al³⁺ and Pb²⁺.

(d)Reaction of cation with sulphate(VI)/SO₄^2-and sulphate(IV)/SO₃^2- ions

All sulphate(VI) and sulphate(IV)/SO₃^2- 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)/SO₄^2- 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

CaSO₃and CaSO₄do not form a thick precipitate as they are sparingly soluble.

Ca²⁺(aq)+ SO₃^2-(aq) -> CaSO₃(s)

Ca²⁺(aq)+ SO₄^2-(aq) -> CaSO₄(s)

Ba²⁺(aq)+ SO₃^2-(aq) -> BaSO₃(s)

Ba²⁺(aq)+ SO₄^2-(aq) -> BaSO₄(s)

Pb²⁺(aq)+ SO₃^2-(aq) -> PbSO₃(s)

Pb²⁺(aq)+ SO₄^2-(aq) -> PbSO₄(s)

(e)Reaction of cation with carbonate(IV)/CO₃^2- ions

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

They dissociate /ionize to release CO₃^2- ions. CO₃^2- 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 andZnSO₄in separate boiling tubes.

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

Observation

Inference

Explanation

Green precipitate

Cu²⁺ ,Fe²⁺,ions

CO₃^2-(aq)

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

Cu²⁺(aq)+ CO₃^2-(aq) -> CuCO₃(s)

Fe²⁺(aq)+ CO₃^2-(aq) -> FeCO₃(s)

When sodium carbonate(IV)is added to CuCO₃(s) the CO₃^2-(aq) ions are first hydrolysed to produce CO₂(g) and OH^–(aq)ions.

CO₃^2-(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)+ CO₃^2-(aq)+2OH^– (aq)

->CuCO₃.Cu(OH)₂(s)

White precipitate

CO₃^2-

White ppt of the carbonate(IV)salt is precipitated

Ca²⁺(aq) + CO₃^2-(aq) -> CaCO₃(s)

Mg²⁺(aq) + CO₃^2-(aq) -> MgCO₃(s)

Pb²⁺(aq) + CO₃^2-(aq) -> PbCO₃(s)

Zn²⁺(aq) + CO₃^2-(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;

BaCO₃ produces CO₂gas. When bubbled/directed into lime water solution,a white precipitate is formed.
I. BaSO₃ produces SO₂gas. 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 / S^2- 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 ZnSO₄in separate boiling tubes.

Add six drops of Potassium /sodium sulphide solution.

Observation

Inference

Explanation

Black ppt

S^2- ions

CuS, FeS,MgS,PbS, ZnS are black insoluble precipitates

Cu²⁺(aq) + S^2-(aq) -> CuS(s)

Pb²⁺(aq) + S^2-(aq) -> PbS(s)

Fe²⁺(aq) + S^2-(aq) -> FeS(s)

Zn²⁺(aq) + S^2-(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 CO₃^2-, SO₄^2-, SO₃^2-, Cl^–,

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

Observation Inference (1mark)

White ppt persist/remains SO₄^2-, Cl^–,

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

Observation Inference (1mark)

White ppt persist/remains SO₄^2-,

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

SO₄^2- 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, “sO₄^2-” 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 80^oC,how much can dissolve in 5cm3 of water at 80^oC .

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 20^oC,how much can saturate 25g of water at 20^oC .

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 10^oC. Calculate the solubility of potassium nitrate(V) at 10^oC.

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 10^oC if the solubility of KNO₃ at 10^oC is 5.0g/100g H₂O?

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

X= 100 – 22 = 78 cm3 of H₂O

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

Saturated solution of KClO₃

NaCl

KClO₃

KNO₃

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

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

At 100^oC = 60.0g

Less at 75^oC = – 39.0g

Mass of crystallized out 21.0g

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

At 100^oC = 60.0g

Less at 35^oC = – 28.0.0g

Mass of crystallized out 32.0g

KNO₃ has a solubility of 42 g/100g water at 20^oC.The salt was heated and added 38g more of the solute which dissolved at100^oC. Calculate the solubility of KNO₃ at 100^oC.

Solubility of KNO₃ at 100^oC = solubility at 20^oC + mass of KNO₃added

=> 42g + 38g = 80g KNO₃ /100g H₂O

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

Mass of solvent at 20^oC = 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

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

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

Study the table below

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

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

(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.

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 H₂O = 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

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:

20^oC

From a correctly plotted graph = 32g

35^oC

From a correctly plotted graph = 57g

III. 55^oC

From a correctly plotted graph = 104g

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

22g

From a correctly plotted graph =13.0^oC

30g

From a correctly plotted graph =17.5^oC

III.100g

From a correctly plotted graph =54.5^oC

(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 80^oC to:

20^oC

Solubility before heating = 180 g

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

Mass of KNO₃crystals = 148 g

35^oC

Solubility before heating = 180 g

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

Mass of KNO₃crystals = 122 g

III. 55^oC

Solubility before heating = 180 g

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

Mass of KNO₃crystals = 78 g

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 = 24^oC

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

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

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

Method I.

Total mass before cooling at 50^oC = 100.0 g

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

Mass of crystals deposited 68.0 g

Method II.

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

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

Mass of crystals that cannot dissolve at 50^oC 15.0 g

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

(From graph) Solubility/mass after cooling at 20^oC = 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 70^oC.The solution is then allowed to cool to 10^oC.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

(Theoretical)Solubility of A before heating = 400 g

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

Mass of crystals that can not dissolve at70^oC = 352 g

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

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

Mass of crystals that crystallize out on cooling to10^oC = 17 g

Mass of crystals that can not dissolve at70^oC = 352 g

Add Mass of crystals that crystallize out on cooling to10^oC = 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

(Theoretical)Solubility of B before heating = 600 g

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

Mass of crystals that cannot dissolve at70^oC = 462 g

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

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

Mass of crystals that crystallize out on cooling to10^oC = 117 g

Mass of crystals that cannot dissolve at70^oC = 462 g

Add Mass of crystals that crystallize out on cooling to10^oC = 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

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 KclO₃at different temperatures.

(b)Plot a graph of mass of KClO₃per 100g water against temperature at which crystals form.

(c)From the graph, show and determine ;

(i)the solubility of KClO₃ at

50^oC

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

35^oC

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

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 80^oC to 40^oC.

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

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

Mass of crystallized/precipitated = 7.2 g

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 (30-40^oC) 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 CaSO₄and 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 /Na₂CO₃.10H₂O precipitates 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

CO₃^2-(aq)+ Ca²⁺ (aq) -> CaCO₃ (s)

CO₃^2-(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

CO₃^2-(aq)+ Ca²⁺ (aq) -> CaCO₃ (s)

CO₃^2-(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)

2NH₃(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

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

Sample II: Uses little sample of soap .

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 /25^oC 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

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.

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 = H_final – H_initial

For chemical reactions:

∆H = H_products – H_reactants

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

H₂H₂SO₄(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

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)

25^oC

time(seconds)

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

Note:

Water boils at 100^oC at sea level/one atmosphere pressure/101300Pa but boils at below 100^oC 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 96^oC.

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

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

69^oC

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(25^oC) 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 ∆Hᶿ_r
Standard enthalpy/heat of combustion ∆Hᶿ_c
Standard enthalpy/heat of displacement ∆Hᶿ_d
Standard enthalpy/heat of neutralization ∆Hᶿ_n
Standard enthalpy/heat of solution/dissolution ∆Hᶿ_s
Standard enthalpy/heat of formation ∆Hᶿ_f

(a)Standard enthalpy/heat of reaction ∆Hᶿ_r

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.

A–B + 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 ∆H_r

1.Calculate ∆H_rfrom the following reaction:

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

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

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) ∆H_r= -93kJ

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

∆H_r=( + 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

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

∆H_r=( + 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 ∆Hᶿ_c

The molar standard enthalpy/heat of combustion(∆Hᶿ_c) 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(∆Hᶿ_c) 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	25.0^oC
Temperature of the water after heating T₂	35.0^oC
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 => (35.0^oC – 25.0^oC) = 10.0^oC

(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

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 ∆H_c = Heat produced ∆H

Number of moles of fuel

=> 0.84 kJ = 96.5517 kJmole^-1

0.0087 /8.7 x 10^-3moles

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 ∆H_c

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 20^oC to 27^oC.(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 ∆H_c = 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

1.0 g of carbon burn in excess air to raise the temperature of 400g of water by 18^oC.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 ∆H_c = 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 ∆Hᶿ_d

The molar standard enthalpy/heat of displacement ∆Hᶿ_d 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(∆Hᶿ_d) 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 0.5^oC- T₂ . Repeat the experiment to complete table 1 below

Table 1

Experiment	I	II
Final temperature of solution(T₂)	30.0^oC	31.0^oC
Final temperature of solution(T₁)	25.0^oC	24.0^oC
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 = 5.5^oC

(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.

Calculate the enthalpy change for one mole of displacement of Cu²⁺ (aq) ions.

Molar heat of displacement ∆H_d = 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

Iron is less reactive than Zinc. Explain the effect of using iron instead of Zinc on the standard molar heat of displacement ∆H_d 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 ∆H_d 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 ∆H_d x Number of moles

=>209kJmole^-1x 0.01 moles = 2.09 kJ

∆T (change in temperature) = Heat produced ∆H Molar heat of displacement ∆H_d 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

The enthalpy of displacement ∆H_d 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 ∆H_d

=>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

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 ∆H_d 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 0.5^oC.

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

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 = 11.5^oC

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 ∆H_d

=>.966 kJ = 0.007728moles

125 moles 7.728 x 10^-3moles

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 25^oC.While that of the mixture was 43^oC.

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 ∆Hᶿ_n

The molar standard enthalpy/heat of neutralization ∆Hᶿ_n 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 ∆Hᶿ_ncan be determined as in the examples below:

To determine the molar enthalpy of neutralization ∆H_n 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 0.5^oC 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 ∆Hᶿ_n.

(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.

∆H_n = 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 ∆H_n

Heat/energy is absorbed by the reaction vessel/calorimeter/plastic cup lowers ∆T and hence ∆H_n

(e)Compare the ∆H_nof 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/∆H_nwould 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 5.0^oC rise in temperature?

Working:

Moles of sodium hydroxide = molarity x volume => 0.5 M x 20cm3 = 0.01 moles

1000 1000

Enthalpy change ∆H = ∆H_n=>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

∆H_n 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 10.0^oC to 16.5^o

∆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

∆H_n

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 13.25^oC 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 = ∆H_nx 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 ∆H_n 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 0.5^oC.

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.

∆H_n = ∆H_n = 1.174845kJ = 35.0701kJ

Number of moles 0.0335

(d)Standard enthalpy/heat of solution ∆Hᶿ_s

The standard enthalpy of solution ∆Hᶿ_sis 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 (∆H_l). Lattice energy /heat/enthalpy of lattice (∆H_l) is an endothermic process (+∆Hl).

The table below shows some ∆H_lin 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(∆H_h).Hydration energy /enthalpy of hydration(∆H_h) is an exothermic process(∆H_h).

The table below shows some ∆H_hin kJ for some ions;

ion	Li⁺	Na⁺	K⁺	Mg²⁺	Ca²⁺	F^–	Cl^–	Br^–
∆H_h	-1091	-406	-322	-1920	-1650	-506	-364	-335

The sum of the lattice energy +∆H_l (endothermic) and hydration energy –∆H_h (exothermic) gives the heat of solution–∆H_s

∆H_s = ∆H_l +∆H_h

Note

Since ∆H_lis an endothermic process and ∆H_h is an exothermic process then ∆H_s is:

(i)exothermic if ∆H_l is less than ∆H_h and hence a solid dissolve easily in water.

(ii)endothermic if ∆H_l is more than ∆H_h 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) ∆H_l =+771 kJ

(ii) Hydrating the ions;

Na ⁺(g) + aq -> Na(aq) ∆Hh = – 406 kJ

Cl^–(g) + aq -> Cl^–(aq) ∆Hh = – 364 kJ

∆H_s =∆H_h +∆H_s -> (- 406 kJ + – 364 kJ) + +771 kJ = + 1.0 kJmole^-1

NaCl does not dissolve easily in water because overall ∆H_s 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) ∆H_l =+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

∆H_s =∆H_h +∆H_s -> (- 1920 kJ + (- 364 x 2 kJ)) + +2493 kJ = –155.0 kJmole^-1

MgCl₂ (s) dissolve easily in water because overall ∆H_s 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) ∆H_l =+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

∆H_s =∆H_h +∆H_s -> (- 1650 kJ + (- 506 x 2 kJ)) + +760 kJ = –1902.0 kJmole^-1

CaF₂ (s) dissolve easily in water because overall ∆H_s 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) ∆H_l =+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

∆H_s =∆H_h +∆H_s -> (- 1920 kJ + (- 335 x 2 kJ)) + +2226 kJ = –364.0 kJmole^-1

MgBr₂ (s) dissolve easily in water because overall ∆H_s 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 ∆H_s 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 0.5^oCafter 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).

∆H_s = ∆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 25^oC was added carefully 3cm3 concentrated sulphuric(VI)acid of density 1.84gcm^-3.The temperature of the mixture rose from 25⁰C to 38^oC.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 13^oC = 5761.392 J = 5.761392 kJ

1000

∆H_s of H₂SO₄= ∆H => 5.761392 kJ = -102.33378kJmoles^-1

Moles of H₂SO₄ 0.0563 moles

(e)Standard enthalpy/heat of formation ∆Hᶿ_f

The molar enthalpy of formation ∆Hᶿ_f is defined as the energy change when one mole of a compound is formed from its elements at 298K(25^oC) and 101325Pa(one atmosphere)pressure. ∆Hᶿ_f 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) ∆H₂B(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+ ∆H₃

(ii) ∆H₂=∆H₁+ -∆H₃

(iii) ∆H₃= – ∆H₁+ ∆H₂

Examples of applying Hess^’ law of constant heat summation

1.Calculate the molar enthalpy of formation of methane (CH₄) given that ∆Hᶿ_cof 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=∆Hᶿ_c =-890.4kJ CH₄(g)+2O₂(g)

∆H₃=∆Hᶿ_c =-393.5kJ ∆H₃=∆Hᶿ_c =-285.7kJ x 2 ∆H₂= ∆Hᶿ_f= 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 ∆Hᶿ_f CH₄ = -74.5 kJmole^-1

Calculate the molar enthalpy of formation of ethyne (C₂H₂) given : ∆Hᶿ_c 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=∆Hᶿ_f =x C₂ H₂+2½O₂(g)

∆H₃=∆Hᶿ_c =-394kJx 2 ∆H₃=∆Hᶿ_c =-286kJ ∆H₂= ∆Hᶿ_c= -1300kJ

2CO₂(g) + H₂O(l)

Substituting:

∆H₃= ∆H₁+ ∆H₂

( -394 x 2) + -286 = -1300kJ + x

x = +244 kJ

Heat of formation ∆Hᶿ_f CH₄ = +244 kJmole^-1

Calculate the molar enthalpy of formation of carbon(II)oxide (CO) given : ∆Hᶿ_c of carbon-graphite = -393.5kJmole^-1, ∆Hᶿ_c 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=∆Hᶿ_f =x CO+½O₂(g)

∆H₃=∆Hᶿ_c =-393.5kJ ∆H₂= ∆Hᶿ_c= -283kJ

CO₂(g)

Substituting:

∆H₃= ∆H₁+ ∆H₂

-393.5kJ = -283kJ + x

x = -110 kJ

Heat of formation ∆Hᶿ_f 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 ∆H₄of ethanol

Energy cycle diagram

2C(s) + 3H₂ (g) +3½O₂(g) ∆H₃=∆Hᶿ_f =-227kJ C₂ H₅OH +3O₂(g)

∆H₂=∆Hᶿ_c =-394kJx 2 ∆H₁=∆Hᶿ_c =-286kJx 3 ∆H₄= ∆Hᶿ_c= x

2CO₂(g) + 3H₂O(l)

Substituting:

∆H₁+ ∆H₂= ∆H₃+ ∆H₄

( -394 x 2) + -286 x 3 = -277 + x

∆H₄ = -1369 kJ

Heat of combustion ∆Hᶿ_c 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 ∆Hfor 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= ∆H₂+ ∆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

(ii)Table II

∆T= T₂ -T₁ => 1.0 +2.0 = 1.5^oC

(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)

∆H₃= ∆H₁+∆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 0.5^oC 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 0.5^oC 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

(ii)Table II

∆T= T₂ -T₁ => 7.0 +7.0 = 7.0^oC

(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∆H_f 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 + ∆H_f

– 40.3kJ + -16.43kJ + -1.47kJ = +0.525kJ + ∆H_f

=>∆H_f = -58.865kJ.

Practice theoretical examples:

Using an energy level diagram calculate the ∆H_sof ammonium chloride crystals given that.

∆H_f of NH₃ (aq) = -80.54kJ mole^-1

∆H_f of HCl (aq) = -164.46kJ mole^-1

∆H_f of NH₄Cl (aq) = -261.7483kJ mole^-1

∆H_s of NH₄Cl (aq) = -16.8517kJ mole^-1

N₂(g) + 1½ H₂(g) + ½ Cl₂ ∆H_f=-261.7483kJ NH₄Cl(s) + aq

x=∆H_s

(aq) (aq)

– 80.54kJ=∆H₁ -164.46kJ=∆H₂

NH₃ (aq) + HCl(aq) 16.8517kJ=∆H₃ NH₄Cl(s)

∆H₁+ ∆H₂+ ∆H₃ = ∆H₄ + ∆H_f

– 80.54kJ + -164.46kJ + -16.8517kJ = -261.7483kJ + ∆H_f

=>∆H_f = -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 (∆H_f)

∆H₂-enthalpy/heat of atomization of sodium (∆H_at)

∆H₃ -enthalpy/heat of ionization/ionization energy of sodium (∆H_i)

∆H₄ -enthalpy/heat of atomization of chlorine (∆H_at)

∆H₅ -enthalpy/heat of electron affinity of chlorine (∆H_e)

∆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)∆H_at of sodium = + 110kJmole^-1

(iii) Electron affinity of chlorine = – 363kJmole^-1

(iv)∆H_at of chlorine = + 120kJmole^-1

(v) ∆H_f of sodium chloride = -411kJ , calculate the lattice energy of sodium chloride using an energy cycle diagram.

Working:

Applying Hess law then:

∆H_f =∆H_a +∆H_i +∆H_a +∆H_e +∆H_l

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(∆H_a) /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

A-A B-B

Reaction path/coordinate/path

Energy

Energy level diagram showing the activation energy for endothermic processes /reactions.

Activated complex

A B

∆Hr

A-B A-B

A-A B-B

Reaction path/coordinate/path

Energy

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.

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.

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

(d)Surface area

(e)Catalyst

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.

(d) correct trend (time increase as more water is added/acid is diluted) in conformity with expected theoretical results.

Sample questions

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

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^-1x 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

Write the equation for the reaction taking place

Na₂S₂O₃ (aq) + 2HCl(aq) -> 2NaCl (aq)+ SO₂ (g) + S(s) + H₂O(l)

Ionically:

S₂O₃^2- (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

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

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

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

Write an ionic equation for the reaction taking place.

Mg²⁺(s) + 2H⁺(aq) -> Mg²⁺(aq) + H₂(g)

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 10^oC 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 30^oC.

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

Plot a graph of temperature(x-axis) against ¹/_t

2(a)From your graph determine the temperature at which:

(i)1/t is ;

0.03

Reading directly from a correctly plotted graph = 32.25 ^oC

0.07

Reading directly from a correctly plotted graph = 48.0 ^oC

(ii) t is;

30 seconds

30 seconds => 1/t =1/30 =0.033

Reading directly from a correctly plotted graph 0.033 => 33.5 ^oC

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 30^oC, 40^oC , 50^oC and 60^oC 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

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

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 C₂and 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

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) + SO₄^2-(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

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(∆H_a)/activation energy (Ea) of the reactants .The catalyst lowers the Enthalpy of activation(∆H_a)/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 ∆H_rbut only lowers the Enthalpy of activation(∆H_a)/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

A-A B-B

Reaction path/coordinate/path

Energy

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

Reaction path/coordinate/path

Energy

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

b) On the same axes, plot a graph of the uncatalysed 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:

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)

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

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)

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)/CrO₄^2- ions in solution are yellow. Dichromate(VI)/Cr₂O₇^2- ions in solution are orange. The two solutions exist in equilibrium as in the equation:

2H⁺ (aq) + 2CrO₄^2-(aq) Cr₂O₇^2- (aq) + H₂O(l)

(Yellow) (Orange)

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 Cr₂O₇^2- ions formed in the solution mixture make it to be more orange in colour.

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 CrO₄^2- ions formed in the solution mixture makes it to be more yellow in colour.

2OH^– (aq) + 2Cr₂O₇^2-(aq) CrO₄^2- (aq) + H₂O(l)

(Orange) (Yellow)

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 Cr₂O₇^2-(aq)ions formed in the solution mixture makes it to be more Orange in colour.

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 Cr₂O₇^2- 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 Cr₂O₇^2- / CrO₄^2- 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:

Cr₂O₇^2-

CrO₄^2-

^OH-

_H+

When an acid is added, the equilibrium shift forward to the right and the mixture become more orange as more Cr₂O₇^2- ions exist.

When a base is added, the equilibrium shift backward to the left and the mixture become more yellow as more CrO₄^2- 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
Indicator	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 450^oC 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 450^oC 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 900^oC 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.

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.

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)50^oC

Reading from a correctly plotted graph =

(ii) 70^oC

Reading from a correctly plotted graph =

(iii) 90^oC 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:

Reduction and oxidation
Electrochemical (voltaic) cell

Electrolysis (electrolytic) cell

(i)REDUCTION AND OXIDATION (REDOX)

In teams of oxygen transfer:
i) Reduction is removal of oxygen.
ii) Oxidation is addition of oxygen.

iii) Redox is simultaneous addition and removal of oxygen.

iv) Reducing agent is the species that undergoes oxidation, therefore gains
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) + H_{2 (g)} -> Cu _(s) + H₂O _(l)

(Oxidising agent) (Reducing agent)

In terms of hydrogen transfer:
i) Oxidation is the removal of hydrogen.
ii) Reduction is the addition of hydrogen.

iii) Redox is simultaneous addition and removal of hydrogen.

iv) Reducing agent is the species that undergoes oxidation, therefore looses/ donates
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.

Cl_{2 (g)} + H₂S_(g) -> S_(S) + 2HCl _(g)

(Oxidizing agent) (Reducing agent)

In terms of electron transfer:
i) Oxidation is donation/ loss/ removal of electrons.
ii) Reduction is gain/ accept/ addition of electrons.

iii) Redox is simultaneous gain/ accept/ addition and donation/ loss/ removal of electrons.

iv) Reducing agent is the species that undergoes oxidation, therefore looses/ donates
v) Oxidizing agent is the species that undergoes reduction, therefore gains/ accepts

Example

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

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)

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)

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)

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)

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)

2Mg(s) -> 2Mg²⁺(aq) + 4e (oxidation/donation of electrons)

O₂(g) + 4e -> 2O^2-(aq) (reduction/gain of electrons)

O₂(g) + 2Mg(s) -> 2Mg²⁺(aq) + 2O^2-(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.

(a)In terms of oxidation number:
i) Oxidation is increase in oxidation numbers.
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.

iv) Reducing agent is the species that undergoes oxidation, therefore increases its oxidation number.
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
O^2-	2	-2	-2
Na⁺	1	+1	+1
Al³⁺	3	+3	+3
P^3-	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:

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

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

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

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;

a) SO₄^2- 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 SO₄^2- = ( +6 + (-2 x 4)) = -2

The chemical name of this radical is thus sulphate(VI) ion

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

Sulphur in;

-SO₃^2- => 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;

-CO₃^2- = 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

-Cr₂O₇^2- => 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

-CrO₄^2- => 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) + Cr₂O₇^2-(aq) -> 6Fe³⁺(aq) + Cr³⁺(aq) + 7H₂O (l) +2 +1 +6 -2 +3 +3 +1 -2

Oxidizing agent:

Cr in Cr₂O₇^2-;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) Cr₂O₇^2-(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 Cr₂O₇^2- 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 Cr₂O₇^2- 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

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)

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

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 1^st metal rod (A)

Changes on the 2^nd metal rod (B)

Changes on the 1^st solution (A(aq))

Changes on the 2^nd 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

Zinc rod ionizes /dissolves to form Zn²⁺ ions at the negative terminal

Zn(s) -> Zn²⁺(aq) + 2e

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

Magnesium rod ionizes /dissolves to form Mg²⁺ ions at the negative terminal

Mg(s) -> Mg²⁺(aq) + 2e

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

Magnesium rod ionizes /dissolves to form Mg²⁺ ions at the negative terminal

Fe(s) -> Fe²⁺(aq) + 2e

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.

Fe++

(iv)Ag/Cu cell

Copper rod ionizes /dissolves to form Cu²⁺ ions at the negative terminal

Cu(s) -> Cu²⁺(aq) + 2e

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++

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 25^oC

(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 Cu^{2+ (}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

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/ SO₄^2- (aq) from sulphuric (VI)acid to form insoluble Lead(II) sulphate (VI).

Pb²⁺ (aq) + SO₄^2- (aq) -> PbSO₄(s)

5.The overall cell reaction is called discharging

PbO₂(s) +Pb(s) + 4H⁺(aq) + 2SO₄^2- (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.

During recharging, the electrode reaction is reversed as below:

2PbSO₄(s) + 2H₂O(l) ->PbO₂(s) +Pb(s) + 4H⁺(aq) + 2SO₄^2- (aq)

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

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)

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

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

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.

The below set up shows an electrolytic cell.

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 / Cl^–donate/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 / Br^–donate/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 / Cl^–donate/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) + 3O^2-(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(+);

6O^2-(l) -> 3O₂ (g) + 12e

(Anion /6O^2- 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.

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:

Position of cations and anions in the electrochemical series

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(SO₄^2- ,SO₃^2- ,CO₃^2- ,HSO₄^– ,HCO₃^–,NO₃^–,PO₄^3-)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

SO₄^2- ,SO₃^2- ,CO₃^2- ,HSO₄^– ,HCO₃^–,NO₃^–,PO₄^3- 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:

Write the equation for the decomposition of the electrolytes during the electrolytic process.

H₂O(l) -> OH^– (aq) + H⁺(aq)

H₂ SO₄(aq) -> SO₄^2-(aq) + 2H⁺(aq)

Name the ions in acidified water that are attracted/move to:

Cathode- H⁺(aq) from either sulphuric(VI) acid (H₂ SO₄) or water (H₂O)

Anode– SO₄^2-(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

(4OH^–ions selectively discharged instead of SO₄^2- ions at the anode)

Name the products of electrolysis of acidified water.

Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound

Anode–Oxygen gas (colourless gas that relights /rekindles glowing splint)

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.

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.”

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:

Write the equation for the decomposition of the electrolytes during the electrolytic process.

H₂O(l) -> OH^– (aq) + H⁺(aq)

Mg SO₄(aq) -> SO₄^2-(aq) + Mg²⁺(aq)

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)

Anode– SO₄^2-(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

(4OH^–ions selectively discharged instead of SO₄^2- ions at the anode)

Name the products of electrolysis of Magnesium sulphate(VI) solution

Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound

Anode–Oxygen gas (colourless gas that relights /rekindles glowing splint)

Explain the difference in volume of products at the cathode and anode.

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.

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.

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

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:

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)

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)

Anode– Cl^–(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

(4OH^–ions selectively discharged instead of Cl^– ions at the anode)

Name the products of electrolysis of dilute sodium chloride solution

Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound

Anode–Oxygen gas (colourless gas that relights /rekindles glowing splint)

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.

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.

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:

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)

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)

Anode– Cl^–(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

(Cl^–ions with a higher concentration block the discharge of OH^– ions at the anode)

Name the products of electrolysis of concentrated sodium chloride solution/brine

Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound

Anode–Chlorine gas(pale green gas that bleaches damp/moist/wet litmus papers)

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.

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

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:

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)

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)

Anode– Cl^–(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

(4OH^–ions selectively discharged instead of Cl^– ions at the anode)

Name the products of electrolysis of dilute Hydrochloric acid

Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound

Anode–Oxygen gas (colourless gas that relights /rekindles glowing splint)

Explain the difference in volume of products at the cathode and anode.

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.

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:

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)

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)

Anode– Cl^–(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

(OH^–ions concentration is low.Cl^– ions concentration is higher at the anode thus cause over voltage/block discharge of OH^–ions)

Name the products of electrolysis of 2M Hydrochloric acid

Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound

Anode–Chlorine gas (Pale green gas that bleaches blue/red moist/wet/damp litmus papers)

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.

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.

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

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:

Write the equation for the decomposition of the electrolytes during the electrolytic process.

H₂O(l) -> OH^– (aq) + H⁺(aq)

CuSO₄(aq) -> SO₄^2-(aq) + Cu²⁺(aq)

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)

Anode– SO₄^2-(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

(OH^–ions ions are higher than SO₄^2-ions in the electrochemical series therefore selectively discharged at the cathode.))

Name the products of electrolysis of 1M copper(II) sulphate(VI) solution

Cathode-2 moles of copper metal as brown solid coat

Anode–Oxygen gas (Colourless gas that relights /rekindles glowing splint)

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

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.

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.

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:

Write the equation for the decomposition of the electrolytes during the electrolytic process.

H₂O(l) -> OH^– (aq) + H⁺(aq)

CuSO₄(aq) -> SO₄^2-(aq) + Cu²⁺(aq)

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)

Anode– SO₄^2-(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 OH^–ions and SO₄^2-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 OH^–ions.

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

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 1^st law of electrolysis.

Faradays 1^st 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

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

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

= 48250 seconds

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

2O^2– 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

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

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

(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

(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

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

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

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

Practically Faraday 1^st law of electrolysis can be verified as below.

Verifying Faraday 1^st 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:

Write the equation for the decomposition of the electrolytes during the electrolytic process.

H₂O(l) -> OH^– (aq) + H⁺(aq)

CuSO₄(aq) -> SO₄^2-(aq) + Cu²⁺(aq)

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)

Anode– SO₄^2-(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 OH^–ions and SO₄^2-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 OH^–ions.

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

(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

(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 2400 = 121920 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

(iii) The faraday constant obtained above is far lower than theoretical.Explain

-high resistance of the wires used.

-temperatures at 25^oC 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

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 ^-3moles

Molar mass 63.5

9.4488 x10 ^-3moles -> 1800 coulombs

1 Mole -> 1 x 1800 coulombs = 190500.381 coulombs

9.4488 x10 ^-3moles

(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

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

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

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)

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 OH^–ions 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.

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

Diagram showing the Manufacture of Sodium hydroxide from the flowing Mercury-cathode cell.

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

Raw materials

(i)Rock salt/solid sodium chloride

(ii)calcium(II)chloride

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 800^oC. A little calcium (II) chloride is added to lower the melting point of the electrolyte to about 600^oC.

The molten electrolyte is the electrolyzed in a carbon graphite anode suspended at the centre and surrounded by steel cathode.

Questions

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.

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

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

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.

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.

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

= 1,930,000,000 / 1.93 x 10⁹C

Time(t) in seconds = Quantity of electricity Current(I) in amperes

Substituting

= 1,930,000,000 / 1.93 x 10⁹C

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 10⁹C->1,930,000,000 / 1.93×10⁹C 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

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 10⁹C ->

1,930,000,000 / 1.93 x 10⁹C 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 = 0.4M / 0.4 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)

Na₂O₂Sodium 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) Al₂O₃(s) + 3H₂O(l)

Pure aluminium (III)oxide (Al₂O₃) has a very high melting point of 2015^oC.

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 800^oC.

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;

6O^2-(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- Fe₂O₃as 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 1000^oC

Cryolite

Electrolysis

Oxygen gas at anode

Pure aluminium sinks in Hall cell

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.

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.

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:

Hydrogen/H₂

Aluminium sulphate(VI)/Al₂(SO₄)₃

Aluminium hydroxide/ Al(OH₄)₃

Tetrahydroxoaluminate(III)/ [Al(OH₄)₃]^–

Aluminium hydroxide/ Al(OH)₃

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 2015^oC to 900^oC 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

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 (827^oC) 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(1650^oC).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(500^oC) 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.

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 .

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;

I) an insoluble green precipitate in excess of 2M sodium hydroxide /2M aqueous ammonia if Fe²⁺ ions are present.
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

Copper pyrites(CuFeS₂) ore with impurities Fe₂O₃ and SiO₂

Froth flotation

Crush (increase surface area)

Concentration chamber

1^st roasting chamber

Silicon(IV)

oxide

Smelting furnace

2^nd roasting furnace

Calcium aluminate (CaAl₂O₄)slag

Limestone

Sulphur(IV)Oxide

Iron Silicate (FeSiO₃)Slag

Anode; Impure Copper eroded.

Cathode; Pure Copper deposited.

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

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.

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.

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;

I) an insoluble blue precipitate in excess of 2M sodium hydroxide if Cu²⁺ ions are present.
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

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

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

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)

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.

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 SO₄^2-, SO₃^2-, CO₃^2-, 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

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

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 800^oC-> 600^oC)		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 2015^oC -> 800^oC)	Cathode: 4Al³⁺(l) + 12e -> 4Al(l) Anode: 6O^2-(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)

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

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 2015^oC while that of sodium is about 98^oC.This is mainly because aluminium has more/three delocalized electrons than sodium/has one.

Aluminium has a boiling point of about 2470^oC while that of sodium is about 890^oC.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 2015^oC to 900^oC

(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)₄and 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 2O^2-(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.

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

Anode used

Aluminium sheet

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

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

HALF-LIFE PERIOD AND DECAY CURVES

D .CHEMICAL vs NUCLEAR REACTIONS

E .APPLICATION OF RADIOACTIVITY AND RADIO ISOTOPES.

DANGERS OF RADIOACTIVITY AND RADIO ISOTOPES.
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 ¹⁶₈O 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 ²³⁷₉₂U 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)

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.

iv) have high ionizing power thus cause a lot of damage to living cells.
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 -> ^x_y W + 10 α

²¹⁰ ₈₄ U -> ¹⁷⁰₆₄ W + 10 α

²¹⁰ ₉₂U -> ^x_y 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.

iv) have medium ionizing power thus cause less damage to living cells than the α particle.
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

²³11 Na -> ²³₁₂Mg + ⁰ _-1e

²³⁴ x Th -> ^y₉₁ Pa + ⁰ _-1e

²³⁴ 90 Th -> ^y₉₁ Pa + ⁰ _-1e

²⁰⁷₇₀Y -> x _y Pb + 3⁰ _-1e

²⁰⁷₇₀Y -> ²⁰⁷ ₇₃Pb + 3⁰ _-1e

x_y C -> ¹⁴₇N + ⁰ _-1e

14₆ C -> ¹⁴₇N + ⁰ _-1e

¹ x n -> ^y₁H + ⁰ _-1e

¹ 0 n -> ¹₁H + ⁰ _-1e

⁴₂He -> 4¹₁H + x ⁰ _-1e

⁴₂He -> 4¹₁H + 2 ⁰ _-1e

²²⁸₈₈Ra -> ²²⁸₉₀Th + x β

²²⁸₈₈Ra -> ²²⁸₉₂Th + 4 β

²³²₉₀Th -> ²¹²₈₂Pb + 2 β + x α

²³²₉₀Th -> ²¹²₈₂Pb + 2 β + 5 α

²³⁸₉₂U -> ²²⁶₈₈ Ra + x β + 3 α

²³⁸₉₂U -> ²²⁶₈₈ Ra + 2 β + 3 α

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..

iv) have very low ionizing power thus cause less damage to living cells unless on prolonged exposure..
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 + ⁴₂ He²⁺

²³⁸ ₉₂ U -> ²³⁴₉₀ T + α

(ii)²³⁸ ₉₂ U to ²²²₈₄ Rn

²³⁸ ₉₂ U -> ²²²₈₄ Rn + 4 ⁴₂ He²⁺

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 ⁴₂ 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 ⁴₂ 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 ⁴₂ He

Using Mass numbers only

238 = 206 + 4y => 4y = 238 – 206 = 32

y = 32 = 8 α

Using atomic numbers only and substituting the 8 α(above)

²³⁸ ₉₂ U -> ²⁰⁶ ₈₂ Pb + 8 ⁴₂ 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 ⁴₂ 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 ⁴₂ He + y ⁰_-1 e

Using Mass numbers only

298 = 214 + 4x => 4x = 298 – 214 = 84

y = 84 = 21 α

Using atomic numbers only and substituting the 21 α (above)

²³⁸ ₉₂ U -> ²¹⁴ ₈₃Bi + 21 ⁴₂ 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 ⁴₂ 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:

a) nuclear fission
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 + ²⁸_a Al -> ^b₁₁ Na + ⁴₂He

^a₀n + ¹⁴₇ N -> ¹⁴_b C + ¹₁ H

¹₀n + ¹₁ H -> ²₁ H + a

¹₀n + ²³⁵ ₉₂ U -> ⁹⁵ ₄₂ Mo + ¹³⁹ ₅₇ La + 2¹₀ n + 7 a

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 -> ^a_bHe + ¹₀ n

²₁H + a -> ³₂He

²₁H + ²₁H -> a + ¹₁ H

4 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

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

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

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 6^th half life 0.03125 g of nitrogen-13 remain.

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 3^rd half-life ¹/₈ of the gas remain

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

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

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 ¹/₂changes 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.

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.

c) Food preservation:

X-rays are used to kill bacteria in tinned food to last for a long time.

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.

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

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 10¹⁸atoms

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—1^st –>5g—2^nd–>2.5—3^rd –>1.25—4^th –>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?

(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 t¹ /₂ in 112 days = 4

t¹ /₂ = 112 = 28 days

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.

Alpha–Mass 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

Beta–Mass 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 ⁴₂He

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 ²¹⁴₈₃Bi and ending at ²⁰⁶₈₂Pb. 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 + ⁴₂ He

(b) step 1 to 3

²¹⁴₈₃Bi ->²¹⁰₈₁Bi + ⁴₂ He + 2 ⁰_-1 e

(c) step 3 to 5

²¹⁰₈₂Pb ->²⁰⁶₈₂Pb + ⁴₂ He + 2 ⁰_-1 e

(c) step 1 to 5

²¹⁴₈₃Bi ->²⁰⁶₈₂Pb + 2 ⁴₂ 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:

Half – life of the Bismuth isotope
Original mass of the Bismuth isotope given that the mass that remained after 70 minutes was 0.16g (2mks)

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 Z₂in 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

Identify the rays S,R and Q

S- Beta ( β )particle/ray

R- Alpha (α )particle/ray

Q- Gamma (y )particle/ray

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

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