Education Cabinet Secretary George Magoha inspects 2019 KCSE Chemistry Practical exams in a Nairobi school. This year's Chemistry practical has been doffed by cheating claims and assertions that one of the reagents used is very poisonous.
Form 4 Chemistry Notes

ORGANIC CHEMISTRY II: ALCOHOLS AND ALKANOIC ACIDS:

1. Alkanols

  • Nomenclature of alkanols
  • Primary secondary and tertiary alcohols
  • Preparation and properties of alcohols
  • Hydrolysis of halogenoalkanes
  • Hydration of alkenes
  • Fermentation of sugars and starches (Ethanol)
  • Physical properties of alcohols
  • Chemical properties of alcohols
  • Combustion
  • Reaction with metals (sodium)
  • Esterification
  • Oxidation
  • Reaction with acidified potassium dichromate
  • Reaction with acidified potassium permanganate
  • Reaction with copper metal
  • Dehydration reactions of alcohols
  • Uses of alkanols

 

  1. Alkanoic acids
  • Nomenclature
  • Preparation and properties of alkanoic acids
  • Physical properties
  • Chemical properties
  • Reaction with sodium carbonate
  • Reaction with sodium hydroxide
  • Reaction with magnesium metal
  • Esterification
  • Uses of ethanoic acid

 

  1. Fats and oils

 

  1. Soaps and soapless detergents
  • Soaps
  • Preparation of soaps
  • Role of soap in cleaning
  • Effect of hard water on soap
  • Removal of hardness
  • Temporary hardness
  • Permanent hardness

 

  • Soapless detergents
  • Preparation
  • Advantages

 

  1. Polymers
  • Natural and synthetic polymers
  • Addition polymerization
  • Condensation polymerization

ALCOHOLS (ALKANOLS):

– They derivatives of alkanes in which a hydrogen has been replaced by a hydroxyl group (OH); which is the functional group.

– They form a homologous series of the general formula CnH2n+1OH, in which the OH can also be denoted as ROH, where R is an alkyl group.

Note: alkanol is the IUPAC name while alcohol is the common name;

 

Nomenclature of alcohols

  1. The ‘e’ of the corresponding alkane molecule is replaced with the suffix – ol;
  2. The parent molecule is the longest chain containing the – OH group;
  3. The numbering of carbon atoms is done such that the carbon atom with the hydroxyl group –OH- attains the lowest possible number
  4. The constituent branch is named accordingly;

 

Examples:                                         

CH3                                           (ii). CH3CH2CH2CH2OH                    (iii). H     OH     H

|                                                 Butan-1- ol;

CH3-C-OH                                                                                          H     C       C       C        H

|

CH3                                                                                                                   H       H      H

2,2 dimethylpropanol-1-ol;                                                                                                      Propan-2-ol;

 

(iv).                  CH3                 (v). H     H      OH     H     H             (vi).          H      H

 

CH3     CH2    CCH3               H    C     C        C      C      C    H               H    C       C      OH

 

OH                         H     CH2   H     H      H                             H       H

2 methyl butan-2-ol;                                                                                                Ethanol;

CH3

3,4 ethyl methylhexan-3-ol;

 

Isomerism in alkanols.

– Alcohols exhibit positional isomerism due to the fact that the position of attachment of the functional group varies within the carbon chain;

 

Examples of isomeric alcohols

(a). Isomers of propanol

(i).      H     OH     H                            (ii).      H     H            H

 

H     C     C         C     H                       H     C     C     O    C     H

 

H     H        H                                       H     H            H

Propan-2-ol                                                             Ethyl methyl ether or methoxyethane;

 

(iii).      H     H     H                             (iv).

 

H     C     C     C     O     H

 

H     H    H

Propan-1-ol;

Note: Ethyl methyl ether is not actually an alcohol as it lacks the –OH group;

– All have the molecular formula: C3H7OH

 

(b). Isomers of butanol

(i). CH3CH2CH2OH                (ii). CH3CH2CHCH3               (iii).            CH3

                  Butan-1-ol;                                           

OH                            CH3     C      OH

                                                            Butan-2-ol;

CH3

                                                                                                2,2-dimethyl propan-2-ol

Note: All have the molecular formula: C4H9OH

 

(c). Draw the Isomers of pentanol, C5H11OH;

(i).       H    H    H          H   H                   (ii).                                   (iii)                               (iv)

 

H – C – C – C – O – C – C – H

 

H    H   H           H    H

                      Ethylpropylether

 

Primary, secondary and tertiary alcohols.

  • Primary alcohols:

– Are alcohols in which the OH group is attached to a carbon atom to which 2 hydrogen atoms are attached;

– Thus they contain a –CH2OH group;

Examples:

H    H    H    H

 

H – C – C – C – COH

                      H     H   H    H

    Butan-1-ol;

 

  • Secondary alcohols:

– Are alcohols in which the hydroxyl group is attached to the carbon atom to which only one other hydrogen atom is attached;

– The carbon atom with the OH group is thus bonded to two carbon atoms;

– They contain a CHOH group;

Example:

H    H    H    H

 

H – C – C – C – C – H

                      H     H   OH H

Butan-2-ol;

 

  • Tertiary alcohols:

– The hydroxyl group is attached to a carbon atom with no hydrogen atoms attached

– The carbon atom with the OH group is bonded to 3 other carbon atoms; hence sorrounded by the methyl groups;

– Tertiary alcohols thus a contain a –COH group;

Examples:

H   OH H

 

H – C – C – C – H

          H    CH3   H

         2-methyl propan-2-ol;

Preparation and properties of alkanols.

– Alkanols are prepared from three main methods.

  • Hydrolysis of halogenoalkanes;
  • Hydration of alkenes
  • Fermentation of starches and sugars (mainly for ethanol)

 

(a). Hydrolysis of halogenoalkanes;

– Halogenoalkanes are compounds in which one or more hydrogen atoms in an alkane are replaced by halogens;

– Addition of aqueous KOH or NaOH to a halogenoalkane and heating results to corresponding alcohol;

– Reaction involves replacement of the halogen atoms with the -OH from the alkali;

 

Examples:

(i). Preparation of methanol

CH3Cl     +       NaOH                  heat              CH3OH    +    NaCl

Chloromethane          sodium hydroxide;                           methanol                sodium chloride

 

(ii). Preparation of propanol;

CH3CH2CH2Br    +      KOH              heat      CH3CH2CH2OH       +     KBr

1-bromopropane                          potassium hydroxide                       Propan-1-ol                                      Potassium bromide

 

Note: –the conversion of a halogenoalkane to an alcohol is known as hydrolysis;

– Reagent in this case is an alkali and condition for reaction is heat;

 

(b). Hydration of alkenes.

– Conversion of an alkene to an alcohol is known as hydration;

– Main reagent for the reaction is water;

– Conditions for the reaction are:

  • An acid catalyst, mainly conc. H2SO4 or phosphoric acid (H3PO4);
  • High temperatures of about 80oC;
  • High pressures of about 25-30 atmospheres;

 

Examples:

(i). Preparation of ethanol from ethene.

 

C2H4   +  H2O      Conc. H2SO4           C2H5OH;

80oC; 25-30 atm;

H     H                                                         H    H

 

 

H     C  = C     H     +     H     OH               H     C     C     H

Ethane                                   Water

H     OH

Ethanol

(ii). Preparation of butanol from butene.

 

CH3CHCHCH3   +    H2O      Conc H3PO4         CH3CH2CH2CH2OH

       Butene                                   80oC, 25-30 atm                              Butanol

 

(c). Preparation (of ethanol) by fermentation.

It is prepared from the fermentation of starches or sugars in the presence of yeast;

Fermentation: Is a chemical decomposition brought by bacteria or yeast (anaerobically) usually accompanied by evolution of carbon (IV) oxide and heat.

 

The chemical process

– Starch is broken into sugars by the action of the enzyme amylase or diastase;

Break up into

Starch molecule + water                                    sucrose molecules

Amylase

(C6H10O5) n  +  nH2O                         nC6H12O6

      Starch                      Water                                  many sucrose molecules

 

– When yeast is added to dilute sucrose solution (ordinary sugar); the enzyme sucrase in yeast catalytically breaks down sugar (sucrose) into the simplest sugars, glucose and fructose i.e.

 

Equation:

C12H22O11(aq) +  H2O(l)        Sucrase             C6H12O6(aq) +  C6H12O6

     Sugar                          Water                                                     glucose                      fructose

 

– Finally the enzyme zymase, also produced by yeast converts glucose and fructose into ethanol and carbon (IV) oxide.

 

Equation:

C6H12O6 (aq)        Zymase         2C2H5OH(aq) + CO2

Glucose/fructose                                    ethanol                         carbon (IV) oxide

 

Optimum conditions for fermentation:

  • Temperatures of 25-30oC;
  • Yeast catalyst;
  • Absence of oxygen (airtight);

 

Note:

– When the reaction mixture contains about 12% by volume of ethanol, the activity of yeast ceases.

– This is because higher ethanol concentrations kill the yeast cells;

– Fermentation provides about 10% alcohol by volume;

– The concentration of resultant ethanol can be increased by fractional distillation.

– During the process, ethanol distills over fast due to its lower boiling point (78°C)

– The distillate at below 95°C is first collected (leaving water behind).

– The resultant fraction will have 95% alcohol by volume; and is called rectified spirit;

– Absolute ethanol; which is 99.5% by volume can be obtained by re-distillation of rectified ethanol between 78-82oC to remove all the water in the mixture;

 

– This can be done in two main ways:

  • Addition of a small amount of benzene to the rectified spirit and then distilling; (benzene dissolves in the water in the alcohol)
  • Distillation of rectified spirit over a suitable drying agent like calcium oxide and then over calcium; (calcium reacts with steam, calcium oxide takes in condensed water)
Properties of alcohols (Ethanol)

(a) Physical properties

(i). It is a colourless, volatile liquid soluble in water in all proportions forming a neutral solution;

(ii). Has a characteristic smell and boils at78.5°C

 

Variation in physical properties of alkanols.

 

Name Molecular formula Molecular mass Boiling point (oC) Melting point (oC) Solubility in 100g of water
Methanol CH3OH 32 64.5 -94 Soluble
Ethanol C2H5OH 46 78.5 -117 Soluble
Propanol C3H7OH 60 97 -127 Soluble
Butanol C4H9OH 74 117 -90 Slightly soluble
Pentanol C5H11OH 88 138 -79 Slightly soluble
Hexanol C6H13OH 102 158 -52 Slightly soluble
Heptanol C7H14OH 116 175 -34.6 Very slightly soluble
Octanol C8H16OH 130 194 -16 Very slightly soluble;

 

Note:

– Solubility of alkanols decreases with increase in molecular mass;

– Both melting and boiling points increases with increase in the relative molecular mass; due to progressive increase in number of van der waals forces;

– Alkanols have higher melting and boiling points than their corresponding alkanes with the same molecular formula;

Reason:

– Alkanols have hydrogen bonding between their molecules, caused by the presence of the OH group; alkanes have van der waals between its molecules; Hydrogen bonds are stronger than weak van der waals;

 

Diagram: Hydrogen bonding between four ethanol molecules.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(b) Chemical properties /main reactions of ethanol.

Note; the main reactions of ethanol are those of its functional group –OH;

 

(iii). Combustion

Procedure;

– A few drops of ethanol are placed in a watch glass and lit.

– A dry gas jar is held over the flame and the gas collected tested with limewater.

 

Observation:

– It burns with a blue flame, which is almost colourless.

– The resultant gas turns limewater into a white precipitate, indicating it is carbon (IV) oxide.

 

Explanations:

– Ethanol (alcohols) burns in air (oxygen) producing carbon (IV) oxide, water and heat energy;

– The lower members of the homologous series burn with a blue or non-luminous flame leaving no residue;

– As the hydrocarbon chain increases the flame becomes more luminous and smoky and a black residue remains;

 

Equation

C2H5OH(l) +  3O2(g)                                         2CO2(g) + 3H2O(l)

 

Note:

– If an alkanol is burnt in a limited supply of oxygen, then the combustion is incomplete and the products include carbon (II) oxide or carbon and water

 

Equations

(i). C2H5OH(l) +  2O2(g)                                   2CO(g) + 3H2O(l)

 

(ii). 2C2H5OH(l) +  2O2(g)                                4C(s) + 6H2O(l)

 

(iv). Reaction with metals (sodium);

Procedure:

– Tiny pieces of sodium one at a time; are added 1cm³ of pure ethanol in a boiling tube.

 

Observation;

– Sodium metal darts on the surface of the ethanol and then dissolves /disappears;

– The beaker becomes warmer indicating an exothermic reaction.

Effervescence occurs and bubbles of a colourless gas are observed; gas burns with a pop sound.

 

Explanation:

– Sodium reacts with alcohol much as it does with water but the reaction is more gentle.

– Sodium reacts with ethanol to produce hydrogen gas, which on testing burns with a pop sound.

– The reaction is exothermic producing heat hence warmer beaker.

– A clear solution of sodium ethoxide is left formed in the boiling tube.

 

Equation;

2C2H5OH + Na (l)                                   2CH3CH2ONa(l)  + H2(g)

 

Note:

Alkanols react with electropositive metals such as sodium, potassium and aluminium to liberate hydrogen gas and form a solution of the metal salt, the metal alkoxide;

 

Examples

(i). 2CH3CH2CH2OH  +  Na                           2CH3CH2CH2ONa  +  H2

       Propanol                                 Sodium metal                           Sodium propoxide                         hydrogen

 

(ii). CH3CH2CH2CH2CHOH  +  K                             2CH3CH2CH2CH2CHOK  + H2

          Pentanol                                        Potassium                                        Potassium pentoxide                       Hydrogen

 

Note: The reactivity of alkanols with metals decreases as the hydrocarbon chain increases;

 

(v). Esterification;

– It is the production of esters (alkyl alkanoates) from the reaction between alcohols and carboxylic acids;

 

Procedure

– 2-3 drops of concentrated sulphuric acid are added to a mixture of equal proportions of ethanol and pure ethanoic acid in an evaporating dish.

– The mixture is warmed gently in a water bath for sometime.

– The mixture is poured into a beaker and smelt;

 

Observation

A fruity sweet smell (of ethyl ethanoate);

 

Explanation

– Ethanol reacts with ethanoic acid in the presence of a few drops of concentrated sulphuric acid to form ethyl ethanoate and water.

– The reaction is very slow and so catalyzed by the hydrogen ions from the sulphuric acid.

 

Equation                                       Conc. H2SO4; warm

C2H5OH(l) +  CH3COOH (aq)                                CH3COOC2H5(l)   +   H2O(l)

Ethanol                     ethanoic acid                                                                 ethyl ethanoate (ester)

 

Structurally;

H     H                          H    O                                 H     O            H     H

Conc. H2SO4

H        C      C     OH  +   H     C    C                         H     C     C     O     C     C    H   +   H2O

                                                                                                                                             Water

H     H                          H    OH                              H                    H      H

Ethanol                                                          Ethanoic acid                                                           Ethylethanoate                     

 

– The alkyl part of the ester is derived from the alkanol, while the alkanoate part is derived from the alkanoic acid;

– The alkanol attaches itself at the group in the carboxylic acid thereby displacing hydrogen atom;

– Under ordinary conditions the reaction takes place slowly; but in presence of concentrated sulphuric acid which act as a catalyst and warm (heat) conditions, the reaction is enhanced;

 

 

 

Further examples:

Write balanced equations for each of the following esterification reactions:

(i). Ethanol and propanoic acid;

 

(ii). Propanol and ethanoic acid;

 

(iii). Ethanol and methanoic acid;

 

Note:

– Generally a reaction between alcohol and a carboxylic acid (-COOH-) produces an ester and water in a process called Esterification;

 

General equation:

Conc. H2SO4, warm

Organic acid +  alcohol                               ester  + water….(Esterification), while;

 

 

Inorganic acid   + alkali                                        salt +  water….(neutralization)

 

Differences between neutralization and Esterification

  1. Esterification is slower than neutralization as the reaction is between molecules and not ions as in neutralization
  2. Esterification is reversible; the forward reaction is esterification and the backward reaction is hydrolysis.
  3. Esterification results to esters which are covalent compound; neutralization forms salts which are electrovalent.

 

Name and formulae of some common esters.

 

Alkanol Alkanoic acid Ester
Methanol

CH3OH

Propanoic acid

O

 

CH3CH2     C     OH

Methylpropanoate

O

 

CH3CH2     C     OCH3

 

Ethanol

C2H5OH

Methanoic acid

O

 

H     C     OH

Ethylmethanoate

O

 

H     C     OCH2CH3

Propan-1-ol Ethanoic acid

 

 

 

Butan-1-ol Ethanoic acid

 

 

 

 

 

 

(vi). Oxidation of primary alkanols;

– On heating in presence of oxidising agents, primary alkanols are oxidised to alkanoic acids;

Note:

– During oxidation of alcohols, they first lose hydrogen to form compounds called aldehydes (compounds ending in –al)

– The resultant aldehydes (alkanal) then gain oxygen to form alkanoic acids.

 

Equations:

(i). Alkanol – hydrogen                 alkanal + water

Then;

(ii). Alkanal +  oxygen                alkanoic acid;

 

General equation:

H                                                         O

 

R     C     OH  +   2 [O]                        R     C     OH   +   H2O

From oxidizing agent

H

 

 

Example: Oxidation of ethanol

– Ethanol like all other alcohols is oxidized by strong oxidizing agents such as potassium dichromate (VI) and potassium manganate (VII) to form ethanoic acid.

 

(a). Reaction with acidified potassium dichromate (VI):

Procedure

– A little solution of acidified potassium dichromate (VI) is added to a little solution of ethanol in a test tube and then warmed gently;

 

Observation:

– The solution (dichromate) changes from yellow to green;

 

Explanation

– The acidified potassium dichromate (VI) oxidizes the ethanol to ethanal then to ethanoic acid, while the dichromate undergoes reduction (chromate (VI) to Cr3+ changing colour from yellow to green;

 

Equations

O

H+ from Conc. H2SO4

C2H5OH(l)    +   [ O ]                                           CH3CH(aq)   +   H2O(l)

Ethanol                           From                        Slow reaction          Ethanal                          water

K2Cr2O7

 

Structurally:

H     H                                                                 H    H

 

H     C      C     OH   +    [ O ]                             H     C     C    O   +   H2O

 

H     H                                                                  H

Then;

O                                                                     O

 

CH3CH(aq)  +   [ O ]                                         CH3C     OH

Ethanal                                       H+ reaction                     Ethanoic acid

General equation

H     H                                                                        O

 

H     C     C    OH(l)   +   2 [ O ]                                   CH3C   OH   +   H2O

 

H     H

(b). Reaction with acidified potassium manganate (VII).

Procedure

Acidified potassium permanganate solution is added to ethanol in a test tube end the mixture is warmed gently.

 

Observation

The  permanganate solution turns from purple to colourless.

The characteristic smell of ethanoic acid is felt.

 

Explanation

The ethanol is oxidized to ethanal then to ethanoic acid. The reduced permanganate decolorizes (turns from purple to colourless);

-The H+/KMnO4 is decolourised due to reduction of manganate (VII) ions to Mn2+

 

General equation:

 

H     H                                                                        O

 

 

H     C     C    OH(l)   +   2 [ O ]                                   CH3C   OH   +   H2O

H+/ KMnO4

H     H

 

 

(c). Catalytic oxidation of alkanols.

– Catalytic oxidation of alkanols results to a dehydrogenation (removal of hydrogen) resulting to the formation of an alkanal (aldehyde);

– These are compounds with the functional group COH;

 

Example: Catalytic oxidation of ethanol with hot copper metal

When ethanol is passed over heated copper at about 300 °C, it is dehydrogenated i.e. hydrogen is removed.

This results into formation of an ethanal; and hydrogen gas is liberated.

 

Equation                       Cu(s)

CH3CH2OH(g)                                    CH3CHO(l)   +   H2(g)

Ethanol                             250oC             Ethanal

(vii). Dehydration reactions of alcohols.

– Dehydration is the removal of water molecules from a compound;

– Excess concentrated sulphuric (VI) acid dehydrates alkanols and forms corresponding alkenes.

 

Conditions required: –

– High temperatures of 140-180°C.

– Catalysts such as conc. sulphuric acid, phosphoric acid and aluminium oxide;

 

Example: Dehydration of ethanol

(i). Apparatus

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(ii). Procedure

– 15cm3 of absolute ethanol are put in around bottomed flask and 5cm3 of con H2SO4 added.

– The contents are mixed thoroughly by swirling the flask.

– The flask is then heated (warmed) gently with little shaking for about 1 minute.

– The gas collected is tested with acidified potassium manganate (VII) and bromine water.

 

(iii). Observations

– Evolution of a colourless gas that decolourises the purple acidified potassium manganate (VII)

– The resultant gas also decolourises the red brown bromine water.

 

(iv). Explanations

– On heating the alkanol (ethanol) undergoes an elimination reaction.

– It loses both a hydrogen (H) and a hydroxyl (OH) from two adjacent carbon atoms.

– The H and OH combine to form water; and the remnants form ethene.

– The sulphuric acid acts as a catalyst.

 

Note: Disadvantage of concentrated sulphuric acid over phosphoric acid.

– Being a strong oxidising agent the concentrated sulphuric acid oxidizes some of the alkanol formed to CO2 and it is itself reduced to SO2;

– This reduces the volume and purity of resultant alkanol; and is also a potential source of pollution;

 

 

Equation

H     H                                                           H    H

Conc. H2SO4

H     C     C     H                                           H     C = C     H   +   H2O

170oC

H     OH

 

Note:

With cold concentrated sulphuric (VI) acid, alkanols react to form alkyl hydrogen sulphates.

 

Example: Ethanol and cold conc. sulphuric acid.

 

C2H5OH   +  H2SO4(l)                         C2H5HSO4(l)  +   H2O(l)

Ethanol                                                                         Ethyl hydrogen sulphate

                                                                                 (Ethoxyethane)

Uses of ethanol

  1. Used as solvents; in the preparation of drugs, perfumes, liquors, vanish and paints.
  2. Source of fuel e.g ethanol when blended with gasoline to form gasohol.

Note:

– Addition of ethanol to petrol improves the antiknock propertied of petrol, due to its low ignition point;

– The alkanol also absorbs any traces of moisture that may enter and damage the petrol system;

– Alkanols also ignite on their own to liberate heat;

Example:

CH3CH2OH(l)   +   3O2(g)                                 2CO2(g)   +   3H2O(l)  ∆H = -1368KjMol

 

  1. Starting material for the manufacture of polyvinyl chloride (P.V.C)
  2. Ethanol is used as a disinfectant (antiseptic) at special concentrations e.g for cleaning tissues and surgical equipments during operations and in dressing wounds;
  3. Manufacture of alkanoic acids e.g ethanoic acid
  4. Ethanol is used as thermometer liquid for measuring low temperatures;
  5. Manufacture of alcoholic drinks
  6. Large amounts of methanol are used in the manufacture of formaldehyde a chemical used in preservation of corpses;
  7. They are used as antifreeze mixtures in car radiators e.g a mixture of ethanol and water freezes at lower temperatures that pure water;
  8. Manufacture of esters to giver fruity flavourings for confectionery and drinks;

 

Tests for primary alkanols.

  1. Reaction with sodium metal

– Alkanols liberate hydrogen gas, a colourless gas that burns with a pop sound;

 

Equation:

2CH3CH2OH(l)  +   2Na(s)                   2CH3CH2ONa   +   H2(g)

 

  1. Reaction with phosphorus (V) chloride:

– Alkanols liberate misty fumes of hydrogen chloride gas;

 

Equation:

PCl5(s)   + CH3CH2OH                                   2CH3CH2OP   +   HCl2

Note:

– Both alkenes and alkanols decolourise the purple acidified potassium manganate (VII);

– However, alkenes decolourise the red bromine water; while alkanols do not.

 

Summary on preparation of alkanols (ethanol).

 

 

Ester                                                                                                        RCOOC2H5

 

 

 

                                                                                            Heat with NaOH(aq)

 

 

 

Ethyl iodide          Boil with KOH(aq)                     Ethanol                                Fermentation         Starch, sugars

(Iodoethane)                                                                                                                         cellulose

CH3CH2I

 

 

                                                                                            H2O + H2SO4

 

 

 

                                                                                Ethene

                                                            CH2CH2

 

 

 

Summary on reactions of alkanols.

 

 

CO2(g)  +  H2O(l)

 

 

 

                                                                                                Combustion

 

 

 

Ester                        Esterification                            Ethanol                  Sodium metal                  Sodium ethoxide   + Hydrogen

RCOOC2H5          H+/ RCOOH                        CH3CH2OH                                           CH3CH2ONa  + H2

 

 

                                Conc. H2SO4 at 170oC                                       Warm with acidified K2Cr2O7 or KMnO4

 

 

 

 

Ethene                                                                            PCl5                                                              Ethanoic acid

CH2CH2                                                                                                                                             CHCOOH

 

 

 

                                                           

                                                            Chloroethane

                                                            CH3CH2Cl

 

 

 

 

 

 

ALKANOIC ACIDS (CARRBOXYLIC ACIDS)

– Also called organic acids and form a homologous series with a general formula of CnH2n-1OOH

– The formula can also be written as CnH2n + 1COOH; in which case n = (no. of carbon atoms – 1)

– Members differ from each other by an additional CH2 group.

– Their functional group is the carboxylic group (-COOH) which is attached to the alkyl group.

– Graphical representation of carboxyl group; O

 

C   OH

Note: All carboxylic acids have the –COOH as the functional group but alkanoic acids are strictly alkanoic acids derived from alkanes

 

Nomenclature of alkanoic acids

– The ending of the corresponding alkane is replaced by ¢oic acid’.

 

Examples;

– Methane to methanoic acid.

– Ethane to ethanoic acid.

 

– They are named as if they are derived from alkanes through replacement of one of the hydrogen atoms by the -COOH group.

 

Note: –Unlike alkanols the functional group (COOH) in alkanoic acids can only be at the end of the carbon chain.

– The C in the COOH is always given the first position, while the substituents are given locants (numbers in reference to the first position).

 

Examples:

IUPAC name Old (traditional ) name Structural formula
Methanoic acid Formic acid              OH

 

H     C = O

Ethanoic acid Acetic acid                OH

 

CH3 – C = O

Propanoic acid Propionic acid                     OH

 

CH3CH2 –  C = O

Ethanedioic acid Oxalic acid  

O                    O

C –   C

HO                  OH

 

Butanedioic acid Succinic acid                  O

CH2C

OH

O

CH2C

OH

Branched alkanoic acids

– The naming of branched alkanoic acids follow the same general rules like that of alkanes; as long as the carbon atom with the –COOH group is given the first position.

– The branch can either be an alkyl group or a halogen other than hydrogen.

 

Examples:

 

Compound IUPAC name
                            O

Cl  –  CH2 –  C

OH

 

2-chloroethanopic acid;

                             O

CH3  –  CH –  C

OH

CH3

 

3-methylpropanoic acid;

                            O

CH3  –  CH –  C

OH

OH

 

2-hydroxypropanoic acid;

 

OH

O

CH3  –  C –  C

OH

CH3

 

 

2-hydroxy, 2-methylpropanoic acid;

                                         O

CH2  – CH  –  CH2 –  C

OH

Br        Cl

 

4-bromo, 3-chlorobutanoic acid;

 

 

Isomerism in alkanoic acids.

– Due to the existence of branched alkanoic acids, it is possible to obtain various isomers for a given alkanoic acid;

 

Example:

Draw all the isomers of pentanoic acid.                                                                              (3 marks)

 

 

 

 

 

 

 

 

 

 

 

Preparation of alkanoic acids

(a). Industrial manufacture

– Is done by the oxidation of primary alkanols using air (oxygen) as the oxidising agent.

 

Conditions:

– Moderate temperatures

– 5 atm pressure

– Hot copper catalyst;

 

Laboratory preparation.

– Is done by the oxidation of primary alkanols using acidified potassium dichromate (VI).

 

Example: Laboratory preparation of ethanoic acid.

(i). Apparatus

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(ii). Procedure

– Acidified potassium dichromate (VI) is heated in a water bath and ethanol added slowly from a water bath.

-The mixture is heated further and then distilled.

– The distillate is collected at about 105oC.

 

(iii). Observations and explanations.

Colour of potassium dichromate (VI) changes from orange to green.

Reason:

The dichromate ions are reduced to chromium (III) ions.

 

Equation:

Cr2O72-(aq) +  14H+(aq)  +  6e-                      2Cr3+(aq)  +   7H2O(l)

 

– The ethanol is oxidised to ethanal (acetaldehyde); which is further oxidised to alkanoic acid.

Equations

O

H+ from Conc. H2SO4

C2H5OH(l)    +   [ O ]                                           CH3CH(aq)   +   H2O(l)

Ethanol                           From                        Slow reaction          Ethanal                          water

K2Cr2O7

 

Structurally:

H     H                                                                 H    H

 

H     C      C     OH   +    [ O ]                             H     C     C    O   +   H2O

 

H     H                                                                  H

Then;

O                                                                     O

 

CH3CH(aq)  +   [ O ]                                         CH3C     OH

Ethanal                                       H+ reaction                     Ethanoic acid

General equation

H     H                                                                        O

 

H     C     C    OH(l)   +   2 [ O ]                                   CH3C   OH   +   H2O

 

H     H

 

Properties of alkanoic acids

 

Gradation in physic al properties of alkanoic acids

 

Name of acid Formula (structural) Molecular formula M.P °C B.P °C Solubility
Methanoic acid

Ethanoic acid

Propanoic acid

Butanoic acid

Pentanoic acid

Hexanoic acid

HC OOH

CH3COOH

CH3CH2COOH

CH3CH2CH2COOH

CH3CH2CH2CH2COOH

CH3(CH2)4COOH

CH2O2

C2H4O3

C3H6O2

C4H8O2

C5H10O2

C6H12O2

8.4

16.6

-20.8

-6.5

-34.5

-1.5

101

118

141

164

186

205

Most Soluble

 

increasing

     solubility

 

 

Least soluble

 

Physical properties of ethanoic acid

– A colourless liquid with a sharp pungent smell.

– B.P is 118oC and freezes at 17°c forming ice like crystals termed as glacial ethanoic acid.

– It is soluble in water and is weakly acidic with a P.H of approximately4.8.

 

Note

– Concentrated ethanoic acid is only slightly ionized and is a poor conductor of electricity

– On dilution its conductance steadily improves as the extent of ionization increases

 

 

 

 

Chemical properties (reactions) of alkanols.

(i). Reaction with carbonates.

– Alkanoic acids react with metal carbonates to form a salt (metal alkanoate), carbon (IV) oxide and water.

 

Examples:

Ethanoic acid and sodium carbonate

Ethanoic acid reacts with sodium carbonate to form sodium ethanoate and water with the liberation of carbon (IV) oxide gas.

 

Equation

Na2CO3  +   2CH3COOH                       2CH3COONa   +   CO2(g)   +   H2O(l)

Sodium carbonate                                                            Sodium ethanoate

 

Zinc carbonate and ethanoic acid.

ZnCO3  +   2CH3COOH                        (CH3COO)2Zn   +   CO2(g)   +   H2O(l)

Zinc carbonate                                                                 Zinc ethanoate

 

(ii). Reaction with metal hydroxides (Neutralization)

Alkanols neutralize alkalis like sodium hydroxide forming a salt (metal alkanoate) and water only.

 

Examples:

Sodium hydroxide and ethanoic acid.

CH3COOH   +   NaOH                      CH3COONa   +   H2O(l)

Ethanoic acid                 Sodium hydroxide                      Sodium ethanoate

 

Potassium hydroxide and methanoic acid

HCOOH   +   KOH                             HCOOK   +   H2O(l)

Methanoic acid         Potassium  hydroxide                      Potassium methanoate

 

(iii). Reaction with metal oxides (neutralization).

– Alkanoic acids react with metal oxides to produce salt (metal alkanoate and water only.

 

Examples:

Ethanoic acid and copper (II) oxide.

2CH3COOH   +   CuO                       (CH3COO)2Cu   +   H2O(l)

Ethanoic acid                     Copper (II) oxide                      Copper ethanoate

 

(iv). Reaction with metals

Alkanoic acids react with reactive metals to form a salt (metal alkanoate) and hydrogen gas;

 

Examples:

Ethanoic acid and sodium metal

2CH3COOH  +  Na(s)                                      2CH3COONa   +   H2(g)

Ethanoic acid                                                                             Sodium ethanoate

 

Propanoic acid and magnesium metal

2CH3CH2COOH  +  Mg                                 (CH3CH2COO)2Mg   +   H2(g)

Propanoic acid                                                                        Magnesium propanoate

 

 

(v). Reaction with alkanols (Esterification)

– Alkanoic acids react with alkanols to form esters;

Conditions:

– Drops of concentrated sulphuric acid.

– Gentle warming.

 

Reaction of ethanoic

– Ethanoic acid reacts with ethanol in the presence of a few drops of concentrated sulphuric acid forming a sweet fruity smelling compound called ester. 

– The process is called esterification.

 

Procedure

– 2-3 drops of concentrated sulphuric acid are added to a mixture of equal proportions of ethanol and pure ethanoic acid in an evaporating dish.

– The mixture is warmed gently in a water bath for sometime.

– The mixture is poured into a beaker and smelt;

 

Observation

A fruity sweet smell (of ethyl ethanoate);

 

Explanation

– Ethanol reacts with ethanoic acid in the presence of a few drops of concentrated sulphuric acid to form ethyl ethanoate and water.

– The reaction is very slow and so catalyzed by the hydrogen ions from the sulphuric acid.

 

Equation                                       Conc. H2SO4; warm

C2H5OH(l) +  CH3COOH (aq)                                CH3COOC2H5(l)   +   H2O(l)

Ethanol                     ethanoic acid                                                                 ethyl ethanoate (ester)

 

Structurally;

H     H                          H    O                                 H     O            H     H

Conc. H2SO4

H        C      C     OH  +   H     C    C                         H     C     C     O     C     C    H   +   H2O

 

H     H                          H    OH                              H                    H      H

 

– The alkyl part of the ester is derived from the alkanol, while the alkanoate part is derived from the acid;

– The alkanol attaches itself at the group in the carboxylic acid thereby displacing hydrogen atom;

– Under ordinary conditions the reaction takes place slowly; but in presence of concentrated sulphuric catalyst and warm (heat) conditions, the reaction is enhanced;

 

If propanol were used in place of ethanol, the reaction would yield the ester propyl ethanoate, according to the following equation;

 

CH3COOH(aq) +  CH3CH2CH2OH(aq)                  CH3COOCH2CH3(aq) + H2O(l)

Ethanoic acid                    propanol                                                    Propylethanoate

 

 

 

Note: –

Esters react with water to form the respective alkanoic acid and alkanol.

This reaction is termed hydrolysis and occurs in presence of concentrated sulphuric acid and heat as conditions.

 

Example:                                Conc. H2SO4

CH3COOCH3   +   H2O                                          CH3COOH   +   CH3OH

Methyl ethanoate                                               Heat                            Ethanoic acid                  Methanol

 

(vi). Reaction with ammonia.

Alkanoic acids react with ammonia to produce the ammonium salt of the acid.

General formula of the ammonium alkanoate salt is RCOONH4

 

Example:

Ethanoic acid and ammonia gas.

CH3COOH   +   NH3                          CH3COONH4

 

Uses of alkanoic acids

  1. In pharmaceuticals, for making medicines e.g ethanoic acid is used in the manufacture of aspirin.
  2. Manufacture of dyes and insecticides
  3. Seasoning food as vinegar
  4. Coagulation of rubber latex
  5. Preparation of polyethenyl ethanoate and cellulose ethanoate which are used are used to make artificial fibres such as rayon.
  6. Manufacture of soaps.
  7. Preparation of perfumes and artificial favours used in food manufacture.

 

Tests for alkanoic acids.

The following tests can be used to test for alkanoic acids.

Reaction with carbonates and hydrogen carbonates

Esterification

 

Summary: Draw a summary flow chart to show all the reactions of a named alkanoic acid.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

FATS AND OILS.

– Are esters of long chain carboxylic acids and glycerol

(a). Oils

– Oils occur naturally in plants and animals.

Examples: Whale oil, groundnut oil, corn oil and castor oil.

– Oils are liquids at room temperature.

Reason:

– They have high proportion of esters derived from the unsaturated oleic acid in them.

 

(b). Fats.

– These occur naturally in animals only.

Examples: Tallow, butter from milk, lard from pigs etc.

– Fats are solids at room temperature

– Oils can be converted to fats /hardened into fats by hydrogenation.

– This is the conversion of oils into fats by use of hydrogen; and forms the basis of margarine manufacture.

– During hydrogenation:

  • Hydrogen is bubbled into oils under high pressure and temperatures of about 400oC in the presence off a nickel catalyst

 

Note

– Fats and oils are important raw materials in the manufacture of soaps.

 

Soaps and soapless detergents

Soaps

Are a variety of compounds produced when oils or fats are reacted with sodium hydroxide.

– They are similar in that they contain a long hydrocarbon chain ending in a carboxylate anion to which is attracted a sodium cation.

– A typical soap is sodium stearate; C17H35COONa+

 

Structurally

                                                                                                                               O

CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2C

ONa+

 Preparation of soaps

– 2cm3 of castor oil and 10 cm3 of 4M sodium hydroxide are poured into a 100cm3 beaker.

– The mixture is then for about 10 minutes, stirring continuously and adding distilled water to make up for evaporation.

 

Explanation:

– On boiling an alkali with fat or oil, a hydrolysis reaction occurs.

 

 

 

 

 

 

Equation:

CH2 – COO – C17H35                                                                                                  CH2OH

 

 

CH –   COO – C17H35   +   3NaOH                            3C17H35COO Na+   +   CHOH

Sodium stearate (soap)

CH2 – COO – C17H35                                                                                                  CH2OH

Fat                                                                                                                                                              Glycerol

 

– When hydrolysis reaction occurs in the presence of an alkali (sodium hydroxide), the process is known as saponification (the chemical reaction between a fat and an alkali)

– In the fat hydrolysis NaOH neutralizes the acid formed (i.e. stearic acid) to form the sodium salt of the acid removing it from the aqueous mixture.

– Thus in excess alkali, all the fat is utilized.

– The sodium salt (sodium stearate) of the acid is termed soap if the number of carbon atoms per molecule is more than eight.

 

Note:

– KOH may be used in place of NaOH as the alkali.

 

  1. To the boiled mixture 3 spatula measures of sodium chloride are added, stirred well and allowed to cool.

Reason:

– The NaCl helps in separating the soap from the glycerol.

– This step is called salting and it reduces the solubility of the soap in the aqueous layer.

– The lower layer consists of glycerol, salts and unused alkali solution.

 

General formula of ordinary soap

CnH2n+1COONa+ where n >8 (n is greater than eight)

 

  1. The solid is filtered off and washed with cold distilled water, to remove impurities like NaCl

 

  1. The solid sample is placed in a test tube with distilled water, then with tap water.

 

Note:

The resultant soap may not have lathered easily with tap water.

Reason:

Some tap water contains a high proportion of calcium or magnesium ions that make water hard.

Summary on soap preparation

 

Fat (oil)                    Step I: saponification        Solution of soap and alcohol

 

 

Add NaOH and boil                                    Step II: salting

Add NaCl, stir well

And allow to cool

 

 

 

Ordinary soap

 

 

 

 

The role of soap in cleaning

Note: functions of soap in water

  1. It makes the water able to wet material more effectively by lowering the surface tension.
  2. Emulsification of oil and grease.

Soap molecules have two dissimilar ends:

  • A hydrocarbon chain which is non-polar and has no attraction for water- hence oil soluble
  • A Carboxylate end, which is polar and is attracted to water; hence is water-soluble.

Note: the Carboxylate end is in fact negatively charged in water because after dissolution, the sodium ion and the carboxylate ion exist as separate entities.

 

Illustration:

O

 

CH3(CH2)16 – C – ONa+

Non polar                Carboxylate (Polar)

 

Effects of soap on oil –water mixture (removal of oils and grease during washing)

Note: schematic representation of a soap molecule

 

 

 

 

 

  • On adding soap into oil water mixture the following build up occurs:

– A molecule of soap has a polar (hydrophilic) and non-polar (hydrophobic) parts;

– The non-polar end dissolves in oil and the polar end dissolves in water.

– When the mixture is agitated (thoroughly shaken) the hydrocarbon chain (tail) dissolves in grease while the carboxylate – sodium end of the soap molecule (the head) remains dissolved in water.

 

Diagrams: role of soap in cleaning.

 

 

 

 

 

 

 

 

– Each oil drop ends up with a large cloud of negative charge around it as the polar heads are negatively charged.

– Consequently the oil drops repel each other, hence preventing them from coalescing.

– The water –soluble sodium “heads” on the surface of the droplets keep the droplets emulsified (suspended) in the water.

– During rinsing the water carries away the oil droplets.

 

 

 

 

Effect of hard water on soap

– The Calcium and magnesium ions in hard water react with soap (sodium stearate) and remove it as an insoluble grey scum of magnesium or calcium stearate.

 

Equations:

(i). 2C17H35COONa+(aq)   +   Ca2+(aq)                           (C17H35COO)2Ca2+(s)   +   2Na+ (aq)

            Sodium stearate (soap)                                                                                               Magnesium stearate

 

(ii). 2C17H35COONa+(aq)   +  Mg2+(aq)                          (C17H35COO)2Mg2+(s)  +  2Na+ (aq)

           Sodium stearate (soap)                                                                                Magnesium stearate

– Soap is wasted in this way until all the calcium (II) and magnesium (II) stearate has been removed.

– The resultant scum is deposited on fabrics, giving them an unsightly dull appearance.

– Thus in hard water districts it is obviously advantageous to remove hardness before washing.

 

Removal of hardness:

Depends on whether the hardness is temporary or permanent

 

(a). Temporary hardness

Cause:

– The presence of calcium hydrogen carbonate or magnesium hydrogen carbonate dissolved in water.

 

Removal

– By boiling the water

– During the process the soluble calcium or magnesium hydrogen carbonate is precipitated out as insoluble calcium or magnesium carbonate.

 

Equation:

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

 

(b). Permanent hardness

Cause:

– Presence of calcium or magnesium chlorides and sulphates

 

Removal

Can be removed using the following methods:

 

(i). Distillation

– The water is distilled and the dissolved substances are left behind as water is evaporated and condensed

 

(i). Addition of washing soda (Na2CO3)

The washing soda reacts with the Mg²+ and Ca²+(aq) ions precipitating them as the insoluble carbonates

Equations:

2Na+(aq)   +   CO32-(aq)   +   2Cl-   +  Ca2+ aq)                         CaCO3(s)   +   2Na+ (aq)  + 2Cl (aq)

 

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

 

Disadvantage of washing soda as a water softener

– It is alkaline and can cause damage to wool and silk.

(iii). Ion exchange process (e.g. in the permutit water softener)

– It involves use of resins and compounds which will exchange their own Na+ for Ca²+or Mg²+ dissolved in hard water

– Thus as the Na²+ go into the water are left in the resin.

 

Equation:

2Na+ (resin)(aq)   +   Ca2+(aq)                             Ca2+ (Resin)2 (aq)   +   Na+(aq)

 

Advantages and disadvantages of hard water

Advantages of hard water

  1. It is good for drinking purposes as calcium contained in it helps to form strong bones and teeth.
  2. When soft water flows in lead pipes some lead is dissolved hence lead poisoning. However when lead dissolves in hard water insoluble PbCO3 are formed, coating the inside of the lead pipes preventing any further reaction
  3. It is good for brewing and the tanning industries;

 

Disadvantages of hard water

  1. Soap forms insoluble salts with magnesium and calcium ions; scum (calcium or magnesium stearate) thereby wasting soap.

Note: For these reason soapless detergents are preferred to ordinary soaps because they do not form scum; but rather form soluble salts with Mg2+ and Ca2+

– Examples of soapless detergents: brand names such as omo, perfix, persil, and fab e.t.c.

 

  1. Deposition of insoluble magnesium and calcium carbonates and sulphates formed from hard water result into blockage of water pips due to the formation of boiler scales

 

3.      Formation of kettle fur which make electrical appliances inefficient and increases running costs.

 

Soap and pollution effects.

  1. The wash water with soap and the dirt (grease) ends up in rivers and lakes thus affecting aquatic life; since plants do not grow well in soapy water.

 

Note: soaps are however biodegradable and so do not persist long in the environment.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Soapless detergents (synthetic detergents)

– Are cleansing agents lacking the carboxylate ions; but also act like soap in the cleaning process

– Instead they have sulphates (-OSO3Na+) or sulphonate groups (-SO3Na+) groups.

Are thus of two main types:

  • Sodium alkyl sulphates
  • Sodium alkylbenzene sulphonate.

 

(i). Sodium alkyl sulphates.

– Are detergents with the formula R – OSO3Na

 

Illustration.

– Consider the structure of sulphuric acid

                      O

 

HO     S      OH

 

O

– Replacing one of the hydrogen atoms with an alkyl group, R results into the compound

R – OSO3H;

– This is known as alkyl hydrogen sulphate;

– If the alkyl group (R) is a long chain such as dodecyl; CH3(CH2)10CH2 ; then the formula of the compound becomes CH3(CH2)10 – OSO3H (alkylhydrogen sulphate)

– Reacting the alkylhydrogen sulphate with an alkali (NaOH) results into a compound with the formula CH3(CH2)10 – OSO3Na.

– This is the soapless detergent, and is known as sodium dodecyl sulphate.

 

(ii). Sodium alkyl benzene sulphonates.

– Are formed when one of the OH groups in sulphuric acid is replaced with an alkyl benzene group.

 

 

– Have the general formula R –           – SO3Na

 

Illustration.

– If the alkyl benzene is of formula CH3(CH2)10CH2 ; then the resultant compound is of formula CH3(CH2)10CH2 –           – SO3H (hydrogen dodecyl benzene sulphonate)

 

– The sodium alkylbenzene sulphonate is neutralized using sodium hydroxide to obtain the detergent, CH3(CH2)10CH2 –           – SO3Na

 

The compound CH3(CH2)10CH2 –           – SO3Na is called sodium dodecyllbenzene sulphonate.

 

Conclusion:

The two main types of soapless detergents are:

  • Sodium alkyl sulphates; CH3(CH2)10 – OSO3Na
  • Sodium alkylbenzene sulphonate; CH3(CH2)10CH2 – – SO3Na

 

Note:

– Due to lack of carboxylate ions, soapless detergents do not form scum with hard water. Advantage over soap

– They are not affected by hard water, as they do not form scum in hard water as follows:

Preparation of soapless detergents

– Most are made from residues from crude oil distillation

– The hydrocarbons are treated with concentrated sulphuric acid  (instead of alkalis in cases of soap)

 

Procedure

– About 10cm³of olive oil in a small beaker, which is then stood in a larger one with ice-cold water.

– While stirring with a glass rod, concentrated sulphuric acid is carefully added to the olive oil using a dropping pipette.

– The acid is added until the yellow oil turns uniformly brown.

– 20cm³of 6M NaOH is then added; to neutralize the acid solution, resulting into a slightly basic product.

The soap is then tested with tap water and distilled water.

 

Explanation

– Some compounds like olive oil contain double bonds which can react concentrated sulphuric acid to form compounds of alkylhydrogen sulphate

 

Example:

 

R – CH = CH – CH – CH2 – Est   +   H2SO4 (l)                  R – CH – CH – CH – CH2 – Est

 

 

H   OSO3H

– In this case; R=alkyl group, either branched or straight; while Est = ester group.

– On adding NaOH, the alkyl hydrogen sulphate is neutralized where the hydrogen of the hydrogen sulphate is replaced by a sodium atom.

– The resultant compound is sodium alkylsulphate; R-CH  – CH-CH2-Est;

|          |

H     OSO3Na

Equation:

 

R – CH – CH – CH – CH2 – Est   +   NaOH(aq)          R – CH – CH – CH – CH2 – Est   +  H2O(l)

 

 

H   OSO3H                                                                     H     OSO3Na

 

– The R – CH  – CH – CH – CH2 – Est is the detergent and lathers easily with both tap and distilled water.

|          |

H     OSO3Na

Note:

– Alkylhydrogen sulphates can also be from alcohols.

 

Example:

 

R – CH – CH3   +   H2O(l)                             R – CH – CH3   +   H2O(l)

 

 

OH                                                                  OSO3H

– Most soapless detergents are sodium alkyl sulphates with a general formula ROSO3Na

 

– Sodium alkylbenzene soapless detergents are industrially manufactured from alkylbenzene;

– In this process alkylbenzene (a petroleum product) reacts with SO3 to form sulphuric acid

– Upon neutralization with NaOH, sulphuric acid forms sodium alkylbenzene sulphonate,

R –            – – SO3Na, a detergent.

 

Summary: manufacture of soapless detergents

 

Dodecene                                              benzene

CH3(CH2)9CH = CH2   +

 

 

                              Chemical reaction

 

                                   

CH3(CH2)11                   Dodecylbenzene

 

 

Conc. sulphuric acid

 

 

 

CH3(CH2)11 –         SO3H

Hydrogen dodecylbenzene sulphonate

 

 

Sodium hydroxide

 

 

CH3(CH2)11            SO3 Na+

 

Sodium dodecylbenzene sulphonate

 

 

Mode of action of soapless detergent.

Soapless detergents have two ends; a long hydrocarbon part, the tail and a short ionic part the head.

 

Simple representation of a detergent molecule.

                                                                        SO3  Na+

 

 

 

Non-polar tail (water hating)                                         Polar head (water loving)

 

Examples:

(i). Sodium lauryl sulphate

 

CH3(CH2)10CH2  – O – SO3Na+

 

 

       Tail                                                head

 

(ii). Sodium alkylbenzene sulphonate

 

CH3(CH2)10CH2  –              SO3Na+

 

 

            Tail                                                     head

– The tail is non-polar and dissolves in oil or grease (waterphobic) while the head is polar and dissolves in water (waterphilic).

– Each oil or grease gets sorrounded by the detergent molecules and hence a cloud of charged heads hence repel each other and do not coalesce.

– The dirt (grease loses its direct contact with the fabric being washed.

– Any agitation at this point then removes the dirt from the object.

 

Advantages of soapless detergents over soap;

  1. They lather easily with hard water since the corresponding calcium and magnesium salts are soluble in water due to lack of the carboxylate ions.
  2. Are mainly prepared from non-food raw materials.
  3. They do not react with acidic water.

 

Note:

– Soapless detergents with branched chain alkyl groups are not easily broken down by bacteria and are therefore the cause of frothing in sewerage plants, rivers etc.

– Consequently modern industry is overcoming this disadvantage by making the detergents from alkylbenzene with straight chain alkyl groups.

 

Pollution effects of soapless detergents.

  1. The active ingredients such as alkylbenzene sulphonates are non-biodegradable, hence accumulate in water sources and end up in human bodies.
  2. Some of the additives such as phosphates cause eutrophication hence excessive build up of algae (algal blooms) which change the water taste and odour, and also reduce oxygen supply hence poor growth of aquatic organisms.
  3. Because of their high lathering tendency, they cause excessive frothing and foaming in water sources especially after heavy rains.

 

Comparisons between soaps and soapless detergents.

 

Detergents Soaps
Have strong cleansing action Have weaker cleansing action
Are highly soluble in water; hence can be used in acidic or hard water. Are not very soluble in water, and tend to be wasted when used in hard water;

They cannot be used in acidic water.

Are made from byproducts of petroleum industry; which helps to conserve edible fats and oils Are made from edible fats and oils;
They cuse water pollution; Are biodegradable and have minimum pollution effects.
They are expensive. Are cheaper than detergents.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

POLYMERS

– A polymer is a macromolecule formed when two or more molecules link together to form a larger unit.

– Polymers have different properties different from those of the monomers.

– The process of polymer formation is called polymerization.

– Two types of polymers exist; natural and synthetic /artificial polymers.

 

(a). Natural polymers and fabrics

– Natural polymers occur naturally in living systems.

 

Examples:

– Rubber (latex), starch, cellulose, wool and proteins.

 

Natural rubber.

– Rubber trees give out a liquid called latex, which is collected from cuts in the trunks of rubber trees.

– Natural rubber is made out of latex from rubber trees.

– Latex is a hydrocarbon C5H8, called isoprene (2,methylbut 1,3 diene).

Formula of isoprene.

 

CH2 = C – CH = CH2

 

CH3

 

– Coagulation of latex leads to formation of a hydrocarbon polymer consisting of isoprene

(2, methylbut-1, 3-diene) units.

– This polymer is called polyisoprene with the formula; [ CH2 – C = CH – CH2 ]

 

CH3

Characteristics of natural rubber.

– Soft and sticky;

– Low resistance and low tensile strength; thus breaks easily upon stretching;

– Loses its rubber like properties at temperatures above 60oC;

 

Note:

– These are not good qualities and for industrial purposes this quality must be improved.

– This is done through vulcanisation of rubber.

 

Vulcanization of rubber.

– Is a chemical reaction in which raw rubber is heated with sulphur and is done purposely to improve the wear quality of rubber.

 

Process of vulcanization;

– Rubber is heated with sulphur

– The sulphur atoms form links between chains of rubber molecules.

– This reduces the number of double bonds in the polymer; making the material tougher, less flexible and less softer.

– During the process the sulphur atoms attach themselves to the rubber molecule in such a way that the molecules become locked in place and are prevented from slipping.

 

Note: –

– Soft rubber has about 2% sulphur, while toughened rubber about 10 % sulphur.

 

– Rubber can also be made artificially in industries and this gives a form of synthetic rubber.

– Other than undergoing vulcanisation it has chemicals that give it desired properties.

– It is made from byproducts of petroleum industry.

– Examples of synthetic rubber include neoprene and thiokol.

 

Note: Neoprene is made by polymerization of chloroprene (2, chlorobut 1, 3 diene), C4H5Cl

 

Equation:

CH2 = C – CH = CH2                                                CH2 – C = CH – CH2

 

Cl                                                                    Cl

Chloroprene                                                                                             Neoprene or polychloroprene

 

 

 

Properties of synthetic rubber.

  1. Are unreactive, hence they don’t react with industrial chemicals like oils, grease, petrol etc.
  2. Are non-inflammable, as they don’t catch fire easily.
  3. Capable of withstanding wide range of temperatures without changing shape.
  4. Have high mechanical strength.

 

Uses of synthetic rubber.

  1. Manufacture of insulating materials for electrical connections.
  2. Making conveyor and seat belts.
  3. Manufacture of car tyres and tubes.
  4. Making gaskets, flexible pipes.

 

Advantages and disadvantages of natural polymers.

– Are biodegradable hence not a likely cause of environmental pollution.

– Are made from renewable resources such as wool and trees; hence not easily exhaustible.

– Most are not easily flammable hence good materials for items like clothing.

 

  • Disadvantages

– Are often very expensive compared to synthetic polymers.

– Some do not last for very long.

– Are easily affected by acids, alkalis, air etc.

 

 

 

 

 

 

 

 

 

 

(b). Synthetic polymers

– They are man-made e.g. polythene, Perspex

– They are of two main types; thermoplastic and thermosetting polymers.

 

  • Thermoplastic;

– Softens on heating and becomes rigid on cooling.

Examples: nylon, polythene, polystyrene, etc.

 

  • Thermosetting:

– Are those that become hard on heating and cannot be softened by heating.

 

Advantages of synthetic polymers and fibres over natural polymers

  1. They can be made into different shapes easily.
  2. They are cheaper.
  3. Are often unaffected by ac ids, alkalis, water and air.
  4. Are usually less denser and yet stronger

 

Disadvantages

  1. Plastics are non-biodegradable hence causes a lot if problems in disposal
  2. Plastics burn more readily than natural material
  3. Some synthetic polymers give off poisonous gases when they burn. E. g. polyurethane gives off cyanide and carbon monoxide

 

Methods of polymerization:

  1. Addition polymerization;

– Occurs when unsaturated molecules (monomers) join to form a long chain molecule (polymer) without the formation of any other product.

– Usually the monomers must have at least a double or triple bond.

– One of the bonds in the double or triple bonds in the monomer opens up, and the unbonded electrons form bonds with neighbouring molecules.

 

Conditions for polymerization.

– High pressures.

 

Examples.

Polymerization of ethene to Polythene

 

CH2 = CH2   +  CH2 = CH2                                         [-CH2 – CH2 – CH2 – CH2 -]

 

Calculations involving polymers.

  1. A polyisoprene molecule is represented as: -[ CH2 – C = CH – CH -]n

 

CH3

– Given that the relative formula mass of polyisoprene is 748 000, calculate the number of isoprene units in the polymer.

 

 

 

 

Summary on common addition polymers.

 

Monomer Polymer properties Uses
H                 H

C = C

H                 H

Ethene

   H     H

 

C –  C

 

H     H

Polythene

 

Light, tough, and durable – Polythene bags, bowls packaging, electrical insulation; plastic pipes etc
       H    H     H

 

H    C    C =  C – H

 

H

Propene

   H       H

 

C       C

 

CH3  H

Polypropylene

 

Light, tough and durable – Making crates, boxes; and plastic ropes
H                 H

C = C

H                 Cl

Chloroethene

   H     H

 

C –  C

 

H     Cl

Polychloroethene(polyvinyl chloride)

 

Strong and hard (not as flexible as polythene) –          Making plastic pipes,

-electrical insulators, floor tiles, credit cards,

                 CH = CH2

 

 

Phenyl ethene (styrene)

         H     H

 

C –   C

 

Cl

 

Polyphenylethene (polystyrene)

 

Light, poor heat conductor, brittle; -Insulation,

-packaging materials and food containers;

F                  F

C = C

F                 F

 

Tetrafluoroethene

      F       F

 

C –    C

 

F      F

Polytetrafluoroethene (Teflon)

 

Non-stick surface;

Withstands high temperatures

– Non-stick coatings on pans; Insulation;
                  O

 

CH2 = C – C – O – CH3

 

CH3

Methylmethacrylate

     H      CH3

 

C  –    C

 

H       C – O – CH3

 

O

Polymethylmethacrylate (Perspex)

  -Optical components

– transparent doors and windows;

– Display signs;

– dental fillings;

 

 

 

  1. Condensation polymerization

– Occurs when monomers (similar or different) combine to form a long chain molecule; with the loss of small molecules like ammonia or water.

– The monomers should have at least two functional groups.

Reason:

– For molecules to join at both ends permitting chain formation;

 

Illustration

– Consider two molecules A and B; each with 2 functional groups:

  • Molecule A; HO – A – OH, with two OH functional groups;
  • Molecule B; HOOC – B – COOH, with two -COOH functional groups

 

On condensation

HO – A – O H +  H O O C – B – COOH + HO – A – OH +…………..

                   

                       Lost to form water

Hence;

HO – A – OH  + HOOC – B – COOH                       [HO-A-OOC-B-CO]n + 2H2O(l)

 

Types of condensation polymers.

– Are of two types:

  • Polyesters

– Are polymerized formed by an ester linkage; usually with the liberation of a water molecule.

 

Examples

O                   O                                                                 O                 O

 

 

HO    C                   C – OH + H – OCH2CH2O – H                             –    C                   C – OH + H – OCH2CH2O –    +  H2O

A diol                                                                                                         n

          A dioic acid                                                                                   Polyester

 

– Are condensation polymer involving monomers that have at least an amine group (in at least one of them); and thus usually result to the evolution of ammonia gas or water.

 

Example:

O                   O                                   H                  H                                        O                   O                  H                   H

 

H – C – (CH2)4 – C – OH   +  H – N – (CH2)6 – N – H                                   – C – (CH2)4 – C – OH   +  H – N – (CH2)6 – N –     +   H2O

n

Examples of condensation polymers and their uses.

 

Polymer Monomer Uses
Polyvinyl chloride Vinyl chloride Making rain coats, plastic discs, plastic water pipes, and electric insulation;
Starch Glucose Laundry
Cellulose. Glucose Paper and clothing manufacture
Silk and wool Proteins (amino acids) Making clothes
Polyester (terylene) Terephthalic acid and Ethan 1,2  diol Clothes and seat belts
Bakelite Urea and methanol Electrical fittings
Perspex Methylmethacrylate Safety belts, windscreen, and plastic lenses;
Nylon Hexane-1,6-dioyl dichloride; and hexane- 1,6- diamine; Used in making safety glass; reflectors; contact lenses; false teeth

Further examples of polymerization reactions.

Formation of nylon.

                                                                                                              H                     H    O                    O

H                                                               H           O                                       O

NCH2CH2CH2CH2CH2CH2N           +            CCH2CH2CH2CH2C                                        – N – (CH2)6 – N – C – (CH2)4 – C –

H                                                     H            Cl                                          Cl                                                                                  n

 

Hexane – 1,6 – diamine                         hexane – 1,6 – dioyl dichloride                                                              Nylon

 

 Formation of Terylene
UNIT 2: ACIDS, BASES AND SALTS

Unit Checklist:

  1. Acids;
  • Meaning
  • Strong and weak acids
  • Concentrated and weak acids
  • Comparing strength of acids
  • Using evolution of hydrogen and carbon (IV) oxides
  • Using electrical conductivity
  • Using PH
  • Role of solvents in acidic properties of solvents
  • Hydrogen chloride in water
  • Hydrogen chloride in methylbenzene

 

  1. Bases:
  • Meaning
  • Strong and weak bases (alkalis)
  • Measuring strength of alkalis
  • Using electrical conductivity
  • Using PH values
  • Effect of solvent type on properties of ammonia solution
  • Ammonia n water
  • Ammonia in methylbenzene

 

  1. Uses of acids and bases

 

  1. Oxides and hydroxides
  • Basic oxides
  • Acidic oxides
  • Neutral oxides
  • Amphoteric oxides
  • Meaning of amphoteric oxides
  • To verify amphoteric oxides
  • Reactions of amphoteric oxides

 

  1. Salts
  • Meaning
  • Preparation methods (summary)
  • Solubility of salts
  • Qualitative tests for cations using NaOH and NH4OH;
  • Effect of heat on metal oxides and hydroxides.
  • Effect of sodium carbonate on salt solutions
  • Properties of cations with sodium chloride, sodium sulphate and sodium sulphite

 

  1. Solubility and solubility curves
  • Definition of solubility
  • Factors affecting solubility
  • Solubility curves and related calculations.
  • Fractional crystallization

 

  1. Water
  • Hardness of water
  • Temporary hardness (Meaning; causes; and removal)
  • Permanent hardness (meaning; cause; and removal)

 

 

  1. Acids

– Are substances whose molecules yield hydrogen ions in water; or

– Are substances, which contain replaceable hydrogen, which can be wholly or partially replaced by a metal.

 

HCl (aq)                      H+(aq)  + Cl(aq)

 

OR: – Acids are proton donors i.e. a substance which provides protons or hydrogen ions.

 

Strength of Acids

– Acids can be categorized as either strong or weak acids;

 

  • Strong acids

– Are those which dissociate or ionize completely to a large extent in water, to yield many hydrogen ions.

– They yield to the solution as many protons as they possibly can.

 

  • Examples

Hydrochloric acid; HCl(aq)                             H+ (aq)  + Cl(aq)

Sulphuric acid; H2SO4(aq)                              2H+ (aq)  + SO4 2-(aq)

Nitric acid; HNO3(aq)                                    H+ (aq)    NO3 aq)

 

  • Weak acids.

– Are acids, which undergo partial dissociation to yield fewer hydrogen ions.

– They do not ionize in water completely or to a large extent i.e. some of their molecules remained unionized in solution.

 

Examples:

Carbonic acid:

H2CO3(aq)              water           H+(aq) +  HCO3(aq)

 

Ethanoic acid

CH3COOH (aq)            water                  H+(aq) + CH3COO(aq)

 

Note: – concentrated acids and dilute acids

Concentrated acids

– Is an acid with a high number of acid molecules per given volume.

 

Dilute acids:

Are acids with a low number of acid molecules per given volume.

 

– Thus there are concentrated strong acids or dilute strong acids; as well as concentrated weak acids and dilute weak acids.

 

 

 

 

Comparing the strength of acids

(i). Using rate of evolution of hydrogen

Apparatus:

– Boiling tubes; 1M HCl/ H2SO4/ HNO3; Methanoic acid/ tartaric acid; magnesium ribbon.

 

Procedure:

– One boiling tube is half filled with 1M HCl; while another is half filled with 1M Ethanoic acid.

– 2 pieces of magnesium ribbons are cleaned to remove a layer of oxide on the surface.

– One of the two pieces is put in each tube of the acid.

 

Observations:

– Hydrochloric acid evolves hydrogen much more quickly than Ethanoic acid yet they were of equal concentration.

 

Conclusion

– Hydrochloric acid is a strong acid;

– Ethanoic acid is a weak acid.

 

Note:

– The same experiment can be repeated with marble chips (CaCO3) in acids of same concentration.

-The marble chips dissolve more quickly in HCl, which is a strong acid.

 

(ii). Using electrical conductivity

Procedure:

– 50cm3 of 2M-hydrochloric acid solution is placed into a beaker and set up apparatus as shown below.

– The switch is closed and the brightness of the bulb noted.

 

Diagram: Electrolytic circuit

 

 

 

 

 

 

 

 

 

 

 

Observations

– Strong acids like HCl, HNO3 and sulphuric acid gave a brighter bulb light than weak acids like ethanoic, carbonic acids e.t.c

 

Explanations

– Strong acids are completely dissociated and have more H+ in solution and hence have got a higher electrical conductivity; than solutions of weak acids which are only partially ionized thus have fewer hydrogen ions in solution

 

 

(iii). Using PH

Procedure

– 2cm3 solutions of different acids of equal concentrations are paired into different test tubes.

– To each test tube 2 drops of universal indicator are added.

– Acids tested: HCl, H2SO4, HNO3; ethanoic acid, carbonic acid, and tartaric acid.

– All acids are of 2M solutions

– The indicator colour and hence the PH number of each is noted; by comparing against the indicator chart.

 

Observations

Substance (1M) Colour of universal indicator PH
Sulphuric acid

Hydrochloric acid

Nitric acid

Ethanoic

Carbonic acid

Tartaric acid

Red

Red

Red

Orange

Yellow

orange

3

3

3

5

6

5

 

Explanations

– Solutions of strong acids contain a higher concentration of hydrogen ions than those of weak acids

– Strong acids have low PH usually less than 3.

– Weak acids have higher PH values usually between 5 – 6.

 

Role of solvents on acidic properties of a solute.

Experiment: – to find out if solutions of HCl in different solvents display acidic properties

Apparatus and reagents

– Hydrogen chloride gas, water and methylbenzene

– Beakers and a funnel

– Blue and red litmus papers.

 

Procedure

– Solutions of hydrogen chloride gas are made by bubbling the dry gas from a generator into water and into methylbenzene contained in separate beakers.

– The hydrochloric gas is passed into the solution using an inverted funnel to prevent sucking back.

 

Apparatus

 

 

 

 

 

 

 

 

 

The resultant solutions are each separately subjected to various tests as shown below and observations recorded

 

Tests and observations

 

Test. Aqueous HCl solution Solution of HCl in methylbenzene
1.       A piece of dry blue litmus paper is dropped into solution

2.       Dry universal indicator paper

3.       Add magnesium ribbon

4.       Add small marble chips

5.       Electrical conductivity

Blue litmus turns red

 

– Turns red (strong acid)

– Evolution of hydrogen

– CO2 evolved

– Good conductor

No effect on litmus

 

– Turns green (neutral)

– No reaction

– No reaction

– Does not conduct

 

Explanation

– The results show that the aqueous solution of hydrogen chloride behaves as an acid; but the solution in methylbenzene lacks acidic properties

– When HCl gas dissolves in water it changes from molecules to ions;

 

Equation:

HCl (aq)        water                 H+(aq) + Cl(aq)

 

– It is the hydrogen ions which give the acidic properties and these can only be formed in the presence of water

– HCl in water conducts electric current due to presence of free ions in solution

– HCl gas in methylbenzene does not conduct electric current because the HCl exists as molecules hence lack free ions

 

Note: – hydrogen chloride gas dissolves in water because both HCl and water polar molecules;

– This causes mutual attraction of both ends of HCl molecule by different water molecules causing the dissociation of HCl molecules into ions.

 

Illustration:

 

 

 

 

 

 

Hence:

HCl (g) + water                       HCl (aq)

HCl (aq)                      H+(aq)  + Cl(aq)

– The presence of hydrogen ions in aqueous solution of hydrogen chloride explains the electrical conductivity and acidic properties of hydrogen chloride

  • Acidic properties: –
  • turns blue litmus paper red;
  • evolves hydrogen gas when reacted with magnesium;
  • evolves carbon dioxide on reaction with CaCO3;

 

2H+ (aq)  +  CaCO3 (s)                           Ca2+(aq)  +  CO2(g) + H2O(l)

 

2H+(aq)  +  Mg(s)                         Mg 2+(aq)   + H2  (g)

 

– Methylbenzene has a weak attraction for hydrogen chloride and hence hydrogen chloride remains as molecules in methylbenzene

 

  1. Bases

– Are substances which accept the protons donated by acids and are hence proton acceptors

NH3 (aq)   +   H+ (aq)                          NH4+(aq)

 

CuO(s) +  2H+ (aq)                         Cu 2+(aq)   +   H2O(l)

 

Alkalis

– An alkali is a soluble base i.e. a base that is soluble in water.

– They are compounds, which produce hydroxyl ions in aqueous solutions.

 

NaOH(aq)                           Na+(aq)   +  OH (aq)

 

Note: –

When an acid proton reacts with a base (hydroxyl ions) in aqueous solution, a neutralization reaction occurs.

 

Strength of an Alkali

– Alkalis can be grouped as either strong or weak alkalis.

 

(a). Strong alkalis

– Are alkalis that undergo complete dissociation in aqueous solution; yielding a large number of hydroxyl (OH-) ions

 

Examples:

– Sodium hydroxide.

– Potassium hydroxide.

 

(b). Weak alkalis

– Are alkalis that undergo only partial dissociation in aqueous solution (water) yielding fewer numbers of hydroxyl ions.

 

Examples

– Calcium hydroxide

– Ammonium hydroxide

 

Measuring the strength of alkalis

(i). Using electrical conductivity

Procedure

50 cm3 of 2 M sodium hydroxide solution is put into a beaker and the apparatus set as shown below

Apparatus

 

 

 

 

 

 

 

 

 

Procedure

– The same procedure is repeated using other alkalis like NH4OH; Ca (OH)2 e.t.c.

 

Observation

– The bulb lights brightly with KOH and NaOH as electrolyte than with NH4OH and Ca (OH)2

 

Explanation

– NaOH and KOH are strong alkalis and are completely dissociated and have more ions in solution and hence have got a higher electrical conductivity than the weak alkalis of NH4OH and Ca(OH)2(aq)

 

(ii). Using PH values

Procedure

– 2 cm3 of NaOH and 2 cm3 of NH4OH are each poured into 2 different test tubes separately

– Into each test tube 2 drops of universal indicator are added.

– The colour change is noted and the corresponding PH scale recorded

 

Observations

 

Alkali Colour of universal indicator PH
– Ammonium hydroxide (1M)

– Calcium hydroxide (1M)

– Sodium hydroxide (0.1M)

– Sodium hydroxide (1M)

– Potassium hydroxide

Blue.

Blue.

Purple.

Purple.

Purple.

11

10

13

14

14

 

 

 

 

 

 

 

Note: the PH scale

– Is a scale which gives a measure of the acidity of alkalinity of a substance.

 

Illustration: a PH scale.

 

1     2     3     4     5     6     7     8     9     10     11     12     13     14

 

 

Strong acid               Weak acids       Neutral   Weak alkali                                    Strong alkali

 

 

Increasing acidity                                                                      Increasing alkalinity

(High H+ ion concentration)                                                   (Low H+ ion concentration)

 

 

Indicator colours:

 

PH 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Colour Red Orange/ red Yellow/ Green Green Green/ Blue Blue/ Purple Purple

 

Effects of type of solvent on the properties of ammonium solution

Procedure

– Ammonium solution is prepared by bubbling the gas from a generator into methylbenzene (toluene) and into water contained in separate beakers

– The solutions are each divided into 3 portions and tested with litmus paper; universal indicator and for electrical conductivity

 

Apparatus

 

 

 

 

 

 

 

 

 

Observations

Test Solution of NH3 in water Solution of NH3 in methylbenzene (toluene)
Dry litmus paper Red litmus paper turns blue; No effect
Dry universal indicator paper Colour turns purple (alkaline PH) Turns green (Neutral PH)
Electrical conductivity Poor conductor Non-conductor

 

Explanations

– When NH3(g) dissolves in water it changes from molecules to ions.

Equation:

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

 

– It is the hydroxide ions that cause alkaline properties.

– Since ammonium hydroxide is a weak alkali, it dissociates partially releasing fewer hydroxide ions hence the poor electric conductivity.

– Ammonium gas in methylbenzene or trichloromethane exists as molecules without free ions hence no alkaline properties and the electrical conductivity.

Uses of acids and bases

  1. Acids

– Refer to the various acids for uses of sulphuric, nitric and hydrochloric acids.

 

  1. Bases/ alkalis

– Some weak bases e.g milk of magnesia, are used to relieve stomach disorders.

 

Amphoteric oxides and hydroxides.

  • Oxides

– An oxide is a binary compound of oxygen and another element.

– Are of four categories:

  • Basic oxides
  • Acidic oxides
  • Neutral oxides
  • Amphoteric oxides

 

(i). Basic oxides

– Are usually oxides of metals (electronegative elements)

– They react with acids to form salt and water only.

Examples

CaO, MgO, CuO etc.

 

(ii). Acidic oxides

– Are usually oxides of non metals (electronegative elements).

– Many of them react with water to form (give) acids and are known as acid anhydrides

Examples

CO2; SO2; SO3; P2O5 and NO2

 

(iii). Amphoteric oxides

– Are oxides, which behave as both bases and acids.

– Are mainly oxides of certain metals in the middle group of the periodic table.

Examples

Oxides of Zn, Al, Pb

 

Experiment: – To verify amphoteric oxides

Procedure

– A small sample of aluminium oxide is placed in a test tube and 5 cm3 of 2M nitric acid added to it and the mixture shaken.

– The procedure is repeated in different test tubes with ZnO, PbO, CuO and CaO.

– The experiments are repeated using excess 2M sodium hydroxide in place of nitric acid

 

Observations

 

Name of solid Observations when
acid is added hydroxide is added
Aluminium oxide

Zinc oxide

Lead II oxide

Zinc hydroxide

Lead hydroxide

Aluminium hydroxide

Oxide dissolves

Oxide dissolves

‘’

Hydroxide dissolves

‘’

Oxide dissolves

‘’

‘’

Hydroxide dissolves

‘’

Explanations

– These oxides are soluble in acids as well as in the alkalis (NaOH)

  • Reaction with acids

– Oxides react with acids to form a salt and water only in a reaction called neutralization reaction.

 

Equations

  • Oxides

(i).PbO(s) + 2H+(aq)                   Pb2+(aq) + H2O(l)

 

(ii). Al2O3(s) + 6H+(aq)                  2Al3+(aq) + 3H2O(l)

 

(iii). ZnO(s) + 2HCl(aq)                ZnCl2(aq) + H2O(l)

 

  • Hydroxides:

(i). Pb(OH)2(s) + 2HCl(aq)                        PbCl2(aq) + 2H2O(l)

 

(ii). Zn(OH)2(s) + 2HCl(aq)                    ZnCl2(aq) + 2H2O(l)

 

(iii). Al(OH)3(s) + 3HCl(aq)                      AlCl3(aq) + H2O(l)

 

Note: in these reactions the metal oxides are reacting as bases

 

  • Reaction with alkalis

– These oxides and hydroxides also react with alkalis e.g sodium hydroxide in which case they are reacting as acids.

– Their reactions with alkalis involve the formation of complex ions; M(OH)2-4

 

Equations

  • Oxides

(i). PbO(s) + 2NaOH(aq) + H2O(l)                   Na2Pb(OH)4(aq)

 

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

 

(ii). Al2O3(s) + 2OH(aq) +  3H2O (l)                   2[Al(OH)4](aq) + 3H2O(l)

 

Ionically: Al2O3(s) + 2OH(aq) + 3H2O(l)                       2[Al(OH)4](aq)

 

(iii). ZnO(s) + 2NaOH(aq) + H2O(l)                     Na2Zn(OH)4(aq)

 

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

 

  • Hydroxides:

(i). Al(OH)3(s) + NaOH(aq)                             NaAl(OH)4(aq)

 

Ionically: Al(OH)3(s) + OH(aq)                     [Al(OH)4](aq)

 

(ii). Zn(OH)2(s) + 2NaOH(aq)                            Na2Zn(OH)4(aq)

 

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

 

(iii). Pb(OH)2(s) + 2NaOH(aq)                            Na2Pb(OH)4(aq)

 

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

 

 

Salts

– Is a compound formed when cations derived from a base combine with anions derived from an acid.

– Salts are usually formed when an acid reacts with a base i.e. when the hydrogen ions in an acid re wholly or partially by a metal ion or ammonium (NH4+) radical.

 

Laboratory preparations of salts

– Salts are prepared in the laboratory using various depending on property of the salt especially solubility

Examples

(a). Preparations by direct synthesis

Equation:

Fe(s) + Cl2(g)                        2FeCl3(s)

 

(b). Reactions of acids with metals, metal oxides, metal hydroxides and metal carbonate

Equations:

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

 

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

 

NaOH(aq) + HCl(aq)                      NaCl(aq) + H2O(l)

 

PbCO3(s) + 2NHO3(aq)                 Pb(NO3)2(aq) + CO2(g) + H2O(l)

 

Note:

– Acid + metal method will not be suitable if:

  • The metal is too reactive e.g. sodium or potassium.
  • The salt formed is insoluble; as it will form an insoluble layer on the metal surface preventing further reaction.
  • The metal is below hydrogen in the reactivity series.

 

(c). Double decomposition/ precipitation

– Mainly for preparations of insoluble salts

– Involves formation (precipitation) of insoluble salts by the reaction between two solutions of soluble salts.

 

Equations:

Pb(NO3)2(aq) + 2NaCl(aq)                      PbCl2(s) + 2NaNO3(aq)

 

AgNO3(aq) + HCl(aq)                   AgCl(s) + HNO3(aq)

 

Types of salts:

– Are categorized into three main categories:

  • Normal salts
  • Acid salts
  • Double salts

 

 

 

 

Solubility of salts: – a summary

– All common salts of sodium, potassium and ammonium are soluble.

– All common nitrates are soluble.

– All chlorides are soluble except silver, mercury and lead chlorides.

– All sulphates are soluble except calcium, barium, lead and stomium sulphates.

– All carbonates are insoluble except sodium, potassium and ammonium carbonates.

– All hydroxides are insoluble except sodium, potassium ammonium and calcium hydroxides is        sparingly soluble.

 

Note:

– Lead(II) chloride is soluble in hot water.

– Calcium hydroxide is sparingly soluble in water.

 

Reactions of some cations with NaOH(aq)  and NH4OH(aq) and solubilities of some salts in water

 

Cation Soluble compounds in water Insoluble compounds in water Reaction with NaOH(aq) Reaction with NH4OH solution
K+ all None No reaction No reaction
Na+ all None No reaction No reaction
Ca2+ Cl; NO3 CO32-; O2-; SO42-; OH; White precipitate insoluble in excess No precipitate
Al3+ Cl; NO3; SO42-; CO32-; O2-; OH; White precipitate

soluble in excess

White precipitate insoluble in excess
Pb2+ NO3; ethanoate; All others White precipitate soluble in excess White precipitate insoluble in excess
Zn2+ Cl; SO42-; NO3; CO32-; OH; White precipitate soluble in excess White precipitate soluble in excess
Mg2+ Cl; SO42-; NO3; CO32-; OH; White precipitate insoluble in excess No precipitate
Fe2+ Cl; SO42-; NO3; CO32-; O2-; OH; (dark) green precipitate insoluble in excess Green precipitate insoluble in water
Fe3+ Cl; SO42-; NO3; CO32-; O2-; OH; (red) brown precipitate insoluble in excess Brown precipitate insoluble in excess
Cu2+ Cl; SO42-; NO3; CO32-; O2-; OH; Pale blue precipitate insoluble in excess Pale blue precipitate soluble in excess forming a deep blue solution
NH4+ all none Ammonium gas on warming Not applicable

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Explanations

– In these experiments NaOH forms insoluble hydroxides with ions of Zn2+, Al3+, Cu2+, Fe2+, Ca2+, Mg2+, Fe3+, and Pb2+.

– These hydroxides have a characteristic appearance, which form the basis of their identification

Examples

Equations:

Zn2+(aq) + 2OH(aq)                         Zn(OH)2(s)

(White ppt).

 

Cu2+(aq) + 2OH(aq)                         Cu(OH)2(s)

(Pale blue ppt).

 

Fe2+(aq) + 2OH(aq)                          Fe(OH)2(s)

(Dirty green ppt).

 

Fe3+(aq) + 2OH(aq)                          Fe(OH)3(s)

(Red-brown ppt).

 

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

(White ppt).

 

– The hydroxides of aluminum, zinc and lead dissolves in excess sodium hydroxide solution because of complexes are formed

Equations:

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

(Tetra-hydroxyl-aluminium (III) ion)

 

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

(Tetra-hydroxyl-lead (II) ion)

 

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

(Tetra-hydroxyl-zinc (II) ion)

 

Note: – in these reactions KOH(aq) may be used instead of sodium hydroxide

 

With ammonia solution

– Insoluble metals hydroxides are similarly formed.

Zn2+(aq) + 2OH(aq)                         Zn(OH)2(s)

(White ppt).

 

Cu2+(aq) + 2OH(aq)                         Cu(OH)2(s)

(Pale blue ppt).

 

Fe2+(aq) + 2OH(aq)                          Fe(OH)2(s)

(Dirty green ppt).

 

Fe3+(aq) + 2OH(aq)                          Fe(OH)3(s)

(Red-brown ppt).

 

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

(White ppt).

 

– However hydroxides of copper and zinc dissolve in excess ammonia solution due to formation of complex ions/ salts

Equations:

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

(White ppt)                                            (Tetra-amine zinc (II) ion; colourless solution)

 

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

(Pale blue ppt)                                          (Tetra-amine copper (II) ion; deep blue solution)

Effects of heat on metal hydroxides

Procedure

– Hydroxides of Zn, Ca, Pb, Cu e.t.c are strongly heated in a test tube each separately

 

Observation

– Most metal hydroxides are decomposed by heat to form metal oxides and water

– Sodium and potassium hydroxides only decompose at very high temperatures.

– Hydroxides of metals lower in the reactivity series are readily decomposed by heat than those metals higher in the series.

 

Examples

Cu(OH)2(s)            Heat                CuO(s) + H2O(l)

(Blue)                                                      (Black)

 

Pb(OH)2(s)             Heat                PbO(s) + H2O(l)

(White)                                                    (Red brown when hot; yellow when cold)

 

Zn(OH)2(s)            Heat                ZnO(s) + H2O(l)

(White)                                                    (Yellow when hot; white when cold)

 

Ca(OH)2(s)            Heat                CuO(s) + H2O(l)

(White)                                                    (White)

 

Note:

– Both iron (II) and iron (III) hydroxides give iron (III) oxide when heated.

Equations:

2Fe(OH)2(s)  + ½ O2(g)          Heat                Fe2O3(s) + 2H2O(l)

(Green)                                                                             (Red-brown)

 

Fe(OH)3(s)             Heat                Fe2O3(s) + 3H2O(l)

(Brown)                                                   (Red-brown)

 

These oxides do not decompose on further heating

 

Effects of sodium carbonate on various salt solutions

Procedure

– 3 drops of NaOH(aq) are added to 2cm3 of 1M solution containing magnesium ions in a test tube

the procedure is repeated with salt solutions containing

 

Solution containing Observations after adding sodium carbonate
Mg2+ A white precipitate is formed
Ca2+ A white precipitate
Zn2+ A white precipitate
Cu2+ A green precipitate
Pb2+ A white precipitate
Fe2+ A green precipitate
Fe3+ A brown precipitate and a colourless gas that forms a white ppt. in lime water;
Al3+ A white precipitate and a colourless gas that forms a white ppt. in lime water;

 

Explanations:

– Sodium carbonate, potassium and ammonium carbonate are soluble in water; all other metal carbonates are insoluble

– Hence their solutions may be used to precipitate the insoluble metal carbonates.

Ionic equations:

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

 

Note:

Iron (III) and Aluminium salts hydrolyse in water giving acidic solutions which react with carbonates to liberate carbon dioxide gas; hence effervescence.

 

Reaction of metal ions in salt solutions with sodium chloride, sodium sulphate and sodium sulphate

(i). Procedure

– 2cm3 of a 0.1M solution containing lead ions is placed in a test tube.

– 2-3 drops of 2M sodium chloride solution are added and the mixture warmed;

– The procedure is repeated using salt solutions containing Ba2+; Mg2+; Ca2+; Zn2+; Cu2+; Fe2+ and Fe3+

– Each experiment (for each salt) is repeated using Na2SO4 and Na2SO3 respectively, in place of sodium chloride.

 

(ii). Observations

 

Solution containing Sodium sulphate Sodium chloride Sodium sulphate
Zn2+ – Colourless solution – Colourless solution Colourless solution
Mg2+ – Colourless solution – Colourless solution Colourless solution
Cu2+ – Blue solution – Blue solution Blue solution
Fe2+ – Greenish solution – Green solution
Fe3+ – Yellow solution – Yellow/ dark brown solution
Pb2+ – White precipitate – White precipitate which dissolve on warming White precipitate
Ba2+ – White precipitate – White precipitate White precipitate

 

Explanations

– All the listed cations soluble salts except Ba2+ and Pb2+

– Lead sulphate and barium sulphate are insoluble in water;

– Lead chloride and barium sulphite are insoluble; however PbCl2(s) dissolves on warming

 

Equations:

Pb2+(aq) + 2Cl(aq)                       PbCl2(s)

 

Pb2+(aq) + SO42-(aq)                     PbSO4(s)

 

Ba2+(aq) + SO42-(aq)                     BaSO4(s)

 

Ba2+(aq) + SO32-(aq)                     BaSO3(s)

 

Note:

– To distinguish the precipitate of barium sulphate from barium sulphite; dilute HNO3 (aq) or HCl(aq) is added to both;

– BaSO3(s) will dissolve in the dilute acid but barium sulphate will not.

 

Uses (importance) of precipitation reactions.

– Precipitation of metal carbonate from aqueous solutions is useful in softening hard water; usually by removing calcium and magnesium ions from water as insoluble carbonate

 

Useful information on salts (qualitative analysis)

Colours of substances in solids and solutions in water.

 

COLOUR  
SOLID AQUESOUS SOLUTION

(IF SOLUBLE)

1. White Colourless Compound of K+; Na+, Ca2+; Mg2+; Al3+; Zn2+; Pb2+; NH4+
2. Yellow Insoluble Zinc oxide, ZnO (turns white on cooling); Lead oxide, PbO (remains yellow on cooling, red when hot)
Yellow Potassium or sodium chromate;
3. Blue Blue Copper (II) compound, Cu2+
4. Pale green

 

Green

Pale green (almost colourless)

Green

Iron (II) compounds,Fe2+

 

Nickel (II) compound, Ni2+; Chromium (II) compounds, Cr3+; (Sometimes copper (II) compound, Cu2+)

5. Brown Brown (sometimes yellow)

 

Insoluble

Iron (III) compounds, Fe3+;

 

Lead (IV) oxide, PbO2

6. Pink Pink (almost colourless)

Insoluble

Manganese (II) compounds, Mn2+;

Copper metal as element (sometimes brown but will turn black on heating in air)

7. Orange Insoluble Red lead, Pb3O4 (could also be mercury (II) oxide, HgO)
8. Black Purple

Brown

Insoluble

Manganate (VII) ions (MnO) as in KMnO4;

Iodine (element)-purple vapour

Manganese (IV) oxide, MnO2

Copper (II) oxide, CuO

Carbon powder (element)

Various metal powders (elements)

 

Reactions of cations with common laboratory reagents and solubilities of some salts in water

 

CATION SOLUBLE COMPOUNDS (IN WATER) INSUOLUBLE COMPOUNDS (IN WATER) REACTION WITH AQUEOUS SODIUM HYDROXIDE REACTION WITH AQUEOUS AMMONIA SOLUTION
Na+ All None No reaction No reaction
K+ All None No reaction No reaction
Ca2+ Cl; NO3; CO32-; O2-; SO42-; OH; White precipitate insoluble in excess White precipitate insoluble in excess, on standing;
Al3+ Cl; NO3; SO42- O2-; OH; White precipitate soluble in excess White precipitate insoluble in excess
Pb2+ NO3; ethanoate; All others; White precipitate soluble in excess White precipitate insoluble in excess
Zn2+ Cl; NO3; SO42- CO32-; O2-; SO42-; OH; White precipitate soluble in excess White precipitate soluble in excess
Fe2+ Cl; NO3; SO42- CO32-; O2-; OH; (Dark) pale green precipitate insoluble in excess (Dark) pale green precipitate insoluble in excess
Fe3+ Cl; NO3; SO42- CO32-; O2-; OH; (Red) brown precipitate insoluble in excess (Red) brown precipitate insoluble in excess
Cu2+ Cl; NO3; SO42- CO32-; O2-; OH; Pale blue precipitate insoluble in excess Pale blue precipitate soluble in excess forming a deep blue solution
NH4+ All None; Ammonias gas on warming Not applicable.

 

 

 

Qualitative analysis for common anions.

 

  SO42-(aq) Cl(aq) NO3(aq) CO32-(aq)
TEST Add Ba2+(aq) ions from Ba(NO3)2(aq); acidify with dilute HNO3(aq) Add Ag+(aq) from AgNO3(aq).

Acidify with dilute HNO3

Alternatively;

Add Pb2+ from Pb(NO3)2 and warm

Add FeSO4(aq);

Tilt the tube and carefully add 1-2 cm3 of concentrated H2SO4(aq)

Add dilute HNO3(aq); bubble gas through lime water;
OBSERVATION The formation of a white precipitate shows presence of SO42- ion; The formation of a white precipitate shows presence of Cl ion;

Formation of a white precipitate that dissolves on warming shown presence of Cl(aq) ions

The formation of a brown ring shows the presence of NO3 ions Evolution of a colourless gas that forma a white precipitate with lime water, turns moist blue litmus paper red; and extinguishes a glowing splint shows presence of CO32- ions
EXPLANATION Only BaSO4 and BaCO3 can be formed as white precipitates.

BaCO3 is soluble in dilute acids and so BaSO4 will remain on adding dilute nitric acid

Only AgCl and AgCO3 can be formed as white precipitates.

AgCO3 is soluble in dilute acids but AgCl is not;

– PbCl2 is the only white precipitate that dissolves on warming

Concentrated H2SO4 forms nitrogen (II) oxide with NO3(aq) and this forms brown ring complex (FeSO4.NO) with FeSO4; All CO32- or HCO3 will liberate carbon (IV) oxide with dilute acids

 

Checklist:

  1. Why is it not possible to use dilute sulphuric acid in the test for SO42- ions;
  2. Why is it not possible to use dilute hydrochloric acid in the test for chloride ions?
  3. Why is it best to use dilute nitric acid instead of the other two mineral acids in the test for CO32- ions?
  4. How would you distinguish two white solids, Na2CO3 and NaHCO3?

 

What to look for when a substance is heated.

 

1. Sublimation White solids on cool, parts of a test tube indicates NH4+ compounds;

Purple vapour condensing to black solid indicates iodine crystals;

2. Water vapour (condensed) Colourless droplets on cool parts of the test tube indicate water of crystallization or HCO3 (see below)
3. Carbon (IV) oxide CO32- of Zn2+; Pb2+; Fe2+; Fe3+; Cu2+;
4. Carbon (IV) oxide and water vapour (condensed) HCO3
5. Nitrogen (IV) oxide NO3of Cu2+; Al3+; Zn2+; Pb2+; Fe2+; Fe3+
6. Oxygen NO3 or BaO2; MnO2; PbO2;

 

 

 

 

 

 

Reduction-oxidation (Redox reactions)

(a). Displacement reactions.

(i) More reactive halogens metals will displace less reactive metals from solutions of their salts in the series:

Zn                               Fe  Pb Cu

More reactive                                Less reactive

 

Example:

– Zinc powder placed in a solution of copper (II) sulphate, which contains Cu2+(aq) ions, will become Zn2+(aq) ions and brown copper solid (metal) will be deposited.

– The Cu2+(aq) is reduced to copper by addition of electrons and the zinc is oxidized to Zn2+(aq) by removal of electrons.

 

(ii). More reactive halogens will displace less reactive halogens from solutions of their salts in series:

Cl2                             Br2 I2

More reactive                              Less reactive.

 

Example:

– Chlorine bubbled into a solution of potassium iodide (colourless), which contains I(aq) ions will turn grey (black) as iodine is liberated.

– The chlorine is reduced to Cl(aq) ions by addition of electrons and the I(aq) ions are oxidized to iodine by removal of electrons.

 

(b). Decolourisation of purple potassium manganate (VII) ions.

When a few drops of purple KMnO4 solution are added to a compound and the purple colour disappears, then this shows that the MnO4(aq) ions have been reduced to almost colourlessMn2+ (aq) ions

The substance in the solution has been oxidized.

 

Example:

– KMnO4 will oxidize Fe2+(aq) ions to Fe3+(aq) ions; pale green solution turns red-brown.

– KMnO4 will oxidize Cl(aq) ions to Cl2(g); colourless solution results to a green gas with a bleaching action;

 

(c). Orange potassium chromate (VI) turning to a green solution.

– Orange solution of dichromate ions, Cr2O72-(aq), changes to green Cr3+(aq) ions when the dichromate is reduced.

– The substance causing this change is oxidized.

 

Example:

– K2Cr2O7 will oxidize Fe2+(aq) to Fe3+(aq)

– K2Cr2O7 will oxidize SO2(g) to SO42-(aq)

– Formation of sulphate ions in solution from sulphur (IV) oxide gas is often used in the test for sulphur (IV) oxide gas.

 

(d). Oxidation of Fe2+(aq) to Fe3+(aq) ions by concentrated nitric acid.

 

 

 

 

Solubility and solubility curves

  • Solubility

– Is the maximum number of grams of a solid which will dissolve in 100g of solvent (usually water) at a particular temperature

– A solution is made up of two parts: – a solute and a solvent.

 

Solute

– The solid part of a solution usually dispersed in the solvent e.g. a salt.

 

Solvent

– The liquid part of the solution into which the solute is dissolved.

 

Experiment: to determine the solubility of potassium nitrate at 20oC.

(i). Materials

– Beakers, evaporating dish, measuring cylinder, burner, scales, thermometer, distilled water and potassium nitrate

 

(ii). Apparatus

 

 

 

 

 

 

 

 

 

 

 

(iii). Procedure

– About 50cm3 of distilled water is placed in a beaker

– Potassium nitrate is added to it a little at a time stirring continuously.

– The nitrate is added until no more will dissolve and there is an excess undissolved salt present. This is the saturated solution of KNO3 at the temperature.

Note:

  • Saturated solution: solution that cannot dissolve any more of the solid/ solute at a particular temperature

– The solution is allowed to settle and it is temperature recorded.

– About 25 cm3 of clear solution is poured in a previously weight evaporating dish.

– The mass of the dish and solution is recorded.

– The dish is then heated in a water bath (to avoid spurting) till the solution is concentrated.

– The concentrated solution is allowed to cool and the dish weighted with its contents.

 

Results and calculations

 

Temperature 20.0oC
Mass of evaporating dish + solution 100.7g
Mass of evaporating dish 65.3g
Mass of solution 35.4g
Mass of evaporating dish + dry salt 73.8g

Calculating:

Mass of salt dissolved = (73.8 – 65.3)g = 8.5g;

Mass of water (solvent) = (35.4 – 8.5)g = 26.9g

Thus:

If 26.9g of water dissolves 8.5g of KNO3 at 20oC;

Then 100g of water will have ? = 100 x 8.5 = 31.6g of salt;

26.9

Therefore the solubility of KNO3 at 20oC = 31.6g per 100g of water

 

Factors affecting solubility.

(i). Temperature

– For most salts solubility increases with rise/ increase in temperature.

Reason

– Increased temperature increases the kinetic energy, and hence the momentum and velocity of the solvent molecules so that they can disintergrate the solute molecules more effectively.

– However solubilities of certain salts remain almost constant with temperature change

– Solubility of gases however decreases with increase in temperature;

Reason:

Increase in temperature causes the gas molecules to expand and hence escape from the solvent.

 

Experiment: To investigate the effect the effect of temperature on solubility.

Requirements: potassium nitrate, distilled water, test tube, thermometer, stirrer, bunsen burner, 250 cm3 glass beaker, 4.5g of potassium nitrate.

 

(ii). Procedure.

Using a 10ml measuring cylinder, measure 5 cm3 of distilled water and add it to the boiling tube containing solid potassium nitrate.  Insert a thermometer into the boiling tube and heat the mixture gently in a water bath or while shaking to avoid spillage. Continue heating until all the solid has dissolved.  Stop heating and allow the solution to cool while gently stirring with a thermometer. Record the temperature at which the crystals of potassium nitrate first appear. Note this in the table below.

Retain the boiling tube and its contents for further experiments.

Measure 2 cm3 of distilled water and add to the mixture in the boiling tube. Heat until the crystals dissolve, then cool while stirring with a thermometer. Record the temperature at which the crystals again first stat to reappear. Repeat this procedure, each time adding more 2 cm3 of distilled water, heating, cooling and recording the crystallization temperature until the table is completely filled.

 

Table 2:

 

Experiment number I II III IV V
Volume of water added 5 7 9 11 13
Temperature at which crystals appear (oC)
Solubility of K in g/100g of water

 

 

 

 

Questions:

(a). Complete the table and calculate the solubility of solid X in g/100g of water at different temperatures.                                                                                                                                        (2 marks)

 

(b). Using the table above, plot a graph of solubility of solid X in g/100g of water against temperature.                                                                                                                                               (5 marks)

(c). From the graph:

(i). calculate the mass of K that would be obtained if the saturated solution is cooled from 60oC to 40oC.                                                                                                                                           (2 marks)

 

(ii). determine the solubility at 70oC.                                                                                              (1mark)

 

(iii). at what temperature would solubility of K be 100g/100g of water?                                      (1mark)

 

(ii). Stirring

– Stirring increases the solubility of a solid

Reason

– Stirring causes the molecules of solvent and solute to move faster causing the solute particles to disintergrate more effectively

 

Solubility curves

– Are curves showing the variation of solubility with temperature.

 

Uses / importance of solubility curves

– Can be used to determine the mass of crystals that would be obtained by cooling a volume of hot saturated solution from one known temperature to another.

– Solubility differences can be used to separate substances i.e. recrystallization or fractional crystallization (refer to separation of mixtures)

– Separation of salts from a mixture of salts with differing solubilities e.g. extraction of sodium carbonate from Trona (refer to carbon and its compounds)

– Manufacture of certain salts e.g. sodium carbonate by the Solvay process (refer to carbon and its compounds)

 

Worked examples

  1. An experiment was carried out to determine the solubility of potassium nitrate and the following results were obtained.

 

Temperature 10 15 30 40 50 60
Mass of KNO3 per 100g of water 20 25 45 63 85 106

 

(a). What is meant by solubility?                                                                                                   (1 mark)

 

(b). Plot a graph of mass of potassium nitrate against temperature.                                            (3 marks)

 

(c). From the graph work out the mass of KNO3 that would crystallize if a solution containing 70g of KNO3 per 100g of water was cooled from 45oC to 25oC.                                                                 (2 marks)

 

(d). Explain what would happen if 100g of KNO3 was put in cold water and heated to 50oC.                                                                                                                                                              (2 marks)

  1. The table below shows the solubility of sulphur (IV) oxide at various temperatures.

 

Temperature (oC) 0 5 10 15 20 25 35 40 45 50 55 60
Mass of SO2 per 100g of water 22 18.4 15.4 13.0 10.8 9.05 7.80 6.80 5.57 4.80 4.20 3.60

 

(a). On the grid provided plot a graph of solubility against temperature.                                    (3 marks)

 

(b). From the graph determine:

(i). The lowest temperature at which 100cm3 of water would contain 11.6g of sulphur dioxide.(1 mark)

 

(ii). The maximum mass of sulphur (IV) oxide that would dissolve in 2 litres of solution at 10oC. (Assume that the density of the solution is 1gcm-3)                                                                          (3 marks)

 

(c) (i). Sulphur (IV) oxide reacts with sodium hydroxide solution to form sodium sulphite and water.                                                                                                                                                 (1 mark)

(ii). Write the equation for this reaction.                                                                                        (1 mark)

 

(iii). Using the information from the graph, determine the volume of the saturated sulphur (IV) oxide solution that can neutralize 153 cm3 of 2M sodium hydroxide solution at 25oC.                     (3 marks)

Water

– Can be pure or impure

Pure water

– Is a pure substance which is a compound of hydrogen and oxygen; that boils at 100oC; melts at 0oC and has a density of 1gcm-3 at sea level.

 

Impure water

– Are the natural waters constituted of dissolved solutes in pure water.

 

Hardness of water

– Water without dissolved substances (salts) hence lathers easily with soap is referred to as soft water while water with dissolved substances that does not lather easily with soap is termed as hard water

 

Experiment: effect of water containing dissolved salts on soap solution

Procedure

– 2 cm3 of distilled water is put in a conical flask.

– Soap solution from a burette is added into the water and shaken until formation of lather is noted.

– If the soap fails to lather more soap solution is added from the burette till it lathers and the volume of the soap required for lathering recorded.

– The procedure is repeated with each of the following: tap water, rain water, dilute solutions of MgCl2, NaCl, CaCl2, a(NO3)2, CaSO4, MgSO4, Mg(HCO3)2, Ca(HCO3)2, ZnSO4 , NaHCO3, and KNO3.

– The procedure is repeated with each of the solutions when boiled.

 

 

 

 

 

 

Observations

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Explanations

– Distilled water requires very little soap to produce lather because it lacks dissolved salts and hence termed soft water.

– Solutions containing NaCl, ZnSO4, KNO3 and NaHCO3 do not require a lot of soap to form lather

Water containing Ca2+ and Mg2+ ions do not lather easily (readily) with soap

Reason:

– These ions react with soap (sodium stearate) to form an insoluble salt (metal stearate) called (Mg and Ca stearate respectively); which is generally termed scum.

 

Equations:

With Ca2+

2C17H35COONa+(aq) + Ca2+(aq)                                  (C17H35COO)2Ca(s) + 2Na+(aq)

Sodium stearate                                                                                             calcium stearate

 

With Mg2+

2C17H35COONa+(aq) + Mg2+(aq)                                 (C17H35COO)2Mg(s) + 2Na+(aq)

Sodium stearate                                                                                             Magnesium stearate

 

– Thus water with Mg and Ca is termed hard water and can only be made soft by removing these ions upon which the water will lather easily with water

– When Ca(HCO3)2(aq) and Mg(HCO3)2(aq) are boiled the amount of soap required for lathering decreases than before boiling

 

Reason

– Boiled decomposes the 2 salts into their respective carbonate s which precipitates from the solution leaving soft water which leathers easily with water

 

– The amount of soap solution used with  solutions containing sulphates and chlorides of calcium and magnesium did not change significantly even after boiling

Reason

– The soluble sulphates and chlorides of Mg and Ca do not decompose upon boiling hence can not be precipitated out.

 

Types of water hard ness

Temporary hardness

– Is hardness due to the presence of CaHCO3 or Mg(HCO3)2 in water; and can usually be removed by boiling.

 

Removal of temporary hardness in water:

(i). Boiling:

– Boiling decomposes and an insoluble chalk of CaCO3 and MgCO3 respectively is deposited in the sides of the vessel.

– This forms an encrustation commonly known as furr the process being furring.

 

Equations:

Ca(HCO3)2(s)           Heat        CaCO3(s) + 2CO2(g) + H2O(l)

 

Mg(HCO3)2(s)           Heat        MgCO3(s) + 2CO2(g) + H2O(l)

 

(ii). Distillation:

– Water containing dissolved salts is heated in a distillation apparatus;

– Pure water distils over first leaving dissolved salts in the distillation flask (refer to separation of mixtures)

– Is of less economic value as it is too expensive hence disadvantageous.

 

(iii). Addition of calcium hydroxide:

– Involves adding correct amount of lime water where CaCO3 is precipitated out.

– This method is cheap and can be used on large scale at water treatment plants.

– However if excess lime (Ca2+) ions is added this will make water hard again.

 

Equation:

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

 

(iv). Addition of ammonia solution:

– Addition of aqueous ammonia to water containing calcium and magnesium hydrogen carbonates (temporary hard) precipitates calcium and magnesium ions as corresponding carbonates.

 

Equations:

Ca(HCO3)2(aq) + 2NH4OH(aq)                          CaCO3(s) +2H2O(l) + (NH4)2CO3(aq)

 

 

(ii). By permutit softener (ion exchange).

– Uses a complex sodium salt (NaX), such as sodium aluminium silicate commonly known as sodium permutit.

– Permutit is a manufactured ion exchange resin.

 

  • Iron exchange resin: materials that will take ions of one element out of it’s compounds and replace it with ions another element

 

 

 

 

Working principle

– The permutit is contained in a metal cylinder

– The hard water is passed through the column of permutit in the cylinder and it emerges softened at the other end

– As hard water passes through the column ion exchange takes place.

– The Ca2+ and Mg2+ remain in the column while sodium ions from the permutit pass into water thus softening it.

 

Diagram: permutit water softener.

 

 

 

 

 

 

 

 

 

 

Equations:

NaX(aq) + Ca2+(aq)                        CaX(s) + 2Na+(aq)

 

NaX(aq) + Mg2+(aq)                       MgX(s) + 2Na+(aq)

 

– When all the Na+ ions in the permutit have been replaced by Ca2+ and Mg2+ ions the permutit can not go on softening water.

– It is then regenerated by washing the column with brine (a strong NaCl solution); during which calcium and magnesium chlorides are washed away.

 

Equation:

CaX(s) + 2NaCl(aq)                        CaCl2(aq) + Na2X(s)

 

MgX(s) + 2NaCl(aq)                       MgCl2(aq) + Na2X(s)

 

Permanent hardness

– Is that due to soluble sulphates and or chlorides of calcium and or magnesium and cannot be removed by boiling

 

Removal of permanent hardness

(i). By the addition of washing soda (sodium carbonate)

– Washing soda softens hard water by causing the formation of insoluble CaCO3 or MgCO3

– The soluble sodium salts left in water do not react with soap.

 

Equations:

Na2CO3(aq) + CaCl2(aq)                 2NaCl(s) + CaCO3(S)

 

Ionically:

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

 

Na2CO3(aq) + MgSO4(aq)                          Na2SO4(s) + CaCO3(S)

 

Ionically:

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

– This method is very convenient and economical on large scale. It softens both temporary and permanent hardness

 

(ii). By permutit softener (ion exchange); explanations as before

(iii). Distillation.

 

Advantages of hard water

(i). It is good for drinking purposes as calcium ions contained in it helps to form strong bones and teeth.

 

(ii). When soft water flows in lead pipes some lead is dissolved hence lead poisoning. However when lead dissolves in hard water insoluble PbCO3 are formed, coating the inside of the lead pipes preventing any further reaction; this reduces any chances of lad poisoning.

 

(iii). It is good for brewing and the tanning industries; it improves wine or beer flavour in brewing industries.

 

Disadvantages of hard water

(i). Soap forms insoluble salts with magnesium and calcium ions; scum (calcium or magnesium stearate) thereby wasting soap.

– For these reason soapless detergents are preferred to ordinary soaps because they do not form scum; but rather form soluble salts with Mg2+ and Ca2+

  • Examples of soapless detergents: omo, perfix, persil, fab e.t.c.

 

(ii). Deposition of insoluble magnesium and calcium carbonates and sulphates formed from hard water result into blockage of water pips due to the formation of boiler scales

(iii). Formation of kettle fur which makes electrical appliances inefficient hence increasing running costs.

 

(iv). Formation of scum on clothing reduces their durability and aesthetic appearance

 

 

UNIT 3: ENERGY CHNAGES IN CHEMICAL AND PHYSICAL PROCESSES (THERMOCHEMISTRY)

 

Unit Checklist.

1. Introduction

2. Specific heat capacity and enthalpy of a system

3. Endothermic and exothermic reactions.

4. Activation energy

5. Determination of heat changes

Ø  Heat of combustion

  • Meaning
  • Determination
  • Calculations

Ø  Fuels

  • Meaning
  • Heating value of a fuel
  • Fuel pollution
  • Choice of a fuel

Ø  Heat of neutralization

  • Meaning
  • Determination
  • Calculations

Ø  Enthalpy of solution

  • Meaning
  • Determination
  • Calculations
  • Enthalpy of displacement
    • Meaning
    • Determination
    • Calculations
  • Enthalpy of precipitation.
    • Meaning
    • Determination
    • Calculations
  1. Energy level diagrams
  2. Thermochemical cycles
  3. Hess’s law
  4. Enthalpy of formation.
  5. Heat of formation and bond energies.
  6. Latent heat
  • Latent heat of fusion
  • Latent heat of vapourisation
  • Relationship between vapourisation, fusion and structure.

 

Introduction:

– Most chemical and physical process are accompanied by energy changes which occur in the form of heat measured in joules (J) and kilojoules (KJ).

– The heat results from the motion of atoms and molecules.

 

Specific heat capacity

– Specific heat capacity of a substance is the number of joules required to raise the temperature of one gram of the substance by one degree Kelvin e.g. – SHC for water is 4.18Jg-1K-1

 

The heat content (Enthalpy) of a system.

– Heat content is denoted by H; while heat change is denoted as ∆H1;

– And; ∆H = HProducts – HReactants

= H2-H1

– Heat changes in chemical reactions can either be exothermic and endothermic.

 

Endothermic and exothermic reactions.

(a). Endothermic reactions

– Is a reaction accompanied by a fall in temperature and energy is absorbed from the surroundings.

– The enthalpy change is normally positive since heat of products (HProducts) is higher than heat of reactants (Hreactants).

– Thus H2-H1 is a positive value; since HI is less than H2.

 

Graphically this can be denoted as follows;-

 

 

 

 

∆H = +Ve; reaction is endothermic

 

 

Reactants; H1

 

 

 

 

Reaction path

 

Examples:

(i). When NH4NO3 dissolves in water, the temperature of the solution drops.

 

Procedure

– A spatula end-full of ammonium nitrate is dropped into a test tube of water.

– The bottom of the test tube is felt with the hand.

 

Observations:

– The hand feels cold.

Reason:

– Energy is absorbed by the products, cooling the test tube.

 

Thus H1 is less than H2 giving a positive ∆ H1 showing an endothermic reaction.

 

 

(ii) N2(g) +  O2 (g)                2NO(g); ∆H=+91KJMol-1

 

(b). Exothermic reactions.

– Are reactions accompanied by a rise in temperature and energy is liberated to the surroundings.

– The enthalpy change is normally negative, since heat of reactants (H1) is normally higher than heat of products (H2).

– Thus, H2-H1 gives a negative value since H1 is higher than H2.

 

Exothermic reactions can be represented graphically as follows;-

 

 

 

 

 

∆H = Ve; reaction is exothermic

 

 

Products; H2

 

 

 

 

Reaction path

 

– In exothermic reactions H2 is lower than H1.

 

Examples:

(i). Manufacture of ammonia in the Haber process;

i.e. N2(g) +  3H2 (g)                     2NH3(g); ∆H=-46Kjmol

 

(ii) Dissolving sodium hydroxide pellets in water,

-The temperature of the resulting NaOH(aq) is higher than the temperature of water at room temperature.

-This implies the internal temperature of products is lower than the reactants’ original temperature.

-Thus (H products – H reactants) = -ve value, an exothermic reaction.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Activation energy.

– Is the energy required to activate the reactants before a reaction can take place.

– Thus activation energy is the energy required to initiate a reaction.

– The size of activation energy will differ from one reaction to another and so will be the gap between the energy of reactants and the energy is products.

 

Examples:

(i). Exothermic reactions

 

 

E1

A               B      Activation energy

Reactants

C(s) + O2(g)

 

∆H = -ve.

E2

 

 

Products

CO2(g)

 

 

Reaction progress (pathway)

 

  • At A; bonds are being broken and the energy is absorbed.
  • At B; bonds are now being formed, and so energy is evolved.

 

Endothermic reactions

 

 

A          B

 

E2

Products

CaO(s) + CO2(g)

 

∆H = +ve

 

 E1

Reactants

CaCO3(s)

 

Reaction progress (pathway)

 

  • At A; bonds are broken and energy is absorbed.
  • At B; bonds are formed and energy is evolved.

 

 

 

 

 

 

 

 

 

 

 

Determination\ measurement of Enthalpy (heat) changes.

Calorimeters are sued and have to be insulated to reduce heat loss to the surrounding.

 

Source of errors:

– Heat loss to the surrounding.

– Absorption of heat by the calorimeter (vessels).

 

Main heat changes under consideration:

  • Enthalpy of combustion (∆Hc)
  • Enthalpy of neutralization (∆Hneut)
  • Enthalpy of solution (∆Hsoln)
  • Enthalpy of precipitation (∆Hprecip)
  • Enthalpy of displacement (∆HDisp)
  • Enthalpy of formation (∆Hf)

 

  1. Heat of combustion ∆Hc

– Is the heat changes when the mole of a substance is completely burned in oxygen, at one atmospheric pressure.

– Since heat is usually evolved and hence ∆H is usually negative.

 

Example;

C(s) +  O2(g)                             CO2(g) ; ∆ H= -394 Kjmol-1

 

 

Apparatus for finding ∆Hc of a fuel.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Heat evolved = Specific heat capacity of water x Mass of water x Temperature rise

 

= CM∆T; joules, where; C = specific heat capacity.

M = mass of water.

∆T= temperature change.

 

 

 

 

Examples

  1. Assume

Volume of water in calorimeter   = 100cm3

Initial temperature of water         = 225oC

Final temperature of water           = 50.5oC

Change in temperature of water   = 28.0oC

Mass of water                              = 100g

Mass of lamp before burning       = 30.46g

Mass of lamp after burning          = 30.06g

Mass of ethanol burnt                  = 30.46 -30.06

                                                     = 0.40g

(a). Determine the heat evolved;

= ∆ H= CM∆ T

=4.2 x 100 x 28

=11760 Joules

=11.760 KJ

 

(b). Hence calculate the molar heat of combustion of ethanol;

 

Moles of ethanol burnt  = Mass of ethanol burnt; 0.4 = 0.008695 moles.

RFM of ethanol            46

Thus if 0.008695 moles = 11.76 kilojoules,

Then 1mole = ?

 

1 mole = 11.76 x 1 = -1352.50 KJMol

0.008695

 

= -1352.5 KJMol-1 (negative value as the reaction is exothermic since heat is evolved).

 

  1. When ethanol was burnt in the apparatus shown (in example 1), the results were: mass of fuel burnt, M1=1.50g; mass of water, M2=500g; ∆T= 13.0oC. (C=12; H=1; O=16; SHC of water=4.18KJKgK). Find the molar enthalpy of combustion of ethanol.

Compare the experimental value with the listed value of –1368 KJMol and explain any difference.

 

Solution:

Heat evolved = CM∆T

= 4.18 x 500 x 13

1000

= 27.17KJ

Molar mass of ethanol, C2H5OH = [(12 x 2) + (1 x 6) + (1 x 16)] = 46

Thus if 1.5g = 27.17KJ

Then 46 g = ?

= 46 x 27.17

1.5

= 833.213KJMol

 

 

  1. In an experiment to determine the heat of combustion of methanol, CH3OH a student used a set up like the one shown in the diagram below. Study the set up and the data and answer the questions that follow.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Data:

Volume of water       =500 cm3

Final temperature of water =27.0oC

Initial temperature of water =20.0oC

Final mass of lamp + menthol = 22.11g

Initial mass of lamp + methanol =22.98g

Density of water- 1gcm-3

(Heat change = mass x temperature change x 4.2Jg-1oC1

 

Questions:

(a). Write an equation for the combustion of methanol.                                                                {1mark}

 

(b). calculate:

(i). The number of moles of methanol used in the experiment                                                    {2marks}

 

(ii). The heat change for this experiment.                                                                                      {1mark}

 

(iii). The heat of combustion per mole of methanol.                                                                   {2marks}

 

(c) Explain why the molar heat of combustion for methanol obtained in this experiment is different from the theoretical value.                                                                                                           {2marks}

 

  1. When 0.6g of element J were completely burned in oxygen, all the heat evolved was used to heat 500 cm3 of water, the temperature of the water rose from 23.0oC to 32.0oC. Calculate the relative atomic mass of element J given that the specific heat capacity of water = 4.2jg-1k-1, density of water=1.0gcm-3 and molar heat of combustion of J is 380JMol                           {3marks}

 

  1. Hydrogen peroxide contained in 100cm3 of solution with water was completely decomposed. The heat evolved was 1380J. Determine the rise in temperature due to the reaction.

(SHC of water = 4.2Jg-1k-1; density of water = 1gcm-3; O =16; H = 1)                                      {3 marks}

 

Fuels:

– Are compounds which produce a high heat of combustion.

– Fuels can be:-

  • Solids; such as charcoal, wood, coal.
  • Liquids; such as ethanol, gasoline
  • Gaseous; such as methane, water, gas etc.

 

Basic concepts:

  • Calorific value

– Is the energy content of a fuel;

– Is the heat evolved when a given mass of fuel is completely burnt in oxygen;

 

Note:

– Sometimes fuels may undergo incomplete combustion.

– Incomplete combustion of fuel is disadvantageous in that:-

  • It reduces the energy content.
  • It leads to pollution.

 

  • Heating value:

– Is the amount of heat energy given out when a unit mass or unit volume of a fuel is completely burnt in oxygen.

 

Fuel pollution.

– Is commonly caused by internal combustion engine.

– Fuel in engine (e.g. petrol) burns completely to water and carbon (IV) oxide, only under ideal conditions.

 

Causes of fuel pollution:

(i). Incomplete combustion which causes production of (produces) CO and unburnt carbon (soot).

 

(ii). Some fuels contain sulphur and nitrogen and on combustion release SO2 and NO2. These gases are acidic, resulting to acidic rain which corrodes buildings and affects trees and animals in various ways.

 

(iii). Fuel additives; e.g. tetraethyl lead, Pb (C2H5)4 added to petrol to enhance burning efficiency produces volatile lead compounds in the exhaust fumes.

– Lead is very poisonous and affects the nervous system and the brain in children.

 

Factors to consider when choosing a fuel.

(i) Heating value,

(ii) Ease and rate of combustion.

(iii) Availability

(iv) Ease of transportation

(v) Ease of storage

(vi) Environment effects.

 

 

 

 

 

  1. Heat of neutralization;

Is the heat evolved when acid and a base react to form and mole of water.

– Alternatively;

– It is the heat evolved when one mole of hydrogen ions from an acid reacts with one mole of hydroxide ions from an alkali to form\ give one mole of water.

 

Equation:

OH(aq) +  H+(aq)                     H2O(l)

 

– Neutralization reactions are exothermic.

– Are determined by:

– measuring the temperature rise produced when a known volume of acid is neutralized by a known volume of alkali.

Examples.

(a). Strong acids reacting with strong alkalis.

The ∆H is always about –57KJMoland higher than that for weak acids – weak alkalis.

Reason:-

– Strong acids and alkalis are already fully ionized in aqueous solution and no heat is lost in ionizing the acid or the alkali.

 

Consider:

– Reaction between sodium hydroxide and hydrochloric acid

 

NaOH(aq)                     Na+(aq) + OH(aq)

HCl(aq)                  H+(aq) +  Cl(aq)

 

On reacting; OH(aq) + H+(aq)                H2O(l)

 

Diagrammatically (energy level diagram)

 

 

 

 

 

 

∆H = -57KJMol-1

 

H2O(l)

 

 

 

 

 

 

Reaction progress (pathway)

 

(b). Neutralization reactions involving weak acids or weak alkalis.

– The ∆H is lower than expected, e.g. only –52Kjmol-; and hence lower than that for strong acids-strong alkalis.

 

Reason:

– Weak acids and weak alkalis are NOT fully ionized in aqueous solutions and some heat is used in ionizing them.

Consider; reaction between ethanoic acid and ammonia solution.

 

 

NH4+(aq) + OH(aq) + CH3COO(aq) + H+(aq)

 

 

 

∆H = +5KJ

 

∆H = -57KJ

NH3(aq) + H2O(l) + CH3COOH(aq)

 

 

 

 

CH3COO(aq) + NH4+(aq) + H2O(l)

 

 

 

Reaction pathway

 

Therefore;

∆H is given by;

∆H = ∆HI + ∆HII

= 5 + (-57)

= -52KJMol-1

 

Note:

– Dibasic acids e.g. H2SO4 contains two replaceable hydrogen atoms hence on incomplete neutralization, they form two moles of water.

 

 

– Therefore, H neutralization =  ½ x ∆H reaction.

 

Experiment: – To determine the heat of neutralization of hydrochloric acid by sodium hydroxide.

(i) Procedure:-

– A clean dry 250 cm3 glass or plastic beaker is wrapped with a newspaper leaf.

– Exactly 50 cm3 of 2.0M hydrochloric acid solution is transferred into the beaker.

– The temperature T1 of the acid solution is noted.

– Using another clean dry measuring cylinder, exactly 50 cm3 of 2.0M NaOH solution is measured and its steady temperature T2 is noted.

– The contents of the beaker (acid), are carefully stirred with a thermometer while adding NaOH.

– The highest temperature T4 attained by the resulting mixture is noted.

 

(ii) Apparatus

 

 

 

 

 

 

 

 

 

 

(iii). Results:

– Temperature of the acid, T1 = oC;

– Temperature of the hydroxide, T2 = oC;

– Average temperature of the two solutions; T1 + T2 = T3 = oC

2

-The highest temperature of the mixture; T4 = oC;

-The temperature change, ∆T= (T4 –T3) = oC;

 

Sample calculations:

Given: Temperature of hydrochloric acid solution T1= 22.75oC;

Temperature of sodium hydroxide solution T2 = 22.80oC;

Average temperature of acid and alkali T1 + T2 = T3 = 22.78oC;

2

Highest temperature of alkali and acid mixture, T4 = 36.40oC;

Temperature change, ∆T= T4 – T3 = (36.40 – 22.78)oC;

=13.62oC;

 

  • Assumption: the specific heat capacity of the solution= 4.2KJKg-1K-1.

 

– In the experiment, 50 cm3 of 2M HCl are neutralized by 50 cm3 of NaOH, thus; volume of the mixture= (50 + 50) cm3 = 100 cm3.

– Taking density of the resultant solution to be 1gcm-3, then;-

Mass of solution M;

=density x volume

=1g/cm3 x 100cm3= 100g.

=0.1 kg i.e. (100/1000)

Thus heat evolved,

Mass of solution x specific heat capacity x temperature change

= MC∆T,

= 0.1Kg x 4.2kj/kg.k x 13.62oC

=- 5.7 KJ.

 

– But 50 cm3 of 2M HCl contains 2 x 50 moles, = 0.1 moles of H+ ions.

1000

– Similarity 50 cm3 of 2M NaOH contains 2 x 50 = 0.1 moles of OH ions.

1000

– This implies that the two solutions neutralize each other completely.

Therefore;

Molar heat of neutralization (heat liberated when one mole of each reagent is used;-

  • Mole = 5.72KJ

1mole = 1 x 5.72 = -57.2 KJMol-1

0.1

– Since heat is evolved, the reaction is exothermic, the molar heat of neutralization= -57.2 KJMol-1.

Thus, enthalpy change;

H+(aq) + OH(aq)                    H2O(l); ∆H(neut) = -57.2 KJMol. (Thermochemical equation)

1mole            1mole                                   1mole

Note:

Thermochemical equation: refers to a chemical equation which shows the enthalpy change.

 

Worked examples:

  1. Given, T1= 21.0oC; T2=22.0oC; T4 = 34.5oC; volume of hydrochloric acid= 100 cm3, volume of NaOH(aq) =100 cm3, molarity of the solutions are each 2M. Calculate the heat of neutralization for the two reagents. (Assumptions, density of the mixture=1gcm-3 and SHC=4.2KJKg-1K-1)

 

2 (a). When 100g of water at 94.0oC were added to a calorimeter at 17.5oC, the temperature rose to 80.5oC. Determine the heat capacity of the calorimeter. What assumption did you make in your calculations?

 

Solution:

Heat given out by water = heat received by calorimeter of heat capacity C.

Heat = MC∆T,

0.100 x 4180 x (94.0 – 80.5) = C (80.5 – 17.5)

C = 0.100 x 4180 x 13.5

62.5

= 90.288Jg-1k-1

Assumption; specific heat capacity of water is 4180jg-1k-1

 

(b). 250 cm3 of sodium hydroxide were added to 250 cm3 of hydrochloric acid in the calorimeter. The temperature of the two solutions was 17.5oC initially and rose to 20.1oC. Calculate the standard enthalpy of neutralization.

 

Solution:

– Assuming the specific heat capacities of the solutions is the same as that of water, 4180jg-1k-1.

Heat from neutralization = heat by calorimeter + solutions.

= (C∆T)   + MC∆T,

Mass of solutions = 250+250) =500 cm3

= density x volume; = 1gcm-3 x 500 cm3=500g

Hat from neutralization = 90(20.1-17.5) + (0.500 x 4180 x (20.1-17.5)

= 5670J.

 

  1. The following results were obtained in an experiment to determine the heat of neutralization of 50 cm3 2M HCl and 50 cm3 2M sodium hydroxide.

Mass of plastic cup = 45.1g

Initial temperature of acid = 27.0oC

Initial temperature of alkali = 23.0oC

Mass of plastic cup + HCl + NaOH = 145.1g

Temperature of the mixture of acid and alkali = 38.5oC.

 

(a) Define heat of neutralization;                                                                                                    {1mark}

Solution:-

Enthalpy change when one mole of water is formed from a reaction between an acid and a base.

 

(b) Write an ionic equation for the neutralization of hydrochloric acid and sodium hydroxide.  {1mark} Solution:

H+(aq) + OH(aq)                      H2O(l)

 

 

 

(c) Calculate;

(i) The amount of heat produced during the experiment.                                                            {3marks}

(Specific heat capacity of solution=4.2kjkg-1k-1, density of solution= 1gcm-3).

 

Solution:

Amount of heat = MC∆T

Mass of solution = (145.1 – 45.1) = 100g

Temperature change; ∆T = (38.5)-(27.0 +23.0) = 38.5 – 25 =13.5oC;

2

Heat produced ∆H = 100g x 4.2kjg-1k-1 x 13.5oC

=5670 joules

=5.67KJ

 

(ii). Molar heat of neutralization for the reaction.                                                                        {3marks}

 

Solution:

Number of moles involved; taking only NaOH or HCl’

1000 cm3 = 2moles

50 cm3= 2 x 50 = 0.1moles

1000

If 0.1 mole produced 5.67 kg; 1 mol = ?

0.1=5.67kg

1mol= (5.67 x 1) = 56.7 KJMol-1

0.1

 

(d). Explain why the molar heat of neutralization of NaOH and ethanoic acid of equal volume and molarity would be less than the value obtained in c (ii) above.                                                     (1mark)

 

Solution:

– Some of the heat produced during neutralization is used up by the weak acid to dissociate fully hence the lower value.

 

(e). Write down the Thermochemical equation for the reaction between NaOH and dilute hydrochloric acid above.

 

Solution

NaOH(aq) + HCl(aq)                     NaCl(aq) + H2O(l); ∆H= -56.7KJMol-1

 

(f). Draw an energy level diagram for the neutralization reaction in 4 (c ) above.

 

Na+(aq) + Cl(aq)

 

 

∆ H=-56.7KJMOL-1

H2O(l)

 

 

 

 

 

Reaction pathway

 

  1. Heat of solution/ enthalpy of solution.

– Is the heat change when a given mass (moles) of a substance is dissolved in a stated amount of solvent (water).

 

Molar enthalpy of solution:

– Is the heat change when one mole of a substance is dissolved in a stated amount of solvent (water).

– Alternatively;

– It is the heat change when one mole of a substance dissolves in water to give an infinitely dilute solution i.e. (a solution which shows no change in its properties when more water is added).

 

– Is determined as;

 

 

∆Hsolution = – ∆Hlattice + ∆Hhydration    ; when ∆Hlattice is given as a negative value;

 

Alternatively;

 

 

 ∆Hsolution = ∆Hlattice + ∆Hhydration   ; where ∆Hlattice is given a s appositive value;

 

Where,

  • Lattice energy

– Is the leaf evolved when one mole of a compound is formed from its separate gaseous ions.

Alternatively;

– It is the energy required to break the ionic bonds within a crystal (solid) lattice.

 

  • Hydration (solvation) energy,

– Is the heat evolved when one mole of ions are hydrated by water molecules.

 

Note:

– Water is a good solvent because it has a high negative(-ve) enthalpy of solvation resulting from powerful interaction between polar molecules and solute ions due to large dipoles on water molecules.

 

– Therefore,

– Heat of solution can either be exothermic or endothermic depending on the magnitude of hydration and lattice energies.

 

– Solution process occurs into two stages.

 

Example:

– Consider the dissolution of sodium chloride solid.

(i). Energy is taken in to break the crystalline lattice.

 

NaCl(s)                       Na+(g) +  Cl(g)

∆H = +∆Hlattice

=+776KJ.

 

(ii). Heat is evolved when one mole of icons are hydrated by water molecules.

Na+(g) + Cl(g)   H2O(l)          Na+(aq) + Cl(aq)

 

∆H hydration=-771KJ

 

Illustrations:

(a) Energy level diagram for an endothermic dissolving process for a solid MX(s)

 

 

M+(g) + X(g) + H2O(l)

 

Step II

Step I                                                            ∆Hhydration

 

Final state

∆Hlattice                                  M+(aq) + X(aq)

 

 

∆Hsolution

 

MX(s) + H2O(l)

 

 

 

 

 

-The lattice energy is larger than the hydration energy hence;-

∆Hsolution = +ve

 

(b) Energy level for the diagram for an exothermic dissolving process for solid MX(s)

 

 

M+(g) + X(g) + H2O(l)

 

Step I                                                           Step II

∆Hlattice                                                          ∆Hhydration

 

Initial state

MX(s) + H2O(l)

 

 

∆Hsolution

 

 

Final state

M+(aq) + X(aq)

 

 

 

 

The lattice energy is smaller than hydration energy hence;

∆H solution = -ve.

 

Note: All gases dissolve with evolution of heat;

Reason:

– There are no intermolecular forces or bonds to break before hydration occurs.

 

 

 

 

 

 

 

 

Worked examples:

  1. The equation below represents changes in physical states for iron metal.

 

Fe(s)                         Fe(l); ∆H = 15.4KJMol

Fe(l)                          Fe(g); ∆H = 354KJMol

 

Calculate the amount of heat required to change 11.2g of solid into iron to gaseous iron. (Fe=56.0)

 

Solution:

Total heat needed to convert 1 mole of iron from solid to gas = (15.4Kj + 354Kj); = 369.4Kj;

1 mole = 56g;

Thus if 56g requires 369.4Kj;

Then 11.2g =?

= 11.2 x 369.4; = 73.88Kj;

56

 

  1. The lattice and hydration enthalpies for lithium chloride and potassium chloride are given below.

 

Salt ∆Hlattice ∆Hhydration.
LiCl -861 -884
KCl -719 -695

 

(a). Calculate the enthalpy of solution, ∆Hsolution; Kj mol for;

  1. Lithium chloride. (2 marks)

 

Solution:

 

Hsolution = -∆Hlattice + ∆Hhydration

= – (-861) + (-884)

= 861- 884;

= -23Kjmol-

 

  1. Potassium chloride: (2 marks)

 

Solution:

 

Hsolution = -∆Hlattice + ∆Hhydration

= – (-719) + (-695)

= 719- 695;

= -24Kjmol-

 

(b) Which of these two salts above has a higher solubility in water? Give reasons for your answer.

 

Solution:

Potassium chloride; the difference in ∆Hsolution is greater;

 

  1. Given that the lattice energy of NaCl(s) is +77kjmol-1 and hydration energies of Na+(g) and Cl(g) are

-406kjmol-1 and -364kjmol-1 respectively. Calculate the heat of solution, ∆Hsoln for one mole of NaCl(s)

 

 

 

 

Experiment: To measure the molar enthalpy change for the dissolution (enthalpy of solution) for various compounds.

(i) Procedure;

– A clean 250 cm3 glass or plastic beaker is wrapped with a newspaper leaf.

– About 50 cm3 of tap water is measured into the beaker and the steady temperature noted.

– The beaker is held in a tilted position and 2 cm3 of and sulphuric acid added into the water.

 

(ii) Apparatus.

 

 

 

 

 

 

 

 

 

 

 

 

Caution:

– Concentrated sulphuric acid should always be added to water and never vise versa.

Reason:

 

 

– The mixture is then carefully but vigorously stirred using a thermometer and the highest temperature of the solution recorded.

– The experiment is repeated with other compounds:

  • Ammonium nitrate solid
  • Potassium nitrate solid
  • Sodium hydroxide pellets.
  • Sodium nitrate solid

 

Results and observations.

 

Concentrated sulphuric acid Ammonium nitrate solid Potassium nitrate solid Sodium hydroxide pellets
Temperature of water(˚c)
Highest temperature of solution (˚c)
Change in temperature(˚c)

 

Calculate the enthalpy change (∆Hsoln) in each experiment done.

 

Additional information;

– Density of concentrated Sulphuric acid =1.84gcm-3

– Density of water =1gcm-3

– No change in volume when any of the solids dissolved in water

– The specific heat capacities of the solutions=4.2kjkg-1k-1

 

 

(a). Sulphuric acid:

– Total volume of solution = (50 +2) =52 cm3

– Temperature change =A rise of 1ºC

– Enthalpy change; ∆Hsoln; =MC∆T

M = 52 cm3 x 1gcm-3 = 52g = 0.52kg.

C = 4.2kjkg-1k-1

∆T = 5K.

Thus enthalpy change = 0.52 x 4.2 x 1

= – 2.184kj

 

Calculating molar enthalpy of solution of concentrated sulphuric acid:

Mass of acid used = density x volume

= 1.84 x 2 = 3.68g;

Formula mass of Sulphuric acid = 98g

3.6g of acid liberated 2.184kj

Thus 98g of acid liberated; 2.198 x 98 =56Kj;

3.68

Because heat was liberated (+ve ∆T), the ∆Hsoln = -56Kjmol-1

 

(b). Sodium nitrate:

Data:

– Volume of water used = 100 cm3

– Mass of sodium nitrate = 2.5g

– Initial temperature of water = 21.0ºC

– Final temperature of water =19.5ºC

– Temperature change; = A fall of 1.5ºC i.e. (19.5 – 21.0) ºC;

– Mass of 100 cm3 of water = 100g = 0.1kg;

 

Enthalpy change i.e. heat absorbed = MC∆T

= 0.1 x 4.2 x 1.5

= 0.63Kj.

Molar heat of dissolution;

– Molar mass of NaNO3 =23 + 14 + 48 = 85g;

– 2.5g of NaNO3 absorbed 0.63Kj;

– 85g of NaNO3 will absorb =?

= 0.63 x 85

2.5

= +21.42Kjmol-1

Since heat is absorbed (-ve heat change), the reaction is endothermic thus ∆Hsoln = +21.42Kjmol-1

 

 

 

 

 

 

 

 

 

 

Note: energy level diagram for the dissolution of sodium nitrate:

 

NaNO3(aq)

 

 

 

 

∆Hsoln

 

 

 

 

 

Reaction path

 

 

  1. Potassium hydroxide;

– Calculate the molar enthalpy of dissolution of potassium hydroxide (based on values obtained).

 

  1. Heat of displacement.

– This refers to the enthalpy change that occurs when one mole of a substance is displaced from a solution of its ions.

 

Experiment: To determine the molar enthalpy change in the reaction between Cu2+ ions and zinc or iron.

 

(i) Procedure:

– A plastic cup or glass beaker is wrapped with a newspaper leaf.

– 25 cm3 of 0.2M copper (II) sulphate solution is transferred into the beaker.

– The steady temperature of the solution is noted.

– 0.5g of zinc powder are carefully transferred into the plastic cup and stirred with a thermometer.

– The highest temperature attained by the solution is recorded.

 

 

Spatula

(ii). Procedure:

 

Thermometer

 

 

 

 

 

 

 

 

 

Observations:

– The blue colors of copper (II) sulphate fades.

– Brown deposits of copper metal are formed in the plastic cup.

 

Explanation:

– Zinc is higher in the electrochemical series than copper;

– Zinc therefore displaces copper ions from its solution.

 

Equations:

Zn(s) + CuSO4(aq)                      ZnSO4(aq) + Cu(s)

 

Ionically:

Zn(s) + Cu2+(aq)                   Zn2+ + Cu(s)

 

– During the reaction the blue Cu2+ in the solution are replaced by the colorless Zn2+.

– Consequently the blue colour of the solution fades, as brown deposits of copper metal are formed in the plastic cup.

– Excess solid (zinc powder) was used to ensure complete displacement of Cu2+.

 

Results:

– Initial temperature of copper sulphate solution, T1 = 23˚C

– Highest temperature of the mixture T2 = 33ºC

– Temperature change, ∆T; = T2-T1 =10ºC

– Volume of copper sulphate solution used = 25.0 cm3.

– Mass of zinc powder taken = 0.6g

– Density of the solution = 1gcm-3.

– Specific heat capacity of the solution = 4.2Kjkg-1k-1.

 

Assumptions:

– The volume of the solution remains unchanged after the reaction.

(Cu = 63.5, S = 32, O =16, Zn = 65, Fe =56)

 

Question:

Using the above data, calculate, calculate the;

(i). Heat change for the above reaction.

(ii). The molar heat of displacement of copper (II) ions.

 

Solutions:

(i) Heat change = MC∆T

Mass = density x volume

1gcm-3 x 25cm3 =25g.

∆H = 25kg x 4.2 kjkg-1k-1x 10k

1000

= 1.050Kj;

 

(ii). Molar heat of displacement of Cu2+ displacement.

1000 cm3 = 0.2 moles.

25 cm3 = 0.2 x 25 = 0.005moles

1000

Thus,

0.005moles =1.050Kj

1mole =1 x 1.050 = 1050 = 210Kj;

0.005          5

– Since the final temperature of the mixture is higher than initial temperature; it means the Hproducts is lower than Hreactants.

– The reaction is thus exothermic.

– Thus molar heat of displacement of Cu2+ = -210Kjmol

Thermochemical equation:

 

Zn(s) + Cu2+(aq)                    Cu(s) + Zn2+(aq); ∆H=-210kJmol-1

 

Note:

– The experimental value for the heat liberated in this reaction is lower than the theoretical value of 216Kjmol-1.

Reasons:

– The heat lost to the surroundings and the heat absorbed by the apparatus is not accounted for in the calculations.

– This reaction is typically a Redox reaction.

 

i.e.                               Oxidation

 

 

Cu2+(aq) + Zn(s)                        Cu(s) +  Zn2+(aq)

 

 

Reduction.

 

  1. Heat of precipitation:

– Is the heat change which occurs when one mole of a substance is precipitated from the solution.

 

Experiment: To determine the heat of precipitation of silver chloride.

Procedure:

– 50 cm3 each of 2M AgNO3 and 2M NaCl are left in separate beakers and their constant temperatures noted and recorded.

– The NaCl solution is added to the silver nitrate solutions.

– The beaker is covered with a card board and shaken gently to allow mixing of the solution.

– The highest temperature of the mixture is noted and recorded.

 

 

(ii). Apparatus:                                                                                                                           50cm3 2M NaCl(aq)

 

Thermometer

 

 

 

 

 

 

 

Beakers

 

 

 

 

 

   50cm3 2M AgNO3(aq                    50cm3 2M HCl(aq)                               50cm3 2M AgNO3(aq)                                                                        Mixture (1ooml)

 

(ii) Data (results)

– Temperature of AgNO3 solution = 21.0ºC

– Temperature of NaCl solution =19.0ºC

– Mean temperature of the solutions = {19+21} = 20.0ºC

2

– Final temperature after mixing = 34.0ºC;

– Temperature change = a rise of 14oC;

– Volume of solution mixed =100 cm3

– Mass of solution mixed =100g (assuming density=1gcm-3)

Questions:

– From the above data, calculate;-

(i) Energy change/ heat change for the reaction

(ii) The molar ∆Hprecipitation for AgCl.

 

Solution:

(i) Heat change, = MC∆T

= 100 x 4.2 x 14

1000

= – 5.88kj

(ii) Equation,

AgNO3(aq) + NaCl(aq)                      AgCl(s) + NaNO3(aq)

Mole ratio                1   :               1       :                       1       :       1

 

Moles of AgNO3;

1000 cm3 = 2 moles

50 cm3 = 2 x 50 = 0.1mole;

 

Since reaction ratio AgNO3: AgCl = 1:1, then moles of AgCl(s) precipitated = 0.1mole

If 0.1mole = 5.88kj

1 mole = (1 x 5.88) = 58.8kJmol

0.1

Reaction being exothermic, ∆Hprecip = -58.8kJmol

 

Thermochemical equation:

 

Ag+(aq) + Cl(aq)             AgCl(s); ∆H= -58.8kjmol-

 

Suitable energy level diagram,

 

 

                        Ag+(aq) + Cl(aq)

 

 

 

∆Hprep = -58.8kJmol

 

                                                                              

AgCl(s)

 

 

 

                        Reaction pathway

 

 

  1. Enthalpy of formation.

– Is the heat change when one mole of a substance is formed from its constituent elements under standard conditions.

Note: this will be dealt with under thermochemical cycles and Hess’s law.

 

 

 

Standard conditions for measuring enthalpy changes.

– Experimental evidence shows that enthalpy changes are affected by physical stats of the reactants, temperature, concentration and pressure.

– Consequently, for comparison purposes certain conditions have been chosen as standard for measurement and determination of enthalpy changes.

– These conditions are:

  • A temperature of 25oC (298K)
  • Pressure of 1 atmosphere (101.325K)

 

Standard enthalpy changes:

– Are values of enthalpy changes that are measured under standard conditions.

– They are denoted by he symbol ∆H0

– A subscript is also added to the symbol to indicate the type of enthalpy change involved.

 

Common standard enthalpy changes.

 

(i). ∆H0f; refers to the standard molar enthalpy change of formation.

 

Example:

H2(g) + ½ O2(g)                       H2O(l); ∆H0f (H2O) = -286KJmol

 

(ii). ∆H0c; refers to the standard molar enthalpy change of combustion;

 

Example:

CH4(g) + 2O2(g)                            CO2(aq) + 2H2O(l); ∆H0c(CH4)  = -890KJmol

 

(iii). ∆H0neut; refers to the standard molar enthalpy change of neutralization.

 

Example:

HCl(aq) + NaOH(aq)                         NaCl(aq) + H2O(l); ∆H0neut = -58KJmol

 

(iv). ∆H0soln; refers to the standard molar enthalpy change of solution (dissolution).

 

Example:

NaNO3(s) + H2O(l)                             NaNO3(aq); ∆H0soln (NaNO3) = +21KJmol

 

(v). ∆H0hyd; refers to the standard molar enthalpy change of hydration.

 

Example:

Na+(s) + Cl(s)    + H2O(l)                        Na+(aq) + Cl(aq); ∆H0hyd (NaCl) = -774KJmol

 

(vi). ∆H0latt; refers to the molar enthalpy change of lattice formation.

 

Example:

Na+(g) + Cl(g)                        NaCl(aq); ∆H0latt (NaCl) = -774KJmol

 

(vi). ∆H0at; refers to the standard molar enthalpy change of atomization.

 

Example:

Na(s)                       Na(g); ∆H0at (Na) = 108.4 KJmol

 

 

 

Energy level diagrams, thermochemical cycles and Hess’s law.

Hess’s law (the law of constant heat summation);

States that:

– If a reaction can occur by more than one route the overall change in enthalpy is the same,           whichever route is followed.

 

Route 1; ∆H1

Illustration:

 

 

 

Route 2

Route 2       ∆H2                              ∆H3

 

B

 

 

 

 

Therefore,∆H1= ∆H2 + ∆H3

 

Worked examples:

  1. Use the information below to determine the enthalpy of combustion of carbon (formation of carbon (IV) oxide).

C(s) + ½O2(g)                    CO(g); ∆H = -110.45KJ

CO(g) + ½O(g)                 CO2(g); ∆H = -282.0KJ

 

Solution:

 

 

 

Route 2

Route 2       ∆H2                              ∆H3

 

CO(g)

 

 

 

 

 

Therefore; ∆H1 = ∆H2 + ∆H3

= -110-45 + (-282.0) = – 392.45kJ

∆H1 = ∆Hcombustion

= -392.45KJ;

 

Note: – Enthalpy of formation, ∆Hf;

– Is the heat absorbed or evolved when one mole of a substance is formed from its elements under standard conditions.

– This can not be determined experimentally because reactions cannot take place under standard conditions.

– In such cases the enthalpy change is determined theoretically using other measurable enthalpies.

– In the case of enthalpy of formation, the enthalpies of combustion of the compound and that of its constituent elements are linked using an energy level or energy cycle diagram.

– The enthalpy of formation is then calculated using the law of constant heat summation (Hess’s law).

 

 

 

 

 

 

Worked examples:


  1. One mole of butane(C4H10) burns completely in oxygen and liberates 2877kj.

(a) Write an equation for the combustion of butane.

2C4H10(g) + 13O2(g)                          8CO2(g) + 10H2O(l)

 

(b) Draw an energy cycle, hence an energy level diagram for the reactions concerned.

 

Solution:

Equations for combustion:

 

C(s) + O2(g)                       CO2(g); then  8C(s) + 😯2(g)                     8CO2(g) (based on balancing for butane)

H2(g) + ½ O2(g)                H2O(g); then  10H2(g) + 5O2(g)                10H2O(g); (based on balancing for butane)

2C4H10(g) + 13O2(g)                          8CO2(g) + 10H2O(l)

 

Then, energy cycle for the changes,

 

 

 

 

 

      ∆Hc;       ∆H1 = ∆H2 + ∆H3                              ∆H3 = Hc        

 

8CO2(g) + 10H2O(l)

 

 

 

 

 

 

Thus energy level diagram for the reactions.

8C(s) + 10H2O(l)

 

 

 

∆H2 = ∆Hf

 

2C4H10(l)

+½13O2(g)

∆H1 = ∆H2 + ∆H3                               ∆H3 = ∆Hc

+13O2(g)

 

8CO2(g) + 10H2O(l)

 

 

Reaction pathway (progress)

 

(c) If the following heats of combustion are given

∆Hºc (graphite) = -393kjmol

∆Hºc (H2(g)) = -286KJmol

∆Hºc (C4H10) = -2877kjmol

 

Calculate the heat of formation of butane, C4H10 from its elements in their normal states at S.T.P.

 

 

 

Solution:

 

∆H1=∆H2 + ∆H3 where:

∆H2 = ∆Hformation of C4H10;

∆H1 =∆Hºc(graphite) + ∆Hºc(hydrogen);

While ∆H3= ∆Hºc(butane)

 

From the equation,

∆H1 = ∆H2 + ∆H3,

∆H2 = ∆H1 – ∆H3

= {8(-393) +10(-286)]-[2(-2877)]

Dividing all through by 2;

= {4(-393) +5(-286)]-2877]; i.e. in order to work with 1 mole of butane;

 

= (-1572) + (-1430) – (-2877)

= -3002 + 2877

= -125KJmol

 

2 (a). Define standard heat of combustion of a substance.

 

Solution:

– The heat change when one mole of a substance is completely burnt in oxygen under standard conditions;

 

(b). Ethanol, CH3CH2OH; cannot be prepared directly from its elements and so its standard heat of formation must be obtained indirectly.

(i) Write an equation for the formation of ethanol from its elements in their normal physical states at standard conditions of temperature and pressure.

 

Solution:

 

2C(s) + 3H2(g) +  ½O2(g)                       C2H5OH(l)

 

(ii). Draw an energy level diagram linking the heat of formation with its heat of combustion and the heats of combustion of its constituent elements.

 

Solution:

Equations

(i). C2H5OH(l) + 3O2(g)                         2CO2(g) + 3H2O(l)

Balancing based on equation for combustion of ethanol

(ii). 2C(s) + 2O2(g)                                2CO2(g)

(iii). 3H2O(g) + 1½O2(g)                          3H2O(l)

 

Then, energy cycle for the changes,

 

 

 

 

+3O2(g)

      ∆Hc;       ∆H1 = ∆H2 + ∆H3                              ∆H3 = Hc        

+3O2(g)

2CO2(g) + 3H2O(l)

 

 

 

Thus energy level diagram for the reactions.

2C(s) + 3H2(l) + ½ O2(g)

 

 

 

∆H2 = ∆Hf (ethanol)

 

C2H5OH(l)

+3O2(g)

∆H1 = ∆H2 + ∆H3                               ∆H3 = ∆Hc (ethanol)

+3O2(g)

 

2CO2(g) + 3H2O(l)

 

 

Reaction pathway (progress)

 

(iii). If the following heats of combustion are given:

∆Hc(graphite) =-393kJmol

∆Hc(hydrogen) =-286kJmol

∆Hc(ethanol) = -1368kJmol

 

Calculate the heat of formation of ethanol, ∆H0f (ethanol)

 

Solution:

∆H1 = ∆H2 + ∆H3

∆H2 = ∆H1 – ∆H3

= [2(-393 + 3(-286)]-[-1368]

= [(-786) + (-856)] – [-1368]

= -164 + 1368

= – 276kJmol

 

  1. Given the standard enthalpies of combustion as

C(s) + O2(g)                CO2(g); ∆H1 = -394kJmol;

H2(g) + ½O2(g)                    H2O(l); ∆H2 = -286kJmol

C2H2(g) +  1½O2(g)                    2CO2(g) + H2O(l); ∆H = -1300kJmol.

 

Find the standard enthalpy of formation of ethyne, C2H2(g).

 

Solution:

– Considering the Hc(ethyne), 2 moles of CO2 and 1 mole of H2O(l) are produced;

– Rebalancing the equations for Hc(carbon) and Hc(hydrogen) likewise:

2C(s) + O2(g)                2CO2(g);

H2(g) + ½O2(g)                    H2O(l);

 

 

 

 

 

 

 

 

 

Then, energy cycle for the changes,

 

 

 

 

+2½O2(g)

      ∆Hc;       ∆H1 = ∆H2 + ∆H3                              ∆H3 = Hc (ethyne)        

+2½O2(g)

     2CO2(g) +  H2O(l)

 

 

 

 

Thus energy level diagram for the reactions.

 

2C(s) + H2(l)

 

 

 

∆H2 = ∆Hf (ethanol)

 

C2H2(l)

+2½O2(g)

∆H1 = ∆H2 + ∆H3                               ∆H3 = ∆Hc (ethanol)

+2½O2(g)

 

2CO2(g) + H2O(l)

 

 

Reaction pathway (progress)

 

Thus ∆H1 = ∆H2 + ∆H3

∆H2 = ∆H1 – ∆H3

= [2(-394) + (-286)]-[-1300]

= (-788-286) + 1300

= -1074 + 1300

= +226kJmol

 

 

Since ∆Hf is +ve, ethyne is described as an endothermic compound.

 

Note:

– In the determination of ∆Hf of ethyne, Hess law is used because attempts to make ethyne from carbon and hydrogen {2C(s) + H2(g)                C2H2(g) }will result to the formation of a mixture of hydrocarbons.

 

 

 

 

 

 

 

 

 

 

 

 

Heat of formation and bond energies

– Chemical reactions involve:

  • Bond breaking; this requires energy.
  • Bond formation; releases energy.

 

Example:

– Formation of methane from carbon and hydrogen.

C(s) + 2H2(g)                CH4(g); ∆H = -74.9kJmol

 

Energy level diagonal for the formation of CH4.

 

 

C(s) + 4H(g)

 

 

II   ∆Hat;

 

C(s) + 2H2(g)

 

 

 

I   ∆Hat = +715KJmol–                    III   ∆H = -4(C-H); bond energy

C(s) + 2H2(g)

 

 

∆HIV = I + II + III

Final enthalpy               = -74.9Kj

CH4(g)

 

 

Reaction pathway (progress)

 

Energy changes involved are:

  • ∆HI; Enthalpy of atomization – this is the energy absorbed when a substance decomposes to form one mole of gaseous atoms.
  • ∆HII; enthalpy of atomization of H2; i.e. dissociation of hydrogen molecules to free hydrogen atoms.
  • ∆HIII; enthalpy of formation of four C – H bonds, during which heat is liberated;

 

Therefore, enthalpy of reaction ∆HR = ∆HIV

∆HR = ∆H1+ ∆HII + ∆HIII

= +715 + 4(218) + -4(C-H)

-74.9 =+715 + 872 +-4 (bond energy)

4 (C-H) =1587 + 74.9

=1661.9

C –H =1661.9 = 415.4kjmol-

4

– This is the amount of energy released when one C-H bond is formed;

 

 

 

 

 

 

 

 

Worked examples.

 

  1. Use the bond energies given below to calculate the heat of reaction for;-

H2(g) + Cl2(g)                     2HCl(g)

 

Bond Bond energy
H-H 435kjmol
Cl-Cl 243kjmol
H-Cl 431kjmol

 

Solution:

 

Bond breaking,

Total energy absorbed = +678 KJ

H2(g) = H-H                  2H(g); ∆H = +435Kj;

Cl2(g) = Cl-Cl                2Cl(g); ∆H= +243Kj

 

Total energy released = -862 KJ

Bond formation,

2H(g) + 2Cl(g)                  2(H-Cl)(g); ∆H=2(-431) = -862Kj;

 

Heat of reaction:

Energy absorbed in bond breakage + energy released in bond formation.

= [+435 + 243] + [-862]

= [+678-862] kJ

= -184kj

Thus, H – H + Cl – Cl                 2[H – Cl]; ∆H=-184Kj

 

Note: energy level diagram.

 

 

2H(g) + 2Cl(g)

 

 

∆HII  ∆Hat (chlorine)

+243KJ

2H(g) + Cl2(g)

 

 

 

∆HI   ∆Hat = +435KJmol–                          ∆HIII = -2 (H-Cl); bond energy

H2(g) + Cl2(g)

 

 

∆HIV = I + II + III

Final enthalpy               = -74.9Kj

2HCl(g)

 

 

Reaction pathway (progress)

 

 

 

 

 

 

 

 

  1. Study the information given in the table below and answer the questions that follow;-

        

Bond Bond energy(kJmol)
C – H 414
Cl – Cl 244
C – Cl 326
H – Cl 431

(a). Calculate the enthalpy change for the reaction

    CH4(g) + Cl2(g)                      CH3Cl(g) + HCl(g)

 

Solution:

 

Bond breaking

(i). 4(C-H)                  4CH(g); ∆H =4(+414) = +1656kj;

(ii). Cl – Cl                2Cl(g); ∆H1 = +244kj

 

Bond formation

(i). 3(CH)           3(C-H); ∆H= 3(-414) =-1242kj

(ii). CCl              (C-Cl); ∆H= -326kj

(iii). HCl             (H-Cl); ∆H= -431kj

 

Heat change = ∆Hbond breakage + ∆Hbond formation

= [(+1656) + (+244)] + [(-1242) + (-326) + (-431)]

= -1900 + (-1999)

∆H = -99kJ

 

(b). Draw an energy level diagram for the reaction

 

 

 

C4H(g) + 2Cl(g)

 

 

∆HII  ∆Hat (chlorine)

+244KJ

C4H(g) + Cl2(g)

 

 

 

∆HI   ∆Hat = +1656Kj                                 ∆HIII = -2 (H-Cl); bond energy

CH4(g) + Cl2(g)                                                          = -99Kj;

 

 

∆HIV = I + II + III

Final enthalpy             

CH3Cl(g) + HCl(g)

 

 

Reaction pathway (progress)

 

 

 

 

 

 

 

  1. Study the information given in the table below and answer the questions that follow.

 

Bond Bond energy
C – C 346
C = C 610
  C – Br 280
  Br – Br 193
C – H 413

 

(i) Calculate the enthalpy change for the reaction.

 CH2=CH2 + Br2                       CH2-CH2

                                                     

                                                  Br     Br

Solution:

 

Bond breaking,

4(C-H)                4CH(g); ∆H = 4 (+413) = +1652kj

Br2                      2Br       ∆H = +193kj

C = C                       C = C; ∆H= + 610kj;

 

Bond formation:

4(C-H),                 4CH; = 4(-413) = -1662kj

2(C-Br),                2CBr; = 2(-280) = -560kj

(C-C )                   2C; = -346kj

 

Heat change:

∆H bond breakage + ∆H bond formation

= [+1652 + 193 + 610] + [-1652 + -560 + -346kJ]

= [+2455] + (-2558)

∆H change = -103kJ;

 

Molar heat of fusion, ∆Hf;

– Is the amount of heat required to change one mole of a solid into a liquid at its melting point.

– It is a measure of intermolecular forces between solid particles since it is the heat energy used to separate the solid particles.

 

Molar heat of vaporization, ∆Hvap;

– Is the amount of heat required to change one mole of liquid into vapor at its boiling point.

– It is a measure of the intermolecular forces between the liquid particles since it is the heat energy used to separate liquid particles.

 

 

 

 

 

 

 

 

Relationship between heats of fusion and vaporization to structure.

They are;-

(a). Higher for substances with giant structures e.g.

  • Ionic solids
  • Metals
  • Giant covalent structures, e.g. diamond, graphite and silicon (IV) oxide

(b). Lower in simple molecular structures whose particles are held together by weak intermolecular forces.

Examples:

  • Iodine crystals
  • Sulphur

 

Examples:

 

NaCl C(graphite) Cu H2O I(s) S
Structure type Giant ionic Giant atomic Giant Metallic Molecular Molecular Molecular
∆Hfusion kjmol 28 13 6 8
∆Hvap kjmol 171 717 305 41 21 10

 

 

UNIT 4: RATES OF REACTION AND REVERSIBLE REACTIONS

UNIT OUTLINE:

  1. Reactions rates:

Ø  Measurements of reaction rates

  • Rate of product yield pre unit time
  • Rate of disappearance of reactants

Ø  Apparatus for measuring reaction rates

  • Collision theory and activation energy
  • Factors affecting reaction rates

§  Temperature

  • Particles size (surface area)
  • Catalysts
  • Pressure

 

  1. Reversible reactants
  • Meaning
  • Chemical equilibrium

Ø  Factors affecting position of equilibrium

§  Temperature

  • Pressure
  • Concentration (of reactants or products)
  • Catalysts
  • Le’Chartaliers principle.

 

Reaction rate:

– The rate of a chemical reaction is taken as the rate at which products are formed or the rate at which reactants disappear.

 

Measurements of reaction rates

Reaction rates are measured in terms of

  • How much product appears in a given time.
  • How much reactant disappears in a given time

– Rates curves are then plotted

– These are plots of concentration or properties, which are functions of concentration against time

 

Graph: concentration of products against time.

 

 

 

 

 

 

 

 

 

 

 

 

Time

 

Graph: concentration of reactants against time.

 

 

 

 

 

 

 

 

 

 

 

 

 

Time

 

 

 

 

 

 

 

Apparatus used to determine rates of reaction

Examples:

– Reaction between dilute HCl and marble chips.

– This can be by:

(a). Measuring the decrease in mass with time

Apparatus:

 

 

 

 

 

 

 

Graph:

 

 

 

 

 

 

 

 

 

 

 

 

Time

 

(b). Measuring the volume of the gas produced with time and then plotting a graph.

Apparatus:

Graph:

 

 

 

 

 

 

 

 

 

 

 

Time

Similar experiments maybe done using zinc granules and zinc powder with an acid

 

Worked example:

Marble chips and dilute hydrochloric acid were mixed. The mass of the reaction mixture was measured and recorded with time. The results are shown in the table below.

 

Time (sec) 0 30 60 90 12 150 180 21 240 270 300
Mass of mixture (g) 42.0 41.5 41.0 40.7 40.4 4.02 40.1 40.0 40.0 40.0 40.0
Loss in mass (g)

 

(a). Complete the table.                                                                                                                 (5 marks)

 

(b). Draw a graph of loss of mass against time.                                                                            (3 marks)

 

(c). On the same axes sketch the curves that would be obtained if:

(i). the acid was more concentrated.                                                                                               (1mark)

 

(ii). larger marble chips were used.                                                                                                 (1mark)

 

Collision theory and activation energy.

The collision theory of reacting particles postulates that:

Ø  Reacting particles must collide before a chemical reaction occurs

  • Not all collisions are effective/ result in chemical reaction
  • Only those particles with sufficient energy result in effective collisions i.e. energy equal to or greater than the activation energy.
  • Any factor, which increases the rate of a chemical reaction, does so by increasing the number of effective collisions;

 

Activation energy (EA):

– Is the minimum amount of energy required by reacting particles to cause a successful collision to form products;

– It refers to the energy an energy barrier that must be overcome by the reactants to be converted to products;

– This energy barrier determines the magnitude of the activation energy of the reactants;

 

Factors determining the value of activation energy:

– Strength of the bonds in the particles of the reactants;

– Whether the reaction is exothermic or exothermic;

– Presence of catalysts;

 

 

 

 

 

 

 

 

 

Graph: activation energy:

 

 

 

 

 

 

 

 

 

 

 

 

 

Factors affecting the rate of reaction

– Factors that increase the rate of a reaction does so by:

  • Increasing number of effective collisions;
  • Lowering the activation energy; since a reaction with high activation energy is slow at ordinary conditions;

 

These factors are:

  • Concentration of reactants
  • Temperature
  • Surface area (size of the particles
  • Catalyst Light
  • Pressure

 

(a).  Concentration of reactants.

– Increase in concentration of one of the reactants increases the rate of the reaction;

Reason:

– Higher concentration results in a greater probability of collision hence a higher rate of reaction

 

Example: reaction between Mg s) and 2M and 4M hydrochloric acid.

 

Graphical representation

 

 

 

 

 

 

 

 

 

 

 

 

 

Time (seconds)

– The maximum volume of a gas collected in both experiments is the same because the number of moles of HCl is the same

– Curve for 4M HCl is steeper indicating a faster rate because the acid is more concentration

– The 4M HCl contains more H+ ions per unit volume than 2M HCl. This reaction takes a shorter time to go to completion

– The start of the flat part of the curve indicates the end of the reaction;

 

Diagram:

Magnesium and hydrochloric acid mixture

Note: Rate of reaction can also be verified by measuring the rate of disappearance of reactants per unit time.

– Reactants with 4M HCl will disappear faster;

– Curve for a faster reaction rate is always to the left of the reference curve;

 

Worked example:

In an experiment to determine the effect of concentration on the rate of a reaction, various concentrations of sodium thiosulphate were reacted with equal volume of hydrochloric acid and the time taken for the precipitate to obscure a cross on paper put under the reaction beaker was determined.

The results are shown in the table below.

 

Volume of S2O32- (cm3) Volume of H2O, cm3 Volume of HCl, cm3 Time for cross to disappear, (s)
50 0 10 20
40 10 10 25
30 20 10 34
20 30 10 52
10 40 10 70

 

(a). Plot a graph of time (vertical axis) against volume of aqueous sodium thiosulphate.             (3marks)

 

(b). From the graph estimate the time for the cross to disappear when 10 cm3 of hydrochloric acid is added to a mixture of 35cm3 of aqueous thiosulphate and 15cm3 of water.                                     (2 marks)

 

(c). What is the effect on the rate of the reaction of adding more water to the aqueous sodium thiosulphate? Explain your answer.                                                                                              (2 marks)

 

(d). On the same graph, plot the curves you would obtain if the experiment were repeated at:

(i). at 45oC;                                                                                                                                      (1mark)

 

(ii). using 10 cm3 of less concentrated hydrochloric acid.                                                              (1mark)

 

 

(b). Effect of temperature.

– An increase in temperature increases the rate of a chemical reaction.

Reasons:

– Increase in temperature increases the kinetic energy of the reacting particles;

– The particles therefore move faster leading to more frequent and more effective collisions.

– The rate of reaction therefore increases;

 

Example: reaction between sodium thiosulphate and dilute HCl(aq)

Stoichiometric equation:

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

 

Ionic equation:

S2O32- (aq)  +   2H+(aq)                    S(s)   +   SO2(g)   +  H2O(l);

 

– The time taken to precipitate the same amount of sulphur by the same volume of acid at different temperatures is measured.

– Concentration of thiosulphate solution and the acid are kept constant.

 

Graph: relationship between time and temperature in a reaction.

 

 

 

 

 

 

 

 

 

 

 

 

 

Temperature (oC)

 

– The rate is proportionate to 1/ t;

  • Hence;

– The higher the temperature the shorter the time for completion of the reaction and the higher the rate of the reaction;

  • Alternatively:

You can plot a graph of 1/ t against temperature;

 

Graphically:

Temperature (oC)

 

 

 

 

 

 

 

 

 

(c). Effect of particle size (surface area).

– A decrease in particle size (surface area) increases the rate of a chemical reaction; and an increase in particle size decreases the rate of a chemical reaction.

Reason:

– A smaller particle size means a larger total surface area and this offers a large surface on which the reacting particles can collide;

– This means more collisions, hence more chances of effective collisions leading to a higher rate of reaction;

 

Example: Reaction between dilute HCl and marble chips

The larger CaCO3 granules undergo a slower rate of reaction than the finely powdered CaCO3

 

Equation:

CaCO3  (S)     +  2H+ aq)                 Ca2+(aq)  +   CO2(g)   +   H2O(l)

 

Graph 1: volume of gas produced against time (for granules and powder)

 

 

 

 

 

 

 

 

 

 

 

 

 

Time (seconds)

 

Graph 2: loss in mass of calcium carbonate against time

 

 

 

 

 

 

 

 

 

 

 

 

 

Time (seconds)

 

 

 

 

 

Worked example:
1. exactly 3.0g of powdered carbonate of metal M of formula MCO3 were mixed with excess dilute hydrochloric acid. The mass of the reaction vessel and its contents were recorded at various times. From these readings, the total loss in mass of the reaction vessel and its contents was calculated and recorded as shown in the table below. The experiment was carried out at room temperature.

 

Time (secs) 0 30 60 90 120 150 180 210
Total loss in mass (g) 0 0.08 0.37 0.90 1.19 1.28 1.32 1.32

 

(a) (i). Plot a graph of total loss in mass against time.                                                                    (1mark)

 

(ii). Determine the total loss in mass after 100seconds.                                                                (2 marks)

 

(b) (i). Write an equation for the reaction that occurs.                                                                    (1mark)

 

(ii). calculate the number of moles of:

  • Carbon (IV) oxide gas produced.       (2 marks)

 

  • MCO3       (2 marks)

 

(iii). Calculate the relative atomic mass of metal M.                                                                     (3 marks)

 

(c). On the same axis sketch a curve that you are likely to obtain if the experiment was repeated:

(i)9. at 50oC. Explain.                                                                                                                   (2 marks)

 

(ii). Using MCO3 granules instead of the powder. Explain.                                                         (2 marks)

 

(d). Effect of catalysts

– Catalysts are substances which alter the rate of a chemical reaction without being consumed, e.g. finely divided iron in the harber process;

– They alter the rate of a chemical reaction but remain unchanged at the end of the reaction;

– They can therefore be reused at the end of the reaction;

 

Modes of action of catalysts:

  • Provision of a surface over which particles can react.

Note: Due to this solid catalysts are more effective when used in powder form;

  • Formation of short-lived intermediate compounds with products which then break up to give the products and the catalyst;
  • By adsorption;

 

Note: – Generally, a catalyst increases the rate of a reaction by providing a different pathway of lower activation energy.

– This means more collisions can overcome this energy barrier and result in a reaction.

– It lowers the activation energy of the reaction;

– Catalysts are also reaction-specific;

 

 

 

Some common catalysts:

 

Catalyst Reaction catalyzed.
Manganese (IV) oxide Dissociation of hydrogen peroxide to give oxygen gas and water;
Vanadium (V) oxide or platinum Contact process for the conversion of sulphur (IV) oxide to sulphur (VI) oxide;
Platinum Manufacture of nitric acid in the Oswald’s process;
Finely divided iron or vanadium Haber process in the manufacture of ammonia;
Nickel Hydrogenation of oils to make fats;

Hydrogenation of unsaturated hydrocarbons

Copper (II) oxide Oxidation  of ethanol to produce ethanal;
Titanium (IV) chloride Ziegler method of polymerizing alkenes

 

Graph: reaction rates with and without a catalyst.

 

 

 

 

 

 

 

 

 

 

 

 

 

Time (seconds)

 

Worked example:

Hydrogen peroxide decomposes slowly to water and oxygen under normal conditions.

When a little manganese (IV) oxide is added to the solution the rate of decomposition is enhanced. The results of an experiment on the decomposition of hydrogen peroxide in presence of manganese (IV) oxide are shown below. Study the results and answer the questions that follow.

 

Volume of oxygen (cm3) 0 19 27 33 36 38 39 40 40
Time (seconds) 0 30 60 90 120 150 180 210 240

 

(a). Plot a graph of volume of oxygen produced (vertical axis) against time.                              (3 marks)

 

(b). What is the rate of reaction:

(i). during the first 30 seconds.                                                                                                     (2 marks)

 

(ii). between 30 seconds and 60 seconds.                                                                                     (2 marks)

 

(c). Explain why: (i). the slope of the curve is steeper at the start of the reaction.                       (2 marks)

 

(ii). the curve goes flat at the end of the reaction.                                           (2 marks)

(d) (i). Suggest the role of manganese (IV) oxide I the decomposition of hydrogen peroxide.     (1mark)

 

(ii). Comment on change in mass of manganese (IV) oxide at the end of the reaction. Explain.(2 marks)

 

(e). Effect of light on the rate of reaction

-The effect of heating and illuminating substances is the same;

– In both cases the constituent particles absorb radiant energy leading to an increase in the number of particles with activation energy resulting in increased rate of reaction;

– Light energizes the particles involved in a reaction;

– This increases the chances of effective collisions per unit time thus increasing the rate of reaction.

– Light of higher frequencies give higher reaction rates e.g. UV Light;

– Examples of reactions affected by light are those involving halogens:

 

Examples of reactions affected by light.

  • Reaction between Cl2(g) and H2(g) does not take place in the dark but is explosive in bright light;

Cl2(g)  +  H2(g)    bright light            2HCl (g)

 

  • For the reaction between methane and bromine; decolourisation of bromine only occurs in presence of light;

CH4(g)  +   Br2(l)           Light        CH3Br(g)  +  HBr(g)

 

Experiment: Effect of light on the decomposition of silver bromide.

  • Procedure:

– About 20cm3 of 0.1M potassium bromide is put in a glass beaker.

– 5cm3 of 0.05M silver nitrate solution is added.

– The resulting pale yellow precipitate is divided into three portions in 3 separate test tubes.

– One of the test tubes in immediately placed in a dark cupboard; the second on a bench and the third is placed in a direct source of light e.g. sunlight.

 

– Formation of a pale yellow precipitate of silver bromide when silver nitrate reacts with potassium bromide.

  • Equation:

KBr (aq) +  AgNO3(aq)                           AgBr(s)   +    KNO3(aq)

Pale yellow

  • Test tube in light: precipitate changes colour from pale yellow to grey.
  • Test tube on the bench: slight change in colour from pale yellow to slight grey.
  • Test tube in dark cupboard: no observable (noticeable) colour change in precipitate.

 

Explanation:

– Light decomposes silver bromide to metallic silver (hence the grey colour) and bromine.

 

Equation:

Light

2AgBr(aq)                       2Ag(s)  +   Br2(g)

 grey

– No observable change in test tube placed in darkness due to lack of light;

– The degree of decomposition and hence change depends on the light intensity falling on the test tubes;

– The rate of decomposition of silver bromide increases with increase in light intensity.

 

 

 

Conclusion:

– Light affects the rate of some chemical reactions by energizing the particles involved in a reaction hence increasing the chances of effective collisions per unit time thus increasing the rate of reaction.

 

Application of effect of light on reaction rate.

– Processing of black and white photographic films is done in dark to prevent the decomposition of the silver bromide that is usually used to coat photographic plates;

 

Further examples of reactions affected by light:

(i). Cl2(g) + H2(g)            UV Light             2HCl(g)

 

CH4(g) + Br2(g)       UV Light     CH3Br(g) + HBr(g)

 

(iii). 6CO2(g) + 6H2O(l)             Light       C6H12O6(aq) + 6O2(g)

 

(f). Pressure

– Increase in pressure increases the rate of reaction involving the formation of small volume of product because of increase in concentration and slight increase in temperature

– In a given volume the higher the number of molecules of a given gas in a container, the greater the pressure;

– Increase in pressure causes the same effect as increase in concentration;

– Thus the rate of reactions involving gases can be increased by increasing the pressure of the gases

 

Note:

From gas laws, decreases in volume results in increase in pressure. Thus, reactions accompanied by decrease in volume move faster to completion.

 

 

 

 

 

 

 

 

 

 

 

Reversible reactions

– Are reactions which can be made to go to either direction (forward or backward) by changing conditions such as temperatures, pressure etc.

– In such cases none of the reactants is completely used up and so the reaction does not go into completion.

 

Note:

Reversible reactions are of two types:

  • Reversible physical changes e.g. heating ice, iodine etc.
  • Reversible chemical changes e.g. heating hydrated copper (II) sulphate, Haber process, decomposition of limestone, heating blue cobalt chloride, ammonium chloride etc.

 

Examples:

– Reaction of nitrogen and hydrogen to give ammonia in the Haber process.

N2(g)  +  3H2(g)                     2NH3 (g)

 

– The reversible sign means the reaction can reach a state of equilibrium if left undisturbed.

 

Explanations of reversible reactions:

Consider the curves below:

 

 

 

 

 

 

 

 

 

 

Time

 

Point (a): forward reaction.

– At the start of the reaction, the rate of the forward reaction is faster.

– The concentration of the reactants is greatest at the beginning.

– The rate of forward reaction and the concentration of reactants decreases with time as the reaction proceeds.

 

Point (b): backward reaction.

– At zero time the rate of the backward reaction is zero.

– The concentration of the products is lowest at this time.

– The rate of the backward reaction and concentration of products increases as the forward reaction proceeds.

 

Point (c): equilibrium.

– It comes to a time (t) when the rate of forward reaction is equal to the rate of backward reaction.

– This balance is called equilibrium.

 

 

 

Equilibrium:

– Is the state at which the rate of forward reaction is equal to the rate of backward reaction.

 

Types of equilibrium:

Static equilibrium:

– refers to a situation when two opposing forces balance each other and whatever was happening before comes to a standstill.

 

Dynamic equilibrium:

– Refers to a situation when two opposing processes, the forward and reverse, continue taking place but at the same rate.

– The equilibrium is said to be dynamic and this state of balance can be reached from either direction.

 

Examples:

3Fe (s)  + 4H2O (g)                    Fe3O4 (s)  + 4H2 (g)

– The Equilibrium state here is established when:

  • Either steam is passed over heated iron in a closed container; or
  • When hydrogen is passed over heated iron oxide;

 

Characteristics of equilibrium systems.

(i) The concentrations of reactants and products do not change after the equilibrium has been reached unless the system is disturbed.

 

Note:

– Observable properties of reactants or products such as colour, mass, volume, PH, temperature etc can be used to detect whether a system is in equilibrium or not.

Examples:

 

Property How it can be used to detect attainment of equilibrium
Colour The colour intensity remains constant at equilibrium
Volume. Total volume of solution remains constant at equilibrium;
Precipitate The height of the precipitate remains constant;

 

(ii). The equilibrium can be reached from either direction.

(iii). All the reactants and products are present in the system.

 

 

 

Factors affecting the position of equilibrium

– Such factors act as a strain on the state of equilibrium and the system reacts in a way to oppose the change.

– The factors are:

  • Temperature
  • Pressure
  • Concentration

 

(a). Temperature

  • Endothermic reactions

– Increase in temperature favours the endothermic reactions;

 

Example:

N2O4 (g)                        2NO2 (g) ; ∆H = + ve

Pale yellow                                dark brown

 

– Increase in temperature shifts the equilibrium to the right; since the reaction is endothermic and hence more yield of nitrogen (IV) oxide; hence dark brown fumes will be observed;

– Decrease in temperature shifts the equilibrium to the left;  since the reaction is endothermic favoured by low temperatures; hence formation of N2O4; and the pale yellow colour is observed;

  • Exothermic reactions:

– Decrease in temperature favours exothermic reactions.

 

Example:

2 SO2(g)   +    O2(g)                    2SO3(g); ∆H = -ve

 

Thus:

– Decrease in temperature causes more yield of sulphur (VI) oxide; because the equilibrium shifts to the right; since the reaction is exothermic favoured by low temperatures;

– Increase in temperature causes SO3 to decompose decreasing its yields. Increase in temperature shifts the equilibrium to the left; because the reaction is exothermic which is favoured by low temperatures;

 

(b). Pressure

-Increase in pressure favours the side with fewer numbers of gaseous molecules since this is the side where pressure is reduced (low).

 

Example:

  • N2 (g) + 3H2 (g) 2NH3 (g)

 

– Increase in pressure favours the production of ammonia; by shifting the equilibrium position to the right; because the volume of gaseous reactants is higher than the volume of gaseous products.

– Decrease in pressure leads to the production of less ammonia (ammonia decomposes); by shifting the equilibrium position to the left; because the volume of gaseous reactants is higher than the volume of gaseous products.

 

 

 

  • N2O4 (g) 2NO2 (g)

Pale yellow                                               dark brown

 

Increase in pressure:

– Shifts the equilibrium to the left; leading to formation of more N2O4 hence mixture turns pale yellow; because the volume of gaseous products (NO2) is higher than the volume of gaseous reactants (N2O4).

Decrease in pressure:

– Shifts the equilibrium position to the right; leading to formation of more NO2 hence colour turns dark brown; because the volume of gaseous reactants  (N2O4) is lower than the volume of gaseous products (NO2);

 

Note: –

  • Change in pressure has no effect on the equilibrium mixture where gaseous molecules on the two sides are equal

Example:

N2 (g)  +  O2 (g)                            2NO(g)

 

– Increase or decrease in pressure has no effect on position of equilibrium; because the volume of gaseous reactants is equal to volume of gaseous products;

 

Note:

– Generally reactions involving only solids and, or liquids are not affected by pressure change because they are not compressible;

 

Summary: effects of pressure on equilibrium:

 

Reaction Effects of pressure change on position of equilibrium
Increase Decrease
N2O4(g)                    2NO2(g) Equilibrium shifts to the left; more N2O4 is formed; Equilibrium shifts to the right; more NO2 is formed;
N2(g) + 3H2(g)                     2NH3(g) – More NH3 is formed; equilibrium shifts to the right; since volume of gaseous reactants is higher than volume of gaseous products – More N2 and H2 are formed; equilibrium shifts to the left; since volume of gaseous reactants is higher than volume of gaseous products
2SO2(g) + O2(g)                   2SO3(g) – More SO3 is formed; equilibrium shifts to the right (forward reaction is favoured); since volume of gaseous reactants is higher than volume of gaseous products – Less SO3 is formed (more SO2 and O2 are formed); equilibrium shifts to the left (backward reaction is favoured); since volume of gaseous reactants is higher than volume of gaseous products
4NH3(g) + 5O2(g)               4NO(g) + 6H2O(g) – More NH3 and O2 is formed; equilibrium shifts to the left; since volume of gaseous reactants is lower than volume of gaseous products – Less NH3 and O2 is formed (more NO and H2O are formed); equilibrium shifts to the right; since volume of gaseous reactants is lower than volume of gaseous products
H2(g) + Cl2(g)                     2HCl(g) – No effect on the equilibrium; because volume of gaseous reactants and volume of gaseous products is the same; – No effect on the equilibrium; because volume of gaseous reactants and volume of gaseous products is the same;

 

 

 

(c). Concentration

– Consider the equilibrium in bromine water system:

 

Br2(aq)  +  H2O(l)                              OBr(aq)+ Br(aq) + 2H+(aq)

Yellow orange                                                            Colourless

 

  • Addition of sodium hydroxide to the equilibrium:

– When NaOH(aq) is added to the equilibrium, the concentration of H+  decrease and the rate of forward reaction are favoured (equilibrium shifts to the right); the reaction of bromine with water increases and there is loss of colour of bromine water.

 

Explanation:

– Addition of sodium hydroxide provides hydroxyl ions into the equilibrium.

– The hydroxyl ions react with the hydrogen ions (on the right of equilibrium) to form water.

 

Equation:

OH(aq) + H+(aq)                         H2O(l)

 

– This process removes hydrogen ions from the equilibrium mixture.

– This shifts the equilibrium to the right hence formation of more products;

– This leads to a change in colour from yellow orange to colourless;

 

  • Addition of hydrochloric acid:

– Addition of HCl(aq) is added concentration of H+ increases; more bromine is formed and the orange-yellow colour of bromine water becomes more intense;

 

Explanation:

– Addition of hydrochloric acid introduces more hydrogen ions into the equilibrium;

– The hydrogen ions react with the colourless bromide and hypobromite ions to form yellow-orange aqueous bromine;

– This shifts the equilibrium to the left hence the increase in the intensity of the yellow-orange colour of bromine water;

 

Further examples:

 

(i). Given the equilibrium:

 

SO2(g) +  O2(g)                  2SO3(g)

 

  • Removal of sulphur (VI) oxide:

– Reducing the concentration of SO3 by removing it causes more sulphur (IV) oxide to be converted to sulphur (IV) oxide.

 

  • Addition of either oxygen or sulphur (IV) oxide.

– Addition of either sulphur (IV) oxide or oxygen to the equilibrium shifts the equilibrium to the right; due to increase of concentration of products hence more yield of sulphur (VI oxide;

 

  • Addition of pyrogallic acid:

– Addition of pyrogallic acid into the equilibrium shifts the equilibrium to the left; since pyrogallic acid dissolves oxygen gas reducing concentration of reactants on the left hence less yield of sulphur (VI) oxide;

(ii). Given the equilibrium:

 

2CrO42-(aq) + 2H+(aq)                          2Cr2O72-(aq) + H2O(l)

Yellow                                                                    Orange

 

  • Addition of sodium hydroxide to the equilibrium:

– When NaOH(aq) is added to the equilibrium, the concentration of H+  decrease and the rate of backward reaction are favoured (equilibrium shifts to the left); the reaction of dichromate ions with water to form chromate ions and hydrogen ions increases and there is change in colour to yellow.

 

Explanation:

– Addition of sodium hydroxide provides hydroxyl ions into the equilibrium.

– The hydroxyl ions react with the hydrogen ions (on the left of equilibrium) to form water.

 

Equation:

OH(aq) + H+(aq)                         H2O(l)

 

– This process removes hydrogen ions from the equilibrium mixture.

– This shifts the equilibrium to the left hence formation of more reactants (chromate and hydrogen ions);

– This leads to a change in colour from orange to yellow;

  • Addition of hydrochloric acid:

– Addition of HCl(aq) is added concentration of H+ increases; more dichromate solution is formed and the orange colour of dichromate ions become more intense;

 

Explanation:

– Addition of hydrochloric acid introduces more hydrogen ions into the equilibrium;

– The hydrogen ions react with the yellow chromate solution to form orange dichromate solution (and water);

– This shifts the equilibrium to the right hence the increase in the intensity of the orange colour of dichromate solution;

 

Generally:

– A change in concentration disturbs the already established equilibrium by making the reaction rate in one direction faster

– The reaction then proceeds predominantly in that direction until equilibrium is re established

 

(d). Effect of catalyst

– Presence of a catalyst has no effect on the position of the equilibrium but alters the rate at which the equilibrium is attained.

– Catalysts usually allow the equilibrium to be reached in a shorter period of time by increasing the rates of the reactions

 

 

 

 

 

 

 

 

Le Chatelier’s Principle

– States that:

 

 

When stress is applied to a system in equilibrium, the system reacts so as to oppose the stress.

 

 

– This implies that when a change in condition is applied to a system in equilibrium, the system moves so as to oppose the change.

Note:

– The effect of different factors on equilibrium was first investigated in 1888 by a French Chemist; Henri Louis Le Chatelier.

– All explanations so far described are based on Le Chatelier’s principle.

 

Industrial applications of chemical equilibrium.

– The ability to change the position of an equilibrium by varying the conditions has been important in industrial processes as industrialists aim at obtaining maximum products at minimum cost and shortest time possible.

– Conditions required to obtain greatest yield of products at minimum costs and shortest time possible are called optimum conditions.

– Such optimum conditions are obtained by continuous removal of products hence reduction in its concentration or varying external factors like temperature and pressure.

 

Summary:

Optimum conditions for common industrial processes.

 

Condition Optimum condition for:
Ø  Haber process

N2(g) + 3H2(g)          2NH3(g); ∆H=- 92KJMol-

Ø  The contact process

2SO3(g) + O2(g)           2SO3(g); ∆H = -197Kjmol-

Temperature – The reaction is exothermic hence favoured by low temperatures; which shift the equilibrium to the right hence more yield of ammonia gas;

– However rate at which the NH3(g) will be produced would be too slow and thus uneconomical; an optimum temperature of about 450oC is thus normally used;

– Forward reaction is exothermic thus increase in temperature favours backward reaction; shifts the equilibrium to the left hence less yield of SO2(g);

– Low temperature will favour forward reaction thus shifts the equilibrium to the right; hence more production of SO2(g); However rate of yield will be too slow and hence optimum temperature for maximum yield are set at about 450oC;

Pressure Increase in pressure favours the forward reaction; shifting the equilibrium to the right hence more yield of ammonia; because the volume of gaseous reactants is higher than the volume of gaseous products;

– However the cost of producing and maintaining high pressures in a system is very high;

– Thus for maximum yield the optimum pressure is 200 atmospheres;

– An increase in pressure will shift the equilibrium to the left leading to  low yield of sulphur (IV) oxide; since the volume of gaseous reactants is lower than the volume of gaseous products;

– Thus the optimum pressure used is atmospheric pressure, which gives a percentage conversion of SO2(g) to SO3(g) of about 96%.

Concentration – Removal of ammonia gas shifts the equilibrium to the right hence more yield of ammonia; since this removal lowers concentration of NH3(g)

– thus for maximum yield ammonia is removed as soon as it is produced; so that more H2(g) and N2(g) can continue reacting;

– Removal of sulphur (VI) oxide gas shifts the equilibrium to the right hence more yield of SO3(g); since this removal lowers its concentration;

– thus for maximum yield SO3(g) is removed as soon as it is produced; so that more SO2(g) and O2(g) can continue reacting;

Catalyst – Main catalyst used is finely divided iron; platinum is a better catalyst but is very expensive and easily poisoned by impurities, hence may increase cost of production; – A catalyst of vanadium (V) oxide is used to increase the rate of reaction;

 

UNIT 4: ELECTROCHEMISTRY

Checklist:

  1. Meaning of electrochemistry
  2. Displacement and Redox reactions.
  3. Oxidation and Redox reactions in terms of electron gain and electron loss
  4. Oxidation numbers (states)
  • Rules of assigning oxidation numbers
  • Calculating oxidation numbers
  1. Redox reactions involving halide ions (halogens)
  • Further examples of Redox reactions.
  1. Tendency of metals to from ions.
  2. Measurements of tendency of metals to ionize
  • Electrochemical cells
  • Salt bridge
  • Cell diagrams
  • Cell equations
  1. Standard electrode potential
  • Definition
  • Standard hydrogen electrode
  • The hydrogen half cell
  • Electrode potentials (negative and positive values)
  • Calculating E0 values from Redox reactions.
  1. Voltaic cells
  • Primary cells
    • Structure
    • Reactions
    • Zinc (negative) terminal
    • Brass (positive) terminal
    • Functions of various components
  • Secondary cells (lead acid accumulators)
    • Structure
    • Reactions
    • During discharge
    • During recharging;
  1. Electrolysis
  • Definition
  • Terminologies used in electrolysis (basic concepts)
  • Preferential discharge of ions
  • Electrolysis of various substances
    • Dilute sulphuric acid
    • Dilute sodium chloride
    • Concentrated sodium chloride (brine)
    • The mercury cathode cell
    • Copper (II) sulphate
      • Using inert electrodes
      • Using copper electrodes
    • Factors affecting electrolysis and electrolysis products
    • Applications of electrolysis
      • Extraction of reactive metals
      • Purification of metals
      • Electroplating
      • Anodizing aluminium utensils
      • Manufacture of sodium hydroxide, chlorine and hydrogen;
        • The mercury cathode cell
        • Diaphragm cell;
        • Membrane cell;
      • Quantitative aspects of electrolysis
        • Basic terminologies
          • The ampere
          • The coulomb
          • The faraday
        • Faradays laws of electrolysis
          • Statements

Definition:     

– Electrochemistry is the chemistry of electrochemical reactions; which deal with the relationship between electrical energy and chemical reactions.

– Electrochemical reactions involve transfer of electrons and are essentially REDOX reactions.

 

Displacement and REDOX reactions

Experiment 1:- Displacement reactions among metals

(i). Procedure

(a). 5 cm3 of 1M CuSO4 (aq) is put in a test-tube and its temperature recorded.

– In the solution, a spatula end-full of iron fillings is added.

– Any observations and temperature change are determined and recorded.

– The procedure is repeated with fresh samples of CuSO4 with Zn, Mg, and Cu powders.

 

(b). The procedure is repeated with 1M magnesium sulphate solution instead of CuSO4 (aq).

 

(ii). Observations:

 

 

 

Metal solid

Copper (II) Sulphate Magnesium Sulphate
Iron fillings – A red brown solid (Cu) is formed

– The blue colour of the solution (Cu2+) fades then changes to green (Fe2+)

– No reaction

 

Zinc powder – A red brown solid, copper metal is deposited.

– The blue colour of the solution (Cu2+) fades then turns colourless;

– No observable reaction (change)

 

Copper powder – No reaction

 

– No reaction

 

 

(iii). Explanations

– Reactions between metals and ions of another metal involve transfer of electrons from the metal to the other metal ion in solution.

 

Examples:

  • Fe(s) and CuSO4 (aq)

– Copper being lower in the electrochemical series accepts electrons easier (than Fe) to form copper atoms (brown solid);

 

Half equations

Fe (s)                          Fe2+(aq) + 2e

 

Cu2+(aq) + 2e                        Cu (s)

 

Overall reaction        

Cu2+(aq)  +   2e-  + Fe(s)                            Fe2+(aq)  + 2e +  Cu(s)

Then;

 

Cu2+(aq)  +  Fe(s)                                  Fe 2+ (aq)  +  Cu  (s)

(Blue)                                                                   (Green)               (Brown solid)

 

 

Oxidation and reduction in terms of electron loss and gain

– The loss of electrons is oxidation and the species that gains electrons (causes electron loss); Cu2+ in this case is called oxidizing agent; and is itself reduced.

– Reduction: refers to gain of electrons, and the species that donates electrons (iron solid in this case) is called a reducing agent and is itself oxidized.

– Displacement reactions generally involve reduction and oxidation simultaneously and are thus termed Redox Reactions.

 

Further examples

(i). Zinc solid and CuSO4(aq)

 

(i). Zn(s)                        Zn2+(aq)  +  2e(Oxidation)

(ii). Cu2+(aq)   +  2e               Cu(s) (Reduction)

 

Oxidation

 

 

Then:   Cu2+(aq) +  Zn(s)             Zn2+(aq) + Cu(s) (Redox)

                Blue                                                           Colourless;       Red brown

 

 

Reduction

 

(ii). Magnesium solid and copper (II) sulphate

 

(i). Mg(s)                       Mg2+(aq)  +  2e(Oxidation)

(ii). Cu2+(aq)   +  2e               Cu(s) (Reduction)

 

                           Oxidation

 

 

Then:   Mg(s) + Cu2+(aq)                         Mg2+(aq) + Cu(s) (Redox)

                                      Blue                                     Colourless;       Red brown

 

 

                                                   Reduction

 

(iii). Silver nitrate and copper solid.

 

(i). Cu(s)                       Cu2+(aq)  +  2e(Oxidation)

(ii). 2Ag+(aq)   +  2e              2Ag(s) (Reduction)

 

                                   Oxidation

 

 

Then:   Cu(s) + 2Ag+(aq)                         Cu2+(aq) + 2Ag(s) (Redox)

                Brown       Colourless                                     Blue                Grey

 

 

                                                   Reduction

 

Note:

– Amount of heat evolved in these redox reactions depends on the position of the metal in the activity series relative to the metal ion in solution.

– The closer the metals are in the activity series; the less readily displacement occurs and the lower the heat evolution during the displacement.

E.g.:    Heat evolved Mg//Cu2+ is higher than that evolved between Fe//Cu2+.

 

Conclusion:

– Metals displace from solutions, those metals lower than themselves in the activity series.

Note:

– The more the reactive a metal is; the stronger a reducing agent it is and the weaker an oxidizing agent it is.

 

Example:

Potassium is stronger reducing agent; but weaker oxidizing agent than silver, gold etc.

 

Summary:

Strength of reducing/ oxidizing agent

 

 

                                         Weakest oxidizing agent Potassium, K        Strongest reducing agent

Sodium, Na

Magnesium, Mg

Aluminium, Al

Zinc; Zn

Iron, Fe

Lead, Pb

Copper, Cu

Silver, Ag

                                       Strongest oxidizing agent Mercury, Hg          Weakest reducing agent

 

Summary on displacement reactions

 

 

 

Metal ion (in solution)

Mg Al Zn Fe Pb Cu
K+ C C C C C C
Na+ C C C C C C
Ca2+ C C C C C C
Mg2+ C C C C C C
Al3+ C C C C C
Zn2+ C C C C
Fe2+ C C C
Pb2+ C C
Cu2+ C
Ag2+

 

Key:

A cross (x) indicates no reaction hence no redox reaction occurs.

A tick indicates redox reaction occurs.

 

 

 

 

 

 

 

 

Oxidation numbers/ oxidation state.

– Is the apparent charge that atoms have in molecules or ions.

– For monoatomic ions, the oxidation number (state) is the magnitude and sign of charge;

 

Example:

Oxidation no of Aluminium in Al3+ is +3;

 

Importance of oxidation numbers:

– Helps in keeping track of electron movement in redox reactions; hence determination of the reduced and oxidized species.

 

Oxidation and reduction in terms of oxidation numbers

  • Oxidation            

– Is an increase in oxidation number.

 

  • Reduction

Refers to a decrease in oxidation number

 

Rules in assigning oxidation numbers

  1. Oxidation number of an uncombined element is zero (0)
  2. The charge on a monoatomic ion is equivalent to the oxidation number of that element;
  3. The oxidation number of hydrogen in all compounds is +1 except in metal hydrides where its –1;
  4. The oxidation number of oxygen in all compounds is –2 except in peroxides where it is –1 and 0F2 where it is +2.
  5. In complex ions the overall charge is equal to the sum of the oxidation states of the constituent elements.
  6. In compounds, the sum of oxidation numbers of all constituent atoms is equal to zero.

 

Worked examples

  1. Calculate the oxidation number of nitrogen in:

(i). NO3

 

Solution:

N + (-2 x 3) = -1

N = -1 + 6

= +5

Note: thus nitric acid with a nitrate ion (NO3) is called nitric (V) acid since the oxidation number of nitrogen in it is +5;

 

(ii). NO2;

 

Solution:

N + (-2 x 2) = 0

N = 0 + 4

= +4

Note: Thus the gas NO2 is referred to as nitrogen (IV) oxide because the oxidation umber of nitrogen in it is +4

 

(iii). NO2;

 

Solution:

N + (-2 x 2) = -1

N = -1 + 4

= +3

Note: thus nitrous acid containing nitrite ion is called nitrous (III) acid since the oxidation number of nitrogen in it is +3.

 

(iii). AgNO3;

 

Solution:

1 + N + (-2 x 3) = 0

1 + N + (-6) = 0

N = 0 – 1 + 6;

N = +5

 

  1. Determine the oxidation number of manganese in each of the following, and hence give the systematic names of the compounds.

(i). MnSO4

 

Solution:

Mn + 6 + (-2 x 4) = 0

Mn = 0 – 6 + 8;

Mn = +2

Systematic name: Manganese (II) sulphate;

 

(ii). Mn2O3;

 

Solution:

2Mn + (3 x -2) = 0;

2Mn = 0 + 6

Mn = ½ x 6;

Mn = +3;

Systematic name: Manganese (III) oxide;

 

(iii). KMnO4

 

Solution:

1 + Mn + (-2 x 4) = 0

Mn = 0 – 1 + 8;

Mn = +7;

Systematic name: Potassium manganate (VII) oxide

 

(iv). MnO3;

 

Solution:

Mn + (-2 x 3) = -1

Mn = -1 + 6;

Mn = +5

Systematic name: Manganese (V) ion;

 

 

 

 

 

 

 

 

 

 

Determination of redox reactions using oxidation numbers.

Worked examples:

 

 

 

 

 

Redox reactions involving Halide ions and halogens

Experiment

(i). Procedure:

– 2 cm3 of chlorine gas are bubbled into each of the following solutions: – KI, KCl, KBr, and KF.

– The observations are made and recorded.

– The procedure is repeated using fluorine, bromine and iodine in place of chlorine.

 

Precaution:   

– Chlorine and bromine are poisonous.

 

(ii). Observations

 

 

 

Halogen

Potassium fluoride Potassium Chloride Potassium Bromide Potassium  Iodide
Fluorine (F2) No visible, change Green-yellow gas is evolved Colourless solution  changes to red brown Colourless solution turns black;
Chlorine (Cl2) No visible colour change; No visible colour change; Colourless solution turns red brown Colourless solution turns  black
Bromine (Br2) No visible colour change; No visible colour change; No visible colour change; Colourless solution turns black;
Iodine (I2) No visible colour change; No visible colour change; No visible colour change; Colourless solution turns black;

 

Note:   Colours of halogens in tetra – chloromethane:-

 

Halogen Colour in tetrachloromethane
Fluorine
Chlorine Yellow
Bromine Red-brown
Iodine Purple

 

(iii). Explanations

– Fluorine displaces all the other halogens; Cl2, Br2 and I2 because it has a greater tendency to accept electrons than all the rest.

– Chlorine displaces both Bromine and Iodine from their halide solutions

– Cl2 takes electrons from the bromide and iodide ions i.e. oxidizes them, to form bromine and iodine respectively.

 

Equations:

(i). Chlorine and potassium bromide:

Cl2 (g) + 2KBr (aq)                           2KCl (aq) + Br2 (l)

 

Ionically:

Cl2 (g) + 2Br(aq)                    2Cl(aq) + Br2 (l)

Green-yellow                                                                           Red brown

 

Redox equation:                    Reduction

 

 

Cl2 (g) + 2Br(aq)                         2Cl(aq) + Br2 (l)

 

 

Oxidation

 

(ii). Chlorine and Potasium iodide:

Cl2 (g) + 2KI (aq)                        2KCl (aq) + I2 (l)

 

Ionically:

Cl2 (g) + 2I(aq)                                  2Cl(aq) + I2 (l)

Green-yellow                                                                           Black

 

Half cell reactions:

  • Oxidation: 2I(aq)        I2(aq) + 2e-
  • Reduction: Cl2 (g) + 2e- 2Cl(aq)

 

Redox equation:                    Reduction

 

 

Cl2 (aq) + 2I(aq)                          2Cl(aq) + I2 (l)

 

 

Oxidation

 

– Bromine takes electrons form iodide ions but not from fluorine and chlorine.

– Iodine is formed i.e. due to oxidation of iodide ions by the Bromine.

 

Equations

(ii). Bromine and Potasium iodide:

Br2 (g) + 2KI (aq)                        2KBr (aq) + I2 (l)

 

Ionically:

Br2 (g) + 2I(aq)                                  2Cl(aq) + I2 (l)

Red brown                                                                              Black

 

Half cell reactions:

  • Oxidation: 2I(aq)        I2(aq) + 2e-
  • Reduction: Br2 (g) + 2e- 2Br(aq)

Redox equation:                    Reduction

 

 

Cl2 (aq) + 2I(aq)                          2Cl(aq) + I2 (l)

 

 

Oxidation

Note: oxidation number of chlorine decreases from 0 to -1 hence reduction; while oxidation number of iodine increases from -1 to 0; hence oxidation;

 

(iv). Conclusion:       

– The stronger the tendency of an element to accept electrons, the stronger is its oxidizing power.

– Fluorine is the strongest oxidizing agent of the 4 halogens considered.

 

 

 

 

 

Order of oxidizing power for halogens.

Fluorine; F2;

Chlorine; Cl2     Increasing oxidizing power.

Bromine; Br2

Iodine; I2

 

Further examples of redox reactions

(a). Action of acid on metals

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

 

Ionically:

Mg (s) + 2H+(aq)                                Mg2+(aq) + H2 (g)

 

Half cell reactions:

  • Oxidation: Mg(s)       Mg2+(aq) + 2e-
  • Reduction: 2H+(g) + 2e- H2(g)

Redox equation:                    Reduction

 

 

2H+ (aq) + Mg(s)                           H2 (aq) + Mg2+ (aq)

 

 

Oxidation

Note: oxidation number of hydrogen decreases from 1 to 0 hence reduction; while oxidation number of magnesium increases from 0 to 2; hence oxidation;

 

(b). Reaction of active metals with water

Example

2Na (s) + 2H2O (l)                           2NaOH (aq) + H2 (g)

 

Ionically:

2Na(s) + 2H2O (l)                                2Na+(aq) + 2OH(aq) + H2 (g)

 

Half cell reactions:

  • Oxidation: 2Na(s)        2Na+(aq) + 2e-
  • Reduction: 2H2O (g) + 2e- 2OH(aq) + H2(g)

Redox equation:                    Oxidation

 

 

2Na(s) + 2H2O (aq)                     2Na+ (aq) + 2Cl(aq) + I2 (l)

 

 

Reduction

Note: oxidation number of water decreases from 0 to -1(total) hence reduction; while oxidation number of sodium increases from 0 to 1; hence oxidation;

 

 

 

 

 

 

(c). Reaction of heated Iron with dry chlorine

2Fe (s) + 3Cl 2(g)                           2FeCl 3(s)

 

Ionically (assumed):

2Fe (s) + 3Cl2(aq)                             2Fe3+ (aq) + 6Cl (g)

 

Half cell reactions:

  • Oxidation: 2Fe(s)       2Fe3+(aq) + 6e
  • Reduction: 3Cl2(g) + 6e H2(g)

Redox equation:                    Oxidation

 

 

2Fe (s) + 3Cl2(g)                         2Fe3+ (aq) + 6Cl (aq)

 

 

Reduction

Note: oxidation number of chlorine decreases from 0 to -1 hence reduction; while oxidation number of iron increases from 0 to 3; hence oxidation;

 

 

(d). Reaction between Bromine and Iron (II) ions

Ionically:

Br2(l) + 2Fe2+(aq)                               2Fe3+(aq) + 2Br (aq)

 

Half cell reactions:

  • Oxidation: Br2(l)       2Br(aq) + 2e-
  • Reduction: 2Fe2+(g) + 2e- 2Fe3+(aq)

Redox equation:                    Reduction

 

 

Br2 (aq) + 2Fe2+(s)                         2Br (aq) + 2Fe3+ (aq)

 

 

Oxidation

 

Note: oxidation number of bromine decreases from 0 to -1 hence reduction; while oxidation number of Fe2+ increases from 2 to 3 (in Fe3+); hence oxidation;

 

(e). Oxidation by potassium Manganate (VII) (KMnO4)

Procedure:    

– Purple Potassium manganate (VII) is added into a solution containing iron (II) ions in a test tube.

– A few drops of concentrated sulphuric (VI) acid are added.

 

Observations:

– The purple solution (containing Manganate (VII) ions) turns to colourless (manganate (II) ions) i.e. the purple solution is decolourised;

 

Explanation:

– The Manganate (VII) ions which give the solution a purple colour are reduced to Manganese (II) ions which appear colourless.  This is a redox reaction.

 

 

Equations

Ionically:

MnO4 (aq) + 8H+(aq) + 5Fe2+(aq)                               Mn2+(aq) + 5Fe3+(aq) + 4H2O (l)

Purple                                                                                               Colourless

 

Half cell reactions:

  • Oxidation: 5Fe2+(s)       5Fe3+(aq) + 5e; (since number of electrons is 5)
  • Reduction: MnO4(aq) + 8H+(aq) + 5e                     Mn2+(aq) + 4H2O(l)

Redox equation:                    Reduction

 

 

MnO4 (aq) + 5Fe2+(aq)                    Mn2+(aq) + 5Fe3+ (aq)

 

 

Oxidation

Note:

  • oxidation number of Manganate ions in KMnO4 decreases from 7 to 2 (in Mn2+) hence reduction; while oxidation number of iron increases from 2 (in Fe2+) to 3 (in Fe3+); hence oxidation;

 

  • The presence of Fe3+ at the end of the reaction can be detected by adding sodium hydroxide solution to form a red brown precipitate of Fe(OH)3;

 

(f). Action of potassium dichromate (VI) on iron (II) ions (Fe2+):

(i). Procedure:

– A solution containing iron (II) ions is added into a solution of oxidized potassium dichromate (VI).

 

(ii). Observations:

– The orange solution of potassium dichromate turns green.

 

(iii). Explanations:    

– The iron (II) ions are oxidized to iron (III) ions

– The chromium (VI) ions (orange) are reduced to chromium (III) ions

– This is thus a REDOX reaction.

 

Equations

Ionically:

Cr2O72- (aq) + 14H+(aq) + 6Fe2+(aq)                               2Cr3+(aq) + 6Fe3+(aq) + 7H2O (l)

Orange                                                                                                  Green

 

Half cell reactions:

  • Oxidation: 6Fe2+(s)       6Fe3+(aq) + 6e; (since number of electrons is 6)
  • Reduction: Cr2O72-(aq) + 14H+(aq) + 5e            2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)

Redox equation:                    Reduction

 

 

Cr2O72- (aq) + 6Fe2+(aq)                  2Cr3+(aq) + 6Fe3+ (aq)

 

 

Oxidation

 

Note:

  • The oxidation number of dichromate ions in K2Cr2O7 decreases from 6 to 3 (in Cr3+) hence reduction; while oxidation number of iron increases from 2 (in Fe2+) to 3 (in Fe3+); hence oxidation;

 

  • The presence of Fe3+ at the end of the reaction can be detected by adding sodium hydroxide solution to form a red brown precipitate of Fe(OH)3;

 

(g). Action of acidified potassium permanganate on Hydrogen Peroxide.

– The overall reaction is a Redox reaction

Redox equation

2MnO4 (aq) + 5H2O (l) + 6H+ (aq)                              2Mn2+ (aq) + 5H2O (l) + 5O2 (g)

Purple                                                                                               Colourless

 

Half cell reactions:

  • Oxidation: 5H2O2(aq)            10H+(aq) + 5O2(g) + 10e;
  • Reduction: 2MnO4(aq) + 16H+(aq) + 10e              2Mn2+(aq) + 8H2O(l)

Redox equation:                    Reduction

 

 

2MnO4 (aq) + 5H2O2 (aq)                 2Mn2+(aq) + 10H+ (aq) + 5O2(g)

 

 

Oxidation

 

Note:

  • The oxidation number of manganate ions in KMnO4 decreases from 7 to 2 (in Mn2+) hence reduction; while oxidation number of hydrogen increases from -1 (in H2O2) to 1 (in H+); hence oxidation;
  • Thus the acidified potassium manganate (VII) oxidizes hydrogen peroxide to water and hydrogen;

 

(h). H2O2 oxidizes Iron (II) salts to Iron (III) salts in acidic medium

Oxidation:

2Fe2+ (aq)                            2Fe3+ (aq) + 2e

 

Reduction:

2H+ (aq) + H2O2(aq) + 2e                     2H2O(l)

 

Overall redox:

2Fe2+ (aq) + H2O2 (aq) + 2H+ (aq)                    2Fe3+ (aq) + 2H2O (l)

 

 

 

 

 

 

 

 

 

Terms used in describing oxidation – reduction

 

Term Electron change Oxidation number change
Oxidation

Reduction

Oxidizing agent

Reducing agent

Substances oxidized

Substance reduced

loss of electrons

gain of electrons

receives electrons

loses electrons

loses electrons

gains electrons

increases

decreases

decreases

increases

increases

decreases

 

The tendency of metals to form ions

– When a  metal  is placed in an aqueous solution of its ions, some of the metal dissolves;

Equation:

M(s)                          Mn+ (aq) + ne

 

– Dissolution of the metal causes electron build up on its surface; making it negatively charged, while the surrounding solution becomes positively charged.

 

Diagram: dissolution of metal and electron build up.

 

 

                                                            Metal rod

 

 

 

 

                                                                 Solution containing metal ions

 

 

 

– The positive charge of the solution increases and some of the cations start recombining with the electrons on the metal surface to form atoms.

 

Equation:

Mn+ (aq) + ne                          M(s)

 

– Consequently, an electric potential difference is created between the metal rod and the positively charged ions in solution.

– This arrangement of a metal rod (electrodes) dipped in a solution of its ions constitutes a half – cell.

 

Note:  

– The tendency of a metal to ionize when in contact with the ions differs form one metal to another.

– This difference can be measured by connecting two different Half – cells to make a full cell.

– The electrodes of the 2 half – cells are connected by a metallic conductor; while the electrolytes (solutions) of the half cells are connected through a salt bridge.

 

 

 

 

 

 

 

 

 

Diagram: Connection of two half cells to a full cell.

Experiment: – to measure the relative tendency of metals to ionize

(i). Procedure:

– A Zinc rod is placed into 50 cm3 of 1M zinc sulphate in a beaker;

– Into another beaker containing 50 cm3 of 1M CuSO4 (aq), a copper rod is dipped;

– The two solutions are connected using a salt bridge.

– The two metal rods are connected through a connecting wire connected to a voltmeter

– The experiment is repeated using the following half-cells instead of the Zinc-half cells:-

  • Mg rod dipped in 1M MgSO4 (aq)
  • Lead dipped in 1M lead Nitrate
  • Copper dipped in 1M CuSO4 (aq)

 

(ii). Apparatus

Diagram:

(iii). Observation

– The zinc rod in the zinc-zinc ions half – cell dissolves;

– The blue colour of the copper (II) Sulphate solution fades/ decrease;

– Red-brown deposits of copper appear on the copper rod in the copper-copper ions half-cell.

– A voltage of 1.10 V is registered in the voltmeter.

 

 

 

(iv). Equations/ reactions, at each half cell

  • In zinc-zinc ions half cell

Zn(s)                             Zn2+ (aq) + 2e

 

  • In the copper-copper ion half cell

Cu2+ (aq) + 2e-                           Cu(s);

 

(v). Explanations:     

– Zinc rod has a higher tendency to ionize than the copper rod, when the metal rods are placed in solutions of their ions.

– Thus the zinc rod has a higher accumulation of electrons than the copper rod.

– This makes it more negative compared to the relatively more positive copper rod, which has a lower accumulation of electrons.

– On connecting the 2 half cells; electrons will flow form the zinc rod to the copper rod through the external wire.

– The copper rod gains the electrons lost by the Zinc rod.

 

Roles of the slat bridge:       

Note: It forms a link between the 2 half cells, thereby completing the circuit

– It compete the circuit by:

  1. Allowing its ions to carry charge from one half – cell to the other.
  2. Maintaining the balance of charge in the two half-cells; by providing the ions which replace those used up at the electrodes.

 

– The overall reaction in the cells is a Redox  reaction

Half cell reactions:

Zn(s)                             Zn2+ (aq) + 2e (oxidation)

 

Cu2+ (aq) + 2e-                           Cu(s); (reduction)

 

Overall redox equation

                                                  Oxidation

 

 

Zn (s) + Cu2+ (aq)                          Zn2+ (aq) + Cu (s)

 

 

Reduction

 

– The voltage of 1.10 V registered in the voltmeter is a measure of the difference between the electrode potential (Eθ) of Zinc and Copper electrodes, i.e. the potential difference/ the Electromotive force.

– Thus: An electrochemical/ voltaic cell;

  • Is the combination of two half –cells to give a full cell capable of generating an electric current from a redox reaction.

 

 

 

 

 

 

Cell diagram for a voltaic cell

Rules/conventions for cell representation

  1. A vertical continuous line (/); represents the metal-metal ion or metal ion-metal interphase.
  2. Vertical broken line ( ­); between the 2 half-cells; represents the salt bridge.

 

Note: The salt bridge may also be represented by two unbroken parallel lines (//).

 

Example: – cell diagram for the above cell;

Zn(s) / Zn2+ (aq)     Cu2+ (aq) / Cu(s);

 

Alternatively:

Zn(s) / Zn2+ (aq) // Cu2+ (aq) / Cu(s);

 

Electrode potential E0, values of other metal – metal ions relative tot he Cu/Cu 2+ half cell

Metal / metal ion half cell Electrode potential Eθ relative to Cu2+ / Cu half cell
Mg(s) / Mg2+ (aq) +2.04
Zn(s) / Zn2+ (aq) +1.10
Pb(s) / Pb2+ (aq) +0.78
Cu(s) / Cu2+ (aq) +0.00
Ag(s) / Ag+ (aq) -0.46

 

Positive and negative E values

 

(i). Positive E values –          

– If the Eθ value for a metal / metal ion is positive then the metal undergoes oxidation (loses electrons) while the reference electrode undergoes reduction (accepts electrons)

 

Example:

– The Eθ value for Zn(s) / Zn2+ (aq) relative Cu2+ (aq) / Cu(s) is positive because the zinc metal is oxidized to zinc ions while the copper ions are reduced to copper metal.

 

(ii). Negative E values:

– Implies that the reference half cell undergoes oxidation (donates electrons) while the other metal ions in the other half cell undergoes reduction (accepts electrons)

 

Example:

– The E value for Ag (s) / Ag+ (aq) is negative because Cu is more reactive than silver and gives out electrons (oxidation); while the less reactive Ag has its ions accepting electrons (reduction) to form Ag solid.

 

(iii). The 0 (Zero) E value:

Always indicate the reference electrode / half-cell; in which case there would be no potential difference (with itself)

 

Example:

– The 2 half  cells of Cu(s) / Cu2+(aq) or Cu2+(aq) / Cu(s)  have no potential difference in between them hence a zero (0) E value.

 

Note:

  1. – Any other element could be chosen as the reference electrode in place of copper Cu and difference electrode potentials values would be obtained for the same elements.

 

  1. – The electrode potential of a single element is usually determined by measuring the difference between the electrode potential of the element and a chosen standard electrode.

 

  1. – This gives the standard electrode potential (E0) of the element.

 

The standard electrode potential (E0)

Definition

– Is the potential difference for a cell comprising of a particular element in contact with 1 molar solution of its ions and the standard hydrogen electrode.

– It is denoted with the symbol E0.

 

Importance

– It is useful in comparing the oxidizing and reducing powers of various substances.

 

The standard Hydrogen Electrode

– Is the hydrogen half-cell, which has been conventionally chosen as the standard reference electrode.

– It has an electrode potential of zero at:-

  • A temperature of 25oC
  • A hydrogen pressure of 1 atmosphere
  • A concentration of 1M hydrogen ions

 

Note:   The ions in the other half-cell must also be at a concentration of 1 molar.

 

Components of the Hydrogen half – cell;

– Consist of an inert platinum electrode immersed in a 1M solution of Hydrogen ions

– Hydrogen gas at 1 atmosphere is bubbled into the platinum electrode.

– The hydrogen is adsorbed into the platinum surface and an equilibrium (state of balance) is established between the adsorbed layer of molecular hydrogen and hydrogen ions in the solution.

Equation:

½ H2 (g)                                   H+ (aq)   + e

 

Platinised platinum

– Is platinum loosely coated with finely-divided platinum.

– This enables it to retain comparatively large quantity of hydrogen due to its porous state.

– Platinised platinum also serves as a route by which electrons leave or enter the electrode.

– The hydrogen electrode is represented as: H2 (g)  / H+(aq); 1M

 

 

 

 

 

 

 

 

 

Diagram: The standard hydrogen electrode:

 

 

 

 

 

 

 

 

 

 

 

– The electrode potential of any metal is taken as the difference in potential between the metal electrode and the standard hydrogen electrode.

 

Negative and positive electrode potentials

(a). Negative electrode potential

– If the metal electrode has a higher/ greater tendency to loose electrons than the hydrogen electrode; then the electrode is negative with respect to hydrogen electrode; and its electrode potential is negative.

 

Examples: Zinc, Magnesium etc.

 

(b). Positive electrode potential

If the tendency of an electrode to loose electrode is lower than the hydrogen electrode, then the electrode is positive with respect to the hydrogen electrode; and its potential is positive.

 

Examples: copper, silver etc

 

Reduction potentials

– Is a standard electrode potential measured when the electrode in question is gaining electrons.

– The lower the tendency of an electrode to accept/ gain electrons; the lower (more negative) the reduction potential and vise versa.

 

Examples:

K+ (aq)   + e (aq)                                      K(s);         E0 = -2.92V

F2 (g)      + e (aq)                                      2F(aq);     E0 = +12.87V

Mg (s)    + 2e                                       Mg2+ (aq); Eθ = -2.71 V

 

– Thus potassium ions with E0 = -2.92V have a lesser tendency to gain electrons than magnesium ions.

– Thus Potassium is the weakest oxidizing agent; but the strongest reducing agent, since it has the greatest tendency to donate electrons.

 

Note:

– Oxidation potentials will be the potentials of electrodes measured when they are losing electrons hence undergoing oxidation.

 

 

 

Standard electrode potentials  reduction potentials) of some elements

 

 

  Reduction equation Eθvolts
Least readily reduced

 

Most readily reduced

 

 

 

 

  

                                – Increasing strength of oxidizing agent

                                – Decreasing strength of oxidizing agent

 

F2 (g)     +  2e                           2F (aq)

+2.87
Cl2 (g)   + 2e                            2Cl (aq) + 2.87
Br2(g)     +  2e                            2Br (aq) + 1.36
Ag+(aq)+ 2e                              Ag (s) + 0.80
I2 (g)      +  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+ (g) +  2e                           Zn (s) – 0.76
Al 3+ (aq)+   3e–                          Al (s) – 1.66
Mg2+ (aq) + 2e                          Mg (s) – 2.71
K+ (aq)    + e                             K (s) – 2.92

 

Note:

– The standard electrode potentials in the above table are reduction potentials.

– The greater the tendency to undergo reduction, the higher (more positive) the Eθ value.

– The reverse reaction (oxidation) would have a potential value equal in magnitude but opposite in sign to the reduction potential.

 

Example:

Zinc

Reduction potential

Zn2+ (aq)    +   2e-                                 Zn (s);   E0 = -0.76V;

 

Oxidation potential

Zn (s)                                    Zn2+ (aq)   +   2e; Eθ = + 0.76V;

 

Uses of E0 values

  1. Comparing the reducing powers and oxidizing powers of various substances;
  2. Predicting whether or NOT a stated REDOX reaction will take place.

 

 

 

 

 

 

The E0 value for a REDOX reaction

– Is usually calculated as the sum of the E0 value for the half cells involved.

 

Note:

– If the sum is positive then the reaction can occur simultaneously;

– If the value of the sum is negative the reaction cannot occur;

 

Sample calculations

  1. Cu 2+ can oxidize Zinc but Zn 2+ cannot oxidize Cu;

(a). Cu 2+/ Cu // Zn / Zn2+

Cu 2+ (aq) + 2e                                Cu (s); E0 = + 0.34 V

Zn (s)                                2e + Zn 2+ (aq); E0 = + 0.76 V

Cu 2+ (aq) + Zn (s)                       Cu (s) + Zn (aq); E0 = – 1.10 V

 

– The overall reaction is positive, hence zinc can be oxidized by copper (II) ions, and hence reaction occurs;

 

(b). Zn 2+ (aq) / Zn (s) // Cu (s) / Cu 2+ (aq)

Zn2+ (aq) + 2e-                             Zn(s);  E0 = – 0.76 V

Cu(s)                             2e + Cu2+ (aq);  E0 = – 0.34 V

 

Zn 2+ (aq) + Cu(s)                   Zn (s) + Cu 2+ (aq) ; E0 = – 1.10 V

 

 

– The overall E0 is negative; thus Zn 2+ cannot oxidize Cu to Cu2+ (Cu cannot reduce Zn 2+ to Zn); and hence the reaction cannot occur.

 

  1. Can chlorine displace bromine form bromide solution?

Cell diagram: Cl2 (g) / Cl(aq) // 2Br(aq) / Br2 (g)

Cl2 (g) + 2e                                 2Cl (aq); E0 = + 1.36 V

2Br (aq)                     2e + Br2 (g); E0 = – 1.09 V

Cl2 (g) + 2Br (aq)                     2Cl(aq) + Br2 (g); E0 = +0.27 V

 

– The overall E0 for the reaction is positive, so chlorine can displace bromine from a bromide solution.

 

 

 

 

 

 

 

 

 

 

 

Worked examples

  1. The diagram below represents part of the apparatus to be used for the determination of the standard electrode potential of Aluminum, Eθ Al 3+(aq) /Al (s)

 

 

 

 

 

 

 

 

 

 

 

 

(a). Name the solutions which  could be placed in beakers A and B; specifying their concentrations.

Answer:

– In beaker A; 1 M HCl (aq) i.e. any solution with 1 M hydrogen ions

– In beaker B; 1 M Al (NO3) (aq); i.e. any aqueous solution with 1M Al 3+.

 

(b). One essential part of the cell has been omitted.  Name the missing part and give its functions.

Answer:-       

Missing part: salt bridge

Function: completes the circuit by;

  1. Allowing its ions to carry charge form one half cell to another.
  2. Providing ions which repose those used up of the electrodes, hence maintaining a balance of charge in the 2 half – cells.

 

(c). The voltmeter reading was found to be –1.66 V.

(i). Give the standard electrode potential for the aluminum electrode.

Solution: it is -1.66 since the Hydrogen-hydrogen ions half-cell = 0.00V.

 

(ii). Show the direction of flow of electrons in the circuit;

Solution: From Al3+ / Al half cell to the H2 / 2H+ (aq) half cell

 

(d). Give the half-cell equations and the overall cell equation

Solution:

3H+ (aq) + 3e                               1½ H2 (g); E0 = + 0.00V;

Al (s)                                     3e + Al3+ (aq); E0 = + 1.66V;

Overall:

Al (s) + 3H+ (aq)                      Al3+ (aq) + 1½H2 (g); E0 = +1.66 V

 

 

 

 

 

  1. You are given the following half-equations;

Mg2+ (aq) + 2e-                   Mg (s);  Eθ = – 2.71 V

Zn2+(g)  + 2e-                     Zn (s);   Eθ = – 0.76 V

 

(i) (a). Obtain an equation for the cell reaction.

 

Mg (s)                              2e + Mg2+ (aq); E0 = +2.71V

Zn2+(aq) + 2e                                 Zn (s); E0 = – 0.76V

Mg (s) + Zn2+ (aq)                    Mg2+ (aq) + Zn(s); E0 = +1.95V

 

Thus equation:

Mg (s) + Zn2+ (aq)                    Mg2+ (aq) + Zn(s); E0 = +1.95V

 

(b). Calculate the E0 value for the cell.

Mg (s)                              2e + Mg2+ (aq); E0 = +2.71V

Zn2+(aq) + 2e                                 Zn (s); E0 = – 0.76V

Mg (s) + Zn2+ (aq)                    Mg2+ (aq) + Zn(s); E0 = +1.95V

 

(c). Give the oxidizing species.

– Reducing species

Magnesium i.e. The species that undergoes oxidation since its oxidation number increases (from 0 to 2); as it reduces the other;

 

– Oxidizing species

Zinc/Zinc ions; – the species that undergoes reduction; since its oxidation number decreases (from 2 to 0) as it oxidizes the other species (Mg).

 

  1. Given the following half-equations

I2 (g) + 2e-                          2I (aq); Eθ = + 0.54V

Br2 (g) + 2e-                        2Br (s); Eθ = +1.09 V

 

(a). Obtain an equation for the all reaction

2I(aq)                               2e + I2 (g); E0 = – 0.54V

Br2 (g) + 2e                           2Br (s); E0 = +1.09V

Br2 (g) + 2I (aq)                       2Br (aq) + I2 (g); E0 = +0.55V

 

(b). Calculate the E0 value for the cell

2I(aq)                               2e + I2 (g); E0 = – 0.54V

Br2 (g) + 2e                           2Br (s); E0 = +1.09V

Br2 (g) + 2I (aq)                       2Br (aq) + I2 (g); E0 = +0.55V

 

 

(c). Give the oxidizing species;

Oxidation

 

Br2 (g) + 2I (aq)                       2Br (aq) + I2 (g); E0 = +0.55V

Reduction               

0             -1                                   -1            0

 

Oxidizing species: Bromine; Br2 (aq)

 

  1. Consider the following list of electrodes and electrode potential values.

 

Electrode reaction Eθ volts
A2+/ A

B2+/ B

C2+/ C

D+/ D

E2+/ E

F2+/ F

+0.34

-0.71

-0.76

+0.80

-2.87

-2.92

 

(a). Which of the ions is the strongest oxidizer?

D+; because it is most readily reduced/ have the highest tendency to accept electrons as evidenced by its highest positive Eθ value when the ions change to element (D+/ D)

 

(b). Which of the ions is the strongest reducer?

– Is least readily reduced hence lowest E0 value (-2.92 V); accepts electrons least readily i.e. shows the lowest E0 when its ions gain electrons/ are reduced (F+/ F, = -2.92 V)

 

  1. The following is a list of reduction standard electrode potentials.

 

Metal Eθ volts(standard electrode potential)
Magnesium

Zinc

Iron

Hydrogen

Copper

Silver

-2.36

-0.76

-0.44

0.00

+0.34

+0.79

 

(a). Which two metals, if used together in a cell would produce the largest e.m.f?

Magnesium-silver cell;

Mg2+ (aq) + 2e                     Mg (s);  Eθ = – 2.36 V

2Ag+(g)  + 2e                    2Ag (s); Eθ = – 0.79 V

 

Thus;

Mg (s)                              2e + Mg2+ (aq); E0 = +2.36V

2Ag+(aq) + 2e                               2Ag (s); E0 = – 0.79V

Mg (s) + 2Ag+ (aq)                   Mg2+ (aq) + 2Ag(s); E0 = +3.15V

(b). What would be the voltage produced by:-

(i). Zinc-copper cell

Cu2+ (aq) + 2e-                     Cu (s);  Eθ = + 0.34 V

Zn2+(g)  + 2e-                      Zn (s);  Eθ = – 0.76 V

 

Thus; Zn(s) / Zn2+ (aq) // Cu2+ (aq) / Cu(s);

 

Zn (s)                               2e + Zn2+ (aq); E0 = + 0.76V

Cu2+(aq) + 2e                                Cu (s); E0 = + 0.34V

Zn (s) + Cu2+ (aq)                     Zn2+ (aq) + Cu (s); E0 = +1.10V

 

(ii). Copper-silver cell

Cu2+ (aq) + 2e-                     Cu (s); Eθ = + 0.34 V

Ag+ (g) + 2e-                       Ag (s); Eθ = + 0.79 V

 

Thus; Cu(s) / Cu2+ (aq) // 2Ag+ (aq) / 2Ag(s);

 

Cu (s)                               2e + Cu2+ (aq); E0 = – 0.34V;

2Ag+(aq) + 2e                               2Ag (s); E0 = + 0.79V;

Cu (s) + 2Ag+ (aq)                   Cu2+ (aq) + 2Ag(s); E0 = + 0.45V

 

(c). Explain the meaning of the positive and negative signs;

 

Positive signs

– The metal in question has a lower tendency to loose electrons than hydrogen hence more relatively positive to hydrogen;

– They are stronger oxidizing agents but weaker reducing agents but weaker reducing agents than hydrogen;

 

Negative signs

– The particular metal has a higher tendency to loose electrons than hydrogen; hence relatively more negative than hydrogen.

– They are weaker oxidizing agents but stronger reducing agents than Hydrogen.

 

 

 

 

 

 

 

 

 

 

 

 

  1. The following are some half-cell electrode potentials of some elements.

 

 

 

 

Reaction

 

(a). Select two half cells which when oxidized give the  burst E value; and fill the cell representation.

Solution:          The silver – copper cell; i.e.

Cell representation

 

(b). Calculate the E0 value

 

(c). Give the strongest reducing  agent and strongest oxidizing agent.

 

Strongest reducing agent

Strongest oxidizer       –           silver    –           has highest reduction potential

 

  1. Study the table below and answer the questions that follow.

(a). Which two metals would form a metallic couple with the highest EMS.

 

(b). Calculate the e.m.f. of the cell that would be produced by (i) above.

 

(c). Write down the cell representation for the cell above.

 

(d). Which metal is the strongest reducing agent in the above list

Metal A           –           have the lowest reduction potential.

 

8 (a). The table below gives reduction potentials obtained when the half-cells for each of the metals represented by letters J, K, L, M and N where connected to a copper half- cell as the reference electrode.

 

 

(i). What is metal L likely to be?  Give a reason

 

Copper            It has an E0 value/ reduction potential of 0.00, with copper as the                                        reference electrode

 

(ii). Which of the metals cannot be displaced from the solution of its salt by any other metal in the table?  Give a reason.

 

Metal J:          Has the lowest reduction potential; meaning it least readily accepts                                                 electrons (most readily donates electrons) than any other metal.

(iii). Metal K and M were connected to form a cell as shown in the diagram below;

 

  1. Indicate on the diagram, the direction of flow of electrons. Explain.

from K to M.

K is a stronger reducing agent than M, as evidenced by its lower reduction potential.

It thus loses electrons faster becoming more Negative than M; hence electrons move from K through  external wires to M.

 

  1. Write the equations for the half-cell reactions that occur at:-

Metal K electrode:-

            Metal M electrode

 

III. If the slat bridge is filled with saturated sodium Nitrate solution, explain how it helps to complete the circuit.

 

Answer

It allows its ions (Na+ (aq) , and No– 3 (aq) ) to carry charge from one half cell to another. Providing ions which replace those used up at the electrodes.

 

VOLTAIC CELLS

– Are also called electrochemical cells.

Are cells in which electrical energy is generated from chemical reactions.

 

Types of electrochemical cells

(i). Primary cells        –          

Electrochemical cells which are not rechargeable

 

(ii). secondary cells-  

– Voltaic/electrochemical cells which are rechargeable.

 

  1. Primary cells/dry cells:

– Are of various types and an example is the Le’ clanche dry cell.

 

The Le’clanche dry cell

 

  1. Structure

 

  • Consist of a Zinc can with carbon rod at the centre.
  • The central graphite/carbon rod is surrounded by powdered Manganese (IV) oxide and carbon; which are inturn surrounded by a paste of NH4Cl (s) and Zinc Chloride.
  • The protruding portion of the carbon rod is covered with a brass cap; and the zinc can covered with a sealing material.

 

Chemical reaction

  • The Zinc can is the negative terminal; while the carbon/graphite rod is the positive terminal;
  1. a the Zinc can/negative terminal
  2. positive terminal /brass cap

 

Hydrogen gas

(NH3 (g) + 2e                         2NH3 (g) + H2 (g)

 

Note:   These gases (NH3 (aq)  and H2 (g) ) are NOT used immediately but are used in more complex reactions.

  • The ammonia gas forms a complex with the zinc chloride in the paste.
  • The hydrogen gas is oxidised to water by the Manganese (IV) oxide.

 

Functions of the various components.

 

  1. a) Brass cap –           Functions as the positive terminal where the reduction reaction

 

  1. b) Zinc can –           Is the Negative terminal; where the oxidation reaction occurs.

 

  1. c) Carbon rod –           It serves as the positive electrode.
  • Acts as the connecting wire between the positive and negative terminal through it electron flow form the Zinc can to the brass cap.
  1. d) Manganese (IV) Oxide –           To oxidise the Hydrogen gas produced at the                                                            anode/positive terminal/brass cap to water.

 

  • A single dry cell can produce a potential of 1.5 V

 

Note:   Dry ammonium chloride does not conduct electric current.  This explains why a paste,       which is a conductor, is used.

 

The dry cells cannot provide a continuous supply of electricity for an undulate period of time.

 

Reason            –           The reactants (electrolytes) are used up and cannot be replaced.

 

  1. Secondary cells

 

Are voltaic/electrochemical cells that are rechargeable.

A common example is the lead – acid accumulator.

 

The lead – acid accumulator.

 

Structure:        –           The positive plate is a lead grill filled with lead (IV)  oxide; while                                       the negative plate  consists of a similar lead grill filled with                                                        spongy lead.

 

–           The grills are immersed in sulphuric acid;  which serves as the          electrolyte.

 

Reactions

 

  1. i) During discharge/when in use

–           At the  negative terminal/lead –           The lead dissolves forming lead (II) ions

 

equation

 

  • At the positive terminal (lead (IV)

Lead (IV) oxide reacts with the                                                                                 Hydrogen ions (H+) in sulphuric acid; also forming lead (II) ions.

Then; the lead (II) ions formed at both electrodes react instantly with the Sulphate ions     to form lead (II) Sulphate.

 

–           The  insoluble lead Sulphate adheres to the electrodes.

 

Overall reaction

 

Note:   The Lead Sulphate should NOT be left for too long to accumulate on the electrodes

 

Reason:           The fine PbSO4 (s)  will charge to coarse non – reversible and inactive form and                  the accumulator will become less efficient.

  • During use/discharge; the lead and the lead (IV) oxide are depleted, and the concentration of sulphric acid declines.

 

  1. ii) During recharging –           Is usually done by applying a suitable voltage to                                                       the terminals of the

–           At the negative  terminal        –           the lead ions                                                                accept electrons to form lead solid.

Overall reaction

 

This process restores the original reactants.

 

 

ELECTROLYSIS

 

Defination

 

Is the decomposition of molten or aqueous solutions by passage of electric current through it.

 

Terminologies used in electrolysis

  1. i) Electrolyte

–           Is a solution which allows electric current to pass through while it gets decomposed.

  • Electric current transfer in electrolyte occur through ions.
  • The electrolyte can be aqueous solutions or molten solutions
  • Electrolyte with may ions are called strong electrolyte; while those with few ions are called weak electrolytes.

 

Examples

 

  1. ii) Electrodes –           Are the solid conductors, usually roots, which usually complete                                           the circuit between electrolytes and cell/battery

–           Are of two types

 

  1. a) Anode – The electrode connected to the positive terminal f a battery/cell
  2. b) Cathode– Electrode connected to the negative terminal of the battery/cell.

 

Note:   Graphite rods are commonly preferred as electrodes in most access.

Reasons:         They are inert/unreactive

Are cheap

  • Platinum is also relatively inert; but not less preferred  to Graphite because its expensive.

 

 

Preferential discharge  of ions

 

  • The products of electrolysis of any given electrolyte depend on the ions present in an electrolyte.
  • Commonly most molten electrolytes have only two ions; a cation and an anion and are termed Binary electrolytes.
  • As the electrolyte decomposes, ions collect/move to the opposite poles.
  • Negatively charged ions move to the Anode; the positive electrode, while the positively charged  ions move to the cathode, the negative  electrode;
  • Regardless of how many ions move to an electrode; only one can be discharged ot give a product.
  • Both cations and anions have a preferential discharge series.

 

  1. Discharge for cations

 

Cations  are discharged by reduction (accepting electrons) to form their respective products.

 

The ease of reduction of cations depends on their position of electrochemical series.

 

Thus Ag+ is most readily discharged as it’s the weakest reducing agent.

 

  1. Discharge for anions

 

Anions are discharged by oxidation (electron loss) to form their respective products.

 

Discharge of anions is viz.

 

 

 

ELECTROLYSIS OF VARIOUS SUBSTANCES

 

  1. Electrolysis of various substances
  2. i)

 

 

 

  1. ii) Procedure –           An electric current is passed through the dilute sulphuric acid.

 

iii)        Observation

 

At the Anode            

–           A colourless gas; collects

  • The gas collected relights a glowing splint; and its volume is half the volume of the gas at cathode.  The gas is oxygen.

 

At the cathode

  • A colourless gas collects
  • The collected gas burns with a pop-sound; and its volume is double the volume of gas at the anode.
  • The gas is Hydrogen gas.

 

  1. Explanations –           Ions present in the electrolyte
  2. i) Hydrogen ions and Sulphate ions form sulphuric acid.
  3. ii) Hydrogen ions and hydroxide ions from water.

 

At the Anode (Positive electrode)

 

–           The negatively charged Sulphate ions and hydroxide ions migrate to the anode.

 

Reason:          OH (aq) ions have a greater tendency to loose electrons than the SO2 – 4 (aq) ions

 

Anode equation

 

 

At the cathode (positive electrodes)   –           The positively charged hydrogen ions migrate to the cathode.

 

Equation

 

 

Note:  

 

  1. The volume of oxygen produced at the anode is half the volume of hydrogen produced at the cathode.

 

Reason:          The 4 electrons lost by the hydroxide ions to form 1 mole (1 volume) of                  oxygen molecules are gained by the four hydrogen ions which form 2                      molecules (2 volumes) of hydrogen  molecules.

 

  1. During the electrolysis, the concentration of the electrolyte (H2SO4 (aq), increases

 

Reasons:         The volumes of hydrogen and oxygen gas liberated are in the same ratio                 as they are combined in water.

 

Thus the amount of water in the electrolyte progressively decrease; hence the increased electrolyte concentration.

 

Conclusion

 

Electrolysis of dilute sulphuric acid is thus the electrolysis of water.

 

Note:   The Hoffmans voltmeter can be used instead of the circuit above.  Viz.

 

 

 

 

Electrolysis of dilute sodium chloride

 

  1. i) Apparatus

 

 

 

  1. ii) Procedure
  • An electric current is passed through dilute sodium chloride solution; with carbon rods     as the electrodes.
  • Gases evolved of each  electrode  are collected and tested.

 

iii)        Observations

  • At the anode:
  • A colourless gas is collected
  • The gas relights a glowing  splint, and its volume is half the volume  of the gas collected at the cathode.
  • The gas is Oxygen, O2

 

  • At the Anode
  • A colourless gas collects
  • The gas burns with a pop sound; and its volume is twice the volume of the gas collected at the anode.
  • The gas is hydrogen gas;

 

  1. iv) Explanations
  • The Ions present in the electrolyte

 

 

 

  1. At the Anode
  • Cl and OH migrate to the anode
  • OH- are preferentially discharged – coz they have greater tendency to lose electrons than the chloride ions;  – the OH (aq)  lose electrons to form water and O2 (g) at anode.

 

Anode equations

 

 

  1. At the Cathode

–           The positively charged Na+ (aq) , and H+ (g)   migrate to the cathode

  • the H+(aq) are preferentially discharged.

 

Reason:          They  H+ (aq) have a greater tendency to gain electrons than Na+ (aq) ions

The H+ (aq) gain elect5rons to form Hydrogen atoms (H) which ten form molecules of hydrogen which bubble off at the electrode.

 

Cathode equations

 

 

  1. Conclusion

Ratio of the volumes of H2 (g) and O2 (g) evolved at cathode and anode is 2:1 respectively.

 

Electrolysis f dilute NaCl is thus the electrolyze of water since only water is decomposed.

 

 

 

  1. Electrolysis of Brine/concentrate sodium chloride.

 

  1. i) Apparatus

 

 

  1. ii) Procedure

An electric current is passed though concentrated sodium chloride/brinc

 

iii)        Observation

  1. a) At the Anode
  • A greenish – yellow gas is evolved.
  • The gas has a pungent irritating smell; and its volume is equal to the volume of the gas evolved at he cathode.
  • The gas is chlorine Cl2 (g)

 

  1. b) At the Cathode
  • A colourless gas is liberated
  • The gas burns with a pop sound; and its volume is equal to volume of gas evolved at the anode.
  • The gas is Hydrogen gas; H2 (g)

 

  1. iv) Explanations
  • The ions present in the electrolyte are:-
  • Na+ (aq) and Cl from sodium chloride
  • H+ (aq) and OH- (aq) from water.

 

  1. At the Anode
  • Cl (aq), and OH (aq), migrate to the anode
  • The chloride ions are preferentially discharged.

 

Reason;           –           OH (aq) have higher tendency to lose electrons  than Cl ions.

  • However coz of the higher concentration Cl (aq) , relative to OH (aq), the Cl (aq), are preferentially discharged hence the evolution of Chlorine gas.

 

  1. A the Cathode
  • Na+ (aq) and H+ (aq) migrate to the cathode.
  • H+ with a higher tendency to gain electrons are preferentially discharged; hence the evolution of hydrogen gas at the cathode.

 

NB:     1.         The pH of the electrolyte becomes alkaline/increases with time.

 

Reason:          The removal of H+ (aq) which come form water leaves excess                                                 hydroxide ions  (OH (aq), hence the alkalinity.

 

  1. Evolution of chlorine gas at anode soon stops after sometime and is replaced by O2 (g)

 

Reason:          Evolution of Cl2 (g) decrease/lowers the concentration of Cl (aq) in                            the electrolyte.

                       

As soon as the Cl (aq) concentration  becomes equal to that of OH (aq)

 

The mercury cathode cell

 

Is an electrolytic arrangement commonly used for the large scale manufacture of chlorine and sodium hydroxide.

 

  1. i) Apparatus
  • Electrolyte in the mercury cell is Brine (concentrated NaCl)
  • Anode is carbon or titanium
  • Cathode is a moving mercury film.

 

 

  1. ii) Reactions
  2. At the Anode
  • Both Chloride and Hydroxide ions are attracted
  • Due to their high concentrations the chloride ions are preferentially discharged.
  • The Cl (aq) lose electrons to form Chlorine gas. (greenish yellow)

 

  1. Cathode (moving mercury)
  • The Na+ (aq) and H+ (aq) are attracted
  • The discharge of H+ (aq) is more difficult than expected
  • Hydrogen has a high over voltage at the moving mercury electrode and so sodium is discharged.

 

Equation

 

  • The discharged sodium atoms combine with mercury to form sodium amalgam

 

Equation

 

  • The sodium amalgam reacts with water to form sodium hydroxide, hydrogen and mercury.

 

Equation

 

  • Hydrogen is pumped out while the Mercury is recycled.
  • The resultant NaOH is of very high party.

 

 

Limitations/disadvantages of the Mercury Cathode cell.

 

  1. Its expensive due to the high cost of mercury.
  2. Pollution form Mercury; i.e. Mercury is poisonous and must be removed from the effluent.

 

 

  1. Electrolysis of Copper (II) Sulphate solution.

 

NOTE:                        The products of electrolysis of copper (II) Sulphate solution depends on the nature of the elctrodes used.

 

  1. i) Apparatus

 

 

 

 

  1. ii) Ions present in the elctrolyte:
  • From copper (II)  Sulphate

 

  • From water

 

  • During electrolysis

 

  1. Using carbon/platinum electrodes

 

iii)        Observations

 

  1. At the Anode:
  • A colourless gas is liberated
  • The gas relights a glowing splint; hence its oxygen.

 

  1. At the cathode
  • A reddish – brown coating (of Cu solid) is deposited.

 

  1. In the electrolyte
  • The blue colour of the solution (CuSO4) (aq) / becomes pale and finally colourless after a long time.

 

Reason:           The blue colour is due to Cu2+.  As the Cu2+ (aq) are continuously being                              discharged at the cathode; the concentration of CU 2+ decreases i.e. decrease in                        the concentration of Cu 2+ (aq) in the solution

 

The electrolyte become acidic/pH decreases (declines)

 

Reason:          Accumulation of H+ (aq) in the solution since only OH (from water) are being                      discharged (at the anode).

 

 

  1. iv) Explanation

 

At the anode

 

  • The SO 2- 4 (aq) and OH (aq) migrate to the anode.
  • The hydroxide ions have a higher tendency to lose electrons than the SO 2- 4 (aq)
  • They (OH) easily loose electrons to form the neutral and unstable hydroxide radical (OH)
  • The hydroxide radical (OH) decomposes to form water and Oxygen.

 

 

At the cathode

  • Copper ions and H+ (aq) migrate to the cathode
  • Cu2+ (aq) have a greater tendency to accept electrons than H+ (aq)
  • The Cu 2+ (aq) are thus reduced to form copper metal which is deposited as a red-brown coating on the cathode.

 

Cathode equation.

 

 

  1. Using copper rods electrodes

 

  1. observations

 

At the anode   –           Mass of the anode (Copper anode) decreases

 

At the cathode            –           reddish – brown deposit

  • cathode increases in mass

 

Electrolyte – no apparent change

 

Note:   The gain in mass of the cathode is equal to the loss in mass of the anode.

 

Explanations

 

At the anode

–           The SO 2- 4(aq) and OH (aq) are attracted to the anode.

  • However, none of them is discharged;
  • Instead; the copper anode itself gradually dissolves; hence the loss in mass of the anode;

 

Reason:          it’s easier to remove electrons form the copper anode itself than format the                       hydroxide ions

 

At the cathode

 

  • H+ (aq) migrate to the cathode
  • The Cu 2+ (aq) are preferentially discharged; because they have a greater tendency to accept electrons
  • The copper cathode is thus coated with a reddish brown deposit of copper metal hence increase in mass.

 

 

Cathode equation

 

Factors affecting electrolysis  and electrolytic products.

 

  1. Electrochemical series
  • Electrolytic products at the anode and cathode during electrolysis depends on its position in the Electrochemical series.

 

Cations:          The higher  the cation in the electrochemical series; the lower the tendency of                     discharge at the cathode.

 

Reason:          Most electropositive cations require more energy in order to be reduced and                       therefore are more difficult to reduce.

 

Reduction order

 

 

Anions:           Discharge is through oxidation ad is as follows.

 

 

  1. Concentration of electrolytes

A cation or anion whose concentration is higher is preferentially discharge if the ions are close in the electrochemical series.

 

Example:  dilute and concentrated NaCl

 

Product at the anode.

 

 

  1. The electrodes used: Products obtained at electrodes depend on the types of electrodes                          used

 

Examples:      in the electrolysis of CuSO4 (aq) using carbon and copper rods separately.

 

 

 

APPLICATIONS OF ELECTROLYSIS

 

  1. Extraction of reactive metals

 

Reactive metals/elements like sodium, magnesium, aluminum are extracted form their compounds by electrolysis.

 

Example:        Sodium is extracted from molten sodium chloride using carbon                               electrodes.

 

  1. Purification of metals

 

It can be used in refining impure metals

 

Examples:      Refining copper

 

  • The impure copper is made of the anode.
  • Their strips of copper are used as the cathode
  • Copper (II) Sulphate are used as the electrolyte.

 

  • During the electrolysis the anode dissolves and pure copper is deposited on the cathode.

 

  • The impurities (including valuable amounts of silver and gold) from the crude copper collect as a sludge become the anode.

 

  1. Electroplating

 

Is the process of coating one metal with another, using electrolysis so as to reduce corrosion or to improve its appearance.

 

During electrolysis:

  • the item to be electroplated is made the cathode
  • the metal to be used in electroplating is used as the anode
  • the electrolyte is made from a solution containing the ions of the metal to be sued in electroplating.

 

Examples

  • Gold-plated watches; silver – plated utensils
  • Steel utensils marked EPNS. I.e. Electroplated Nickel Silver.

 

  1. Anodizing Aluminum

 

Is the reinforcement of the oxide coating on Aluminum utensils/articles

Is done by electrolysis of dilute sulphuric acid using Aluminum articles as anode.

 

Importance:   Prevention arrosion of Aluminum articles

 

  1. Manufacture of sodium hydroxide, chlorine and hydrogen

 

  • Sodium hydroxide is prepared by the electrolysis of brine, for which 3 methods are available
  • The method depends o the type of electrolytic cell.
  • These cells are
  • The mercury cell
  • The diaphragm cell
  • The membrane cell

 

  1. The mercury cell

 

  • Components
  • The electrolyte is concentrated sodium chloride
  • The anodes are made of graphite or titanium, which are placed above the cathode.
  • The cathode consists of mercury, which flows along the bottom of the cell.

 

 

 

Chemical reactions

 

Anode

 

  • Both chloride and hydroxide ions are attracted.
  • Chloride ions are preferentially discharged due to their high concentration
  • The chloride ions undergo oxidation to form green – yellow chlorine gas.

 

Equation

 

 

 

At the cathode (flowing mercury)

  • Na+ (aq) migrate to the cathode
  • Sodium ions are preferentially discharged.
  • They undergo reduction to form sodium solid.

 

Equation

 

  • the discharged sodium atoms combine  with mercury to form sodium diagram

 

Equation

 

 

  • The sodium amalgam is then passed into another reactor containing water.
  • The amalgam reacts with water forming hydrogen and sodium hydroxide.

 

The mercury is regenerated and it is recycled into the main cell.

–           Main product:                       Sodium hydroxide

–           By products:                          Sodium and chlorine.

 

Advantages of mercury cathode cell

  • The resultant sodium hydroxide is very pure; as it has no contamination from sodium chloride.
  • It is highly concentrated; e. about 50%.

 

Disadvantages

  • Some of the mercury said its way into the environment leading to mercury pollution; a common case of brain damage in humans.
  • At the operating temperatures (700C – 800C), mercury vapours escape into the atmosphere and cause irritation and destruction of lungs tissues.
  • Its operation requires highly skilled man power.

 

  1. Diaphragm cell

 

Components

  • An asbestos diaphragm; to separate the electrolytic cell into two compartments; thus preventing mixing of H2 and Cl2 molecules
  • The anode compartment contains a graphite rod.
  • The cathode compartment contains a stainless steel cathode.

 

 

Diagram

 

 

Chemical reactions

  • The asbestos diaphragm is permeable only to ions, but not to the hydrogen or chlorine molecules.
  • It thus prevents H2 (g) and Cl (g) form mixing and reacting to yield HCl (g)
  • It also separates NaOH and Cl2 which would otherwise react.

 

Chemical reactions

 

At the anode

  • Chloride ions undergo oxidation to form chlorine gas.

 

Equation

 

–           At the cathode

  • H= and Na+ (a) migrate to the cathode compartment.
  • The H+ are preferentially discharged.
  • They (H+ (aq) undergo reduction to form hydrogen gas.

 

 

Equation

 

  • the discharge of H+ causes more water molecules to dissociate, thus increasing the concentration of OH- in the solution.
  • therefore, the Na+ and OH- ions also react in the cathode compartment to form sodium hydroxide.

 

Equation

 

Advantage

  • Does not result into pollution

 

Disadvantage

  • The resultant NaOh is dilute (12% NaOH)
  • It is also not pure due to contamination with NaCl – (12% NaOH + 15% NaCl by mass.

 

Note:   –           The concentration of the NaOH can be increased by evaporating excess water,                   during             which NaCl with a lower solubility crystallizes out first, leaving NaOH at               a higher concentration.

–           The solid NaCl (Crystals) are then filtered off.

  • This is a case of fractional crystallization.

 

 

 

 

  1. The membrane cell
  • Is divided into 2 compartments by a membrane
  • Most commonly used type of Membrane is the cation – exchange membrane.
  • This membrane type allows only cations to pass through it.

 

Components

  • A cation exchange that divides the cell into 2 compartments; an anode and a cathode compartments.
  • Both electrodes are made of graphite
  • The electrolyte in the anode compartment is purified brine
  • The electrolyte in the cathode compartment is pure water.

 

Diagram

 

Chemical reactions

 

The anode

 

Chloride ions undergo oxidation to form green – yellow chlorine gas.

 

The cathode

 

As current passes through the cell H+ and Na+ pass across the membrane to the cathode

  • H+ are preferentially discharged.
  • They undergo reduction to liberate hydrogen gas.

 

  • Continuos discharge of H+ leaves the OH- at a higher concentration

 

  • The OH- react with Na+ to form sodium hydroxide

 

Advantages

 

  1. Resultant sodium hydroxide is very pure, since it has no contamination from NaCl.
  2. The sodium hydroxide has a relatively high concentration; at about 30 – 35% NaOH by mass.

 

 

Uses of sodium hydroxide, chlorine, and hydrogen

 

  1. Sodium Hydroxide

 

  • React with chlorine to form sodium chlorate sodium hypochlorite or I, NaOCl. This is a powerful oxidising agent which is used for sterilization and bleaching in textiles, paper and textile industries.

 

i.e. 2NaOH (aq)  +   Cl2 (g)                                NaOCl (aq)  +  NaCl (g)  + H2O (l)

 

  • Manufacture of sodas, detergents and cosmetics.
  • Neutralization of acidic solutions in the laboratories.

 

  1. Hydrogen

 

  • For hydrogenation in the manufacture of margarine
  • Manufacture of ammonia
  • Production of hydrochloric acid

 

  1. Chlorine
  • Formation of sodium chlorate I; for bleaching in pulp, textile and paper industries.
  • Sewage and water treatment
  • Manufacture of polymers such as polyvinyl chloride.
  • Manufacture of pesticides.

 

QUANTITATIVE ASPECTS OF ELECTROLYSIS

 

Basic terminologies and concepts

 

  • Ampere

Is the standard unit used to measure an electric  current; the flow of electrons

Is usually abbreviated as amps.

 

  1. Coulomb

Is the quantity of electricity, when a current of 1 ampere flows for one second. I.e.  1 Coulomb = 1 Ampere x 1 second

Generally:

Quantity of     Electricity        =          current x Time in seconds

A         =          It; Where

Q         =          Quantity of electricity in coulombs

I           =          Current in Amperes

T          =          time in seconds

 

  1. Faraday

Is the quantity of electricity produced by one mole of electrons; and is usually a constant equivalent to 96487 (approx. 96500) coulombs

 

Faradays laws of Electrolysis.

First law;

The mass of substance liberated during electrolysis is directly proportional to the quantity of electricity passed.

 

Worked examples

  1. A current of 2.0 Amperes was passed through dilute potassium sulphate solution for  two minutes             using  inert electrodes.
  2. Write the equation for the reaction at anode.

 

 

  1. Work out the mass of the product formed at the cathode.  (H = 1.0, Faraday = 96,000 C)

Solution;          quantity of electricity  =          current x time

=          2 x 2x 60

=          240 coulombs

cathode reaction

 

1 mole of electrons      =          96000 C

4 moles of e-   =          4 x 96000 C

=          384,000 C

 

 

  1. What mass of copper would be coated on the cathode from a solution of copper (II) Sulphate  by a current of 1 amp flowing for 30 minutes.

(Cu = 63.5;      Faraday constant = 96487 Cuo/-)

 

solution

 

Cathode  reaction

 

Cu 2+ (aq)       +          2e-                               Cu (s)

 

–           1 mole of Cu requires 2 moles of electrons

  • Quantity of electricity passed; = 1 x 30 x 60 coulombs; = 1800C
  • 1 mole of electrons carriers a charge of 96487 coulombs = 192974 coulombs
    • coulombs deposit 63.5 g of Cu.

 

Thus 1800 C will deposit   63.5 x 1800          =          0.592 grams

192974

 

  1. An element x has relative atomic mass of 88g. when a current of 0.5 amperes was passed through a solution of x chloride for 32 minutes, 10 seconds;  44 g  of x was deposited at the cathode.  (1 faraday = 96500 c)

Calculate  the charge on the ion of x.

 

 

  1. In the electrolysis of dil CuSO4 solution, a steady current of 0.20 Amperes was passed for 20 minutes. (1 Faraday = 96, 500 C Mol-, Cu = 64)

Calculate

  1. The number of Coulombs of electricity used

 

  1. The mass of the substance formed at the cathode.

2 moles of electrons liberate 1 mole of Cu.  i.e.  Cu 2+ + 2e                           Cu (s)

 

  1. An element p has a relative atomic mass of 44. When a current of 0.5 Amperes was possed through a fused chloride of p for 32 minutes and 10 seconds; 0.22g of p were deposit

 

 

 

 

 

 

UNIT 5: METALS: EXTRACTION PROPERTIES AND USES.

Introduction:

  1. Introduction
  2. Extraction methods
  3. Concentration of the ores
  4. Metal extraction
  • Sodium metal
  • Main ores
  • Extraction process
  • Properties of sodium
  • Uses of sodium

 

  • Aluminium metal
  • Main ores
  • Qualitative analysis
  • Extraction from bauxite.
  • Electrolysis of purified bauxite
  • Properties of aluminium
  • Uses of aluminium

 

  • Zinc metal
  • Main ores
  • Qualitative analysis
  • Extraction process
  • By oxidation
  • By electrolysis
  • Properties of zinc
  • Uses of zinc

 

  • Iron metal
  • Main ores
  • Qualitative analysis
  • Extraction from haematite
  • Properties of iron
  • Uses of iron

 

  • Copper metal
  • Main ores
  • Qualitative analysis
  • Extraction process from copper pyrites
  • Properties of copper
  • Uses of copper

 

  • Lead metal
  • Main ores
  • Extraction process
  • Properties of lead
  • Uses of lead

Introduction:

Only most unreactive metals occur naturally in their elementary form.

Examples: – Gold, Silver, Platinum.

– Other elements occur as ores i.e. metal – bearing rocks.

Examples:

– Oxides

– Sulphides

– Carbonates

– Chlorides.

 

Note:
An ore is a mineral deposit with reasonable composition of a desired metal.

 

Methods of Extraction

Depend on position of the metal in the reactivity series.

– Main methods are:

  1. Electrolytic Method:

– Used for metals high up in the reactivity series

E.g. – Sodium and Potassium

– Calcium and Magnesium

– Aluminium.

– These metals occur in very stable ores

 

  1. Reduction method:

– For less reactive metals.

E.g. Iron, Zinc, and Copper.

 

– Is achieved using;

(i) Carbon in form of coke.

(ii) Carbon (II) oxide

(iii) Hydrogen

– Oxidation is also used followed by reduction.

 

Preliminary steps before extraction.

– Minerals (mineral) are usually mined with several impurities which lower the concentration of the metal per given mass or volume.

– Thus the ore is first concentrated before the actual extraction.

– Concentration is possible due to difference in properties between the mineral compound and the earthy materials.

 

Methods of ore concentration.

  1. Physical methods.

(a). Optical sorting.

– Used to separate ore particles that have sufficiently different colours to be detected by the naked eye.

– It involves physical handpicking of the desired particles.

– Mainly used for minerals containing transition elements such as chromium.

 

 

 

(b). Hydraulic washing.

– Also called sink and float separation.

– Utilizes the difference in density between the minerals and the unwanted materials

– The ore is washed with streams of water.

– The denser ore particles will sink to the bottom of the washing container and can then be collected.

– Examples in ores of tin and lead.

 

(c). Magnetic separation.

– Is used when either the ore particles or the earthy materials (unwanted materials) are magnetic.

– A strong magnet is used to attract the magnetic components and leaving the non-magnetic materials behind.

– Examples: in ores like magnetite (Fe3O4) and chromite which are magnetic.

 

(d). Electrostatic separation.

– Used to separate particles which have different electric charges.

– The particles are subjected into an electric field.

– The oppositely charged particles follow different paths and can then be separated.

 

(e). Froth floatation

– Is mainly use for sulphide ores.

– Takes advantage of two facts.

  • Oil can wet the surfaces of ores.
  • Oil floats on water

The process:

– The ore is ground into a fine powder; to increase the surface area for upcoming reactions.

– It is then mixed with water and a suitable oil detergent e.g pine or eucalyptus;

– The mixture is then agitated by blowing compressed air through it;

– Small air bubbles attach to the oiled ore particles; which are thenn buoyed up and carried to the surface where they float.

– A froth rich in mineral is formed at the top while impurities sink at the bottom.

– The froth is skimmed off and dried.

– Froth floatation process is used for copper, lead and zinc metals;

 

Diagram: froth floatation apparatus.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. Chemical concentration.

– Involves the use of chemical reactions to concentrate the ores.

Examples:

– Bauxite, the main aluminium ore is chemically concentrated by a process known as Bayer’s process.

– This takes advantage of the amphoteric nature of aluminium oxide, which can thus react with both acids and bases.

– Chemical concentration can also be done by leaching.

– This involves reacting the ore with a compound such as sodium cyanide;

– The cyanide ions form complex ions with the metal.

– The complex ions formed are water soluble, and can be separated by filtration, leaving the unwanted materials in the residue.

 

 

The metals

  1. Sodium

Main ores;

  1. Rock salt / sodium chloride; NaCl
  2. Chile saltpetre / sodium Nitrate; NaNO3

iii. Soda ash/sodium carbonate; Na2CO3.

 

Other ores include;

(i). Borax; Na2B4O7.10H2O

(ii). Sodium Sulphate, Na2SO4;

 

Extraction;

– Sodium is obtained by the electrolysis of fused sodium chloride in the electrolytic cell.

– Calcium chloride and calcium fluoride are added to the electrolyte.

Reasons;

– To lower the melting point of sodium chloride from 800oC to 600oC;

 

– Once molten, the electrical resistance within the cell is sufficient to maintain the temperature without external heating.

Steel or iron is used as the cathode, while carbon/graphite is used as the anode.

– Thus steel is not used as the anode.

Reason;

– At high temperatures, steel would react with chloride formed at the anode, but graphite is inert even at high temperatures.

 

– Steel wire gauze separates the electrodes.

Reason;

– To prevent products of electrolysis (sodium and chlorine) from mixing and reacting to form sodium chloride.

 

– The electrolytic apparatus used in sodium extraction is called the Downs cell.

 

Diagram: The Downs cell.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

– During electrolysis, fused sodium chloride dissociates according to the equation;

 

NaCl(l)                        Na+(l) +Cl(l)

 

At the cathode:

  • Observation;

– Soft silvery metal

 

Explanation

– Na+ ions are attracted and undergo reduction (accept electrons) to form/ produce molten sodium metal.

 

Equation;

Na+(l) + e                      Na(l)

 

– Molten sodium is lighter than fused sodium chloride and floats on the surface where it overflows into a separate container / sodium reservoir.

 

Note;

– The resultant sodium is usually collected in liquid / molten state, floating on top of the electrolyte.

Reasons;

– Less dense than molten sodium chloride

– Has a low melting point.

 

At the anode;

Observations;

– Evolution of a green-yellow gas.

 

Explanation:

– Chlorine gas is evolved as a by – product and collected separately.

– Negatively charged Cl ions migrate to the positive anode and undergo oxidation to form chlorine gas;

 

Equation:

2Cl(l)                    Cl2(g) + 2e-

 

 

Properties of Sodium;

– Is a soft silvery metal with low density; 0.979gcm-3

– Has a low melting point, 97oC, and a low boiling point of 883oC.

 

Chemical reactions

(a) With air

– Na is very reactive and tarnishes in moist air to form an oxide layer.

4Na(s)+ O2(g)                2Na2O(s);

 

– The oxide layer reacts with more air moisture to form hydroxide

Na2O(s) + CO2 (g)                  Na2CO3(s) + H2O(l)

 

Note

– Due to those series of reactions sodium is stored under a liquid hydrocarbon e.g. petroleum, kerosene.

 

– Sodium burns in oxygen with a golden yellow flame to form sodium peroxide

Equation:

2Na(s) + O2(g)                         Na2O2(s)

(White)

 

(b). With water

– Na reacts vigorously with water to form NaOH and Hydrogen.

Equation:

2Na(s) + 2H2O(l)                       2NaOH(aq) + H2(g)

 

– The resulting solution is highly alkaline with a PH of 14.

– Sodium is stored under oil to prevent contact with moisture from the atmosphere.

Note:

– The reaction between Na and dilute acids would be explosive and not safe to investigate

 

(c). With chlorine

– Sodium burns in chlorine gas;

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

 

(d) With ammonia gas;

– Sodium forms hydrogen and a solid;

2Na(s) + NH3(g)                    2NaNH2(s) + H2(g)

            Sodamide

And;

NaNH2(s) + H2O(l)                    NaOH(aq) + NH3(g)

 

Uses of sodium;

  1. Is alloyed with lead in the preparation tetraethyl (IV) lead, which is added to petrol as an anti-knock.
  2. Provides the glow in sodium vapours lamps, for street lighting (orange-yellow street lights).
  3. Is an excellent conductor of heat and electricity with low melting point hence used;
  • In nuclear reactors to conduct away heat.
  • Modern aeroplane engines.
  1. Manufacture of sodium peroxide, and sodium cyanide used in the extraction of silver and gold.

 

Question;

– Although electrolysis is an expensive way of obtaining metals, it must be used for some metals. Explain.

Solution;

– Group 1 and 2 metals together with Al are themselves such powerful reducing agents that their oxides cannot be reduced by chemical reducing agents.

 

 

 

 

 

 

Worked example

  1. Below is a simplified diagram of the Downs cell in which sodium metal is manufactured.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(a) (i) Identify; –

Electrolyte X: – Molten sodium chloride

Gas Y: -Chlorine gas

 

(ii) Write an equation for the reaction at the cathode.

Na+(l) + e-                      Na(l).

 

(iii). In what state is sodium collected?

– Molten state/liquid state.

 

(iv). Give two properties of Na that makes it possible to be collected as in (b) (iii) above.

– Its less dense than molten sodium chloride.

– Has a low melting point.

 

(v). The cathode is made of steel but the anode is made of graphite.

Why is this yet steel is a better conductor?

– At high temperature steel would react with chlorine formed but graphite is inert even at high temperatures.

 

(vi). In this process, the naturally occurring, raw material is usually mixed with another compound. Identify the compound and state its use.

Compound; – Calcium chloride

Use;-To lower melting point of Nacl2 from 800oc to 600oc

 

(vii). What is the function of the steel gauze cylinder?

– Prevents sodium reacting with chlorine forming NaCl

 

(viii). Give one industrial use of sodium

– A coolant in nuclear reactors;

– Alloy with lead in tetraethyl (IV) lead;

 

(ix). Explain why sodium metal is stored under paraffin;

– Keep it out of air; reacts very fast with air forming a dull surface.

– Can react with water.

 

(b). State an industry that can be built next to a sodium extracting plant.

 

(c). A current of 100 Amperes flows through an electrolyte of molten sodium chloride for 15 hours. Calculate the mass of sodium produced in kg (Na = 23; 1F = 96500C)

Solution:

Q = It

=100 x 15 x 60 x 60

=5400000C.

Cathode equation

Na+(l) + e                     Na(l)

 

96500C = 23g of Na

5400000 C =23 x 5400000 =1287.04g

96500

=1.287kg

 

(d) .For the same quantity of electricity as in (c) above ; calculate the volume of the gaseous product produced in the cell at 150c and 800mmhg.(Molar gas volume at s.t.p = 22.4dm3)

 

 

  1. Aluminium

– Forms 7% of the earth’s crust and is the most common metal.

 

Main ores;

– Bauxite; Al2O3.H2O

– Mica; K2Al2S6016.

– China clay;Al2S2O72H2O

– Corundum;Al2O3.

 

Chemical test;

– Crush the ore into a fine powder;

– Add dilute nitric (V) acid to the powder

– Filter to obtain a solution of the ore;

– To a solution of the ore add NaOH(aq) dropwise till in excess, and then repeat the same procedure using Ammonia solution, NH4OH.

 

Observations:

  • With NaOH(aq):

– White precipitate in soluble in excess;

 

  • With NH4OH(aq):

– White precipitate insoluble in excess;

 

Extraction from Bauxite;

-Involves two main processes;-

  • Purification of Bauxite.
  • Electrolysis of purified bauxite (alumina)

 

  1. Purification of bauxite

– Chief impurities are small quantities of silica and iron (III) oxide.

– The oxide ore is ground and treated under pressure/ dissolved in hot aqueous sodium hydroxide.

 

During the process;

– The amphoteric bauxite dissolves in NaOH forming sodium aluminate;

Equation:

2NaOH(aq) + Al2O3.3H2O(s)                      2NaAl(OH)4(aq).

 

Ionically:

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

 

– Silica impurities also dissolve forming sodium silicate

Equation:

SIO2(s) + 2NaOH(aq)                       Na2SIO3(aq) + H2O(l)

 

– The iron impurities (mainly iron (III) oxide) DO NOT dissolve.

– This mixture is then filtered, during which iron (III) oxide remain as residue of red mud while a filterate of sodium aluminate and sodium silicate is collected.

– Carbon (IV) oxide is bubbled through the filterate, followed by dilution then addition of a little aluminium hydroxide to cause precipitation (seeding) of Aluminium hydroxide.

Ionically:

2[Al(OH)4](aq) + CO2(g)                          2Al(OH)3(s) + H2O(l);

 

Seeding

Al(OH)3

Alternatively:

Al(OH)4(aq)                                2Al(OH)3(s) + OH(aq);

 

General equation:

NaAlO2(aq) + 2H2O(l)    hydrolysis  NaOH(aq) + Al(OH)3(s).

 

– The precipitated Aluminium hydroxide is then filtered off, washed and ignited to give pure aluminium oxide (Alumina);

Equation:

2Al(OH)3(s)                        Al2O3(s) + 3H2O(l)

Alumina

 

  1. Electrolysis of purified bauxite (alumina)

The Alumina (Al2O3), has a high melting point, 2015oC and a lot of heat would be required to melt it.

– Additionally the molten compound is a very poor conductor of electricity.

– Consequently, cryolite (Na3AlF6) is mixed with the oxide.

Reason;

– To lower the melting temperature of Al2O3 from 2015oC to around 800oC;

– At this lower temperature the molten oxide also conducts well.

 

– The molten alumina mixed with bauxite is then electrolysed in a steel cell lined with carbon graphite as the cathode.

Note;

– Other than being an electrolyte the graphite cathode lining also prevents alloy formation, as it ensures no contact between the resultant aluminium and the steel tank;

-The anodes also made of Graphite dip into the steel tank at intervals.

Diagram: electrolytic steel cell for the extraction of Aluminium.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Electrolytic reactions;

– The Aluminium oxide dissociates to give constituent ions;

Equation:

Al2O3(l)                          2Al3+(l)  +3O2-(l)

At the cathode;

Observation;

– A silvery white metal which quickly becomes dulled.

 

Explanation:

– Aluminium ions move to the cathode and are reduced to form aluminium metal.

 

Equation;

2Al3+(l) + 6e-                2Al(l)

 

At the anode;

Observation;

– Effervescence of a colourless gas.

 

Explanations:

– Oxygen ions migrate to the anode and get oxidized to form oxygen gas.

– The resultant oxygen gas reacts further with the graphite anode to form carbon (IV) oxide.

– This is due to the high temperatures involved during the process.

Note;

– Consequently the carbon anode should be replaced from time to time.

 

Equations;

3O2-(l)                3O2(g) + 6e-

Then;

C(s) (anode) + O2(g)                       CO2(g)

 

Note:

– Cryolite usually adds Na+; and F ions into the electrolyte.

– Thus the anions are O2- and F ions into the electrolyte.

– However oxygen is discharged in preference to Fluorine.

Reason;

– Fluorine is a stronger oxidizing agent than oxygen. Thus oxygen easily gives electrons than fluorine, hence discharge.

– Aluminium is discharged in preference to sodium.

 

 

 

 

 

 

 

 

 

 

Summary: – Flow chart on the Extraction of Aluminium from bauxite.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Properties of Aluminium

Physical properties

– Is a silvery white metal which quickly becomes dulled with a thin oxide layer.

– Has very low density (2.7gcm-3), with ability to be rolled into wires / foil.

– Is a good conductor of heat and electricity.

 

Chemical properties.

  1. Reaction with air;

– In air it acquires a continuous very thin coating of oxide, which resists further reaction.

– Removal of this protective cover renders the metal reactive.

– Consequently steel wool or wood ash should NOT be used in aluminium utensils.

– Usually, salty water attacks the oxide film allowing the aluminium to corrode and for this reason, ordinary aluminium is not used for marine purposes.

– Aluminium will burn in air at 800oC to form is oxide and nitrate.

Equations:

4Al(s) + 3O2(g)                    2Al2O3(s);

2Al(s)­ + N2(g)                    2AlN(s)

 

  1. Reaction with Acids.

Note:

– The protective Aluminium oxide (being covalent and insoluble) layer makes its reactivity with acids less than expected.

 

  • With nitric (V) acid;

– Has hardly any effect on the metal, at any concentration.

Reason:

– Being a powerful oxidizing agent, it simply thickens the oxide layer thereby preventing further reaction.

 

  • With sulphuric (VI) acid;

– Only hot concentrated sulphuric (VI) acid breaks down the oxide layer and reacts with the metal.

Equation:

2Al(s) + 6H2SO4(l)                  Al2(SO4)3(aq) + 6H2O(l)  + 3SO2(g)

 

  • With Hydrochloric acid;

– Dilute HCl dissolves aluminium slowly; liberating hydrogen.

Equation:

2Al(s) + 6HCl(l)                        2AlCl3(aq) + 3H2(g)

With concentrated HCl the rate of reaction is increased.

 

  1. Reaction with chlorine;

– Hot aluminium burns in chlorine gas with a white light, forming dense white fumes of Aluminium (III) Chloride.

-The white fumes cool and collect on the cooler parts of the apparatus as a white solid.

Equation:

2Al(s) + 3Cl2(g)                     2AlCl3(g)

 

Note:

– The apparatus for the preparation of AlCl3 is kept very dry.

Reason:

– Aluminium chloride is readily/easily hydrolysed by water/moisture, and so it fumes in damp air with the evolution of hydrogen chloride gas.

Equation:

AlCl3(s) + 3H2O(l)                     Al(OH)3(s)+3HCl(g)

 

  1. Reaction with water.

– Aluminium does not react with cold water, due to the formation of an insoluble coating of Aluminium oxide.

Note;

– If the oxide film is removed, the metal reacts slowly with cold water.

 

  1. Reaction with caustic soda.

– The metal, especially in powder form, reacts with caustic soda solution, liberating hydrogen and leaving sodium aluminate in solution.

– The reaction is exothermic and once started, it is very vigorous.

Equation:

2NaOH(aq) + 2Al(s) + 2H2O(s)                       2NaAlO2(aq) + 3H2(g)

 

Ionically:

2Al(s) + 2OH(aq) + 2H2O(l)                         2AlO-2(g)  + 3H2(g)

Note:

– Thus aluminium has an amphoteric nature as it reacts with both acids and alkalis.

Uses of aluminium

  1. Making parts of airplanes, railway, trucks, buses, tankers, furniture, and car e.t.c.

Reason;

– It is Very light due to a very low density.

 

  1. Making cooking vessels/ utensils such as sufurias.

Reason:

– It is a good conductor of heat and electricity.

– It is not easily corroded by cooking liquids due to the unreactive coating of aluminium oxide.

 

  1. Making overhead cables

Reason:

– It is a good conductor of electricity.

– It is light hence can easily be supported by poles and ductile to be rolled into wires (cables).

 

  1. Aluminium powder mixed with oil is used as a protective paint.

 

  1. Making Aluminium foils due to its high malleability. The foil is used in cooking; packaging and for milk bottle tops.

 

  1. Making alloys, which have high tensile strength and yet light.

Examples:

 

Alloy Component
Duralumin Aluminium, copper, manganese and magnesium
Magnalium Aluminium (70%) and magnesium (30%)

 

  1. As a reducing agent in the Thermite process in the production of some elements such as chromium, cobalt manganese and titanium.

Example:

Cr2O3(s) + 2Al(s)                       2Cr(s) + Al2O3(s)

 

Note:

  • The thermite process

Is a process of reducing oxides of metals which are ordinarily difficult to reduce using Aluminium powder.

 

Examples:

  • Iron (III) Oxide (Fe2O3)
  • Chromium (III) Oxide (Cr2O3)
  • Compound oxide of manganese (Mn3O4)

 

– The oxide and the Al powder are well mixed together, forming Thermite.

– The thermite is ignited using magnesium ribbon fuse, since the reaction will not start at low temperatures.

– The high heat of formation of Aluminium oxide , results into a vigorous exothermic reaction that leads to a molten metal.

 

Example: -In the reduction of chromium (III) Oxide.

Cr2O3(s) + 2Al(s)                         2Cr(l) + Al2O3(s) + Heat.

 

Sample question:

 

  1. Zinc

Main ores;

(i). Zinc blende; ZnS

(ii). Calamine; ZnCO3.

 

Extraction:

– Is done by electrolysis or reduction of its oxide using carbon.

 

Preliminary steps:

– The ore is first concentrated by froth floatation.

– The ore is roasted in air to convert it to the oxide.

Equations:

  • From Zinc blende:

2ZnS(s) + 3O2(g)                 2ZnO(s) + 2SO2(g)

 

  • From Calamine:

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

 

– After obtaining the oxide the metal is extracted by either reduction or reduction:

 

(a). The reduction method.

– The oxide is mixed with coke and limestone and heated in a furnace.

 

Diagram: Furnace for zinc extraction by reduction:

 

 

– The limestone (CaCO3) decomposes to liberate CO2 which is then reduced by coke to form carbon (II) oxide.

 

Equations:

Heat

 

 

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

Then:

CO2 (g) + C(s)                 2CO (g)

 

– The resultant carbon (II) oxide and coke are the reducing agents in the furnace, at about 1400oC.

– They reduce the oxide to the metal; which is liberated in vapour form.

 

Equations:

  • ZnO(s) + C(s)                      Zn (g)  + CO (g)
  • ZnO(s) + CO(s)                       Zn (g)  + CO2 (g)

– At the furnace temperatures zinc exists in vapour form, and leaves at the top of the furnace.

– Liquid zinc being lighter settles above molten lead and is run off;

– The vapour is condensed in a spray of molten lead to prevent re-oxidation of zinc.

– The resultant zinc is 98-99% pure and can be further purified by distillation.

– SO2 is a by-product and is the main source of pollution in the extraction of zinc.

– Usually it is channeled to a contact process plant for the manufacture of sulphuric acid.

– Alternatively it can be scrubbed off to prevent pollution of the environment.

– Less volatile impurities remain in the furnace.

– The silica impurities combine with the quicklime/ calcium oxide (CaO) from limestone to form calcium silicate.

 

Equation:

CaO(s)  + SiO(s)                             CaSiO3(s)

 

– The silicates together with other less volatile impurities form slag, at the bottom of the furnace from where it is run off.

 

Summary: Flow chart and the extraction of zinc

 

                            Sulphur (IV) oxide                  Coke and limestone                       CO2 and excess CO(g)

ZnO(s)

 

 

 

 

ZnS (ore)

 

 

 

                                                                                                                                                                       Zinc + impurities

Separation chamber
      Slag

 

 

                                     Air

 

 

 

                                                                                                                                                                       Zinc liquid (pure)

 

 

 

 

 

(b). Electrolytic extraction of zinc.

– Zinc metal is obtained from the oxide via a series of steps:

 

Step I: Preparation of electrolyte:

– The ZnO obtained from roasting the ore is converted to zinc sulphate by reacting it with dilute sulphuric (VI) acid.

 

Equation:

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

 

– Any lead (II) oxide impurity present in the zinc oxide reacts with the acid to form lead (II) sulphate.

Equation:

PbO(s) + H2SO4(aq)                          PbSO4(s) + H2O(l)

 

– The insoluble lead (II) sulphate is then precipitated and separated by filtration;

– The zinc sulphate is then dissolved in water and the solution electrolysed.

 

Step II: The electrolytic process:

Electrolyte:

– Zinc (II) sulphate solution;

 

Ions present:

– Zn2+ and H+ as cations; and SO42- and OH as anions;

 

Cathode:

Lead containing 1% silver.

 

Anode:

– Aluminium sheets;

 

Chemical reactions:

Cathode:

Observations:

– Deposits of a grey solid.

 

Explanations:

– Zn2+ and H+ migrate to the cathode.

– The Zn2+ are discharged in preference to H+;

Reason:

– The cathode is relatively reactive. Thus since zinc is more reactive thn hydrogen, its ions undergo reduction faster;

 

Equation:

Zn2+(aq) + 2e-                      Zn(s);

 

Note:

If graphite electrodes were used, hydrogen gas would have been evolved instead;

 

 

Anode:

Observations:

– Evolution of a colourless gas that relights a glowing splint;

 

Explanations:

– OH and SO42- migrate to the cathode.

– The OH are discharged in preference to SO42-; giving off oxygen gas

Reason:

The OH ions have a higher oxidation potential than SO42- and therefore easily giving electrons for reduction at the cathode

 

Equation:

4OH(aq)                              2H2O(l) + O2(g) + 4e-

 

Note:

– Over 80% of zinc is extracted by the electrolytic method.

– Zinc extracted by the electrolytic method is much more pure.

 

Note: – Industrial plants that can be set up near the zinc extraction plant.

– Contact process plant, to make use of the SO2 by-product.

– Lead accumulators factories, to utilize the zinc produced.

– Paper factory using, SO3 and hence SO2 in bleaching.

– Brass factory for alloying zinc and copper.

– Steel factory to use zinc in galvanization.

 

Properties of zinc

Physical properties

– Is a blue-grey lustrous metal with:

  • Density of 7.1gcm-3
  • Melting point of 420oC
  • Boiling point of 907oC

 

Chemical reactions

(a). with air;-

– Zinc tarnishes slowly forming a protective layer which prevents further reaction i.e. oxide layer or basic carbonate.

 

Equations:

2Zn(s) + O2(g)                           2ZnO(s)

Then;

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

 

– It burns with a blue-green flame when strongly heated in air to form an oxide which is yellow when hot and white when cold.

Equation:

2Zn(s) + O2(g)                          2ZnO(s)

 

 

 

(b). With water

– Zinc does not react with water

– Steam reacts with red-hot zinc, forming zinc oxide and liberating hydrogen gas.

Equation:

Zn(s) + H2O(g)                          ZnO(s) + H2(g)

 

(c). with dilute acids

– Zinc is above hydrogen in the reactivity series hence displaces hydrogen from steam (water) and dilute acids like H2SO4 and HCl.

Equation:

Zn(s) + 2H+(aq)                            Zn2+(aq) + H2(g)

 

– Pure zinc reacts slowly while impure zinc reacts faster/ more quickly.

– Copper (II) sulphate is used as a catalyst to speed up the reaction.

 

(d). Concentrated acids.

(i). Concentrated sulphuric (VI) acid.

Equation:

Zn(s) + 2H2SO4(l)                        ZnSO4(aq) + 2H2O(l) + SO2(g)

 

(ii). 50% concentrated nitric (v) acid.

– It reacts with 50% concentrated nitric (V) acid to liberate nitrogen (II) oxide.

Equation:

3Zn(s) + 8HNO3(l)                        Zn(NO3)2(aq) + 2H2O(l) + 2NO(g)

 

(iii). Concentrated nitric (V) acid.

– It reduces concentrated nitric (V) acid to nitrogen (IV) oxide.

Equation:

Zn(s) + 4HNO3(l)                        Zn(NO3)2(aq) + 4H2O(l) + 2NO2(g)

 

(e). with alkalis.

– Zinc is amphoteric and dissolves in hot alkalis to give the zincate ion and hydrogen gas.

Equation:

Zn(s) + 2OH(aq) + 2H2O(l)                          H2(g) + [Zn(OH)4]2-(aq)

 (Zincate ion)

(f). Other reactions.

(i). Zinc burns in chlorine to give zinc chloride

Zn(s) + Cl2(g)                       ZnCl2(s)

(ii). Zinc combines with sulphur

Zn(s) + S(s)                      2ZnS(s)

 

 

 

 

 

 

Uses of zinc:

  1. Galvanization of iron sheets to prevent corrosion and rusting.

Note:

Rusting does not occur even when galvanized iron sheets are scratched and exposed.

Reason:

– The rest of the zinc protects the iron from rusting. This is because zinc being more reactive gets oxidized in preference to iron, and is hence “sacrificed” in the protection of iron.

– This method is referred to as cathodic or sacrificial protection.

  1. Making brass; an alloy of copper and zinc.
  2. Making outer casings of dry batteries;
  3. Die-castings contain zinc and aluminium, and are used for making radio and car parts;
  4. Zinc cyanide is used for refining silver and gold;

 

Sample question:

 

  1. Iron

– Is the second most abundant metal after aluminium, forming about 7% of the earth’s crust.

 

Main ores

– Haematite, Fe2O3;

– Magnetite, Fe3O4;

– Siderite, FeCO3;

 

Qualitative analysis for presence copper in an ore sample.

– Crush the ore into fine powder;

– Add dilute nitric (V) acid to the ore, to dissolve the oxide filter to obtain the filtrate.

– To the filtrate add aqueous sodium hydroxide / ammonium hydroxide dropwise till in excess.

– Formation of a red brown / brown precipitate in both cases indicates presence of Fe3+

 

Extraction from haematite (Fe2O3).

Summary of the process

The ore-haematite is crushed and mixed with coke and limestone.

– The mixture is called charge.

– The charge is loaded into the top of a tall furnace called blast furnace.

– Hot air-the blast is pumped into the lower part of the furnace.

– The ore is reduced to iron as the charge falls through the furnace.

– A waste material called slag is formed at the same time.

– The slag floats on the surface of the liquid iron produced.

– Each layer can be tapped off separately.

 

Details of the extraction process.

(i) The blast furnace

Is a tall, somewhat conical furnace usually made of silica and lined on the inside with firebrick.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Details of the extraction process

Raw materials.

– Iron ore; i.e. haematite

– Coke; C

– Limestone; CaCO3;

– Hot air;

 

Conditions

– Temperature at the bottom of furnace, 1400-1600oC

– Temperature at the top of the furnace, 400oC

 

Reactions and processes

Step-1 – crushing and loading

– The ore is crushed into powder form, to increase the surface area for the upcoming reduction/ redox reactions.

– It is then mixed with coke and limestone and then fed at the top of the furnace using the double bell (double-cone devise) changing system

Note:

– The double bell charging system ensures that the furnace can be fed continuously from the top with very little heat loss, by preventing any escape of hot gases.

– This in turn reduces production costs.

 

Step 2: -Pre heating of the blast furnace.

– Air that has been preheated to about 700oC is blown/ fed into the base of the blast furnace through small pipes called tuyers.

– This provides the required temperatures for the reactions in the blast furnace.

– This results into highest temperatures, about 1600oC at the hearth (bottom of the furnace) which then decreases upwards the furnace.

 

Step 3: -Generation of reducing agents.

– Two reducing agents are used in this process: Coke and carbon (II) oxide; with carbon (II) oxide being the main reducing agent.

 

(i). Oxidation of coke;

– Coke burns in the blast at the bottom of the furnace.

– The reaction temperatures is about 1600oC and the product is Carbon (IV) oxide gas

– This reaction is exothermic, producing a lot of heat in the blast furnace.

 

Equation:

C(s) + O2(g)                CO2(g)

 

(ii). Decomposition of limestone;

– The limestone in the charge decomposes in the blast furnace to calcium oxide (Quicklime) and carbon dioxide.

 

Heat

Equation:

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

 

 

– The calcium oxide will be used in the removal of the main ore impurity/ silicates/ silica in the form of silicon (IV) oxide.

– The CO2 then moves up the blast furnace to regenerate carbon (II) oxide, the chief reducing agent.

 

(iii). Production of carbon monoxide

– The CO2 from oxidation of coke and decomposition of limestone (calcium carbonate) react with (excess) coke, to form carbon (II) oxide

– The reaction occurs higher up in the blast furnace at about 700oC;

 

Equation:

CO2(g) + C(s)               2CO(g)

 

Step 4: The actual reduction process

– Reduction of the ore is by either CO or coke, depending on temperatures.

 

(i). Reduction by coke

– This occurs much lower down the furnace at higher temperatures of about 800oC and above.

– This reaction is ordinarily slow and thus serves to only reduce the part of the ore reduced by CO at lower temperatures in the upper parts of the furnace.

 

Equation:

2Fe2O3(s) + 3C(s)                   4Fe(s) + CO2(g)

 

Note:

– The resultant CO2 is quickly reduced to CO by the white-hot coke to more carbon (II) oxide as per step 3(iii) above.

 

(ii). Reduction by carbon (II) oxide

– This is the main reducing agent.

– The reaction between CO and Fe2O3 is relatively faster and occurs at lower temperatures of 500oC-700oC, higher up the furnace.

 

Equation:

Fe2O3(s) + 3CO (g)                           2Fe(s) + CO2 (g)

 

– The resultant carbon (IV) oxide is also quickly recycled by being reduced to CO by coke to from more reducing agent

 

(iii). Melting

– The iron produced in both of the reduction processes is in solid state.

– As the iron drops / falls down the furnace, it melts as it passes through the melting zone/ molten zone (1500oC-1800oC)

– The molten iron runs to the bottom of the furnace.

– Temperatures at the hearth (bottom of the furnace) is maintained at approx. 1400oC and yet pure iron melts at about 1525oC.

– Consequently the molten iron would easily solidify at the base (Temp =1400oC)

– However this is not usually the case;

Reason:

-Impurities absorbed by iron during melting (mainly carbon) reducing the melting point to below 1400oC.

– The molten iron is then easily tapped off.

 

Step 5: -Removal of earthy impurities.

– The earthy impurities in the ore (mainly silica) react with calcium oxide from decomposition of limestone to form calcium silicate.

 

Equation:

CaO(s) + SiO2(s)                     CaSiO3(s)

 

– These earthy impurities form molten slag whose main component is calcium silicate.

– The slag does not mix with iron but rather floats on top of it, at the base of the furnace.

 

Importance of the slag

– As it floats on top of molten iron it protects it from being re-oxidized by the incoming hot air.

 

Uses/application of the slag

  1. Light-weight building material.
  2. Manufacture of cement.
  3. Road building material.

 

Step 6:- Removal of furnace (waste) gases.

– Hot unreacted/waste gases leave at the top of the furnace.

– Main components include Nitrogen, unreacted CO2, unreacted CO, oxygen and Argon (Noble gases)

– Additionally they contain dust particles.

Note:

– Upon removal of dust particles, the furnace gases, being hot can be used to pre-heat the air blown in at the base.

 

Summary: flow chart for the extraction of iron.

 

Properties of iron:

Physical properties:

– It has a melting point of 420oC and a boiling point of 907oC;

– Have a good thermal and electrical conductivity;

– It is ductile and malleable;

 

Chemical properties.

(i). Reaction with air.

– It readily rusts in presence of moist air hydrated brown iron (III) oxide; Fe2O3.H2O(s)

Equation:

4Fe(s) + 2H2O (l) + 3O2 (g)                          2Fe2O3.H2O (l)

 

– When heated it reacts with oxygen to form tri-iron tetroxide; Fe3O4;

Equation:

 

3Fe(s) + 2O2 (g)                             Fe3O4(s)

 

(ii). Reaction with water.

– It does not readily react with cold water.

– It however reacts with steam liberating hydrogen gas and forming tri-iron tetroxide.

Equation:

3Fe(s) + 4H2O(g)                          Fe3O4(s) + 4H2(g)

 

(iii). Reaction with chlorine.

– Hot iron glows in chlorine without further heating, forming black crystals of iron (III) chloride;

– Iron (III) chloride sublimes on heating and will thus collect on the cooler parts of the apparatus;

Equation:

2Fe(s) + 3Cl2(g)                           2FeCl3(s)

 

Note:
– Iron (III) chloride fumes when it is exposed to damp (moist) air;

Reason:

– It is readily hydrolysed by water with evolution of hydrogen chloride gas;

 

Equation:
FeCl3(s) + 3H2O(l)                       Fe(OH)3(s) + 3HCl(g)

 

(iv). Reaction with acids:

  • Hydrochloric acid:

– Iron reacts with hydrochloric acid to liberate hydrogen gas.

Equation:

2Fe(s) + HCl(aq)                         FeCl2(aq) + H2(g)

 

  • Sulphuric (VI) acid:

Fe(s) + H2SO4 (aq)                        FeSO4 (aq) + H2 (g)

 

Note: With hot concentrated H2SO4;

– The iron reduces hot concentrated H2SO4 to sulphur (IV) oxide and it is itself oxidized to iron (III) sulphate.

Equation:

2Fe(s) + 6H2SO4 (l)                              Fe2 (SO4)3(aq) + 6H2O (l) + 3SO2 (g)

 

  • Nitric (V) acid.

– Iron reacts with dilute nitric (V) acid to form nitrogen (IV) oxide and ammonia which then forms ammonium nitrate.

Equation:

10HNO3 (aq) + 4Fe(s)                          4Fe(NO3)2(aq) + NH4NO3(aq) + 3H2O(l)

 

– Warm dilute nitric (V) acid gives iron (II) nitrate.

– Concentrated nitric (V) cid renders the iron unreactive.

Reason:

– Formation of iron oxide as a protective layer on the metal surface.

 

(vi). Reaction with sulphur.

– Iron when heated in sulphur forms iron (II) sulphide.

Equation:

Fe(s) + S(s)                      FeS(s)

 

Uses of iron.

– Iron exists in different types and alloys, depending on percentage composition of iron, and other elements.

– Each type of alloy of iron has different uses depending on properties.

 

Iron alloy or type Properties. Uses
Cast iron – Refers to iron just after it has been produced in the blast furnace;

– Contains 3-5% carbon, 1% silicon, and 2% phosphorus;

Disadvantage: very brittle hence easily breaks;

Advantage: It is extremely very hard;

– Making:

Ø  Furnaces;

Ø  Grates;

Ø  Railings;

Ø  Drainage pipes;

Ø  Engine blocks;

Ø  Iron boxes;

Note: This is due to its very hard nature;

– Manufacture of wrought iron and steel;

Wrought iron – Refers to cast iron with 0.1% carbon.

– It is malleable hence can easily be moulded or welded;

– making iron nails; horse shoes; agricultural implements like pangas;

Note:

Its use is declining due to increased use of mild steel

Steel – Are alloys whose main component is iron;

– Other components may be carbon; vanadium; manganese; tungsten; nickel and chromium;

Examples:

Mild steel: has about 0.3% carbon, 99.75% iron;

Special steel: has a small percentage of carbon together with other small substances;

 

Stainless steel:

-Contains 74% iron, 18% chromium, and 8% nickel;

 

 

 

Cobalt steel:

– Contains about 97.5% iron and 2.5% cobalt;

– Very tough and hard;

– Slightly magnetic;

– Mild steel is used for making:

Ø  Nails; Car bodies;

Ø  Railway lines; Ship bodies;

Ø  Rods for reinforced concrete, pipes;

Note: Advantage of mild steel:

– It is easy to work on;

 

 

– That with 10-12% chromium and some nickel is used to make: cutlery; sinks; vats;

– Steel containing 5-18% tungsten is used for: making high speed cutting and drilling tools;

 

– For making electromagnets;

  1. Copper

Description: – A red brown metal.

Distribution: – Canada, USA, Zambia, and Tanzania.

 

Main ores;

– Copper pyrites, CuFeS2;

– Cuprite, CuO

– Chalcocite, Cu2S

– Malachite, CuCO3.Cu (OH)2

 

Qualitative analysis / test for presence in an ore sample.

– Crush the ore and then add dilute nitric or hydrochloric or sulphuric acid to dissolve the ore.

– Filter to obtain Cu2+ filtrate.

– Divide filtrate into 2 different test tubes.

– To one sample add aqueous Ammonium hydroxide dropwise till in excess formation of a pale blue precipitate soluble in excess NaOH to form a deep blue solution.

 

Equations

With little NaOH:-

Cu2+(aq) + 2OH(aq)                        Cu(OH)2(s)

Pale blue ppt.

In excess:

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

Deep blue solution

 

-To the second portion add sodium hydroxide solution dropwise till in excess, formation of a pale blue precipitate insoluble in excess confirms presence of Cu2+.

 

Extraction- from copper pyrites.

  1. Crushing the ore

– The ore is crushed to increase the surface area for the succeeding chemical reactions.

– The ore is then concentrated.

 

  1. Concentration of the ore.

– The ore is concentrated by froth floatation.

– The fine ore powder is mixed with water and oil, after which air is blown into the mixture, usually from below.

– Bubbles of the air forms froth, resulting to concentration of the ore.

– The lighter oil floats on top of the water, with the ore floating on top of the oil.

– The denser water sediments the earthy impurities like soil particles.

– The concentrated ore is then tapped off.

– This process involves formation of an oil froth onto which the ore floats hence the name froth formation.

 

  1. First Roasting

-The concentrated copper pyrite, CuFeS2 is then roasted in air to remove some of the sulphur impurities as sulphur (IV) oxide.

Equation:

2CuFeS2(s) + 4O2 (g)                 3SO2 (g) + 2FeO(S) + Cu2S(s);

Note:

– During 1st roasting limestone and silica (SiO2) are added to the roasted ore and the mixture heated in the absence of air.

 

Importance

– Removal of iron impurities.

– The iron (II) so formed during roasting is converted to iron (II) silicate, FeSiO3.

– The iron (II) silicate constitutes the major portion/component of the slag.

 

Equation:

FeO(s) + SiO2(g)                 FeSiO3(s)

 

– The slag separates itself from the copper (I) sulphide.

– The sulphur (IV) oxide escapes into the atmosphere and is the major pollutant in this process.

 

Pollution control mechanisms.

– Scrubbing the gas using calcium hydroxide;

 

Equation:

SO2(g) + Ca (OH)2(s)                CaSO3(s) + H2O(l)

 

– Construction of a contact process nearby.

 

  1. Second Roasting.

– The CuS is heated in a regulated supply of air where some of it is converted to Cu2O

 

Equation:

2Cu2S(s) + 3O2(g)                      Cu2O(s) + 2SO2(g)

 

  1. Reduction of copper (II) oxide

Note: – Not all the Cu2S was oxidized to copper (I) oxide Cu2O.

– The unreacted (unoxidized) Cu2S serves as the reducing agent in this step.

i.e. The copper (I) oxide formed in step 4 is reduced to copper metal by the (unreacted) copper (I) sulphide.

– This is called blister copper.

 

Equation:

Cu2S(s) + 2Cu2O(s)                6Cu(s) + SO2 (g);

 

  1. Electrolysis

-The copper metal from reduction in step 6 is impure and is thus purified by electrolysis.

 

Main impurities

– Traces of gold

– Traces of silver

– Iron

– Sulphur

 

 

 

Electrolytic apparatus

Anode: Impure copper

Cathode: Pure copper plates/ sheets;

Electrolyte: Dilute copper (II) sulphate solution (containing Cu2+; H+; SO42- and OH)

 

Diagram of electrolytic apparatus.

 

 

 

 

 

 

 

 

 

 

 

 

Electrolytic reactions;

At the cathode.

Observations:

– Deposition of a brown solid.

 

Explanations:

– The copper (II) ions, Cu2+ move to the cathode, where they accept electrons and undergo reduction.

– Cations in the electrolyte are Cu2+ and H+ but Cu2+ are preferentially discharged due to their easy tendency to undergo reduction.

 

Equation:

Cu2+(aq) + 2e-                     Cu(s);

 

At the anode

Observations:

– Dissolution of the anode, hence the impure copper rod decreases in size.

 

Explanation

– Since the metal rod is dipped into a solution of its ions, the copper solid undergoes oxidation, losing electrons to form copper ions, Cu2+

– Consequently as more copper ions, Cu2+ get reduced at the cathode; more are released by the dissolving anode.

 

Equation:

Cu(s)                          Cu2+(aq) + 2e-

 

Overall reaction

Cu(s) + Cu2+(aq)                     Cu2+(aq) + Cu(s)

 

– The electrolytic product is 99.98% copper.

– Traces of silver and gold collect as sludge at the bottom of the cell.

Note:-To improve purity of the product of electrolysis, the following steps are advisable;

(i). Increase the dilution of the electrolyte/ use a very dilute electrolyte.

(ii). Reduce the amount of current / use a low current.

 

Summary of extraction of copper from copper pyrites.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Uses of copper

  1. Making copper wires and contacts in switches, plugs and sockets

Reason:-Copper is a good conductor of electricity

Note:-For this purpose, pure copper is necessary, since impurities increase electrical resistance.

 

  1. Making soldering instruments.

Reason:-Copper has a high thermal conductivity.

 

  1. Manufacture of alloys.

Examples

Alloy Components
Brass Copper and zinc
Bronze Copper and tin
German silver Copper, zinc and nickel;

 

  1. Making coins and ornaments.

Reason:-it is durable and aesthetic.

 

 

Properties of copper.

  1. Physical properties

– Soft red-brown metal.

– Melting point of 1083oC and a boiling point of 2595oC

– Density is 8.92gcm-3 and electrical conductivity is about 5.93 x 10-9 Ώ-1m-1

 

  1. Chemical properties

(i). It does not react with cold water or steam.

 

(ii). Heating in air

– When heated in air it forms a black layer of copper (II) oxide on its surface.

– Finely divided copper burns with a blue flame.

 

Equation:

2Cu(s) + O2(g)                     2CuO(s)

(Red brown)                                           (Black)

 

(iii). Reaction with chlorine

– Cu reacts with chlorine in presence of heat to form green copper (II) chloride.

 

Equation:

Cu(s) + Cl2(g)             CuCl2(s)

 

(iv). Reaction with Acids.

– Copper does not react with dilute hydrochloric, nitric and sulphuric acids.

– However it reacts with 50% Nitric acid, concentrated Nitric acid and concentrated sulphuric acid.

 

  • With 50% Nitric (V) acid

– Copper reduces the nitric acid to nitrogen monoxide.

 

Equation;

3Cu(s) + 8HNO3(l)                    3Cu(NO3)2 (aq) + 4H2O(l) + 2NO(g);

 

  • With concentrated Nitric (V) acid

– Copper reduces the acid to brown nitrogen (IV) oxide/Nitrogen dioxide gas.

 

Equation:

Cu(s) + 4HNO3(l)               Cu (NO3)2(aq) + 2H2O(l) + 2NO2(g)

 

  • With concentrated sulphuric acid.

-The sulphuric acid is reduced to sulphur (IV) oxide.

 

Equation:

Cu(s) + 2H2SO4(aq)                 CuSO4 (aq) + 2H2O(l) + SO2 (g)

 

 

Worked example.

The flow chart below outlines some of the processes involved in the extraction of copper from copper pyrites. Study it and answer the questions that follow.

 

Hot air                                          Air

(a) (i). Write an equation for the reaction in the first roasting chamber.                                       (1mark)

 

(ii). Name gas K.                                                                                                                           (1mark)

 

 

(iii). Give the name and formula of slag M.                                                                                    (1mark)

 

(iv). Give the name of the reaction in chamber N.                                                                          (1mark)

 

(v). Name the impure copper X.                                                                                                      (1mark)

 

(b). Pure copper is obtained from impure copper by electrolysis.

(i). Name the anode the cathode and the electrolyte.                                                                    (3 marks)

 

(ii). Write equations for the reactions at the anode and cathode.                                            (2 marks)

 

(iii). calculate the time taken for a current of 10 amperes to deposit 32kg of pure copper. (Cu = 64, 1F = 96000C)                                                                                                                                    (3 marks)

(c). Draw a diagram to show how you would plate an aluminium spoon with copper

 

6. Lead.

– Is a transition element that combines with other elements to form compounds with 2 oxidation states.

– It is among the group 4 elements;

 

Main ores:

– Galena, PbS (lead sulphide); the main ore;

– Cerrusite, PbCO3 (lead carbonate)

– Anglesite, PbSO4 (lead II) sulphate);

Qualitative analysis / test for presence of Zn2+ in an ore sample.

– The ore is crushed and then dilute nitric or hydrochloric or sulphuric acid added to dissolve the ore.

– It is then filtered to obtain Zn2+ filtrate.

– The filtrate is divided into 2 different test tubes.

– To the first portion sodium hydroxide solution is added dropwise till in excess, formation of a white precipitate soluble in excess confirms presence of either Zn2+; Al3+ or Pb2+.

– To the second sample aqueous ammonium hydroxide is added dropwise till in excess; formation of a white precipitate soluble in excess NH4OH(aq) confirms presence of Zn2+ only;

 

Equations

With little NH4OH:-

Zn2+(aq) + 2OH(aq)                        Zn(OH)2(s)

White ppt.

In excess:

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

Colourless solution

Extraction of lead:

– Occur in three main steps:

Ø  Ore concentration;

Ø  Extraction by reduction;

Ø  Purification (refining) by electrolysis;

1. Ore concentration:

– Is done by selective froth floatation;

– The ore is ground into a fine powder, then water and a suitable oil added;

– Air is then blown into the mixture; facilitating formation of a low density froth that floats on top;

– Additionally, chemicals such as sodium cyanide and zinc sulphate are added to facilitate separation of zinc sulphide present in the ore.

– The separated PbS is then dried and broken into smaller pieces, then subjected to reduction;

2. Reduction:

Step I: Roasting the ore:

– The crushed and concentrated ore is roasted in a furnace to convert it to lead (II) oxide;

Equation:

2PbS(s) + 3O2 (g)                       2PbO(s) + 2SO2 (g)

– During roasting some of the lead (II) sulphide is converted to lead (II) sulphate;

Equation:

PbS(s) + 2O2(g)                          PbSO4(s);

– Any lead sulphate formed is converted to lead silicate by silicon (IV) oxide;

– The fate of lead (II) silicate;

Note:

– Additionally the lead (II) sulphate may further react with lead sulphide to form lead metal;

Step II: Ore reduction:

– The lead oxide obtained is mixed with coke, limestone and silica and some scrap iron;

– The mixture is fed into the top of the Imperial smelting furnace (ISF); where it is melted using hot air blasts introduced near the bottom of the furnace;

 

Diagram: The Imperial Smelting furnace for Lead extraction.

Main reactions:

(i). The lead (II) oxide is reduced to lead by the coke.

Equation:
PbO(s) + C(s)                              Pb(s) + CO(s);

 

(ii). The resultant carbon (IV) oxide produced in reaction (i) above further reduces any remaining lead (II) oxide;

 

Equation:
PbO(s) + CO(s)                              Pb(s) + CO2(g);

 

(iii). The scrap iron is added so as to react with any lead sulphide that may be present.

Equation:

Fe(s) + PbS(s)                      Pb(s) + FeS(s)

 

 

 

(iv). The limestone undergoes decomposition to give calcium oxide and liberate carbon (IV) oxide;

Equation:

CaCO3(s)                       CaO(s) + CO2 (g);

 

– The carbon (IV) oxide gets reduced by coke to form more carbon (IV) oxide for reduction as in reaction (ii);

 

Equation:
CO2 (g) + C(s)                    2CO (g);

 

(v). The calcium oxide reacts with silica in form of SiO2 to form calcium silicate;

 

Equation:

CaO(s) + SiO2 (g)                       CaSiO3 (l);

 

Waste gases and residues.

– The iron sulphide and calcium silicate form a molten slag which is less dense and floats on top of molten lead at the bottom of the furnace;

– From here the slag is separately tapped off;

– Excess gases and air that did not react in the blast furnace escape through outlets at the top of the furnace;

– These waste gases can be trapped and recycled;

– These gases include: excess CO; excess CO2; oxygen; nitrogen; some SO2; and argon;

 

Pollution effects:
– Main pollutant is sulphur (IV) oxide from roasting of the ore.

 

Pollution control:
– It is directly fed into a contact process plant or scrubbed using calcium hydroxide forming calcium sulphite;

 

  1. Purification (refining) of lead:
    – The molten lead obtained in this process contains impurities such as gold, silver, copper, arsenic, tin and sulphur;

– The impure lead is refined by electrolysis.

 

Electrolysis of molten lead.

(i). Electrolyte:

Any aqueous solution containing lead ions;

 

(ii). The anode:

– Impure lead;

 

(iii). The cathode:

– Pure lead;

 

 

 

 

Electrolytic reactions;

At the cathode.

Observations:

– Deposition of a grey solid.

 

Explanations:

– The lead (II) ions, Pb2+ move to the cathode, where they accept electrons and undergo reduction.

– Cations in the electrolyte are Pb2+ and H+ but Pb2+ are preferentially discharged due to their easy tendency to undergo reduction.

 

Equation:

Pb2+(aq) + 2e-                     Pb(s);

 

At the anode

Observations:

– Dissolution of the anode, hence the impure lead rod decreases in size.

 

Explanation

– Since the metal rod is dipped into a solution of its ions, the impure lead solid undergoes oxidation, losing electrons to form lead (II) ions, Pb2+

– Consequently as more lead (II) ions, Pb2+ get reduced at the cathode; more are released by the dissolving anode.

 

Equation:

Pb(s)                          Pb2+(aq) + 2e-

 

Overall reaction

                       Oxidation at anode;

 

 

Pb(s) + Pb2+(aq)                     Pb2+(aq) + Pb(s)

 

 

Reduction at cathode;

 

Summary: flow chart on extraction of lead.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Properties of lead:

Physical properties;

– Has a low melting point but a high density;

– The unusually low melting point of lead is difficult to explain using simple metallic bonding theory;

– It is rather soft and pliable;

– Relatively malleable;

 

Chemical properties.

Note: – Lead is fairly unreactive to most other metals;

 

  1. Uses of lead:
  1. It is used in several alloys e.g. solder and also added to bronze alloys to make them stronger;
  2. Lead ingots are used in the manufacture of accumulators;
  3. Being so malleable and so chemically inert lead sheeting was used for roofing (look at roofs of old churches and cathedral)-cost and pollution effects have however brought this to a stop;
  4. Making lead pipes for water supply; this is also discouraged particularly in soft water areas due to threat of lead poisoning;
  5. Making tetraethyl lead (IV) which for many years was used as a fuel additive to increase octane rating of fuels;
  6. Used in weights, clock pendulums, plumb bobs etc; due to its high density;
  7. It absorbs X-rays and hence lead aprons and lead glass are used to shield hospital radiographers;
  8. Used for safe disposal or storage of radioactive substances since no radioactive emission has been known to pass through thick lead blocks;

 

Sample question:

 

 

 

 

UNIT 6: RADIOACTIVITY

Checklist

  1. Meaning of radioactivity
  2. Natural and artificial radioactivity
  3. Nuclear equations and chemical equations.
  4. Types of radiations
  • Alpha particles
  • Beta particles
  • Gamma rays
  1. Half life of a radioisotope
  2. Radioactive decay curves and half life
  3. Radioactive disintegration and nuclear equations.
  4. Radioactive decay series, nuclear fission and nuclear fusion;
  5. Uses of radioactivity;
  6. Dangers and pollution effects of radioactivity;

 

 

Definitions:

  • Radioactivity is an automatic spontaneous disintegration of nuclei of some heavy elements emitting some kinds of radiant energy
  • Elements which exhibit this are said to be radioactive and emit different types of radiation

Types of radiation from radioactive elements

They can be identified by

  1. Measuring their penetration power
  • They were directed to thin leaves of paper, aluminium, lead plates of varying thickness.
  • Observations:
  1. One penetrated only thin foils of paper but not aluminium of 0.05mm thick and lead
  2. One penetrated both paper and Al but not lead
  3. One penetrated paper Al and lead
  4. Directing them to air
  • Observations
  • Penetrated less through air but caused a lot of ionization to the air molecule
  • Penetrated more through air but caused less ionization
  • Penetrated most through air but caused less ionization

These penetrated just as far as X-rays

Directing through magnetic /electric fields

 

 

 

 

 

 

                                                                                                                     S                      N

–           +

 

 

 

 

 

 

The radioactive element radiation emits all the 3types of radiation when they were directed to poss through electric magnetic fields to the fields

Were slightly deflected towards the negative
Were strongly deflected towards the positive
  1. Were not deflected

Therefore

A and B –are charged particles

-Have opposite charges

A is positive, B is negative

And

Deflections obeyed through L.H Rule

  • A were deflected less because they were moving with higher momentum (MV) because of high mass

Experimental evidence

  • Experiments carried out by Rattlerford revealed
  1. Were positively charged and had a mass of 4 units and thus a charge of +2 these are Alpha- particles therefore double charged helium ions,
  2. Were negatively charged and had similar properties to cathode rays. Measurement of charge/ mass confirmed they were electrons and carries a unit of –1

They are beta particles

  1. Were carrying no charge. They are electromagnetic waves similar to light rays and X- rays with a wave length of only 10 metres

Their emission enables a nucleus to lose surplus energy

They are gamma rays

Evidence for the nature of gamma rays

  • Gamma rays
  • Are unaffected by an electric field
  • Are an affected by a magnetic field
  • Can penetrate several centimeters of lead
  • Can be diffracted by the lattice of a crystal
  • Have no change in atomic number mass of the atoms emitting them

 

NATURE AND PROPERTY OF α , β ,AND γ

 

Nature α- particles helium nuclei

4He (He )

Β-particles electrons Rays electromagnetic waves radiation
Relative penetrating power Least 5cm in air stopped by paper and Al foils Several metres of air thin Al foil (100) Penetrate air, Al and many mm of Pb (10000)
Range in air

 

 

 

A few cm A few m A few km
Effect of electric and magnetic fields Small deflection Large deflection No deflection
Ionization of gases Cause much ionization

Discharges electroscopes rapidly

Cause less ionization

Discharges electroscopes slowly

Negligible ionization

 

Radioactive decay curve

Is always a symptotic to the x-axis

 

 

 

 

 

Number of atoms of the

Radioactive element

 

 

 

 

– The number of atoms disintegrating per unit of time /second is always proportional to the number of atoms, N at that time

This number N decreases slowly / exponentially with time

  • At any time to the number of atoms of a radioactive element is N/ No and at time T½, only ½ the total number of atoms of the original radioactive element will be present i.e.

T½ = ½N

This time is referred to as the half-life period of a radioactive element

Therefore

Half-life period T½ of a radioactive element is defined as the time taken for ½ the atoms to disintergrate

Thus

In T½ the radioactivity of the element diminishes to half its value

Example

Radioactive decay curve of radon

With half life of 4 days

 

No = 6 X 103 atoms

T½ = 4days

 

No. Of atoms

103 x 6

 

 

103 x 5

 

 

103 x 4

 

 

103 x 3

 

 

 

 

 

 

 

Radon emits α- particles

Initial no of radon atoms = 103 X 6

After 4 days the number of atoms present=103 X 3

“ “ “ “8 days “ “ “ “ “ “ “ “ “ “ “ “ “ “ “ “ =103 X  1.5

“ “ “ “12 days “” “ “ “ “ “ “ “ “ “ “ “  “ “ = 103 X 0.75

It is impossible to predict the atoms that will disintergrate next

 

The half lives of some radioactive isotopes

Radioactive isotopes Half life
Uranium-238 4.5 X 109  years
Radium-226 1.6 X 103 years
Carbon-14 5.7 X 103 years
Strontium-90 28 years
Iodine-131 8.1days
Radon-222 4days
Bismuth-214 19.7minutes
Polonium-218 3minutes
Polonium-214 1.5 X 10 -4sec

-The half-life of a radioactive isotope provides aquantitave measure of its stability

Thus

-The shorter the half-life the faster the isotope decays and the more unstable it is

-The longer the half-life the slower the decay process and the more stable the isotope

RADIOACTVE DISINTERGRATION EQUATIONS

-Different radioactive elements disintergrate to emit different radiations

  1. Emission of α particles

-Occurs in isotopes with an atomic number greater than 83 (Z> 83) because they are unstable since their nuclei are so heavy and their atomic mass is too large

-They attain stability by ejecting an alpha particle containing two protons and two neutrons

Examples

  • 238U-α partially-

 

(b) Emission of β particles

  • Occur in isotopes with more neutrons than stable isotopes of the same element, therefore heaviest isotopes of an element  are likely to emit β particles to attain stability
  • During the β-decay process, a neutron splits up forming a proton &an electron

 

 

The proton remains in the nucleus while the electron /β particles is ejected.

Result

Number of neutrons in the isotope decreases by one while number of protons increase by one.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

© Emissions of α-& β alternatively

 

 

 

 

 

 

 

 

(2) Emissions of beta particles in two –stage

 

Application of Radioactivity /Using radioactive isotopes.

  1. Carbon dating
  • Radio active Carbon –14 nuclide produced when Nitrogen –14 is bombarded with radiocum from the sum is used.
  • When they die carbon-14 resent starts to reduce as decay takes place by emission of β particles.
  • Using the decay curve of carbon –14 it is possible to estimate age of the animals/plants since the carbon –14 is present in their tissues.

 

2.Medication/treatment of cancer

  • Gamma rays are used to treat /kill cancer cells when the tumour is subjected to the radiations

E.G Gamma rays /penetrating from CO are used in treating inaccessible growths.

Superficial /skin cancers can be treated by less penetrating radiation from or in plastic sheets strapped on the affected

 

3.Studying metabolic pathways

-Radioactive Isotopes can be used to trace the uptake of metabolism of various elements by animals /plants.

E.g Uptake of phosphate & metabolism of phosphorus by plants can be studied using a fertilizer containing

Radioactive tracer studies using have helped in the exudation of photosynthesis and protein synthesis

I have been used in the diagnosis & treatment of thyroid diseases &in research into thyroid gland functioning.

 

4.Thickness gauge and empty packet detectors

-Radiation passing through a material decreases as the material gets thicker

Hence:

Amount of penetrating Beta –or-gamma- radiation can be used to estimate the thickness of various materials like paper, metal or plastic

-Radiation thickness gauges can be used to control the thickness of sheet steel emerging from a high-speed rolling mill.

-β –Rays measure thickness upto ~ 0.2cm of steel γ-rays can be used with steel upto 10cm thick.

-Level gauges are used to measure amount of liquid in fire extinguisher &gas cylinders

-Empty –packet detectors can be set to reject empty/insufficiently filled packets filled packets of biscuits/cigarettes.  \

5.decting pipe bursts

  • Can be underground pipes carrying water /oil
  • If the water / oil is mixed with radioactive substances form the mixture will leak at the point where there is a burst and the radiations can be detected if a detector is passed
  1. Effect on static electricity
  • In textile industry the presence of static charges can attract dust and cause fires
  • When a radioactive element is placed in such industries the radiations emitted will ionize air and ions formed will attract the static charges : this minimises problems due to static charges

Hazards of radioactive

  • Arise from
  1. Exposure of the body to external radiation
  2. Ingestion/ inhalation of the radioactive matter
  • They damage body cells/ tissues
  • Cause mutation/deformities

Precautions

  • Protect the body with lad/ concrete shielding
  • Never pick/ hold radioactive elements with bare hands ; use forceps and well protected tongs
  • Use radiation absorbers

SAMPLE QUESTIONS

  • Particles from a radioactive source move through 7cm in air at ordinary pressure
  • The radioactive emission of radium are α-, β- and γ. Draw labelled diagram to show how the rays can be separated
  • the table below shows nuclides which are radioactive products of   Their ½ lives and K.E during decay are shown
nucleide Half life energy
Th 1.39 X 1010 3.98
Th 1.9yrs 5.42
Ra 3.64days 5.66
Rn 54.5 sec 6.28
Po 0.16 sec 6.77
At 3 X 10-4 sec 7.64
Bi 60.5 min X
Po 2.9 X 10-7 sec 8.78

 

 

  1. Identify pairs of isotopes of same element
  2. Identify 2 nuclides by in this table, which have been produced directly by α- decay of other nuclides in the table. Use equations
  • Identify 2 nuclides in this table, which have been produced by β- decay of other nuclides in the table. Use equations
  1. Suggest a series of decays for formation of
  2. Deduce how ½ life varies with energy of emitted rays

Suggest the value of X for