The e of the corresponding alkane molecule is replaced with the suffix ol;
The parent molecule is the longest chain containing the – OH group;
The numbering of carbon atoms is done such that the carbon atom with the hydroxyl group OH- attains the lowest possible number
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 C – OH

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 80^oC;
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 80^oC, 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:

C₁₂H₂₂O_11(aq) + H₂O_(l) Sucrase C₆H₁₂O_6(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) ^Zymase2C₂H₅OH_(aq) + CO₂

Glucose/fructose ethanol carbon (IV) oxide

Optimum conditions for fermentation:

Temperatures of 25-30^oC;
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 78-82^oC 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) + 3O_2(g) 2CO_2(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) + 2O_2(g) 2CO_(g) + 3H₂O_(l)

(ii). 2C₂H₅OH_(l) + 2O_2(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) + H_2(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) CH₃COOC₂H_5(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

Esterification is slower than neutralization as the reaction is between molecules and not ions as in neutralization
Esterification is reversible; the forward reaction is esterification and the backward reaction is hydrolysis.
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

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

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) + H_2(g)

Ethanol 250^oC 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

170^oC

H OH

Note:

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

Example: Ethanol and cold conc. sulphuric acid.

C₂H₅OH + H₂SO_4(l) C₂H₅HSO_4(l) + H₂O_(l)

Ethanol Ethyl hydrogen sulphate

(Ethoxyethane)

Uses of ethanol

Used as solvents; in the preparation of drugs, perfumes, liquors, vanish and paints.
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) + 3O_2(g) 2CO_2(g) + 3H₂O_(l) ∆H = -1368KjMol^–

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

Tests for primary alkanols.

Reaction with sodium metal

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

Equation:

2CH₃CH₂OH_(l) + 2Na_(s) 2CH₃CH₂ONa + H_2(g)

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.

CO_2(g) + H₂O_(l)

Combustion

Ester ^{Esterification} Ethanol ^{Sodium metal} Sodium ethoxide + Hydrogen

RCOOC₂H₅ H⁺/ RCOOH CH₃CH₂OH CH₃CH₂ONa + H₂

Conc. H₂SO₄ at 170^oC Warm with acidified K₂Cr₂O₇or 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 C_nH_2n-1OOH

– The formula can also be written as C_nH_2n _{+ 1}COOH; 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 105^oC.

(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:

Cr₂O₇^2-_(aq) + 14H⁺_(aq) + 6e- 2Cr³⁺_(aq) + 7H₂O_(l)

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

Equations

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 118^oC 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 + CO_2(g) + H₂O_(l)

Sodium carbonate Sodium ethanoate

Zinc carbonate and ethanoic acid.

ZnCO₃ + 2CH₃COOH (CH₃COO)₂Zn + CO_2(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 + H_2(g)

Ethanoic acid Sodium ethanoate

Propanoic acid and magnesium metal

2CH₃CH₂COOH + Mg (CH₃CH₂COO)₂Mg + H_2(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) CH₃COOC₂H_5(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) CH₃COOCH₂CH_3(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

In pharmaceuticals, for making medicines e.g ethanoic acid is used in the manufacture of aspirin.
Manufacture of dyes and insecticides
Seasoning food as vinegar
Coagulation of rubber latex
Preparation of polyethenyl ethanoate and cellulose ethanoate which are used are used to make artificial fibres such as rayon.
Manufacture of soaps.
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 400^oC 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; C₁₇H₃₅COO^–Na⁺

Structurally

CH₃CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂C

O^–Na⁺

Preparation of soaps

– 2cm³of 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.

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

C_nH_2n+1COO^–Na⁺where n >8 (n is greater than eight)

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

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

It makes the water able to wet material more effectively by lowering the surface tension.
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:

CH₃(CH₂)₁₆– C O^–Na⁺

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). 2C₁₇H₃₅COO^–Na⁺_(aq) + Ca²⁺_(aq) (C₁₇H₃₅COO^–)₂Ca²⁺_(s) + 2Na⁺_(aq)

Sodium stearate (soap) Magnesium stearate

(ii). 2C₁₇H₃₅COO^–Na⁺_(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 CaCO_3(s) + CO_2(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) + CO₃^2-_(aq) + 2Cl- + Ca²⁺_aq) CaCO_3(s) + 2Na⁺_(aq)+ 2Cl^– _(aq)

CO₃^2- _(aq) + Mg²⁺ MgCO_3(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

It is good for drinking purposes as calcium contained in it helps to form strong bones and teeth.
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
It is good for brewing and the tanning industries;

Disadvantages of hard water

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.

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.

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 (-OSO₃^–Na⁺) or sulphonate groups (-SO₃^–Na⁺) 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

HO S OH

– 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 + H₂SO_{4 (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 SO₃^–Na⁺

Tail head

(ii). Sodium alkylbenzene sulphonate

CH₃(CH₂)₁₀CH₂ – SO₃^–Na⁺

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;

They lather easily with hard water since the corresponding calcium and magnesium salts are soluble in water due to lack of the carboxylate ions.
Are mainly prepared from non-food raw materials.
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.

The active ingredients such as alkylbenzene sulphonates are non-biodegradable, hence accumulate in water sources and end up in human bodies.
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.
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.

Are unreactive, hence they dont react with industrial chemicals like oils, grease, petrol etc.
Are non-inflammable, as they dont catch fire easily.
Capable of withstanding wide range of temperatures without changing shape.
Have high mechanical strength.

Uses of synthetic rubber.

Manufacture of insulating materials for electrical connections.
Making conveyor and seat belts.
Manufacture of car tyres and tubes.
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

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

Disadvantages

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

Methods of polymerization:

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.

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;

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

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

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:

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

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

Uses of acids and bases

Oxides and hydroxides

Basic oxides
Acidic oxides
Neutral oxides
Amphoteric oxides
Meaning of amphoteric oxides
To verify amphoteric oxides
Reactions of amphoteric oxides

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

Solubility and solubility curves

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

Water

Hardness of water
Temporary hardness (Meaning; causes; and removal)
Permanent hardness (meaning; cause; and removal)

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; H₂SO_4(aq) 2H⁺_(aq) + SO₄^2-_(aq)

Nitric acid; HNO_3(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:

H₂CO_3(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

Sulphuric acid

Hydrochloric acid

Nitric acid

Ethanoic

Carbonic acid

Tartaric acid

Red

Orange

Yellow

orange

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

– 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) + CO_2(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

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

– Ammonium hydroxide (1M)

– Calcium hydroxide (1M)

– Sodium hydroxide (0.1M)

– Sodium hydroxide (1M)

– Potassium hydroxide

Blue.

Purple.

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 NH_3(g) dissolves in water it changes from molecules to ions.

Equation:

NH_3(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

Acids

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

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

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). Al₂O_3(s) + 6H⁺_(aq) 2Al³⁺_(aq) + 3H₂O_(l)

(iii). ZnO_(s) + 2HCl_(aq) ZnCl_2(aq) + H₂O_(l)

Hydroxides:

(i). Pb(OH)_2(s) + 2HCl_(aq) PbCl_2(aq) + 2H₂O_(l)

(ii). Zn(OH)_2(s) + 2HCl_(aq) ZnCl_2(aq) + 2H₂O_(l)

(iii). Al(OH)_3(s) + 3HCl_(aq) AlCl_3(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). Al₂O_3(s) + 2OH^–_(aq)+3H₂O_(l) 2[Al(OH)₄]^–_(aq) + 3H₂O_(l)

Ionically: Al₂O_3(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) + Cl_2(g) 2FeCl_3(s)

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

Equations:

Zn_(s) + H₂SO_4(aq) ZnSO_4(aq) + H_2(g)

CuO_(s) + H₂SO_4(aq) CuSO_4(aq) + H₂O_(l)

NaOH_(aq) + HCl_(aq) NaCl_(aq) + H₂O_(l)

PbCO_3(s) + 2NHO_3(aq) Pb(NO₃)_2(aq) + CO_2(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) PbCl_2(s) + 2NaNO_3(aq)

AgNO_3(aq)+ HCl_(aq) AgCl_(s) + HNO_3(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^–	CO₃^2-; O^2-; SO₄^2-; OH^–;	White precipitate insoluble in excess	No precipitate
Al³⁺	Cl^–; NO₃^–; SO₄^2-;	CO₃^2-; O^2-; 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^–; SO₄^2-; NO₃^–;	CO₃^2-; OH^–;	White precipitate soluble in excess	White precipitate soluble in excess
Mg²⁺	Cl^–; SO₄^2-; NO₃^–;	CO₃^2-; OH^–;	White precipitate insoluble in excess	No precipitate
Fe²⁺	Cl^–; SO₄^2-; NO₃^–;	CO₃^2-; O^2-; OH^–;	(dark) green precipitate insoluble in excess	Green precipitate insoluble in water
Fe³⁺	Cl^–; SO₄^2-; NO₃^–;	CO₃^2-; O^2-; OH^–;	(red) brown precipitate insoluble in excess	Brown precipitate insoluble in excess
Cu²⁺	Cl^–; SO₄^2-; NO₃^–;	CO₃^2-; O^2-; 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)} + 4NH_3(aq) [Zn(NH₃)₄]²⁺_(aq)+ 2OH^–_(aq)

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

Cu(OH)_2(s) + 4NH_3(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 O_2(g) ^Heat Fe₂O_3(s) + 2H₂O_(l)

(Green) (Red-brown)

Fe(OH)_3(s) ^Heat Fe₂O_3(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) + CO₃^2-_(aq) CaCO_3(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 PbCl_2(s) dissolves on warming

Equations:

Pb²⁺_(aq) + 2Cl^–_(aq) PbCl_2(s)

Pb²⁺_(aq) + SO₄^2-_(aq) PbSO_4(s)

Ba²⁺_(aq) + SO₄^2-_(aq) BaSO_4(s)

Ba²⁺_(aq) + SO₃^2-_(aq) BaSO_3(s)

Note:

– To distinguish the precipitate of barium sulphate from barium sulphite; dilute HNO_{3 (aq)} or HCl_(aq) is added to both;

– BaSO_3(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)
2. Yellow	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^–;	CO₃^2-; O^2-; SO₄^2-; OH^–;	White precipitate insoluble in excess	White precipitate insoluble in excess, on standing;
Al³⁺	Cl^–; NO₃^–; SO₄^2-	O^2-; 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^–; SO₄^2-	CO₃^2-; O^2-; SO₄^2-; OH^–;	White precipitate soluble in excess	White precipitate soluble in excess
Fe²⁺	Cl^–; NO₃^–; SO₄^2-	CO₃^2-; O^2-; OH^–;	(Dark) pale green precipitate insoluble in excess	(Dark) pale green precipitate insoluble in excess
Fe³⁺	Cl^–; NO₃^–; SO₄^2-	CO₃^2-; O^2-; OH^–;	(Red) brown precipitate insoluble in excess	(Red) brown precipitate insoluble in excess
Cu²⁺	Cl^–; NO₃^–; SO₄^2-	CO₃^2-; O^2-; 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.

	SO₄^2-_(aq)	Cl^–_(aq)	NO₃^–_(aq)	CO₃^2-_(aq)
TEST	Add Ba²⁺_(aq) ions from Ba(NO₃)_2(aq); acidify with dilute HNO_3(aq)	Add Ag⁺_(aq) from AgNO_3(aq). Acidify with dilute HNO₃ Alternatively; Add Pb²⁺ from Pb(NO₃)₂ and warm	Add FeSO_4(aq); Tilt the tube and carefully add 1-2 cm³ of concentrated H₂SO_4(aq)	Add dilute HNO_3(aq); bubble gas through lime water;
OBSERVATION	The formation of a white precipitate shows presence of SO₄^2- 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 CO₃^2- 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 CO₃^2- or HCO₃^– will liberate carbon (IV) oxide with dilute acids

Checklist:

Why is it not possible to use dilute sulphuric acid in the test for SO₄^2- ions;
Why is it not possible to use dilute hydrochloric acid in the test for chloride ions?
Why is it best to use dilute nitric acid instead of the other two mineral acids in the test for CO₃^2-ions?
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	CO₃^2- of Zn²⁺; Pb²⁺; Fe²⁺; Fe³⁺; Cu²⁺;
4. Carbon (IV) oxide and water vapour (condensed)	HCO₃^–
5. Nitrogen (IV) oxide	NO₃^–of 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 Cl_2(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, Cr₂O₇^2-_(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 SO_2(g) to SO₄^2-_(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 20^oC.

(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	20.0^oC
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 20^oC = 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 60^oC to 40^oC. (2 marks)

(ii). determine the solubility at 70^oC. (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

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 45^oC to 25^oC. (2 marks)

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

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 SO₂per 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 10^oC. (Assume that the density of the solution is 1gcm^-3) (3 marks)

(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 25^oC. (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 100^oC; melts at 0^oC 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, CaCl_2,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⁺

2C₁₇H₃₅COO^–Na⁺_(aq) + Ca²⁺_(aq) (C₁₇H₃₅COO^–)₂Ca_(s) + 2Na⁺_(aq)

Sodium stearate calcium stearate

With Mg²⁺

2C₁₇H₃₅COO^–Na⁺_(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 CaCO_3(s) + 2CO_2(g) + H₂O_(l)

Mg(HCO₃)_2(s) ^Heat MgCO_3(s) + 2CO_2(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) 2CaCO_3(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)₂CO_3(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 its 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) CaCl_2(aq) + Na₂X_(s)

MgX_(s) + 2NaCl_(aq) MgCl_2(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:

Na₂CO_3(aq) + CaCl_2(aq) 2NaCl_(s) + CaCO_3(S)

Ionically:

Ca²⁺_(aq) + CO₃^2-_(aq) CaCO_3(s)

Na₂CO_3(aq) + MgSO_4(aq) Na₂SO_4(s) + CaCO_3(S)

Ionically:

Mg²⁺_(aq) + CO₃^2-_(aq) MgCO_3(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

Energy level diagrams
Thermochemical cycles
Hesss law
Enthalpy of formation.
Heat of formation and bond energies.
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 = H_Products– H_Reactants

= 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 (H_Products) is higher than heat of reactants (H_reactants).

– Thus H₂-H₁ is a positive value; since H_I 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) N_2(g) + O_{2 (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. N_2(g) + 3H_{2 (g)} 2NH_3(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) + O_2(g)

∆H = -ve.

E₂

Products

CO_2(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) + CO_2(g)

∆H = +ve

E₁

Reactants

CaCO_3(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 (∆H_c)
Enthalpy of neutralization (∆H_neut)
Enthalpy of solution (∆H_soln)
Enthalpy of precipitation (∆H_precip)
Enthalpy of displacement (∆H_Disp)
Enthalpy of formation (∆H_f)

Heat of combustion ∆H_c

– 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) + O_2(g) CO_2(g) ; ∆ H= -394 Kjmol^-1

Apparatus for finding ∆H_c 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

Assume

Volume of water in calorimeter = 100cm³

Initial temperature of water = 225^oC

Final temperature of water = 50.5^oC

Change in temperature of water = 28.0^oC

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).

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= 13.0^oC. (C=12; H=1; O=16; SHC of water=4.18KJKg^–K^–). 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^–

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 =27.0^oC

Initial temperature of water =20.0^oC

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^-1oC^–1

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}

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 23.0^oC to 32.0^oC. 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}

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.

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 –57KJMol^–and 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

NH_3(aq) + H₂O_(l)+ CH₃COOH_(aq)

CH₃COO^–_(aq) + NH₄⁺_(aq) + H₂O_(l)

Reaction pathway

Therefore;

∆H is given by;

∆H = ∆H_I + ∆H_II

= 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

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

-The temperature change, ∆T= (T₄ –T₃) = ^oC;

Sample calculations:

Given: Temperature of hydrochloric acid solution T₁= 22.75^oC;

Temperature of sodium hydroxide solution T₂= 22.80^oC;

Average temperature of acid and alkali T₁ + T₂= T₃= 22.78^oC;

Highest temperature of alkali and acid mixture, T₄= 36.40^oC;

Temperature change, ∆T= T₄ – T₃ = (36.40 – 22.78)^oC;

=13.62^oC;

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 13.62^oC

=- 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:

Given, T₁= 21.0^oC; T2=22.0^oC; T₄ = 34.5^oC; 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 94.0^oC were added to a calorimeter at 17.5^oC, the temperature rose to 80.5^oC. 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 17.5^oC initially and rose to 20.1^oC. 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.

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 = 27.0^oC

Initial temperature of alkali = 23.0^oC

Mass of plastic cup + HCl + NaOH = 145.1g

Temperature of the mixture of acid and alkali = 38.5^oC.

(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 =13.5^oC;

Heat produced ∆H = 100g x 4.2kjg^-1k^-1 x 13.5^oC

=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

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;

∆H_solution= – ∆H_lattice + ∆H_hydration ; when ∆H_lattice is given as a negative value;

Alternatively;

∆H_solution = ∆H_lattice + ∆H_hydration; where ∆H_lattice 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 = +∆H_lattice

=+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 ∆H_hydration

Final state

∆H_lattice M⁺_(aq) + X^–_(aq)

∆H_solution

MX_(s) + H₂O_(l)

-The lattice energy is larger than the hydration energy hence;-

∆H_solution = +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

∆H_lattice ∆H_hydration

Initial state

MX_(s) + H₂O_(l)

∆H_solution

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:

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;

The lattice and hydration enthalpies for lithium chloride and potassium chloride are given below.

Salt	∆H_lattice	∆H_hydration.
LiCl	-861	-884
KCl	-719	-695

(a). Calculate the enthalpy of solution, ∆H_solution; Kj mol^– for;

Lithium chloride. (2 marks)

Solution:

H_solution = -∆H_lattice + ∆H_hydration

= – (-861) + (-884)

= 861- 884;

= -23Kjmol-

Potassium chloride: (2 marks)

Solution:

H_solution = -∆H_lattice + ∆H_hydration

= – (-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 ∆H_solution is greater;

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, ∆H_soln 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 (∆H_soln) 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; ∆H_soln; =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 ∆H_soln = -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 ∆H_soln = +21.42Kjmol^-1

Note: energy level diagram for the dissolution of sodium nitrate:

NaNO_3(aq)

∆H_soln

Reaction path

Potassium hydroxide;

– Calculate the molar enthalpy of dissolution of potassium hydroxide (based on values obtained).

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) + CuSO_4(aq) ZnSO_4(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 H_products is lower than H_reactants.

– 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.

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 AgNO_3(aq50cm³ 2M HCl_(aq)50cm³ 2M AgNO_3(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

– Final temperature after mixing = 34.0ºC;

– Temperature change = a rise of 14^oC;

– 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 ∆H_{precipitation} for AgCl.

Solution:

(i) Heat change, = MC∆T

= 100 x 4.2 x 14

1000

= – 5.88kj

(ii) Equation,

AgNO_3(aq) + NaCl_(aq) AgCl_(s) + NaNO_3(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, ∆H_precip = -58.8kJmol^–

Thermochemical equation:

Ag⁺_(aq) + Cl^–_(aq) AgCl_(s); ∆H= -58.8kjmol-

Suitable energy level diagram,

Ag⁺_(aq) + Cl^–_(aq)

∆H_prep= -58.8kJmol^–

AgCl_(s)

Reaction pathway

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 Hesss 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 25^oC (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). ∆H⁰_f; refers to the standard molar enthalpy change of formation.

Example:

H_2(g) + ½ O_2(g) H₂O_(l); ∆H⁰_{f (H2O)} = -286KJmol^–

(ii). ∆H⁰_c; refers to the standard molar enthalpy change of combustion;

Example:

CH_4(g) + 2O_2(g) CO_2(aq)+ 2H₂O_(l); ∆H⁰_c(CH4)= -890KJmol^–

(iii). ∆H⁰_neut; refers to the standard molar enthalpy change of neutralization.

Example:

HCl_(aq) + NaOH_(aq) NaCl_{(aq) +}H₂O_(l); ∆H⁰_neut = -58KJmol^–

(iv). ∆H⁰_soln; refers to the standard molar enthalpy change of solution (dissolution).

Example:

NaNO_3(s) + H₂O_(l) NaNO_3(aq); ∆H⁰_{soln (NaNO3)} = +21KJmol^–

(v). ∆H⁰_hyd; refers to the standard molar enthalpy change of hydration.

Example:

Na⁺_(s) + Cl^–_(s) + H₂O_(l) Na⁺_(aq) + Cl^–_(aq); ∆H⁰_{hyd (NaCl)} = -774KJmol^–

(vi). ∆H⁰_latt; refers to the molar enthalpy change of lattice formation.

Example:

Na⁺_(g) + Cl^–_(g) NaCl_(aq); ∆H⁰_{latt (NaCl)} = -774KJmol^–

(vi). ∆H⁰_at; refers to the standard molar enthalpy change of atomization.

Example:

Na_(s) Na_(g); ∆H⁰_{at (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

Therefore,∆H₁= ∆H₂ + ∆H₃

Worked examples:

Use the information below to determine the enthalpy of combustion of carbon (formation of carbon (IV) oxide).

C_(s) + ½O_2(g) CO_(g); ∆H = -110.45KJ

CO_(g) + ½O_(g) CO_2(g); ∆H = -282.0KJ

Solution:

Route 2

Route 2 ∆H₂∆H₃

CO_(g)

Therefore; ∆H₁ = ∆H₂ + ∆H₃

= -110-45 + (-282.0) = – 392.45kJ

∆H₁= ∆H_combustion

= -392.45KJ;

Note: – Enthalpy of formation, ∆H_f;

– 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 (Hesss law).

Worked examples:

One mole of butane(C₄H₁₀) burns completely in oxygen and liberates 2877kj.

(a) Write an equation for the combustion of butane.

2C₄H_10(g)+ 13O_2(g) 8CO_2(g) + 10H₂O_(l)

(b) Draw an energy cycle, hence an energy level diagram for the reactions concerned.

Solution:

Equations for combustion:

C_(s) + O_2(g) CO_2(g); then 8C_(s) + 😯_2(g) 8CO_2(g)(based on balancing for butane)

H_2(g) + ½ O_2(g) H₂O_(g); then 10H_2(g) + 5O_2(g) 10H₂O_(g); (based on balancing for butane)

2C₄H_10(g)+ 13O_2(g) 8CO_2(g) + 10H₂O_(l)

Then, energy cycle for the changes,

∆H_c; ∆H₁ = ∆H₂ + ∆H₃ ∆H₃ = H_c

8CO_2(g) + 10H₂O_(l)

Thus energy level diagram for the reactions.

8C_(s) + 10H₂O_(l)

∆H₂ = ∆H_f

2C₄H_10(l)

+½13O_2(g)

∆H₁= ∆H₂ + ∆H₃ ∆H₃ = ∆H_c

+13O_2(g)

8CO_2(g) + 10H₂O_(l)

Reaction pathway (progress)

(c) If the following heats of combustion are given

∆Hº_c (graphite) = -393kjmol^–

∆Hº_c (H_2(g)) = -286KJmol^–

∆Hº_c (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= ∆H_formation of C₄H₁₀;

∆H₁=∆Hº_c(graphite) + ∆Hº_c(hydrogen);

While ∆H₃= ∆Hº_c(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) + 3H_2(g) + ½O_2(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) + 3O_2(g) 2CO_2(g) + 3H₂O_(l)

Balancing based on equation for combustion of ethanol

(ii). 2C_(s) + 2O_2(g) 2CO_2(g)

(iii). 3H₂O_(g) + 1½O_2(g) 3H₂O_(l)

Then, energy cycle for the changes,

+3O_2(g)

∆H_c; ∆H₁ = ∆H₂ + ∆H₃ ∆H₃ = H_c

+3O_2(g)

2CO_2(g) + 3H₂O_(l)

Thus energy level diagram for the reactions.

2C_(s) + 3H_2(l)+ ½ O_2(g)

∆H₂ = ∆H_{f (ethanol)}

C₂H₅OH_(l)

+3O_2(g)

∆H₁= ∆H₂ + ∆H₃ ∆H₃ = ∆H_{c (ethanol)}

+3O_2(g)

2CO_2(g) + 3H₂O_(l)

Reaction pathway (progress)

(iii). If the following heats of combustion are given:

∆H_c(graphite)=-393kJmol^–

∆H_c(hydrogen)=-286kJmol^–

∆H_c(ethanol)= -1368kJmol^–

Calculate the heat of formation of ethanol, ∆H⁰_{f (ethanol)}

Solution:

∆H₁ = ∆H₂+ ∆H₃

∆H₂= ∆H₁– ∆H₃

= [2(-393 + 3(-286)]-[-1368]

= [(-786) + (-856)] – [-1368]

= -164 + 1368

= – 276kJmol^–

Given the standard enthalpies of combustion as

C_(s) + O_2(g) CO_2(g); ∆H₁= -394kJmol^–;

H_2(g) + ½O_2(g) H₂O_(l); ∆H₂= -286kJmol^–

C₂H_2(g) + 1½O_2(g) 2CO_2(g) + H₂O_(l); ∆H = -1300kJmol^–.

Find the standard enthalpy of formation of ethyne, C₂H_2(g).

Solution:

– Considering the H_c(ethyne), 2 moles of CO₂ and 1 mole of H₂O_(l) are produced;

– Rebalancing the equations for H_c(carbon) and H_c(hydrogen) likewise:

2C_(s) + O_2(g) 2CO_2(g);

H_2(g) + ½O_2(g) H₂O_(l);

Then, energy cycle for the changes,

+2½O_2(g)

∆H_c; ∆H₁ = ∆H₂ + ∆H₃ ∆H₃ = H_c_(ethyne)

+2½O_2(g)

2CO_2(g) + H₂O_(l)

Thus energy level diagram for the reactions.

2C_(s) + H_2(l)

∆H₂ = ∆H_{f (ethanol)}

C₂H_2(l)

+2½O_2(g)

∆H₁= ∆H₂ + ∆H₃ ∆H₃ = ∆H_{c (ethanol)}

+2½O_2(g)

2CO_2(g) + H₂O_(l)

Reaction pathway (progress)

Thus ∆H₁ = ∆H₂ + ∆H₃

∆H₂ = ∆H₁– ∆H₃

= [2(-394) + (-286)]-[-1300]

= (-788-286) + 1300

= -1074 + 1300

= +226kJmol^–

Since ∆H_f is +ve, ethyne is described as an endothermic compound.

Note:

– In the determination of ∆H_f of ethyne, Hess law is used because attempts to make ethyne from carbon and hydrogen {2C_(s) + H_2(g) C₂H_2(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) + 2H_2(g) CH_4(g); ∆H = -74.9kJmol^–

Energy level diagonal for the formation of CH₄.

C_(s) + 4H_(g)

II ∆H_at;

C_(s) + 2H_2(g)

I ∆H_at = +715KJmol^–III ∆H = -4(C-H); bond energy

C_(s) + 2H_2(g)

∆H_IV = I + II + III

Final enthalpy = -74.9Kj

CH_4(g)

Reaction pathway (progress)

Energy changes involved are:

∆H_I; Enthalpy of atomization – this is the energy absorbed when a substance decomposes to form one mole of gaseous atoms.
∆H_II; enthalpy of atomization of H₂; i.e. dissociation of hydrogen molecules to free hydrogen atoms.
∆H_III; enthalpy of formation of four C – H bonds, during which heat is liberated;

Therefore, enthalpy of reaction ∆H_R = ∆H_IV

∆H_R= ∆H₁+ ∆H_II + ∆H_III

= +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-

– This is the amount of energy released when one C-H bond is formed;

Worked examples.

Use the bond energies given below to calculate the heat of reaction for;-

H_2(g) + Cl_2(g) 2HCl_(g)

Bond	Bond energy
H-H	435kjmol^–
Cl-Cl	243kjmol^–
H-Cl	431kjmol^–

Solution:

Bond breaking,

Total energy absorbed = +678 KJ

H_2(g) = H-H 2H_(g); ∆H = +435Kj;

Cl_2(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)

∆H_II ∆H_{at (chlorine)}

+243KJ

2H_(g) + Cl_2(g)

∆H_I ∆H_at = +435KJmol^– ∆H_III= -2 (H-Cl); bond energy

H_2(g) + Cl_2(g)

∆H_IV = I + II + III

Final enthalpy = -74.9Kj

2HCl_(g)

Reaction pathway (progress)

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

CH_4(g) + Cl_2(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 = ∆H_{bond breakage} + ∆H_{bond formation}

= [(+1656) + (+244)] + [(-1242) + (-326) + (-431)]

= -1900 + (-1999)

∆H = -99kJ

(b). Draw an energy level diagram for the reaction

C4H_(g) + 2Cl_(g)

∆H_II ∆H_{at (chlorine)}

+244KJ

C4H_(g) + Cl_2(g)

∆H_I ∆H_at = +1656Kj ∆H_III= -2 (H-Cl); bond energy

CH_4(g) + Cl_2(g) = -99Kj;

∆H_IV = I + II + III

Final enthalpy

CH₃Cl_(g) + HCl_(g)

Reaction pathway (progress)

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, ∆H_f;

– 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, ∆H_vap;

– 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
∆H_fusionkjmol^–	28	–	13	6	8	–
∆H_vap kjmol^–	171	717	305	41	21	10

UNIT 4: RATES OF REACTION AND REVERSIBLE REACTIONS

UNIT OUTLINE:

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

Reversible reactants

Meaning
Chemical equilibrium

Ø Factors affecting position of equilibrium

§ Temperature

Pressure
Concentration (of reactants or products)
Catalysts
LeChartaliers 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 (E_A):

– 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 S₂O₃^2- (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 45^oC; (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:

Na₂S₂O_3(aq) + 2HCl_(aq) 2NaCl_(aq) + H₂O_(l) + S_(s) + SO_2(g)

Ionic equation:

S₂O₃^2- _(aq) + 2H⁺_(aq) S_(s) + SO_2(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:

CaCO_{3 (S)}+ 2H⁺ _aq) Ca²⁺_(aq) + CO_2(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 50^oC. 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 Oswalds 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 Cl_2(g) and H_2(g) does not take place in the dark but is explosive in bright light;

Cl_2(g) + H_2(g) ^{bright light} 2HCl _(g)

For the reaction between methane and bromine; decolourisation of bromine only occurs in presence of light;

CH_4(g) + Br_2(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) + AgNO_3(aq) AgBr_(s) + KNO_3(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) + Br_2(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). Cl_2(g) + H_2(g) ^UV ^Light 2HCl_(g)

CH_4(g) + Br_2(g) ^{UV Light} CH₃Br_(g) + HBr_(g)

(iii). 6CO_2(g) + 6H₂O_(l) ^Light C₆H₁₂O_6(aq) + 6O_2(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.

N_2(g) + 3H_2(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) + 4H_{2 (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 SO_2(g) + O_2(g) 2SO_3(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
Reaction	Increase	Decrease
N₂O_4(g) 2NO_2(g)	Equilibrium shifts to the left; more N₂O₄ is formed;	Equilibrium shifts to the right; more NO₂is formed;
N_2(g) + 3H_2(g) 2NH_3(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
2SO_2(g) + O_2(g) 2SO_3(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
4NH_3(g) + 5O_2(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
H_2(g) + Cl_2(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:

Br_2(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:

SO_2(g) + O_2(g) 2SO_3(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:

2CrO₄^2-_(aq) + 2H⁺_(aq) 2Cr₂O₇^2-_(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 Chateliers 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 Chateliers 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:
Condition	Ø Haber process N_2(g) + 3H_2(g) 2NH_3(g); ∆H=- 92KJMol-	Ø The contact process 2SO_3(g) + O_2(g) 2SO_3(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 NH_3(g) will be produced would be too slow and thus uneconomical; an optimum temperature of about 450^oC 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 SO_2(g); – Low temperature will favour forward reaction thus shifts the equilibrium to the right; hence more production of SO_2(g); However rate of yield will be too slow and hence optimum temperature for maximum yield are set at about 450^oC;
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 SO_2(g) to SO_3(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 NH_3(g) – thus for maximum yield ammonia is removed as soon as it is produced; so that more H_2(g) and N_2(g) can continue reacting;	– Removal of sulphur (VI) oxide gas shifts the equilibrium to the right hence more yield of SO_3(g); since this removal lowers its concentration; – thus for maximum yield SO_3(g) is removed as soon as it is produced; so that more SO_2(g) and O_2(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:

Meaning of electrochemistry
Displacement and Redox reactions.
Oxidation and Redox reactions in terms of electron gain and electron loss
Oxidation numbers (states)

Rules of assigning oxidation numbers
Calculating oxidation numbers

Redox reactions involving halide ions (halogens)

Further examples of Redox reactions.

Tendency of metals to from ions.
Measurements of tendency of metals to ionize

Electrochemical cells
Salt bridge
Cell diagrams
Cell equations

Standard electrode potential

Definition
Standard hydrogen electrode
The hydrogen half cell
Electrode potentials (negative and positive values)
Calculating E⁰ values from Redox reactions.

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;

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 CuSO_{4 (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 CuSO_4(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

Oxidation number of an uncombined element is zero (0)
The charge on a monoatomic ion is equivalent to the oxidation number of that element;
The oxidation number of hydrogen in all compounds is +1 except in metal hydrides where its 1;
The oxidation number of oxygen in all compounds is 2 except in peroxides where it is 1 and 0F₂ where it is +2.
In complex ions the overall charge is equal to the sum of the oxidation states of the constituent elements.
In compounds, the sum of oxidation numbers of all constituent atoms is equal to zero.

Worked examples

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

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:

Cl_{2 (g)} + 2KBr_(aq) 2KCl_(aq) + Br_{2 (l)}

Ionically:

Cl_{2 (g)} + 2Br^–_(aq) 2Cl^–_(aq) + Br_{2 (l)}

Green-yellow Red brown

Redox equation: Reduction

Cl_{2 (g)} + 2Br^–_(aq) 2Cl^–_(aq) + Br_{2 (l)}

Oxidation

(ii). Chlorine and Potasium iodide:

Cl_{2 (g)} + 2KI_(aq) 2KCl_(aq) + I_{2 (l)}

Ionically:

Cl_{2 (g)} + 2I^–_(aq) 2Cl^–_(aq) + I_{2 (l)}

Green-yellow Black

Half cell reactions:

Oxidation: 2I^–_(aq) I_2(aq) + 2e-
Reduction: Cl_{2 (g)} + 2e- 2Cl^–_(aq)

Redox equation: Reduction

Cl_{2 (aq)} + 2I^–_(aq) 2Cl^–_(aq) + I_{2 (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:

Br_{2 (g)} + 2KI_(aq) 2KBr_(aq) + I_{2 (l)}

Ionically:

Br_{2 (g)} + 2I^–_(aq) 2Cl^–_(aq) + I_{2 (l)}

Red brown Black

Half cell reactions:

Oxidation: 2I^–_(aq) I_2(aq) + 2e-
Reduction: Br_{2 (g)} + 2e- 2Br^–_(aq)

Redox equation: Reduction

Cl_{2 (aq)} + 2I^–_(aq) 2Cl^–_(aq) + I_{2 (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) + H_{2 (g)}

Ionically:

Mg_(s) + 2H⁺_(aq) Mg²⁺_(aq) + H_{2 (g)}

Half cell reactions:

Oxidation: Mg_(s) Mg²⁺_(aq) + 2e-
Reduction: 2H⁺_(g) + 2e- H_2(g)

Redox equation: Reduction

2H⁺_(aq) + Mg_(s) H_{2 (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) + H_{2 (g)}

Ionically:

2Na_(s) + 2H₂O_(l) 2Na⁺_(aq) + 2OH^–_(aq) + H_{2 (g)}

Half cell reactions:

Oxidation: 2Na_(s) 2Na⁺_(aq) + 2e-
Reduction: 2H₂O_(g) + 2e- 2OH^–_(aq) + H_2(g)

Redox equation: Oxidation

2Na_(s) + 2H₂O_(aq) 2Na⁺_(aq) + 2Cl^–_(aq) + I_{2 (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) + 3Cl_2(aq) 2Fe³⁺_(aq) + 6Cl^–_(g)

Half cell reactions:

Oxidation: 2Fe_(s) 2Fe³⁺_(aq) + 6e^–
Reduction: 3Cl_2(g) + 6e^– H_2(g)

Redox equation: Oxidation

2Fe_(s) + 3Cl_2(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:

Br_2(l) + 2Fe²⁺_(aq) 2Fe³⁺_(aq) + 2Br^–_(aq)

Half cell reactions:

Oxidation: Br_2(l) 2Br^–_(aq) + 2e-
Reduction: 2Fe²⁺_(g) + 2e- 2Fe³⁺_(aq)

Redox equation: Reduction

Br_{2 (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:

Cr₂O₇^2-_(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: Cr₂O₇^2-_(aq) + 14H⁺_(aq)+ 5e^– 2Cr³⁺_(aq) + 7H₂O_(l) + 6Fe³⁺_(aq)

Redox equation: Reduction

Cr₂O₇^2-_(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) + 5O_{2 (g)}

Purple Colourless

Half cell reactions:

Oxidation: 5H₂O_2(aq) 10H⁺_(aq) + 5O_2(g) + 10e^–;
Reduction: 2MnO₄^–_(aq) + 16H⁺_(aq)+ 10e^– 2Mn²⁺_(aq) + 8H₂O_(l)

Redox equation: Reduction

2MnO₄^–_(aq) + 5H₂O_{2 (aq)} 2Mn²⁺_(aq) + 10H⁺_(aq) + 5O_2(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). H₂O₂oxidizes Iron (II) salts to Iron (III) salts in acidic medium

Oxidation:

2Fe²⁺_(aq) 2Fe³⁺_(aq) + 2e^–

Reduction:

2H⁺_(aq) + H₂O_2(aq) + 2e^– 2H₂O_(l)

Overall redox:

2Fe²⁺_(aq) + H₂O_{2 (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

gains electrons

increases

decreases

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 CuSO_{4 (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:

Allowing its ions to carry charge from one half – cell to the other.
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 (E^θ) 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

A vertical continuous line (/); represents the metal-metal ion or metal ion-metal interphase.
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 E^θ 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 E^θ 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 E^θ 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:

– 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.

– 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.

– 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 25^oC
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 H_{2 (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: H_{2 (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

F_{2 (g)}+ e^– _(aq) 2F^–_(aq); E⁰ = +12.87V

Mg _(s) + 2e^– Mg²⁺_(aq); E^θ = -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

E^θvolts

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

Br_2(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 E^θ 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^–; E^θ = + 0.76V;

Uses of E⁰ values

Comparing the reducing powers and oxidizing powers of various substances;
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

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.

Can chlorine displace bromine form bromide solution?

Cell diagram: Cl_{2 (g)}/ Cl^–_(aq) // 2Br^–_(aq) / Br_{2 (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

The diagram below represents part of the apparatus to be used for the determination of the standard electrode potential of Aluminum, E^θ 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;

Allowing its ions to carry charge form one half cell to another.
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 H_{2 (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

You are given the following half-equations;

Mg²⁺_(aq) + 2e- Mg _(s); E^θ = – 2.71 V

Zn²⁺_(g) + 2e- Zn _(s); E^θ = – 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).

Given the following half-equations

I_{2 (g)} + 2e- 2I^– _(aq); E^θ= + 0.54V

Br_{2 (g)} + 2e- 2Br^– _(s); E^θ = +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) + I_{2 (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) + I_{2 (g)}; E⁰ = +0.55V

(c). Give the oxidizing species;

Oxidation

Br₂ _(g) + 2I^– _(aq) 2Br^–_(aq) + I_{2 (g)}; E⁰ = +0.55V

Reduction

0 -1 -1 0

Oxidizing species: Bromine; Br_{2 (aq)}

Consider the following list of electrodes and electrode potential values.

Electrode reaction

E^θ 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 E^θvalue 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)

The following is a list of reduction standard electrode potentials.

Metal

E^θ 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); E^θ = – 2.36 V

2Ag⁺_(g) + 2e^– 2Ag _(s); E^θ = – 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); E^θ = + 0.34 V

Zn²⁺_(g) + 2e- Zn _(s); E^θ = – 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); E^θ = + 0.34 V

Ag⁺_(g) + 2e- Ag _(s); E^θ = + 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.

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

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;

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.

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.

Primary cells/dry cells:

– Are of various types and an example is the Le clanche dry cell.

The Leclanche dry cell

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;

a the Zinc can/negative terminal
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.

a) Brass cap – Functions as the positive terminal where the reduction reaction

b) Zinc can – Is the Negative terminal; where the oxidation reaction occurs.

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.

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.

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

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 PbSO_{4 (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.

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

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

ii) Electrodes – Are the solid conductors, usually roots, which usually complete the circuit between electrolytes and cell/battery

– Are of two types

a) Anode – The electrode connected to the positive terminal f a battery/cell
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.

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 its the weakest reducing agent.

Discharge for anions

Anions are discharged by oxidation (electron loss) to form their respective products.

Discharge of anions is viz.

ELECTROLYSIS OF VARIOUS SUBSTANCES

Electrolysis of various substances
i)

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.

Explanations – Ions present in the electrolyte
i) Hydrogen ions and Sulphate ions form sulphuric acid.
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:

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.

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

i) Apparatus

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;

iv) Explanations

The Ions present in the electrolyte

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

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

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.

Electrolysis of Brine/concentrate sodium chloride.

i) Apparatus

ii) Procedure

An electric current is passed though concentrated sodium chloride/brinc

iii) Observation

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)

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)

iv) Explanations

The ions present in the electrolyte are:-
Na⁺ _(aq) and Cl^– from sodium chloride
H+ _(aq) and OH- _(aq) from water.

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.

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.

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.

i) Apparatus

Electrolyte in the mercury cell is Brine (concentrated NaCl)
Anode is carbon or titanium
Cathode is a moving mercury film.

ii) Reactions
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)

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.

Its expensive due to the high cost of mercury.
Pollution form Mercury; i.e. Mercury is poisonous and must be removed from the effluent.

Electrolysis of Copper (II) Sulphate solution.

NOTE: The products of electrolysis of copper (II) Sulphate solution depends on the nature of the elctrodes used.

i) Apparatus

ii) Ions present in the elctrolyte:

From copper (II) Sulphate

From water

During electrolysis

Using carbon/platinum electrodes

iii) Observations

At the Anode:

A colourless gas is liberated
The gas relights a glowing splint; hence its oxygen.

At the cathode

A reddish brown coating (of Cu solid) is deposited.

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).

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.

Using copper rods electrodes

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: its 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.

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.

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.

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

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.

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.

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.

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

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

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.

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 H_{2 (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.

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

Resultant sodium hydroxide is very pure, since it has no contamination from NaCl.
The sodium hydroxide has a relatively high concentration; at about 30 35% NaOH by mass.

Uses of sodium hydroxide, chlorine, and hydrogen

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.

Hydrogen

For hydrogenation in the manufacture of margarine
Manufacture of ammonia
Production of hydrochloric acid

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.

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

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

A current of 2.0 Amperes was passed through dilute potassium sulphate solution for two minutes using inert electrodes.
Write the equation for the reaction at anode.

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

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

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.

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

The number of Coulombs of electricity used

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)

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:

Introduction
Extraction methods
Concentration of the ores
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:

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

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.

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.

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 Bayers 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

Sodium

Main ores;

Rock salt / sodium chloride; NaCl
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 800^oC to 600^oC;

– 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) Cl_2(g) + 2e-

Properties of Sodium;

– Is a soft silvery metal with low density; 0.979gcm^-3

– Has a low melting point, 97^oC, and a low boiling point of 883^oC.

Chemical reactions

(a) With air

– Na is very reactive and tarnishes in moist air to form an oxide layer.

4Na_(s)+ O_2(g) 2Na₂O_(s);

– The oxide layer reacts with more air moisture to form hydroxide

Na₂O_(s) + CO_{2 (g)} Na₂CO_3(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) + O_2(g) Na₂O_2(s)

(White)

(b). With water

– Na reacts vigorously with water to form NaOH and Hydrogen.

Equation:

2Na_(s) + 2H₂O_(l) 2NaOH_(aq) + H_2(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) + Cl_2(g) 2NaCl_(s)

(d) With ammonia gas;

– Sodium forms hydrogen and a solid;

2Na_(s) + NH_3(g) 2NaNH_2(s) + H_2(g)

Sodamide

And;

NaNH_2(s) + H₂O_(l) NaOH_(aq) + NH_3(g)

Uses of sodium;

Is alloyed with lead in the preparation tetraethyl (IV) lead, which is added to petrol as an anti-knock.
Provides the glow in sodium vapours lamps, for street lighting (orange-yellow street lights).
Is an excellent conductor of heat and electricity with low melting point hence used;

In nuclear reactors to conduct away heat.
Modern aeroplane engines.

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

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 800^oc to 600^oc

(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)

Aluminium

– Forms 7% of the earths 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)

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:

Al₂O_3(s) + 2OH^–_(aq) + 3H₂O_(l) 2[Al(OH)₄]^–_(aq);

– Silica impurities also dissolve forming sodium silicate

Equation:

SIO_2(s) + 2NaOH_(aq) Na₂SIO_3(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) + CO_2(g) 2Al(OH)_3(s) + H₂O_(l);

Seeding

Al(OH)₃

Alternatively:

Al(OH)₄^–_(aq) 2Al(OH)_3(s) + OH^–_(aq);

General equation:

NaAlO_2(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) Al₂O_3(s) + 3H₂O_(l)

Alumina

Electrolysis of purified bauxite (alumina)

– The Alumina (Al₂O₃), has a high melting point, 2015^oC 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 2015^oC to around 800^oC;

– 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:

Al₂O_3(l) 2Al³⁺_(l) +3O^2-_(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;

3O^2-_(l) 3O_2(g) + 6e-

Then;

C_(s) (anode) + O_2(g) CO_2(g)

Note:

– Cryolite usually adds Na⁺; and F^– ions into the electrolyte.

– Thus the anions are O^2- 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.

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 800^oC to form is oxide and nitrate.

Equations:

4Al_(s) + 3O_2(g) 2Al₂O_3(s);

2Al_(s) + N_2(g) 2AlN_(s)

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)+ 6H₂SO_4(l) Al₂(SO₄)_3(aq)+ 6H₂O_(l)+ 3SO_2(g)

With Hydrochloric acid;

– Dilute HCl dissolves aluminium slowly; liberating hydrogen.

Equation:

2Al_(s) + 6HCl_(l) 2AlCl_3(aq)+ 3H_2(g)

With concentrated HCl the rate of reaction is increased.

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)+ 3Cl_2(g) 2AlCl_3(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:

AlCl_3(s)+ 3H₂O_(l) Al(OH)_3(s)+3HCl_(g)

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.

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) 2NaAlO_2(aq)+ 3H_2(g)

Ionically:

2Al_(s)+ 2OH^–_(aq)+ 2H₂O_(l) 2AlO^-2_(g)+ 3H_2(g)

Note:

– Thus aluminium has an amphoteric nature as it reacts with both acids and alkalis.

Uses of aluminium

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.

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.

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).

Aluminium powder mixed with oil is used as a protective paint.

Making Aluminium foils due to its high malleability. The foil is used in cooking; packaging and for milk bottle tops.

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%)

As a reducing agent in the Thermite process in the production of some elements such as chromium, cobalt manganese and titanium.

Example:

Cr₂O_3(s) + 2Al_(s) 2Cr_(s) + Al₂O_3(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.

Cr₂O_3(s) + 2Al_(s) 2Cr_(l) + Al₂O_3(s) + Heat.

Sample question:

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) + 3O_2(g) 2ZnO_(s) + 2SO_2(g)

From Calamine:

ZnCO_3(s) ZnO_(s) + CO_2(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

CaCO_3(s) CaO_(s) + CO_{2 (g)}

Then:

CO_{2 (g)} + C_(s) 2CO_(g)

– The resultant carbon (II) oxide and coke are the reducing agents in the furnace, at about 1400^oC.

– 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) CaSiO_3(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) + H₂SO_4(aq) ZnSO_4(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) + H₂SO_4(aq) PbSO_4(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 SO₄^2- 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 SO₄^2-migrate to the cathode.

– The OH^– are discharged in preference to SO₄^2-; giving off oxygen gas

Reason:

The OH^– ions have a higher oxidation potential than SO₄^2- and therefore easily giving electrons for reduction at the cathode

Equation:

4OH^–_(aq) 2H₂O_(l) + O_2(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 420^oC
Boiling point of 907^oC

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) + O_2(g) 2ZnO_(s)

Then;

ZnO_(s) + CO_2(g) ZnCO_3(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) + O_2(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) + H_2(g)

(c). with dilute acids

– Zinc is above hydrogen in the reactivity series hence displaces hydrogen from steam (water) and dilute acids like H₂SO₄and HCl.

Equation:

Zn_(s) + 2H⁺_(aq) Zn²⁺_(aq) + H_2(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) + 2H₂SO_4(l) ZnSO_4(aq) + 2H₂O_(l) + SO_2(g)

(ii). 50% concentrated nitric (v) acid.

– It reacts with 50% concentrated nitric (V) acid to liberate nitrogen (II) oxide.

Equation:

3Zn_(s) + 8HNO_3(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) + 4HNO_3(l) Zn(NO₃)_2(aq) + 4H₂O_(l) + 2NO_2(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) H_2(g) + [Zn(OH)₄]^2-_(aq)

(Zincate ion)

(f). Other reactions.

(i). Zinc burns in chlorine to give zinc chloride

Zn_(s) + Cl_2(g) ZnCl_2(s)

(ii). Zinc combines with sulphur

Zn_(s) + S_(s) 2ZnS_(s)

Uses of zinc:

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.

Making brass; an alloy of copper and zinc.
Making outer casings of dry batteries;
Die-castings contain zinc and aluminium, and are used for making radio and car parts;
Zinc cyanide is used for refining silver and gold;

Sample question:

Iron

– Is the second most abundant metal after aluminium, forming about 7% of the earths 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, 1400-1600^oC

– Temperature at the top of the furnace, 400^oC

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 700^oC 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 1600^oC 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 1600^oC and the product is Carbon (IV) oxide gas

– This reaction is exothermic, producing a lot of heat in the blast furnace.

Equation:

C_(s) + O_2(g) CO_2(g)

(ii). Decomposition of limestone;

– The limestone in the charge decomposes in the blast furnace to calcium oxide (Quicklime) and carbon dioxide.

Heat

Equation:

CaCO_3(s) CaO_(s) + CO_2(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 700^oC;

Equation:

CO_2(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 800^oC 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:

2Fe₂O_3(s) ₊ 3C_(s) 4Fe_(s) + CO_2(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 500^oC-700^oC, higher up the furnace.

Equation:

Fe₂O_3(s) + 3CO_(g) 2Fe_(s) + CO_{2 (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 (1500^oC-1800^oC)

– The molten iron runs to the bottom of the furnace.

– Temperatures at the hearth (bottom of the furnace) is maintained at approx. 1400^oC and yet pure iron melts at about 1525^oC.

– Consequently the molten iron would easily solidify at the base (Temp =1400^oC)

– However this is not usually the case;

Reason:

-Impurities absorbed by iron during melting (mainly carbon) reducing the melting point to below 1400^oC.

– 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) + SiO_2(s) CaSiO_3(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

Light-weight building material.
Manufacture of cement.
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 907^oC;

– 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) + 3O_{2 (g)} 2Fe₂O₃.H₂O_(l)

– When heated it reacts with oxygen to form tri-iron tetroxide; Fe₃O₄;

Equation:

3Fe_(s) + 2O_{2 (g)} Fe₃O_4(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) Fe₃O_4(s) + 4H_2(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) + 3Cl_2(g) 2FeCl_3(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:
FeCl_3(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) FeCl_2(aq) + H_2(g)

Sulphuric (VI) acid:

Fe_(s) + H₂SO_{4 (aq)} FeSO4 _(aq) + H_{2 (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) + 6H₂SO_{4 (l)} Fe₂ (SO₄)_3(aq) + 6H₂O_(l) + 3SO_{2 (g)}

Nitric (V) acid.

– Iron reacts with dilute nitric (V) acid to form nitrogen (IV) oxide and ammonia which then forms ammonium nitrate.

Equation:

10HNO_{3 (aq)} + 4Fe_(s) 4Fe(NO₃)_2(aq) + NH₄NO_3(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;

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) + 4NH_3(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.

Crushing the ore

– The ore is crushed to increase the surface area for the succeeding chemical reactions.

– The ore is then concentrated.

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.

First Roasting

-The concentrated copper pyrite, CuFeS₂ is then roasted in air to remove some of the sulphur impurities as sulphur (IV) oxide.

Equation:

2CuFeS_2(s) + 4O_{2 (g)} 3SO_{2 (g)} + 2FeO_(S) + Cu₂S_(s);

Note:

– During 1^st 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) + SiO_2(g) FeSiO_3(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:

SO_2(g) + Ca (OH)_2(s) CaSO_3(s) + H₂O_(l)

– Construction of a contact process nearby.

Second Roasting.

– The CuS is heated in a regulated supply of air where some of it is converted to Cu₂O

Equation:

2Cu₂S_(s) + 3O_2(g) Cu₂O_(s) + 2SO_2(g)

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) + SO_{2 (g)};

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⁺; SO₄^2- 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

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.

Making soldering instruments.

Reason:-Copper has a high thermal conductivity.

Manufacture of alloys.

Examples

Alloy	Components
Brass	Copper and zinc
Bronze	Copper and tin
German silver	Copper, zinc and nickel;

Making coins and ornaments.

Reason:-it is durable and aesthetic.

Properties of copper.

Physical properties

– Soft red-brown metal.

– Melting point of 1083^oC and a boiling point of 2595^oC

– Density is 8.92gcm^-3 and electrical conductivity is about 5.93 x 10^-9Ώ^-1m^-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) + O_2(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)+ Cl_2(g) CuCl_2(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) + 8HNO_3(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) + 4HNO_3(l) Cu (NO3)_2(aq) + 2H₂O_(l) + 2NO_2(g)

With concentrated sulphuric acid.

-The sulphuric acid is reduced to sulphur (IV) oxide.

Equation:

Cu_(s) + 2H₂SO_4(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) + 4NH_3(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) + 3O_{2 (g)} 2PbO_(s) + 2SO_{2 (g)}

– During roasting some of the lead (II) sulphide is converted to lead (II) sulphate;

Equation:

PbS_(s) + 2O_2(g) PbSO_4(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) + CO_2(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:

CaCO_3(s) CaO_(s) + CO_{2 (g)};

– The carbon (IV) oxide gets reduced by coke to form more carbon (IV) oxide for reduction as in reaction (ii);

Equation:
CO_{2 (g)} + C_(s) 2CO_(g);

(v). The calcium oxide reacts with silica in form of SiO₂ to form calcium silicate;

Equation:

CaO_(s) + SiO_{2 (g)} CaSiO_{3 (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;

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;

Uses of lead:

It is used in several alloys e.g. solder and also added to bronze alloys to make them stronger;
Lead ingots are used in the manufacture of accumulators;
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;
Making lead pipes for water supply; this is also discouraged particularly in soft water areas due to threat of lead poisoning;
Making tetraethyl lead (IV) which for many years was used as a fuel additive to increase octane rating of fuels;
Used in weights, clock pendulums, plumb bobs etc; due to its high density;
It absorbs X-rays and hence lead aprons and lead glass are used to shield hospital radiographers;
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

Meaning of radioactivity
Natural and artificial radioactivity
Nuclear equations and chemical equations.
Types of radiations

Alpha particles
Beta particles
Gamma rays

Half life of a radioisotope
Radioactive decay curves and half life
Radioactive disintegration and nuclear equations.
Radioactive decay series, nuclear fission and nuclear fusion;
Uses of radioactivity;
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

Measuring their penetration power

They were directed to thin leaves of paper, aluminium, lead plates of varying thickness.
Observations:

One penetrated only thin foils of paper but not aluminium of 0.05mm thick and lead
One penetrated both paper and Al but not lead
One penetrated paper Al and lead
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

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

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,
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

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

10³x 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 10^{9 years}
Radium-226	1.6 X 10^{3 years}
Carbon-14	5.7 X 10^{3 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

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.

(2) Emissions of beta particles in two stage

Application of Radioactivity /Using radioactive isotopes.

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

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

Exposure of the body to external radiation
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

Identify pairs of isotopes of same element
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

Suggest a series of decays for formation of
Deduce how ½ life varies with energy of emitted rays

Suggest the value of X for

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ORGANIC CHEMISTRY II: ALCOHOLS AND ALKANOIC ACIDS:

1. Alkanols

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ALCOHOLS (ALKANOLS):

Nomenclature of alcohols

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

H H H H

H H OH H

H CH3 H

2-methyl propan-2-ol;

Preparation and properties of alkanols.

Starch molecule + water sucrose molecules

Properties of alcohols (Ethanol)

Then;

Ethanal H+ reaction Ethanoic acid

General equation

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

General equation:

Example: Catalytic oxidation of ethanol with hot copper metal

Then;

Ethanal H+ reaction Ethanoic acid

General equation

(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.

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

Soaps and soapless detergents

– 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; C17H35COO–Na+

Structurally

Preparation of soaps

Summary on soap preparation

Effect of hard water on soap

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

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

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

Measuring the strength of alkalis

(ii). Using PH values

Laboratory preparations of salts

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

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

Ø Fuels

Ø Heat of neutralization

Ø Enthalpy of solution

UNIT 4: RATES OF REACTION AND REVERSIBLE REACTIONS

Ø Measurements of reaction rates

Ø Apparatus for measuring reaction rates

§ Temperature

Ø Factors affecting position of equilibrium

§ Temperature

Measurements of reaction rates

Ø Reacting particles must collide before a chemical reaction occurs

Factors affecting the rate of reaction

Reversible reactions

– The Equilibrium state here is established when:

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.

(ii). The equilibrium can be reached from either direction.

Factors affecting the position of equilibrium

Pale yellow dark brown

– Decrease in temperature favours exothermic reactions.

Example:

Pale yellow dark brown

Decrease in pressure:

Experiment 1:- Displacement reactions among metals

Oxidation and reduction in terms of electron loss and gain

QUANTITATIVE ASPECTS OF ELECTROLYSIS

H CH₃ H

Ethanal H⁺reaction Ethanoic acid

Ethanal H⁺reaction Ethanoic acid

– A typical soap is sodium stearate; C₁₇H₃₅COO^–Na⁺

– Cerrusite, PbCO₃ (lead carbonate)

– Anglesite, PbSO₄(lead II) sulphate);