263REDOX REACTIONS

Chemistry deals with varieties of matter and change of one

kind of matter into the other. T ransformation of matter from

one kind into another occurs through the various types of

reactions. One important category of such reactions is

Redox Reactions. A number of phenomena, both physical

as well as biological, are concerned with redox reactions.

These reactions find extensive use in pharmaceutical,

biological, industrial, metallurgical and agricultural areas.

The importance of these reactions is apparent from the fact

that burning of different types of fuels for obtaining energy

for domestic, transport and other commercial purposes,

electrochemical processes for extraction of highly reactive

metals and non-metals, manufacturing of chemical

compounds like caustic soda, operation of dry and wet

batteries and corrosion of metals fall within the purview of

redox processes. Of late, environmental issues like

Hydrogen Economy (use of liquid hydrogen as fuel) and

development of ‘Ozone Hole’ have started figuring under

redox phenomenon.

8.1 CLASSICAL IDEA OF REDOX REACTIONS –

OXIDATION AND REDUCTION REACTIONS

Originally, the term oxidation was used to describe the

addition of oxygen to an element or a compound. Because

of the presence of dioxygen in the atmosphere (~20%),

many elements combine with it and this is the principal

reason why they commonly occur on the earth in the

form of their oxides. The following reactions represent

oxidation processes according to the limited definition of

oxidation:

2 Mg (s) + O

2

(g) → 2 MgO (s) (8.1)

S (s) + O

2

(g) → SO

2

(g) (8.2)

After studying this unit you will be

able to

••

• identify redox reactions as a class

of reactions in which oxidation

and reduction reactions occur

simultaneously;

••

• define the terms oxidation,

reduction, oxidant (oxidising

agent) and reductant (reducing

agent);

••

• explain mechanism of redox

reactions by electron transfer

process;

••

• use the concept of oxidation

number to identify oxidant and

reductant in a reaction;

••

• classify redox reaction into

combination (synthesis),

decomposition, displacement

and disproportionation

reactions;

••

• suggest a comparative order

among various reductants and

oxidants;

••

• balance chemical equations

using (i) oxidation number

(ii) half reaction method;

••

• learn the concept of redox

reactions in terms of electrode

processes.

UNIT 8

REDOX REACTIONS

Where there is oxidation, there is always reduction –

Chemistry is essentially a study of redox systems.

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264 CHEMISTRY

In reactions (8.1) and (8.2), the elements

magnesium and sulphur are oxidised on

account of addition of oxygen to them.

Similarly, methane is oxidised owing to the

addition of oxygen to it.

CH

4

(g) + 2O

2

(g) → CO

2

(g) + 2H

2

O (l) (8.3)

A careful examination of reaction (8.3) in

which hydrogen has been replaced by oxygen

prompted chemists to reinterpret oxidation in

terms of removal of hydrogen from it and,

therefore, the scope of term oxidation was

broadened to include the removal of hydrogen

from a substance. The following illustration is

another reaction where removal of hydrogen

can also be cited as an oxidation reaction.

2 H

2

S(g) + O

2

(g) → 2 S (s) + 2 H

2

O (l) (8.4)

As knowledge of chemists grew, it was

natural to extend the term oxidation for

reactions similar to (8.1 to 8.4), which do not

involve oxygen but other electronegative

elements. The oxidation of magnesium with

fluorine, chlorine and sulphur etc. occurs

according to the following reactions :

Mg (s) + F

2

(g) → MgF

2

(s) (8.5)

Mg (s) + Cl

2

(g) → MgCl

2

(s) (8.6)

Mg (s) + S (s) → MgS (s) (8.7)

Incorporating the reactions (8.5 to 8.7)

within the fold of oxidation reactions

encouraged chemists to consider not only the

removal of hydrogen as oxidation, but also the

removal of electropositive elements as

oxidation. Thus the reaction :

2K

4

[Fe(CN)

6

](aq) + H

2

O

2

(aq) →2K

3

[Fe(CN)

6

](aq)

+ 2 KOH (aq)

is interpreted as oxidation due to the removal

of electropositive element potassium from

potassium ferrocyanide before it changes to

potassium ferricyanide. To summarise, the

term “oxidation” is defined as the addition

of oxygen/electronegative element to a

substance or removal of hydrogen/

electropositive element from a substance.

In the beginning, reduction was

considered as removal of oxygen from a

compound. However, the term reduction has

been broadened these days to include removal

of oxygen/electronegative element from a

substance or addition of hydrogen/

electropositive element to a substance.

According to the definition given above, the

following are the examples of reduction

processes:

2 HgO (s)

2 Hg (l) + O

2

(g) (8.8)

(removal of oxygen from mercuric oxide )

2 FeCl

3

(aq) + H

2

(g) →2 FeCl

2

(aq) + 2 HCl(aq)

(8.9)

(removal of electronegative element, chlorine

from ferric chloride)

CH

2

= CH

2

(g) + H

2

(g) → H

3

C – CH

3

(g) (8.10)

(addition of hydrogen)

2HgCl

2

(aq) + SnCl

2

(aq) → Hg

2

Cl

2

(s)+SnCl

4

(aq)

(8.11)

(addition of mercury to mercuric chloride)

In reaction (8.11) simultaneous oxidation

of stannous chloride to stannic chloride is also

occurring because of the addition of

electronegative element chlorine to it. It was

soon realised that oxidation and reduction

always occur simultaneously (as will be

apparent by re-examining all the equations

given above), hence, the word “redox” was

coined for this class of chemical reactions.

Problem 8.1

In the reactions given below, identify the

species undergoing oxidation and

reduction:

(i) H

2

S (g) + Cl

2

(g) → 2 HCl (g) + S (s)

(ii) 3Fe

3

O

4

(s) + 8 Al (s) → 9 Fe (s)

+ 4Al

2

O

3

(s)

(iii) 2 Na (s) + H

2

(g) → 2 NaH (s)

Solution

(i) H

2

S is oxidised because a more

electronegative element, chlorine is added

to hydrogen (or a more electropositive

element, hydrogen has been removed

from S). Chlorine is reduced due to

addition of hydrogen to it.

(ii) Aluminium is oxidised because

oxygen is added to it. Ferrous ferric oxide

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265REDOX REACTIONS

(Fe

3

O

4

) is reduced because oxygen has

been removed from it.

(iii) With the careful application of the

concept of electronegativity only we may

infer that sodium is oxidised and

hydrogen is reduced.

Reaction (iii) chosen here prompts us to

think in terms of another way to define

redox reactions.

8.2 REDOX REACTIONS IN TERMS OF

ELECTRON TRANSFER REACTIONS

We have already learnt that the reactions

2Na(s) + Cl

2

(g) → 2NaCl (s) (8.12)

4Na(s) + O

2

(g) → 2Na

2

O(s) (8.13)

2Na(s) + S(s) → Na

2

S(s) (8.14)

are redox reactions because in each of these

reactions sodium is oxidised due to the

addition of either oxygen or more

electronegative element to sodium.

Simultaneously, chlorine, oxygen and sulphur

are reduced because to each of these, the

electropositive element sodium has been

added. From our knowledge of chemical

bonding we also know that sodium chloride,

sodium oxide and sodium sulphide are ionic

compounds and perhaps better written as

Na

+

Cl

–

(s), (Na

+

)

2

O

2–

(s), and (Na

+

)

2

S

2–

(s).

Development of charges on the species

produced suggests us to rewrite the reactions

(8.12 to 8.14) in the following manner :

For convenience, each of the above

processes can be considered as two separate

steps, one involving the loss of electrons and

the other the gain of electrons. As an

illustration, we may further elaborate one of

these, say, the formation of sodium chloride.

2 Na(s) → 2 Na

+

(g)

+ 2e

–

Cl

2

(g) + 2e

–

→ 2 Cl

–

(g)

Each of the above steps is called a half

reaction, which explicitly shows involvement

of electrons. Sum of the half reactions gives

the overall reaction :

2 Na(s) + Cl

2

(g) → 2 Na

+

Cl

–

(s) or 2 NaCl (s)

Reactions 8.12 to 8.14 suggest that half

reactions that involve loss of electrons are

called oxidation reactions. Similarly, the

half reactions that involve gain of electrons

are called reduction reactions. It may not

be out of context to mention here that the new

way of defining oxidation and reduction has

been achieved only by establishing a

correlation between the behaviour of species

as per the classical idea and their interplay in

electron-transfer change. In reactions (8.12 to

8.14) sodium, which is oxidised, acts as

a reducing agent because it donates electron

to each of the elements interacting with it and

thus helps in reducing them. Chlorine, oxygen

and sulphur are reduced and act as oxidising

agents because these accept electrons from

sodium. To summarise, we may mention that

Oxidation: Loss of electron(s) by any species.

Reduction: Gain of electron(s) by any species.

Oxidising agent : Acceptor of electron(s).

Reducing agent : Donor of electron(s).

Problem 8.2 Justify that the reaction :

2 Na(s) + H

2

(g) → 2 NaH (s) is a redox

change.

Solution

Since in the above reaction the compound

formed is an ionic compound, which may

also be represented as Na

+

H

–

(s), this

suggests that one half reaction in this

process is :

2 Na (s) → 2 Na

+

(g) + 2e

–

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266 CHEMISTRY

and the other half reaction is:

H

2

(g) + 2e

–

→ 2 H

–

(g)

This splitting of the reaction under

examination into two half reactions

automatically reveals that here sodium is

oxidised and hydrogen is reduced,

therefore, the complete reaction is a redox

change.

8.2.1 Competitive Electron Transfer

Reactions

Place a strip of metallic zinc in an aqueous

solution of copper nitrate as shown in Fig. 8.1,

for about one hour. You may notice that the

strip becomes coated with reddish metallic

copper and the blue colour of the solution

disappears. Formation of Zn

2+

ions among the

products can easily be judged when the blue

colour of the solution due to Cu

2+

has

disappeared. If hydrogen sulphide gas is

passed through the colourless solution

containing Zn

2+

ions, appearance of white zinc

sulphide, ZnS can be seen on making the

solution alkaline with ammonia.

The reaction between metallic zinc and the

aqueous solution of copper nitrate is :

Zn(s) + Cu

2+

(aq) → Zn

2+

(aq) + Cu(s) (8.15)

In reaction (8.15), zinc has lost electrons

to form Zn

2+

and, therefore, zinc is oxidised.

Evidently, now if zinc is oxidised, releasing

electrons, something must be reduced,

accepting the electrons lost by zinc. Copper

ion is reduced by gaining electrons from the zinc.

Reaction (8.15) may be rewritten as :

At this stage we may investigate the state

of equilibrium for the reaction represented by

equation (8.15). For this purpose, let us place

a strip of metallic copper in a zinc sulphate

solution. No visible reaction is noticed and

attempt to detect the presence of Cu

2+

ions by

passing H

2

S gas through the solution to

produce the black colour of cupric sulphide,

CuS, does not succeed. Cupric sulphide has

such a low solubility that this is an extremely

sensitive test; yet the amount of Cu

2+

formed

cannot be detected. We thus conclude that the

state of equilibrium for the reaction (8.15)

greatly favours the products over the reactants.

Let us extend electron transfer reaction now

to copper metal and silver nitrate solution in

water and arrange a set-up as shown in

Fig. 8.2. The solution develops blue colour due

to the formation of Cu

2+

ions on account of the

reaction:

Fig. 8.1 Redox reaction between zinc and aqueous solution of copper nitrate occurring in a beaker.

(8.16)

Here, Cu(s) is oxidised to Cu

2+

(aq) and

Ag

+

(aq) is reduced to Ag(s). Equilibrium greatly

favours the products Cu

2+

(aq) and Ag(s).

By way of contrast, let us also compare the

reaction of metallic cobalt placed in nickel

sulphate solution. The reaction that occurs

here is :

(8.17)

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267REDOX REACTIONS

Fig. 8.2 Redox reaction between copper and aqueous solution of silver nitrate occurring in a beaker.

At equilibrium, chemical tests reveal that both

Ni

2+

(aq) and Co

2+

(aq)

are present at moderate

concentrations. In this case, neither the

reactants [Co(s) and Ni

2+

(aq)] nor the products

[Co

2+

(aq) and Ni (s)] are greatly favoured.

This competition for release of electrons

incidently reminds us of the competition for

release of protons among acids. The similarity

suggests that we might develop a table in

which metals and their ions are listed on the

basis of their tendency to release electrons just

as we do in the case of acids to indicate the

strength of the acids. As a matter of fact we

have already made certain comparisons. By

comparison we have come to know that zinc

releases electrons to copper and copper

releases electrons to silver and, therefore, the

electron releasing tendency of the metals is in

the order: Zn>Cu>Ag. We would love to make

our list more vast and design a metal activity

series or electrochemical series. The

competition for electrons between various

metals helps us to design a class of cells,

named as Galvanic cells in which the chemical

reactions become the source of electrical

energy. We would study more about these cells

in Class XII.

8.3 OXIDATION NUMBER

A less obvious example of electron transfer is

realised when hydrogen combines with oxygen

to form water by the reaction:

2H

2

(g) + O

2

(g) → 2H

2

O (l) (8.18)

Though not simple in its approach, yet we

can visualise the H atom as going from a

neutral (zero) state in H

2

to a positive state in

H

2

O, the O atom goes from a zero state in O

2

to a dinegative state in H

2

O. It is assumed that

there is an electron transfer from H to O and

consequently H

2

is oxidised and O

2

is reduced.

However, as we shall see later, the charge

transfer is only partial and is perhaps better

described as an electron shift rather than a

complete loss of electron by H and gain by O.

What has been said here with respect to

equation (8.18) may be true for a good number

of other reactions involving covalent

compounds. Two such examples of this class

of the reactions are:

H

2

(s) + Cl

2

(g) → 2HCl(g) (8.19)

and,

CH

4

(g) + 4Cl

2

(g) → CCl

4

(l) + 4HCl(g) (8.20)

In order to keep track of electron shifts in

chemical reactions involving formation of

covalent compounds, a more practical method

of using oxidation number has been

developed. In this method, it is always

assumed that there is a complete transfer of

electron from a less electronegative atom to a

more electonegative atom. For example, we

rewrite equations (8.18 to 8.20) to show

charge on each of the atoms forming part of

the reaction :

0 0 +1 –2

2H

2

(g) + O

2

(g) → 2H

2

O (l) (8.21)

0 0 +1 –1

H

2

(s) + Cl

2

(g) → 2HCl(g) (8.22)

–4+1 0 +4 –1 +1 –1

CH

4

(g) + 4Cl

2

(g) → CCl

4

(l) +4HCl(g) (8.23)

It may be emphasised that the assumption

of electron transfer is made for book-keeping

purpose only and it will become obvious at a

later stage in this unit that it leads to the simple

description of redox reactions.

Oxidation number denotes the

oxidation state of an element in a

compound ascertained according to a set

of rules formulated on the basis that

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268 CHEMISTRY

electron pair in a covalent bond belongs

entirely to more electronegative element.

It is not always possible to remember or

make out easily in a compound/ion, which

element is more electronegative than the other.

Therefore, a set of rules has been formulated

to determine the oxidation number of an

element in a compound/ion. If two or more

than two atoms of an element are present in

the molecule/ion such as Na

2

S

2

O

3

/Cr

2

O

7

2–

, the

oxidation number of the atom of that element

will then be the average of the oxidation

number of all the atoms of that element. We

may at this stage, state the rules for the

calculation of oxidation number. These rules are:

1. In elements, in the free or the uncombined

state, each atom bears an oxidation

number of zero. Evidently each atom in H

2

,

O

2

, Cl

2

, O

3

, P

4

, S

8

, Na, Mg, Al has the

oxidation number zero.

2. For ions composed of only one atom, the

oxidation number is equal to the charge

on the ion. Thus Na

+

ion has an oxidation

number of +1, Mg

2+

ion, +2, Fe

3+

ion, +3,

Cl

–

ion, –1, O

2–

ion, –2; and so on. In their

compounds all alkali metals have

oxidation number of +1, and all alkaline

earth metals have an oxidation number of

+2. Aluminium is regarded to have an

oxidation number of +3 in all its

compounds.

3. The oxidation number of oxygen in most

compounds is –2. However, we come across

two kinds of exceptions here. One arises

in the case of peroxides and superoxides,

the compounds of oxygen in which oxygen

atoms are directly linked to each other.

While in peroxides (e.g., H

2

O

2

, Na

2

O

2

), each

oxygen atom is assigned an oxidation

number of –1, in superoxides (e.g., KO

2

,

RbO

2

) each oxygen atom is assigned an

oxidation number of –(½). The second

exception appears rarely, i.e. when oxygen

is bonded to fluorine. In such compounds

e.g., oxygen difluoride (OF

2

) and dioxygen

difluoride (O

2

F

2

), the oxygen is assigned

an oxidation number of +2 and +1,

respectively. The number assigned to

oxygen will depend upon the bonding state

of oxygen but this number would now be

a positive figure only.

4. The oxidation number of hydrogen is +1,

except when it is bonded to metals in binary

compounds (that is compounds containing

two elements). For example, in LiH, NaH,

and CaH

2

, its oxidation number is –1.

5. In all its compounds, fluorine has an

oxidation number of –1. Other halogens (Cl,

Br, and I) also have an oxidation number

of –1, when they occur as halide ions in

their compounds. Chlorine, bromine and

iodine when combined with oxygen, for

example in oxoacids and oxoanions, have

positive oxidation numbers.

6. The algebraic sum of the oxidation number

of all the atoms in a compound must be

zero. In polyatomic ion, the algebraic sum

of all the oxidation numbers of atoms of

the ion must equal the charge on the ion.

Thus, the sum of oxidation number of three

oxygen atoms and one carbon atom in the

carbonate ion, (CO

3

)

2–

must equal –2.

By the application of above rules, we can

find out the oxidation number of the desired

element in a molecule or in an ion. It is clear

that the metallic elements have positive

oxidation number and nonmetallic elements

have positive or negative oxidation number.

The atoms of transition elements usually

display several positive oxidation states. The

highest oxidation number of a representative

element is the group number for the first two

groups and the group number minus 10

(following the long form of periodic table) for

the other groups. Thus, it implies that the

highest value of oxidation number exhibited

by an atom of an element generally increases

across the period in the periodic table. In the

third period, the highest value of oxidation

number changes from 1 to 7 as indicated below

in the compounds of the elements.

A term that is often used interchangeably

with the oxidation number is the oxidation

state. Thus in CO

2

, the oxidation state of

carbon is +4, that is also its oxidation number

and similarly the oxidation state as well as

oxidation number of oxygen is – 2. This implies

that the oxidation number denotes the

oxidation state of an element in a compound.

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269REDOX REACTIONS

The oxidation number/state of a metal in a

compound is sometimes presented according

to the notation given by German chemist,

Alfred Stock. It is popularly known as Stock

notation. According to this, the oxidation

number is expressed by putting a Roman

numeral representing the oxidation number

in parenthesis after the symbol of the metal in

the molecular formula. Thus aurous chloride

and auric chloride are written as Au(I)Cl and

Au(III)Cl

3

. Similarly, stannous chloride and

stannic chloride are written as Sn(II)Cl

2

and

Sn(IV)Cl

4

. This change in oxidation number

implies change in oxidation state, which in

turn helps to identify whether the species is

present in oxidised form or reduced form.

Thus, Hg

2

(I)Cl

2

is the reduced form of Hg(II) Cl

2

.

Problem 8.3

Using Stock notation, represent the

following compounds :HAuCl

4

, Tl

2

O, FeO,

Fe

2

O

3

, CuI, CuO, MnO and MnO

2

.

Solution

By applying various rules of calculating

the oxidation number of the desired

element in a compound, the oxidation

number of each metallic element in its

compound is as follows:

HAuCl

4

→ Au has 3

Tl

2

O → Tl has 1

FeO → Fe has 2

Fe

2

O

3

→ Fe has 3

CuI → Cu has 1

CuO → Cu has 2

MnO → Mn has 2

MnO

2

→ Mn has 4

Therefore, these compounds may be

represented as:

HAu(III)Cl

4

, Tl

2

(I)O, Fe(II)O, Fe

2

(III)O

3

,

Cu(I)I, Cu(II)O, Mn(II)O, Mn(IV)O

2.

The idea of oxidation number has been

invariably applied to define oxidation,

reduction, oxidising agent (oxidant), reducing

agent (reductant) and the redox reaction. To

summarise, we may say that:

Oxidation: An increase in the oxidation

number of the element in the given substance.

Reduction: A decrease in the oxidation

number of the element in the given substance.

Oxidising agent: A reagent which can

increase the oxidation number of an element

in a given substance. These reagents are called

as oxidants also.

Reducing agent: A reagent which lowers the

oxidation number of an element in a given

substance. These reagents are also called as

reductants.

Redox reactions: Reactions which involve

change in oxidation number of the interacting

species.

Problem 8.4

Justify that the reaction:

2Cu

2

O(s) + Cu

2

S(s) → 6Cu(s) + SO

2

(g)

is a redox reaction. Identify the species

oxidised/reduced, which acts as an

oxidant and which acts as a reductant.

Solution

Let us assign oxidation number to each

of the species in the reaction under

examination. This results into:

+1 –2 +1 –2 0 +4 –2

2Cu

2

O(s) + Cu

2

S(s) → 6Cu(s) + SO

2

We therefore, conclude that in this

reaction copper is reduced from +1 state

to zero oxidation state and sulphur is

oxidised from –2 state to +4 state. The

above reaction is thus a redox reaction.

Group 1 2 13 14 15 16 17

Element Na Mg Al Si P S Cl

Compound NaCl MgSO

4

AlF

3

SiCl

4

P

4

O

10

SF

6

HClO

4

Highest oxidation +1 +2 +3 +4 +5 +6 +7

number state of

the group element

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270 CHEMISTRY

Further, Cu

2

O helps sulphur in Cu

2

S to

increase its oxidation number, therefore,

Cu(I) is an oxidant; and sulphur of Cu

2

S

helps copper both in Cu

2

S itself and Cu

2

O

to decrease its oxidation number;

therefore, sulphur of Cu

2

S is reductant.

8.3.1 Types of Redox Reactions

1. Combination reactions

A combination reaction may be denoted in the

manner:

A + B → C

Either A and B or both A and B must be in the

elemental form for such a reaction to be a redox

reaction. All combustion reactions, which

make use of elemental dioxygen, as well as

other reactions involving elements other than

dioxygen, are redox reactions. Some important

examples of this category are:

0 0 +4 –2

C(s) + O

2

(g)

CO

2

(g) (8.24)

0 0 +2 –3

3Mg(s) + N

2

(g) Mg

3

N

2

(s) (8.25)

–4+1 0 +4 –2 +1 –2

CH

4

(g) + 2O

2

(g) CO

2

(g) + 2H

2

O (l)

2. Decomposition reactions

Decomposition reactions are the opposite of

combination reactions. Precisely, a

decomposition reaction leads to the breakdown

of a compound into two or more components

at least one of which must be in the elemental

state. Examples of this class of reactions are:

+1 –2 0 0

2H

2

O (l) 2H

2

(g) + O

2

(g) (8.26)

+1 –1 0 0

2NaH (s) 2Na (s) + H

2

(g) (8.27)

+1 +5 –2 +1 –1 0

2KClO

3

(s) 2KCl (s) + 3O

2

(g) (8.28)

It may carefully be noted that there is no

change in the oxidation number of hydrogen

in methane under combination reactions and

that of potassium in potassium chlorate in

reaction (8.28). This may also be noted here

that all decomposition reactions are not redox

reactions. For example, decomposition of

calcium carbonate is not a redox reaction.

+2 +4 –2 +2 –2 +4 –2

CaCO

3

(s)

CaO(s) + CO

2

(g)

3. Displacement reactions

In a displacement reaction, an ion (or an atom)

in a compound is replaced by an ion (or an

atom) of another element. It may be denoted

as:

X + YZ → XZ + Y

Displacement reactions fit into two categories:

metal displacement and non-metal

displacement.

(a) Metal displacement: A metal in a

compound can be displaced by another metal

in the uncombined state. We have already

discussed about this class of the reactions

under section 8.2.1. Metal displacement

reactions find many applications in

metallurgical processes in which pure metals

are obtained from their compounds in ores. A

few such examples are:

+2 +6 –2 0 0 +2 +6 –2

CuSO

4

(aq) + Zn (s) → Cu(s) + ZnSO

4

(aq)

(8.29)

+5 –2 0 0 +2 –2

V

2

O

5

(s) + 5Ca (s)

2V (s) + 5CaO (s)

(8.30)

+4 –1 0 0 +2 –1

TiCl

4

(l) + 2Mg (s) Ti (s) + 2 MgCl

2

(s)

(8.31)

+3 –2 0 +3 –2 0

Cr

2

O

3

(s) + 2 Al (s) Al

2

O

3

(s) + 2Cr(s)

(8.32)

In each case, the reducing metal is a better

reducing agent than the one that is being

reduced which evidently shows more capability

to lose electrons as compared to the one that

is reduced.

(b) Non-metal displacement: The non-metal

displacement redox reactions include

hydrogen displacement and a rarely occurring

reaction involving oxygen displacement.

2019-20

271REDOX REACTIONS

All alkali metals and some alkaline earth

metals (Ca, Sr, and Ba) which are very good

reductants, will displace hydrogen from cold

water.

0 +1 –2 +1 –2 +1 0

2Na(s) + 2H

2

O(l) → 2NaOH(aq) + H

2

(g)

(8.33)

0 +1 –2 +2 –2 +1 0

Ca(s) + 2H

2

O(l) → Ca(OH)

2

(aq) + H

2

(g)

(8.34)

Less active metals such as magnesium and

iron react with steam to produce dihydrogen gas:

0 +1 –2 +2 –2 +1 0

Mg(s) + 2H

2

O(l)

Mg(OH)

2

(s) + H

2

(g)

(8.35)

0 +1 –2 +3 –2 0

2Fe(s) + 3H

2

O(l)

Fe

2

O

3

(s) + 3H

2

(g) (8.36)

Many metals, including those which do not

react with cold water, are capable of displacing

hydrogen from acids. Dihydrogen from acids

may even be produced by such metals which

do not react with steam. Cadmium and tin are

the examples of such metals. A few examples

for the displacement of hydrogen from acids

are:

0 +1 –1 +2 –1 0

Zn(s) + 2HCl(aq) → ZnCl

2

(aq) + H

2

(g)

(8.37)

0 +1 –1 +2 –1 0

Mg (s) + 2HCl (aq) → MgCl

2

(aq) + H

2

(g)

(8.38)

0 +1 –1 +2 –1 0

Fe(s) + 2HCl(aq) → FeCl

2

(aq) + H

2

(g)

(8.39)

Reactions (8.37 to 8.39) are used to

prepare dihydrogen gas in the laboratory.

Here, the reactivity of metals is reflected in the

rate of hydrogen gas evolution, which is the

slowest for the least active metal Fe, and the

fastest for the most reactive metal, Mg. Very

less active metals, which may occur in the

native state such as silver (Ag), and gold (Au)

do not react even with hydrochloric acid.

In section (8.2.1) we have already

discussed that the metals – zinc (Zn), copper

(Cu) and silver (Ag) through tendency to lose

electrons show their reducing activity in the

order Zn> Cu>Ag. Like metals, activity series

also exists for the halogens. The power of these

elements as oxidising agents decreases as we

move down from fluorine to iodine in group

17 of the periodic table. This implies that

fluorine is so reactive that it can replace

chloride, bromide and iodide ions in solution.

In fact, fluorine is so reactive that it attacks

water and displaces the oxygen of water :

+1 –2 0 +1 –1 0

2H

2

O (l) + 2F

2

(g) → 4HF(aq) + O

2

(g) (8.40)

It is for this reason that the displacement

reactions of chlorine, bromine and iodine

using fluorine are not generally carried out in

aqueous solution. On the other hand, chlorine

can displace bromide and iodide ions in an

aqueous solution as shown below:

0 +1 –1 +1 –1 0

Cl

2

(g) + 2KBr (aq) → 2 KCl (aq) + Br

2

(l)

(8.41)

0 +1–1 +1 –1 0

Cl

2

(g) + 2KI (aq) → 2 KCl (aq) + I

2

(s)

(8.42)

As Br

2

and I

2

are coloured and dissolve in CCl

4

,

can easily be identified from the colour of the

solution. The above reactions can be written

in ionic form as:

0 –1 –1 0

Cl

2

(g) + 2Br

–

(aq) → 2Cl

–

(aq) + Br

2

(l) (8.41a)

0 –1 –1 0

Cl

2

(g) + 2I

–

(aq) → 2Cl

–

(aq) + I

2

(s) (8.42b)

Reactions (8.41) and (8.42) form the basis

of identifying Br

–

and I

–

in the laboratory

through the test popularly known as ‘Layer

Test’. It may not be out of place to mention

here that bromine likewise can displace iodide

ion in solution:

0 –1 –1 0

Br

2

(l) + 2I

–

(aq) → 2Br

–

(aq) + I

2

(s) (8.43)

The halogen displacement reactions have

a direct industrial application. The recovery

of halogens from their halides requires an

oxidation process, which is represented by:

2X

–

→ X

2

+ 2e

–

(8.44)

here X denotes a halogen element. Whereas

chemical means are available to oxidise Cl

–

,

Br

–

and I

–

, as fluorine is the strongest oxidising

2019-20

272 CHEMISTRY

agent; there is no way to convert F

–

ions to F

2

by chemical means. The only way to achieve

F

2

from F

–

is to oxidise electrolytically, the

details of which you will study at a later stage.

4. Disproportionation reactions

Disproportionation reactions are a special type

of redox reactions. In a disproportionation

reaction an element in one oxidation state is

simultaneously oxidised and reduced. One of

the reacting substances in a

disproportionation reaction always contains

an element that can exist in at least three

oxidation states. The element in the form of

reacting substance is in the intermediate

oxidation state; and both higher and lower

oxidation states of that element are formed in

the reaction. The decomposition of hydrogen

peroxide is a familiar example of the reaction,

where oxygen experiences disproportionation.

+1 –1 +1 –2 0

2H

2

O

2

(aq) → 2H

2

O(l) + O

2

(g) (8.45)

Here the oxygen of peroxide, which is present

in –1 state, is converted to zero oxidation state

in O

2

and decreases to –2 oxidation state in

H

2

O.

Phosphorous, sulphur and chlorine

undergo disproportionation in the alkaline

medium as shown below :

0 –3 +1

P

4

(s) + 3OH

–

(aq)+ 3H

2

O(l) → PH

3

(g) + 3H

2

PO

2

–

(aq)

(8.46)

0 –2 +2

S

8

(s) + 12 OH

–

(aq) → 4S

2–

(aq) + 2S

2

O

3

2–

(aq)

+ 6H

2

O(l)

(8.47)

0 +1 –1

Cl

2

(g) + 2 OH

–

(aq) → ClO

–

(aq) + Cl

–

(aq) +

H

2

O (l)

(8.48)

The reaction (8.48) describes the formation

of household bleaching agents. The

hypochlorite ion (ClO

–

) formed in the reaction

oxidises the colour-bearing stains of the

substances to colourless compounds.

It is of interest to mention here that whereas

bromine and iodine follow the same trend as

exhibited by chlorine in reaction (8.48),

fluorine shows deviation from this behaviour

when it reacts with alkali. The reaction that

takes place in the case of fluorine is as follows:

2 F

2

(g) + 2OH

–

(aq) → 2 F

–

(aq) + OF

2

(g) + H

2

O(l)

(8.49)

(It is to be noted with care that fluorine in

reaction (8.49) will undoubtedly attack water

to produce some oxygen also). This departure

shown by fluorine is not surprising for us as

we know the limitation of fluorine that, being

the most electronegative element, it cannot

exhibit any positive oxidation state. This

means that among halogens, fluorine does not

show a disproportionation tendency.

Problem 8.5

Which of the following species, do not

show disproportionation reaction and

why ?

ClO

–

, ClO

2

–

, ClO

3

–

and ClO

4

–

Also write reaction for each of the species

that disproportionates.

Solution

Among the oxoanions of chlorine listed

above, ClO

4

–

does not disproportionate

because in this oxoanion chlorine is

present in its highest oxidation state that

is, +7. The disproportionation reactions

for the other three oxoanions of chlorine

are as follows:

+1 –1 +5

3ClO

–

→ 2Cl

–

+ ClO

–

3

+3 +5 –1

6 ClO

2

–

4ClO

3

–

+ 2Cl

–

+5 –1 +7

4ClO

–

3

→ Cl

–

+ 3 ClO

4

–

Problem 8.6

Suggest a scheme of classification of the

following redox reactions

(a) N

2

(g) + O

2

(g) → 2 NO (g)

(b) 2Pb(NO

3

)

2

(s) → 2PbO(s) + 4 NO

2

(g) +

O

2

(g)

(c) NaH(s) + H

2

O(l) → NaOH(aq) + H

2

(g)

(d) 2NO

2

(g) + 2OH

–

(aq) → NO

2

–

(aq) +

NO

3

–

(aq)+H

2

O(l)

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273REDOX REACTIONS

Solution

In reaction (a), the compound nitric oxide

is formed by the combination of the

elemental substances, nitrogen and

oxygen; therefore, this is an example of

combination redox reactions. The

reaction (b) involves the breaking down

of lead nitrate into three components;

therefore, this is categorised under

decomposition redox reaction. In reaction

The Paradox of Fractional Oxidation Number

Sometimes, we come across with certain compounds in which the oxidation number of a

particular element in the compound is in fraction. Examples are:

C

3

O

2

[where oxidation number of carbon is (4/3)],

Br

3

O

8

[where oxidation number of bromine is (16/3)]

and Na

2

S

4

O

6

(where oxidation number of sulphur is 2.5).

We know that the idea of fractional oxidation number is unconvincing to us, because

electrons are never shared/transferred in fraction. Actually this fractional oxidation state is

the average oxidation state of the element under examination and the structural parameters

reveal that the element for whom fractional oxidation state is realised is present in different

oxidation states. Structure of the species C

3

O

2

, Br

3

O

8

and S

4

O

6

2–

reveal the following bonding

situations:

+2 0 +2

O = C = C*= C = O

Structure of C

3

O

2

(carbon suboxide)

Structure of Br

3

O

8

(tribromooctaoxide) Structure of S

4

O

6

2–

(tetrathionate ion)

The element marked with asterisk in each species is exhibiting the different oxidation

state (oxidation number) from rest of the atoms of the same element in each of the species.

This reveals that in C

3

O

2

, two carbon atoms are present in +2 oxidation state each, whereas

the third one is present in zero oxidation state and the average is 4/3. However, the realistic

picture is +2 for two terminal carbons and zero for the middle carbon. Likewise in Br

3

O

8

, each

of the two terminal bromine atoms are present in +6 oxidation state and the middle bromine

is present in +4 oxidation state. Once again the average, that is different from reality, is

16/3. In the same fashion, in the species S

4

O

6

2–

, each of the two extreme sulphurs exhibits

oxidation state of +5 and the two middle sulphurs as zero. The average of four oxidation

numbers of sulphurs of the S

4

O

6

2–

is 2.5, whereas the reality being + 5,0,0 and +5 oxidation

number respectively for each sulphur.

We may thus, in general, conclude that the idea of fractional oxidation state should be

taken with care and the reality is revealed by the structures only. Further, whenever we come

across with fractional oxidation state of any particular element in any species, we must

understand that this is the average oxidation number only. In reality (revealed by structures

only), the element in that particular species is present in more than one whole number oxidation

states. Fe

3

O

4

, Mn

3

O

4

, Pb

3

O

4

are some of the other examples of the compounds, which are

mixed oxides, where we come across with fractional oxidation states of the metal atom. However,

the oxidation states may be in fraction as in O

2

+

and O

2

–

where it is +½ and –½ respectively.

(c), hydrogen of water has been displaced

by hydride ion into dihydrogen gas.

Therefore, this may be called as

displacement redox reaction. The reaction

(d) involves disproportionation of NO

2

(+4 state)

into NO

2

–

(+3 state) and NO

3

–

(+5 state). Therefore reaction (d) is an

example of disproportionation redox

reaction.

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274 CHEMISTRY

Problem 8.7

Why do the following reactions proceed

differently ?

Pb

3

O

4

+ 8HCl → 3PbCl

2

+ Cl

2

+ 4H

2

O

and

Pb

3

O

4

+ 4HNO

3

→ 2Pb(NO

3

)

2

+ PbO

2

+

2H

2

O

Solution

Pb

3

O

4

is actually a stoichiometric

mixture of 2 mol of PbO and 1 mol of

PbO

2

. In PbO

2

, lead is present in +4

oxidation state, whereas the stable

oxidation state of lead in PbO is +2. PbO

2

thus can act as an oxidant (oxidising

agent) and, therefore, can oxidise Cl

–

ion

of HCl into chlorine. We may also keep in

mind that PbO is a basic oxide. Therefore,

the reaction

Pb

3

O

4

+ 8HCl → 3PbCl

2

+ Cl

2

+ 4H

2

O

can be splitted into two reactions namely:

2PbO + 4HCl → 2PbCl

2

+ 2H

2

O

(acid-base reaction)

+4 –1 +2 0

PbO

2

+ 4HCl → PbCl

2

+ Cl

2

+2H

2

O

(redox reaction)

Since HNO

3

itself is an oxidising agent

therefore, it is unlikely that the reaction

may occur between PbO

2

and HNO

3

.

However, the acid-base reaction occurs

between PbO and HNO

3

as:

2PbO + 4HNO

3

→ 2Pb(NO

3

)

2

+ 2H

2

O

It is the passive nature of PbO

2

against

HNO

3

that makes the reaction different

from the one that follows with HCl.

8.3.2 Balancing of Redox Reactions

Two methods are used to balance chemical

equations for redox processes. One of these

methods is based on the change in the

oxidation number of reducing agent and the

oxidising agent and the other method is based

on splitting the redox reaction into two half

reactions — one involving oxidation and the

other involving reduction. Both these methods

are in use and the choice of their use rests with

the individual using them.

(a) Oxidation Number Method: In writing

equations for oxidation-reduction reactions,

just as for other reactions, the compositions

and formulas must be known for the

substances that react and for the products that

are formed. The oxidation number method is

now best illustrated in the following steps:

Step 1: Write the correct formula for each

reactant and product.

Step 2: Identify atoms which undergo change

in oxidation number in the reaction by

assigning the oxidation number to all elements

in the reaction.

Step 3: Calculate the increase or decrease in

the oxidation number per atom and for the

entire molecule/ion in which it occurs. If these

are not equal then multiply by suitable

number so that these become equal. (If you

realise that two substances are reduced and

nothing is oxidised or vice-versa, something

is wrong. Either the formulas of reactants or

products are wrong or the oxidation numbers

have not been assigned properly).

Step 4: Ascertain the involvement of ions if

the reaction is taking place in water, add H

+

or

OH

–

ions to the expression on the appropriate

side so that the total ionic charges of reactants

and products are equal. If the reaction is

carried out in acidic solution, use H

+

ions in

the equation; if in basic solution, use OH

–

ions.

Step 5 : Make the numbers of hydrogen atoms

in the expression on the two sides equal by

adding water (H

2

O) molecules to the reactants

or products. Now, also check the number of

oxygen atoms. If there are the same number

of oxygen atoms in the reactants and

products, the equation then represents the

balanced redox reaction.

Let us now explain the steps involved in

the method with the help of a few problems

given below:

Problem 8.8

Write the net ionic equation for the

reaction of potassium dichromate(VI),

K

2

Cr

2

O

7

with sodium sulphite, Na

2

SO

3

,

in an acid solution to give chromium(III)

ion and the sulphate ion.

2019-20

275REDOX REACTIONS

Solution

Step 1: The skeletal ionic equation is:

Cr

2

O

7

2–

(aq) + SO

3

2–

(aq) → Cr

3+

(aq)

+ SO

4

2–

(aq)

Step 2: Assign oxidation numbers for Cr

and S

+6 –2 +4 –2 +3 +6 –2

Cr

2

O

7

2–

(aq) + SO

3

2–

(aq) → Cr(aq)+SO

4

2–

(aq)

This indicates that the dichromate ion is

the oxidant and the sulphite ion is the

reductant.

Step 3: Calculate the increase and

decrease of oxidation number, and make

them equal: from step-2 we can notice that

there is change in oxidation state of

chromium and sulphur. Oxidation state

of chromium changes form +6 to +3. There

is decrease of +3 in oxidation state of

chromium on right hand side of the

equation. Oxidation state of sulphur

changes from +4 to +6. There is an increase

of +2 in the oxidation state of sulphur on

right hand side. To make the increase and

decrease of oxidation state equal, place

numeral 2 before cromium ion on right

hand side and numeral 3 before sulphate

ion on right hand side and balance the

chromium and sulphur atoms on both the

sides of the equation. Thus we get

+6 –2 +4 –2 +3

Cr

2

O

7

2–

(aq) + 3SO

3

2–

(aq) → 2Cr

3+

(aq) +

+6 –2

3SO

4

2–

(aq)

Step 4: As the reaction occurs in the

acidic medium, and further the ionic

charges are not equal on both the sides,

add 8H

+

on the left to make ionic charges

equal

Cr

2

O

7

2–

(aq) + 3SO

3

2–

(aq)+ 8H

+

→ 2Cr

3+

(aq)

+ 3SO

4

2–

(aq)

Step 5: Finally, count the hydrogen

atoms, and add appropriate number of

water molecules (i.e., 4H

2

O) on the right

to achieve balanced redox change.

Cr

2

O

7

2–

(aq) + 3SO

3

2–

(aq)+ 8H

+

(aq) →

2Cr

3+

(aq) + 3SO

4

2–

(aq) +4H

2

O (l)

Problem 8.9

Permanganate ion reacts with bromide ion

in basic medium to give manganese

dioxide and bromate ion. Write the

balanced ionic equation for the reaction.

Solution

Step 1: The skeletal ionic equation is :

MnO

4

–

(aq) + Br

–

(aq)

→→

→ MnO

2

(s) + BrO

3

–

(aq)

Step 2: Assign oxidation numbers for Mn

and Br

+7 –1 +4 +5

MnO

4

–

(aq) + Br

–

(aq) →MnO

2

(s) + BrO

3

–

(aq)

this indicates that permanganate ion is

the oxidant and bromide ion is the

reductant.

Step 3: Calculate the increase and

decrease of oxidation number, and make

the increase equal to the decrease.

+7 –1 +4 +5

2MnO

4

–

(aq)+Br

–

(aq) → 2MnO

2

(s)+BrO

3

–

(aq)

Step 4: As the reaction occurs in the basic

medium, and the ionic charges are not

equal on both sides, add 2 OH

–

ions on

the right to make ionic charges equal.

2MnO

4

–

(aq) + Br

–

(aq) → 2MnO

2

(s) +

BrO

3

–

(aq) + 2OH

–

(aq)

Step 5: Finally, count the hydrogen atoms

and add appropriate number of water

molecules (i.e. one H

2

O molecule) on the

left side to achieve balanced redox change.

2MnO

4

–

(aq) + Br

–

(aq) + H

2

O(l) → 2MnO

2

(s)

+ BrO

3

–

(aq) + 2OH

–

(aq)

(b) Half Reaction Method: In this method,

the two half equations are balanced separately

and then added together to give balanced

equation.

Suppose we are to balance the equation

showing the oxidation of Fe

2+

ions to Fe

3+

ions

by dichromate ions (Cr

2

O

7

)

2–

in acidic medium,

wherein, Cr

2

O

7

2–

ions are reduced to Cr

3+

ions.

The following steps are involved in this task.

Step 1: Produce unbalanced equation for the

reaction in ionic form :

Fe

2+

(aq) + Cr

2

O

7

2–

(aq) → Fe

3+

(aq) + Cr

3+

(aq)

(8.50)

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276 CHEMISTRY

Step 2: Separate the equation into half-

reactions:

+2 +3

Oxidation half : Fe

2+

(aq) → Fe

3+

(aq) (8.51)

+6 –2 +3

Reduction half : Cr

2

O

7

2–

(aq) → Cr

3+

(aq)

(8.52)

Step 3: Balance the atoms other than O and

H in each half reaction individually. Here the

oxidation half reaction is already balanced with

respect to Fe atoms. For the reduction half

reaction, we multiply the Cr

3+

by 2 to balance

Cr atoms.

Cr

2

O

7

2–

(aq) → 2 Cr

3+

(aq) (8.53)

Step 4: For reactions occurring in acidic

medium, add H

2

O to balance O atoms and H

+

to balance H atoms.

Thus, we get :

Cr

2

O

7

2–

(aq) + 14H

+

(aq) → 2 Cr

3+

(aq) + 7H

2

O (l)

(8.54)

Step 5: Add electrons to one side of the half

reaction to balance the charges. If need be,

make the number of electrons equal in the two

half reactions by multiplying one or both half

reactions by appropriate number.

The oxidation half reaction is thus rewritten

to balance the charge:

Fe

2+

(aq) → Fe

3+

(aq) + e

–

(8.55)

Now in the reduction half reaction there are

net twelve positive charges on the left hand side

and only six positive charges on the right hand

side. Therefore, we add six electrons on the left

side.

Cr

2

O

7

2–

(aq) + 14H

+

(aq) + 6e

–

→ 2Cr

3+

(aq) +

7H

2

O (l) (8.56)

To equalise the number of electrons in both

the half reactions, we multiply the oxidation

half reaction by 6 and write as :

6Fe

2+

(aq) → 6Fe

3+

(aq) + 6e

–

(8.57)

Step 6: We add the two half reactions to

achieve the overall reaction and cancel the

electrons on each side. This gives the net ionic

equation as :

6Fe

2+

(aq) + Cr

2

O

7

2–

(aq) + 14H

+

(aq) → 6 Fe

3+

(aq) +

2Cr

3+

(aq) + 7H

2

O(l) (8.58)

Step 7: Verify that the equation contains the

same type and number of atoms and the same

charges on both sides of the equation. This last

check reveals that the equation is fully

balanced with respect to number of atoms and

the charges.

For the reaction in a basic medium, first

balance the atoms as is done in acidic medium.

Then for each H

+

ion, add an equal number of

OH

–

ions to both sides of the equation. Where

H

+

and OH

–

appear on the same side of the

equation, combine these to give H

2

O.

Problem 8.10

Permanganate(VII) ion, MnO

4

–

in basic

solution oxidises iodide ion, I

–

to produce

molecular iodine (I

2

) and manganese (IV)

oxide (MnO

2

). Write a balanced ionic

equation to represent this redox reaction.

Solution

Step 1: First we write the skeletal ionic

equation, which is

MnO

4

–

(aq) + I

–

(aq) → MnO

2

(s) + I

2

(s)

Step 2: The two half-reactions are:

–1 0

Oxidation half : I

–

(aq) → I

2

(s)

+7 +4

Reduction half: MnO

4

–

(aq) → MnO

2

(s)

Step 3: To balance the I atoms in the

oxidation half reaction, we rewrite it as:

2I

–

(aq) → I

2

(s)

Step 4: To balance the O atoms in the

reduction half reaction, we add two water

molecules on the right:

MnO

4

–

(aq) → MnO

2

(s) + 2 H

2

O (l)

To balance the H atoms, we add four H

+

ions on the left:

MnO

4

–

(aq) + 4 H

+

(aq) → MnO

2

(s) + 2H

2

O (l)

As the reaction takes place in a basic

solution, therefore, for four H

+

ions, we

add four OH

–

ions to both sides of the

equation:

MnO

4

–

(aq) + 4H

+

(aq) + 4OH

–

(aq)

→

MnO

2

(s) + 2 H

2

O(l) + 4OH

–

(aq)

Replacing the H

+

and OH

–

ions with water,

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277REDOX REACTIONS

the resultant equation is:

MnO

4

–

(aq) + 2H

2

O (l) → MnO

2

(s) + 4 OH

–

(aq)

Step 5 : In this step we balance the charges

of the two half-reactions in the manner

depicted as:

2I

–

(aq) → I

2

(s) + 2e

–

MnO

4

–

(aq) + 2H

2

O(l) + 3e

–

→ MnO

2

(s)

+ 4OH

–

(aq)

Now to equalise the number of electrons,

we multiply the oxidation half-reaction by

3 and the reduction half-reaction by 2.

6I

–

(aq) → 3I

2

(s) + 6e

–

2 MnO

4

–

(aq) + 4H

2

O (l) +6e

–

→ 2MnO

2

(s)

+ 8OH

–

(aq)

Step 6: Add two half-reactions to obtain

the net reactions after cancelling electrons

on both sides.

6I

–

(aq) + 2MnO

4

–

(aq) + 4H

2

O(l) → 3I

2

(s) +

2MnO

2

(s) +8 OH

–

(aq)

Step 7: A final verification shows that the

equation is balanced in respect of the

number of atoms and charges on both

sides.

8.3.3 Redox Reactions as the Basis for

Titrations

In acid-base systems we come across with a

titration method for finding out the strength of

one solution against the other using a pH

sensitive indicator. Similarly, in redox systems,

the titration method can be adopted to

determine the strength of a reductant/oxidant

using a redox sensitive indicator. The usage of

indicators in redox titration is illustrated below:

(i) In one situation, the reagent itself is

intensely coloured, e.g., permanganate ion,

MnO

4

–

. Here MnO

4

–

acts as the self indicator.

The visible end point in this case is

achieved after the last of the reductant (Fe

2+

or C

2

O

4

2–

) is oxidised and the first lasting

tinge of pink colour appears at MnO

4

–

concentration as low as 10

–6

mol dm

–3

(10

–6

mol L

–1

). This ensures a minimal

‘overshoot’ in colour beyond the

equivalence point, the point where the

reductant and the oxidant are equal in

terms of their mole stoichiometry.

(ii) If there is no dramatic auto-colour change

(as with MnO

4

–

titration), there are

indicators which are oxidised immediately

after the last bit of the reactant is

consumed, producing a dramatic colour

change. The best example is afforded by

Cr

2

O

7

2

–

, which is not a self-indicator, but

oxidises the indicator substance

diphenylamine just after the equivalence

point to produce an intense blue colour,

thus signalling the end point.

(iii) There is yet another method which is

interesting and quite common. Its use is

restricted to those reagents which are able

to oxidise I

–

ions, say, for example, Cu(II):

2Cu

2+

(aq) + 4I

–

(aq) → Cu

2

I

2

(s) + I

2

(aq) (8.59)

This method relies on the facts that iodine

itself gives an intense blue colour with starch

and has a very specific reaction with

thiosulphate ions (S

2

O

3

2–

), which too is a redox

reaction:

I

2

(aq) + 2 S

2

O

3

2–

(aq)→2I

–

(aq) + S

4

O

6

2–

(aq) (8.60)

I

2

, though insoluble in water, remains in

solution containing KI as KI

3

.

On addition of starch after the liberation of

iodine from the reaction of Cu

2+

ions on iodide

ions, an intense blue colour appears. This

colour disappears as soon as the iodine is

consumed by the thiosulphate ions. Thus, the

end-point can easily be tracked and the rest

is the stoichiometric calculation only.

8.3.4 Limitations of Concept of Oxidation

Number

As you have observed in the above discussion,

the concept of redox processes has been

evolving with time. This process of evolution

is continuing. In fact, in recent past the

oxidation process is visualised as a decrease

in electron density and reduction process as

an increase in electron density around the

atom(s) involved in the reaction.

8.4 REDOX REACTIONS AND ELECTRODE

PROCESSES

The experiment corresponding to reaction

(8.15), can also be observed if zinc rod is

dipped in copper sulphate solution. The redox

reaction takes place and during the reaction,

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278 CHEMISTRY

zinc is oxidised to zinc ions and copper ions

are reduced to metallic copper due to direct

transfer of electrons from zinc to copper ion.

During this reaction heat is also evolved. Now

we modify the experiment in such a manner

that for the same redox reaction transfer of

electrons takes place indirectly. This

necessitates the separation of zinc metal from

copper sulphate solution. We take copper

sulphate solution in a beaker and put a

copper strip or rod in it. We also take zinc

sulphate solution in another beaker and put

a zinc rod or strip in it. Now reaction takes

place in either of the beakers and at the

interface of the metal and its salt solution in

each beaker both the reduced and oxidized

forms of the same species are present. These

represent the species in the reduction and

oxidation half reactions. A redox couple is

defined as having together the oxidised and

reduced forms of a substance taking part in

an oxidation or reduction half reaction.

This is represented by separating the

oxidised form from the reduced form by a

vertical line or a slash representing an

interface (e.g. solid/solution). For example

in this experiment the two redox couples are

represented as Zn

2+

/Zn and Cu

2+

/Cu. In both

cases, oxidised form is put before the reduced

form. Now we put the beaker containing

copper sulphate solution and the beaker

containing zinc sulphate solution side by side

(Fig. 8.3). We connect solutions in two

beakers by a salt bridge (a U-tube containing

a solution of potassium chloride or

ammonium nitrate usually solidified by

boiling with agar agar and later cooling to a

jelly like substance). This provides an electric

contact between the two solutions without

allowing them to mix with each other. The

zinc and copper rods are connected by a metallic

wire with a provision for an ammeter and a

switch. The set-up as shown in Fig.8.3 is known

as Daniell cell. When the switch is in the off

position, no reaction takes place in either of

the beakers and no current flows through the

metallic wire. As soon as the switch is in the

on position, we make the following

observations:

1. The transfer of electrons now does not take

place directly from Zn to Cu

2+

but through

the metallic wire connecting the two rods

as is apparent from the arrow which

indicates the flow of current.

2. The electricity from solution in one beaker

to solution in the other beaker flows by the

migration of ions through the salt bridge.

We know that the flow of current is possible

only if there is a potential difference

between the copper and zinc rods known

as electrodes here.

The potential associated with each

electrode is known as electrode potential. If

the concentration of each species taking part

in the electrode reaction is unity (if any gas

appears in the electrode reaction, it is confined

to 1 atmospheric pressure) and further the

reaction is carried out at 298K, then the

potential of each electrode is said to be the

Standard Electrode Potential. By

convention, the standard electrode potential

(E



 of hydrogen electrode is 0.00 volts. The

electrode potential value for each electrode

process is a measure of the relative tendency

of the active species in the process to remain

in the oxidised/reduced form. A negative E



means that the redox couple is a stronger

Fig.8.3 The set-up for Daniell cell. Electrons

produced at the anode due to oxidation

of Zn travel through the external circuit

to the cathode where these reduce the

copper ions. The circuit is completed

inside the cell by the migration of ions

through the salt bridge. It may be noted

that the direction of current is opposite

to the direction of electron flow.

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279REDOX REACTIONS

Reaction (Oxidised form + ne

–

→ →

→

Reduced form) E





/ V

F

2

(g) + 2e

–

→ 2F

–

2.87

Co

3+

+ e

–

→ Co

2+

1.81

H

2

O

2

+ 2H

+

+ 2e

–

→ 2H

2

O 1.78

MnO

4

–

+ 8H

+

+ 5e

–

→ Mn

2+

+ 4H

2

O 1.51

Au

3+

+ 3e

–

→ Au(s) 1.40

Cl

2

(g) + 2e

–

→ 2Cl

–

1.36

Cr

2

O

7

2–

+ 14H

+

+ 6e

–

→ 2Cr

3+

+ 7H

2

O 1.33

O

2

(g) + 4H

+

+ 4e

–

→ 2H

2

O 1.23

MnO

2

(s) + 4H

+

+ 2e

–

→ Mn

2+

+ 2H

2

O 1.23

Br

2

+ 2e

–

→ 2Br

–

1.09

NO

3

–

+ 4H

+

+ 3e

–

→ NO(g) + 2H

2

O 0.97

2Hg

2+

+ 2e

–

→ Hg

2

2+

0.92

Ag

+

+ e

–

→ Ag(s) 0.80

Fe

3+

+ e

–

→ Fe

2+

0.77

O

2

(g) + 2H

+

+ 2e

–

→ H

2

O

2

0.68

I

2

(s) + 2e

–

→ 2I

–

0.54

Cu

+

+ e

–

→ Cu(s) 0.52

Cu

2+

+ 2e

–

→ Cu(s) 0.34

AgCl(s) + e

–

→ Ag(s) + Cl

–

0.22

AgBr(s) + e

–

→ Ag(s) + Br

–

0.10

2H

+

+ 2e

–

→ H

2

(g) 0.00

Pb

2+

+ 2e

–

→ Pb(s) –0.13

Sn

2+

+ 2e

–

→ Sn(s) –0.14

Ni

2+

+ 2e

–

→ Ni(s) –0.25

Fe

2+

+ 2e

–

→ Fe(s) –0.44

Cr

3+

+ 3e

–

→ Cr(s) –0.74

Zn

2+

+ 2e

–

→ Zn(s) –0.76

2H

2

O + 2e

–

→ H

2

(g) + 2OH

–

–0.83

Al

3+

+ 3e

–

→ Al(s) –1.66

Mg

2+

+ 2e

–

→ Mg(s) –2.36

Na

+

+ e

–

→ Na(s) –2.71

Ca

2+

+ 2e

–

→ Ca(s) –2.87

K

+

+ e

–

→ K(s) –2.93

Li

+

+ e

–

→ Li(s) –3.05

Increasing strength of oxidising agent

Increasing strength of reducing agent

1. A negative E



means that the redox couple is a stronger reducing agent than the H

+

/H

2

couple.

2. A positive E



means that the redox couple is a weaker reducing agent than the H

+

/H

2

couple.

Table 8.1 The Standard Electrode Potentials at 298 K

Ions are present as aqueous species and H

2

O as liquid; gases and

solids are shown by g and s respectively.

reducing agent than the H

+

/H

2

couple. A

positive E



means that the redox couple is a

weaker reducing agent than the H

+

/H

2

couple.

The standard electrode potentials are very

important and we can get a lot of other useful

information from them. The values of standard

electrode potentials for some selected electrode

processes (reduction reactions) are given in

Table 8.1. You will learn more about electrode

reactions and cells in Class XII.

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280 CHEMISTRY

SUMMARY

Redox reactions form an important class of reactions in which oxidation and reduction

occur simultaneously. Three tier conceptualisation viz, classical, electronic and oxidation

number, which is usually available in the texts, has been presented in detail. Oxidation,

reduction, oxidising agent (oxidant) and reducing agent (reductant) have been viewed

according to each conceptualisation. Oxidation numbers are assigned in accordance

with a consistent set of rules. Oxidation number and ion-electron method both are

useful means in writing equations for the redox reactions. Redox reactions are classified

into four categories: combination, decomposition displacement and disproportionation

reactions. The concept of redox couple and electrode processes is introduced here.

The redox reactions find wide applications in the study of electrode processes and cells.

EXERCISES

8.1 Assign oxidation number to the underlined elements in each of the following

species:

(a) NaH

2

PO

4

(b) NaHSO

4

(c) H

4

P

2

O

7

(d) K

2

MnO

4

(e) CaO

2

(f) NaBH

4

(g) H

2

S

2

O

7

(h) KAl(SO

4

)

2

.12 H

2

O

8.2 What are the oxidation number of the underlined elements in each of the

following and how do you rationalise your results ?

(a) K

I

3

(b) H

2

S

4

O

6

(c) Fe

3

O

4

(d) CH

3

CH

2

OH (e) CH

3

COOH

8.3 Justify that the following reactions are redox reactions:

(a) CuO(s) + H

2

(g) → Cu(s) + H

2

O(g)

(b) Fe

2

O

3

(s) + 3CO(g) → 2Fe(s) + 3CO

2

(g)

(c) 4BCl

3

(g) + 3LiAlH

4

(s) → 2B

2

H

6

(g) + 3LiCl(s) + 3 AlCl

3

(s)

(d) 2K(s) + F

2

(g) → 2K

+

F

–

(s)

(e) 4 NH

3

(g) + 5 O

2

(g) → 4NO(g) + 6H

2

O(g)

8.4 Fluorine reacts with ice and results in the change:

H

2

O(s) + F

2

(g) → HF(g) + HOF(g)

Justify that this reaction is a redox reaction.

8.5 Calculate the oxidation number of sulphur, chromium and nitrogen in H

2

SO

5

,

Cr

2

O

7

2–

and NO

3

–

. Suggest structure of these compounds. Count for the fallacy.

8.6 Write formulas for the following compounds:

(a) Mercury(II) chloride (b) Nickel(II) sulphate

(c) Tin(IV) oxide (d) Thallium(I) sulphate

(e) Iron(III) sulphate (f) Chromium(III) oxide

8.7 Suggest a list of the substances where carbon can exhibit oxidation states from

–4 to +4 and nitrogen from –3 to +5.

8.8 While sulphur dioxide and hydrogen peroxide can act as oxidising as well as

reducing agents in their reactions, ozone and nitric acid act only as oxidants.

Why ?

8.9 Consider the reactions:

(a) 6 CO

2

(g) + 6H

2

O(l) → C

6

H

12

O

6

(aq) + 6O

2

(g)

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281REDOX REACTIONS

(b) O

3

(g) + H

2

O

2

(l) → H

2

O(l) + 2O

2

(g)

Why it is more appropriate to write these reactions as :

(a) 6CO

2

(g) + 12H

2

O(l) → C

6

H

12

O

6

(aq) + 6H

2

O(l) + 6O

2

(g)

(b) O

3

(g) + H

2

O

2

(l) → H

2

O(l) + O

2

(g) + O

2

(g)

Also suggest a technique to investigate the path of the above (a) and (b) redox

reactions.

8.10 The compound AgF

2

is unstable compound. However, if formed, the compound

acts as a very strong oxidising agent. Why ?

8.11 Whenever a reaction between an oxidising agent and a reducing agent is carried

out, a compound of lower oxidation state is formed if the reducing agent is in

excess and a compound of higher oxidation state is formed if the oxidising agent

is in excess. Justify this statement giving three illustrations.

8.12 How do you count for the following observations ?

(a) Though alkaline potassium permanganate and acidic potassium

permanganate both are used as oxidants, yet in the manufacture of benzoic

acid from toluene we use alcoholic potassium permanganate as an oxidant.

Why ? Write a balanced redox equation for the reaction.

(b) When concentrated sulphuric acid is added to an inorganic mixture

containing chloride, we get colourless pungent smelling gas HCl, but if the

mixture contains bromide then we get red vapour of bromine. Why ?

8.13 Identify the substance oxidised reduced, oxidising agent and reducing agent for

each of the following reactions:

(a) 2AgBr (s) + C

6

H

6

O

2

(aq) → 2Ag(s) + 2HBr (aq) + C

6

H

4

O

2

(aq)

(b) HCHO(l) + 2[Ag (NH

3

)

2

]

+

(aq) + 3OH

–

(aq) → 2Ag(s) + HCOO

–

(aq) + 4NH

3

(aq)

+ 2H

2

O(l)

(c) HCHO (l) + 2 Cu

2+

(aq) + 5 OH

–

(aq) → Cu

2

O(s) + HCOO

–

(aq) + 3H

2

O(l)

(d) N

2

H

4

(l) + 2H

2

O

2

(l) → N

2

(g) + 4H

2

O(l)

(e) Pb(s) + PbO

2

(s) + 2H

2

SO

4

(aq) → 2PbSO

4

(s) + 2H

2

O(l)

8.14 Consider the reactions :

2 S

2

O

3

2–

(aq) + I

2

(s) → S

4

O

6

2–

(aq) + 2I

–

(aq)

S

2

O

3

2–

(aq) + 2Br

2

(l) + 5 H

2

O(l) → 2SO

4

2–

(aq) + 4Br

–

(aq) + 10H

+

(aq)

Why does the same reductant, thiosulphate react differently with iodine and

bromine ?

8.15 Justify giving reactions that among halogens, fluorine is the best oxidant and

among hydrohalic compounds, hydroiodic acid is the best reductant.

8.16 Why does the following reaction occur ?

XeO

6

4–

(aq) + 2F

–

(aq) + 6H

+

(aq) → XeO

3

(g)+ F

2

(g) + 3H

2

O(l)

What conclusion about the compound Na

4

XeO

6

(of which XeO

6

4–

is a part) can be

drawn from the reaction.

8.17 Consider the reactions:

(a) H

3

PO

2

(aq) + 4 AgNO

3

(aq) + 2 H

2

O(l) → H

3

PO

4

(aq) + 4Ag(s) + 4HNO

3

(aq)

(b) H

3

PO

2

(aq) + 2CuSO

4

(aq) + 2 H

2

O(l) → H

3

PO

4

(aq) + 2Cu(s) + H

2

SO

4

(aq)

(c) C

6

H

5

CHO(l) + 2[Ag (NH

3

)

2

]

+

(aq) + 3OH

–

(aq) → C

6

H

5

COO

–

(aq) + 2Ag(s) +

4NH

3

(aq) + 2 H

2

O(l)

(d) C

6

H

5

CHO(l) + 2Cu

2+

(aq) + 5OH

–

(aq) → No change observed.

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282 CHEMISTRY

What inference do you draw about the behaviour of Ag

+

and Cu

2+

from these

reactions ?

8.18 Balance the following redox reactions by ion – electron method :

(a) MnO

4

–

(aq) + I

–

(aq) → MnO

2

(s) + I

2

(s) (in basic medium)

(b) MnO

4

–

(aq) + SO

2

(g) → Mn

2+

(aq) + HSO

4

–

(aq) (in acidic solution)

(c) H

2

O

2

(aq) + Fe

2+

(aq) → Fe

3+

(aq) + H

2

O (l) (in acidic solution)

(d) Cr

2

O

7

2–

+ SO

2

(g) → Cr

3+

(aq) + SO

4

2–

(aq) (in acidic solution)

8.19 Balance the following equations in basic medium by ion-electron method and

oxidation number methods and identify the oxidising agent and the reducing

agent.

(a) P

4

(s) + OH

–

(aq) → PH

3

(g) + HPO

2

–

(aq)

(b) N

2

H

4

(l) + ClO

3

–

(aq) → NO(g) + Cl

–

(g)

(c) Cl

2

O

7

(g) + H

2

O

2

(aq) → ClO

2

–

(aq) + O

2

(g) + H

+

8.20 What sorts of informations can you draw from the following reaction ?

(CN)

2

(g) + 2OH

–

(aq) → CN

–

(aq) + CNO

–

(aq) + H

2

O(l)

8.21 The Mn

3+

ion is unstable in solution and undergoes disproportionation to give

Mn

2+

, MnO

2

, and H

+

ion. Write a balanced ionic equation for the reaction.

8.22 Consider the elements :

Cs, Ne, I and F

(a) Identify the element that exhibits only negative oxidation state.

(b) Identify the element that exhibits only postive oxidation state.

(c) Identify the element that exhibits both positive and negative oxidation states.

(d) Identify the element which exhibits neither the negative nor does the positive

oxidation state.

8.23 Chlorine is used to purify drinking water. Excess of chlorine is harmful. The

excess of chlorine is removed by treating with sulphur dioxide. Present a balanced

equation for this redox change taking place in water.

8.24 Refer to the periodic table given in your book and now answer the following

questions:

(a) Select the possible non metals that can show disproportionation reaction.

(b) Select three metals that can show disproportionation reaction.

8.25 In Ostwald’s process for the manufacture of nitric acid, the first step involves

the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam.

What is the maximum weight of nitric oxide that can be obtained starting only

with 10.00 g. of ammonia and 20.00 g of oxygen ?

8.26 Using the standard electrode potentials given in the Table 8.1, predict if the

reaction between the following is feasible:

(a) Fe

3+

(aq) and I

–

(aq)

(b) Ag

+

(aq) and Cu(s)

(c) Fe

3+

(aq) and Cu(s)

(d) Ag(s) and Fe

3+

(aq)

(e) Br

2

(aq) and Fe

2+

(aq).

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283REDOX REACTIONS

8.27 Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO

3

with silver electrodes

(ii) An aqueous solution AgNO

3

with platinum electrodes

(iii) A dilute solution of H

2

SO

4

with platinum electrodes

(iv) An aqueous solution of CuCl

2

with platinum electrodes.

8.28 Arrange the following metals in the order in which they displace each other

from the solution of their salts.

Al, Cu, Fe, Mg and Zn.

8.29 Given the standard electrode potentials,

K

+

/K = –2.93V, Ag

+

/Ag = 0.80V,

Hg

2+

/Hg = 0.79V

Mg

2+

/Mg = –2.37V. Cr

3+

/Cr = –0.74V

arrange these metals in their increasing order of reducing power.

8.30 Depict the galvanic cell in which the reaction Zn(s) + 2Ag

+

(aq) → Zn

2+

(aq) +2Ag(s)

takes place, Further show:

(i) which of the electrode is negatively charged,

(ii) the carriers of the current in the cell, and

(iii) individual reaction at each electrode.

2019-20