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Study notes on REDOX CHEMISTRY, Study notes of Chemistry

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Redox Reactions 541
Chemical reactions involve transfer of electrons from one
chemical substance to another. These electron transfer
reactions are termed as oxidation-reduction or redox-
reactions.
Molecular and Ionic equations
(1) Molecular equations : When the reactants and
products involved in a chemical change are written in molecular
forms in the chemical equation, it is termed as molecular
equation.
Example :
2222 24 ClOHMnClHClMnO
In above example the reactants and products have been
written in molecular forms, thus the equation is termed as
molecular equation.
(2) Ionic equations : When the reactants and products
involved in a chemical change are ionic compounds, these will
be present in the form of ions in the solution. The chemical
change is written in ionic forms in chemical equation, it is
termed as ionic equation. Example,
ClHMnO 44
2
22
222 ClOHClMn
In above example the reactants and products have been
written in ionic forms, thus the equation is termed as ionic
equation.
(3) Spectator ions : In ionic equations, the ions which
do not undergo any change and equal in number in both
reactants and products are termed as spectator ions and are not
included in the final balanced equations. Example,
ClHZn 22
ClHZn 2
2
2
(Ionic
equation)
HZn 2
2
2HZn
(Final ionic
equation)
In above example, the
Cl
ions are the spectator ions
and hence are not included in the final ionic balanced equation.
Oxidation-reduction and Redox reactions
(1) Oxidation : Oxidation is a process which involves;
addition of oxygen, removal of hydrogen, addition of non-metal,
removal of metal, Increase in +ve valency, loss of electrons and
increase in oxidation number.
(i) Addition of oxygen : 2Mg + O2
2MgO
(ii) Removal of hydrogen : H2S+Cl2
2HCl + S
(iii) Addition of Non-metal : Fe + S
FeS
(iv) Removal of metal : 2KI+H2O2
2KOH+I2
(v) Increase in +ve valenc y :
eFeFe 32
(vi) Loss of electrons (also known as de-electronation)
4
M
3
M
2
M
1
M
0
M
1
M
2
M
3
M
4
M
e e e e e e e e
Loss of electrons
(a)
eHH 0
(Formation of proton)
(b)
eMnOMnO 4
2
4
(De-electronation of
2
4
MnO
)
(c)
eFeFe 622 30
(De-electronation of iron)
(vii) Increase in oxidation number
(a)
20 MgMg
(From 0 to +2)
(b)
3
6
3
4
6
2)()( CNFeCNFe
(From +2 to +3)
Redox Reactions
Chapter
13
pf3
pf4
pf5
pf8
pf9
pfa

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Chemical reactions involve transfer of electrons from one

chemical substance to another. These electron – transfer

reactions are termed as oxidation-reduction or redox-

reactions.

Molecular and Ionic equations

(1) Molecular equations : When the reactants and

products involved in a chemical change are written in molecular

forms in the chemical equation, it is termed as molecular

equation.

Example :

2 2 2 2

MnO  4 HClMnCl  2 HOCl

In above example the reactants and products have been

written in molecular forms, thus the equation is termed as

molecular equation****.

(2) Ionic equations : When the reactants and products

involved in a chemical change are ionic compounds, these will

be present in the form of ions in the solution. The chemical

change is written in ionic forms in chemical equation, it is

termed as ionic equation. Example ,

 

MnO  4 H  4 Cl

2

2 2

2

Mn  2 Cl  2 HOCl

 

In above example the reactants and products have been

written in ionic forms, thus the equation is termed as ionic

equation****.

(3) Spectator ions : In ionic equations, the ions which

do not undergo any change and equal in number in both

reactants and products are termed as spectator ions and are not

included in the final balanced equations. Example,

 

Zn  2 H  2 Cl

 

ZnH  2 Cl

2

2

( Ionic

equation )

Zn  2 H

2

2

ZnH

( Final ionic

equation )

In above example, the

Cl ions are the spectator ions

and hence are not included in the final ionic balanced equation.

Oxidation-reduction and Redox reactions

(1) Oxidation : Oxidation is a process which involves;

addition of oxygen, removal of hydrogen, addition of non-metal,

removal of metal, Increase in + ve valency, loss of electrons and

increase in oxidation number.

(i) Addition of oxygen : 2 Mg + O 2 

2 MgO

(ii) Removal of hydrogen : H 2 S+Cl 2  2 HCl + S

(iii) Addition of Non-metal : Fe + SFeS

(iv) Removal of metal : 2 KI+H 2 O 2  2 KOH+I 2

(v) Increase in +ve valency :

  

FeFee

2 3

(vi) Loss of electrons (also known as de-electronation)

 4

M

 3

M

 2

M

 1

M

0

M

 1

M

 2

M

 3

M

 4

M

  • e

     - e - - e - - e - - e - - e - - e - - e - 

Loss of electrons

(a)

 

HHe

0

( Formation of proton )

(b)

  

MnOMnOe

4

2

4

( De-electronation of

2 

4

MnO )

(c)

 

2 Fe  2 Fe  6 e

0 3

( De-electronation of iron )

(vii) Increase in oxidation number

(a)

0 2

Mg Mg ( From 0 to +2)

(b)   

3

6

3

4

6

2

Fe ( CN ) Fe ( CN ) (From +2 to +3)

Redox Reactions

Chapter

(c)

0

2

2 ClCl

(From – 1 to 0)

(2) Reduction : Reduction is just reverse of oxidation.

Reduction is a process which involves; removal of oxygen,

addition of hydrogen, removal of non-metal, addition of metal,

decrease in + ve valency, gain of electrons and decrease in

oxidation number.

(i) Removal of oxygen : CuOCCuCO

(ii) Addition of hydrogen : Cl H 2 HCl

2 2

(iii) Removal of non-metal

2 2 2 2 4

2 HgClSnClHgClSnCl

(iv) Addition of metal :

2 2 2

HgClHgHgCl

(v) Decrease in +ve valency

(a)

 

3 2

Fe Fe (+ ve valency decreases )

(b)

 

4

6

3

6

[ Fe ( CN )] [ Fe ( CN )] (– ve valency increase s)

(vi) Gain of electrons (also known as electronation)

 4

M

 3

M

 2

M

 1

M

0

M

 1

M

 2

M

 3

M

 4

M

+e

+e

+e

+e

+e

+e

+e

+e

Gain of electrons

(a) ( ) 2 ()

2

Zn aqeZnS

 

( Electronation of

2 

Zn )

(b)

2 0

Pb  2 ePb

 

( Electronation of

2 

Pb )

(c)

  

 

4

6

3

6

[ Fe ( CN )] e [ Fe ( CN )]

( Electronation of

3 

6

[ Fe ( CN )] )

(vii) Decrease in oxidation number

(a)

2 0

MgMg

( From +2 to 0)

(b)   

 

4

6

3

6

Fe ( CN ) Fe ( CN ) ( From +3 to +2)

(c)

Cl  2 Cl

0

2

( From 0 to – 1)

(3) Redox-reactions

(i) An overall reaction in which oxidation and reduction

takes place simultaneously is called redox or oxidation-

reduction reaction****. These reactions involve transfer of

electrons from one atom to another. Thus every redox reaction

is made up of two half reactions ; One half reaction represents

the oxidation and the other half reaction represents the

reduction.

(ii) Types of redox reaction

(a) Direct redox reaction : The reactions in which

oxidation and reduction takes place in the same vessel are called

direct redox reactions.

(b) Indirect redox reaction : The reactions in which

oxidation and reduction takes place in different vessels are

called indirect redox reactions. Indirect redox reactions are the

basis of electro-chemical cells.

(c) Intermolecular redox reactions : In which one

substance is oxidised while the other is reduced.

For example, 2 Al FeO AlO 2 Fe

2 3 2 3

Here, Al is oxidised to

2 3

AlO while

2 3

FeO is reduced to

Fe.

(d) Intramolecular redox reactions : In which one

element of a compound is oxidised while the other is reduced.

For example,

3 2

2 KClO  2 KCl  3 O

Here,

 5

Cl in

3

KClO

is reduced to

 1

Cl in KCl while

2 

O in

3

KClO

is oxidised to

0

2

O.

Oxidising and Reducing agents

(1) Definition : The substance (atom, ion or molecule)

that gains electrons and is thereby reduced to a low valency state

is called an oxidising agent , while the substance that loses

electrons and is thereby oxidised to a higher valency state is

called a reducing agent****.

Or

An oxidising agent is a substance, the oxidation

number of whose atom or atoms decreases while a reducing

agent is a substance the oxidation number of whose atom

increases.

(2) Important oxidising agents

(i) Molecules made up of electronegative elements.

Example : O 2 , O 3 and X 2 (halogens).

(ii) Compounds containing an element which is in the

highest oxidation state.

Example : , ,

4 2 2 7

KMnO KCrO

2 2 7

NaCrO, ,

3 , 2 4

CrO HSO

3 3 3 2 4 3 2 2 2

HNO , NaNO , FeCl , HgCl , KClO , SO , CO , HO etc.

(iii) Oxides of elements, , , , , ,

3 2 4 10

MgOCuOCrO CO PO etc.

(iv) Fluorine is the strongest oxidising agent.

(3) Important reducing agents

(i) All metals e.g. Na, Zn, Fe, Al, etc.

(ii) A few non-metals e.g. C, H 2

, S etc.

(iii) Hydracids : HCl, HBr, HI, H 2 S etc.

(iv) A few compounds containing anelement in the lower

oxidation state (ous).

Example : , ,

2 4

FeCl FeSO SnCl HgCl CuO

2 2 2 2

, , etc.

(v) Metallic hydrides e.g. NaH, LiH etc.

(vi) Organic compounds like HCOOH and ( COOH ) 2

and

their salts, aldehydes, alkanes etc.

(vii) Lithium is the strongest reducing agent in solution.

(viii) Cesium is the strongest reducing agent in absence of

water. Other reducing agents are

2 2 3

NaSO and KI.

(ix) Hypo prefix indicates that central atom of compound

has the minimum oxidation state so it will act as a reducing

agent.

Example :

3 2

HPO (hypophosphorous acid).

(4) Substances which act as oxidising as well as

reducing agents

Examples : H 2 O 2 , SO 2 , H 2 SO 3 , HNO 2 , NaNO 2 , Na 2 SO 3 , O 3

etc_._

(5) Tips for the identification of oxidising and

reducing agents

(i) If an element is in its highest possible oxidation state

in a compound, the compound can function as an oxidising

agent.

Example :

4 2 2 7 3 2 4 4

KMnO , KCrO , HNO , HSO , HClO etc.

V 2 O 5 Vanadium (V)

oxide

CuO Copper (II) oxide

SnO 2 Tin (IV) oxide FeCl 3 Iron (III) chloride

(4) Rules for the determination of oxidation

number of an atom : The following rules are followed in

ascertaining the oxidation number of an atom,

(i) If there is a covalent bond between two same atoms

then oxidation numbers of these two atoms will be zero. Bonded

electrons are symmetrically distributed between two atoms.

Bonded atoms do not acquire any charge. So oxidation numbers

of these two atoms are zero.

A : A or AAA

  • A

For example, Oxidation number of Cl in Cl 2

, O in O 2

and

N in N 2

is zero.

(ii) If covalent bond is between two different atoms then

electrons are counted towards more electronegative atom. Thus

oxidation number of more electronegative atom is negative and

oxidation number of less electronegative atom is positive. Total

number of charges on any element depends on number of

bonds.

A – B

A

  • B

:

A – B  A

  • : B

  • 2

:

The oxidation number of less electronegative element (A)

is + 1 and + 2 respectively.

(iii) If there is a coordinate bond between two atoms then

oxidation number of donor atom will be + 2 and of acceptor

atom will be – 2.

A B

A

2+

  • :B
  • 2

:

(iv) The oxidation number of all the atoms of different

elements in their respective elementary states is taken to be

zero. For example, in N , Cl , H , P , S , O , Br , Na , Fe , Ag 2 2 2 4 8 2 2

etc. the oxidation number of each atom is zero.

(v) The oxidation number of a monoatomic ion is the

same as the charge on it. For example, oxidation numbers of

 2 

Na , Mg and

3 

Al ions are + 1, + 2 and + 3 respectively while

those of

 2 

Cl , S and

3 

N ions are – 1, – 2 and – 3 respectively.

(vi) The oxidation number of hydrogen is + 1 when

combined with non-metals and is – 1 when combined with active

metals called metal hydrides such as LiH, KH, MgH 2 , CaH 2 etc.

(vii) The oxidation number of oxygen is – 2 in most of its

compounds, except in peroxides like

2 2 2

HO , BaO etc. where it

is – 1. Another interesting exception is found in the compound

OF

2

( oxygen difluoride ) where the oxidation number of oxygen

is + 2. This is due to the fact that fluorine being the most

electronegative element known has always an oxidation number

of – 1.

(viii) In compounds formed by union of metals with non-

metals, the metal atoms will have positive oxidation numbers

and the non-metals will have negative oxidation numbers.

For example,

(a) The oxidation number of alkali metals ( Li, Na, K etc.)

is always +1 and those of alkaline earth metals ( Be, Mg, Ca etc) is +

(b) The oxidation number of halogens ( F, Cl, Br, I ) is

always – 1 in metal halides such as KF, AlCl 3 , MgBr 2 , CdI 2. etc.

(ix) In compounds formed by the union of different

elements, the more electronegative atom will have negative

oxidation number whereas the less electronegative atom will have

positive oxidation number.

For example,

(a) N is given an oxidation number of – 3 when it is

bonded to less electronegative atom as in NH 3

and NI 3

, but is

given an oxidation number of + 3 when it is bonded to more

electronegative atoms as in NCl 3.

(b) Since fluorine is the most electronegative element

known so its oxidation number is always – 1 in its compounds

i.e. oxides, interhalogen compounds etc.

(c) In interhalogen compounds of Cl, Br, and I; the more

electronegative of the two halogens gets the oxidation number

of – 1. For example, in BrCl 3 , the oxidation number of Cl is – 1

while that of Br is +3.

(x) For neutral molecule, the sum of the oxidation

numbers of all the atoms is equal to zero. For example, in NH 3

the sum of the oxidation numbers of nitrogen atom and 3

hydrogen atoms is equal to zero. For a complex ion, the sum of

the oxidation numbers of all the atoms is equal to charge on the

ion. For example, in

2 

4

SO ion, the sum of the oxidation

numbers of sulphur atom and 4 oxygen atoms must be equal to

(xi) It may be noted that oxidation number is also

frequently called as oxidation state. For example, in H 2

O , the

oxidation state of hydrogen is +1 and the oxidation state of

oxygen is – 2. This means that oxidation number gives the

oxidation state of an element in a compound.

(xii) In the case of representative elements, the highest

oxidation number of an element is the same as its group number

while highest negative oxidation number is equal to (8 – Group

number) with negative sign with a few exceptions. The most

common oxidation states of the representative elements are

shown in the following table,

Group Outer shell

configuration

Common oxidation numbers

(states)

except zero in free state

I A ns

1

II A ns

2

III A ns

2

np

1

+3, +

IV A ns

2

np

2

+4,+3,+2,+1, – 1, – 2, – 3, – 4

V A ns

2

np

3

+5,+3,+1, – 1, – 3

VI A ns

2

np

4

+6,+4,+2,– 2

VII A ns

2

np

5

+7,+5,+3, +1, – 1

(xiii) Transition metals exhibit a large number of

oxidation states due to involvement of ( n – 1 ) d electron

besides ns electron.

(xiv) Oxidation number of a metal in carbonyl complex is

always zero.

Example : Ni has zero oxidation state in   

4

NiCO.

(xv) Those compounds which have only C , H and O the

oxidation number of carbon can be calculated by following

formula,

C

O H

n

n n

C

Oxidation numberof' '

Where,

O

n is the number of oxygen atom,

H

n is the number

of hydrogen atom,

C

n is the number of carbon atom.

For example, (a)

3

CHOH ;  4 ,  1 ,  1

H C O

n n n

Oxidation number of ‘ C ’ = 2

1

( 1 2 4 )



 

(b)

HCOOH ;  2 ,

H

n  2 ,  1

O c

n n

Oxidation number of carbon = 2

1

( 2 2 2 )



 

(5) Procedure for calculation of oxidation number

: By applying the above rules, we can calculate the oxidation

numbers of elements in the molecules/ions by the following

steps.

(i) Write down the formula of the given molecule/ion

leaving some space between the atoms.

(ii) Write oxidation number on the top of each atom. In

case of the atom whose oxidation number has to be calculated

write x.

(iii) Beneath the formula, write down the total oxidation

numbers of each element. For this purpose, multiply the

oxidation numbers of each atom with the number of atoms of

that kind in the molecule/ion. Write the product in a bracket.

(iv) Equate the sum of the oxidation numbers to zero for

neutral molecule and equal to charge on the ion.

(v) Solve for the value of x.

Table : 13.2 Oxidation number of some elements in compounds, ions or chemical species

Element Oxidation

Number

Compounds, ions or chemical species

Sulphur (S) – 2 H 2 S , Zn S , NaH S , ( Sn S 3 )

2 –

, Ba S , C S 2

0 S , S 4 , S 8 , S CN

-

  • 1 S 2 F 2 , S 2 Cl 2

+ 4 S O

2

, H

2

S O

3

, ( S O

3

2 –

, S OCl 2

, NaH S O 3

, Ca [ H S O 3

]

2

, [ H S O

3

]

,

S F

4

+ 6 H

2

S O

4

, ( S O

4

2 -

, [ H S O 4

]

,

Ba S O 4,

KH S O

4,

S O

3

, S F

6

, H

2

S

2

O

7

, ( S

2

O

7

2 –

Nitrogen ( N ) – 3 N H 3 , ( N H 4 )

, Al N, Mg 3 N 2 , ( N )

3 –

, Ca 3 N 2 , C N

-

– 2 N 2 H 4 , ( N 2 H 5 )

– 1 N H 2 OH

  • 1/3 N aN 3

, N

3

H

0 N

2

+ 1 N 2 O

+ 2 N O

+ 3 H N O

2

, ( N O

2

, NaN O 2

, N

2

O

3

, N F

3

+ 4 N O

2

+ 5 H N O 3 , ( N O 3 )

, K N O 3 , N 2 O 5

Chlorine ( Cl ) – 1 H Cl , Na Cl , Ca Cl 2 , Al Cl 3 , I Cl , I Cl 5 , SO Cl 2 , CrO 2 Cl 2 , K Cl , K 2 Pt Cl 6 , HAu Cl 4 , C Cl 4

0 Cl, Cl 2

  • 1 HO Cl , NaO Cl , ( O Cl )

, Cl 2 O

  • 3 K Cl O 2

, ( Cl O 2

, H Cl O 2

  • 4 Cl O 2

  • 5 ( Cl O 3

, K Cl O 3

, Na Cl O 3

, H Cl O 3

  • 7 H Cl O 4 , Cl 2 O 7 , K Cl O 4 , ( Cl O 4 )

Xenon ( Xe ) + 6 Xe O 3

, Xe F 6

(6) Exceptional cases of evaluation of oxidation

numbers : The rules described earlier are usually helpful in

determination of the oxidation number of a specific atom in

simple molecules but these rules fail in following cases. In these

cases, the oxidation numbers are evaluated using the concepts of

chemical bonding involved.

Type I****. In molecules containing peroxide linkage in

addition to element-oxygen bonds. For example,

(i) Oxidation number of S in H 2

SO

5

(Permonosulphuric acid or Caro's acid)

By usual method;

2 5

HSO

2  1  x  5 ( 2 ) 0 or x  8

But this cannot be true as maximum oxidation number

for S cannot exceed + 6. Since S has only 6 electrons in its

valence shell. This exceptional value is due to the fact that two

oxygen atoms in

2 5

HSO

shows peroxide linkage as shown

below,

O

H O S O O H

O

Therefore the evaluation of o.n. of sulphur here should be

made as follows,

2 × (+1) + x + 3 × (–2) + 2 × (–1)

(for H) (for S) (for O) (for O–O)

or 2 + x – 6 – 2 = 0 or x = + 6.

(ii) Oxidation number of S in H 2 S 2 O 8

(Peroxidisulphuric acid or Marshall's acid)

By usual method ;

2 2 8

HSO

1 × 2 + 2x + 8 (–2) = 0

2x = + 16 – 2 = 14 or x = + 7

Similarly Caro's acid, Marshall's acid also has a peroxide

linkage so that in which S shows +6 oxidation state.

O O

H O S O O S O H

O O

Therefore the evaluation of oxidation state of sulphur

should be made as follow,

2 × (+1) + 2 × (x) + 6 × (–2) + 2 × (–1) = 0

(for H) (for S) (for O) (for O–O)

or 2 + 2x – 12 – 2 = 0 or x = + 6.

(iii) Oxidation number of Cr in CrO 5

(Blue perchromate)

By usual method

5

CrO ; x – 10 = 0 or x = + 10

This cannot be true as maximum O. N. of Cr cannot be

more than + 6. Since Cr has only five electrons in 3d orbitals

and one electron in 4s orbital. This exceptional value is due to

the fact that four oxygen atoms in CrO 5 are in peroxide linkage.

The chemical structure of CrO 5

is

O

O O

Cr

O O

Therefore, the evaluation of o.n. of Cr should be made as

follows

x + 1 × (– 2) + 4 (–1) = 0

(for Cr ) (for O ) (for O–O )

or x – 2 – 4 = 0 or x = + 6.

Type II****. In molecules containing covalent and

coordinate bonds, following rules are used for evaluating the

oxidation numbers of atoms.

(i) For each covalent bond between dissimilar atoms the

less electronegative element is assigned the oxidation number of

  • 1 while the atom of the more electronegative element is

assigned the oxidation number of – 1.

(ii) In case of a coordinate-covalent bond between similar

or dissimilar atoms but the donor atom is less electronegative

than the acceptor atom, an oxidation number of +2 is assigned

to the donor atom and an oxidation number of – 2 is assigned to

the acceptor atom.

Conversely, if the donor atom is more electronegative

than the acceptor atom, the contribution of the coordinate bond

is neglected. Examples,

(a) Oxidation number of C in HCN and HN

 C

The evaluation of oxidation number of C cannot be made

directly by usual rules since no standard rule exists for oxidation

numbers of N and C.

In such cases, evaluation of oxidation number should be

made using indirect concept or by the original concepts of

chemical bonding.

(b) Oxidation number of carbon in H – N

 C

The contribution of coordinate bond is neglected since the

bond is directed from a more electronegative N atom (donor) to

a less electronegative carbon atom (acceptor).

Therefore the oxidation number of N in HN C

 remains

  • 3 as it has three covalent bonds.

1 × (+ 1) + 1 × (– 3) + x = 0

(for H ) (for N ) (for C )

or 1 + x – 3 = 0 or x = + 2.

(c) Oxidation number of carbon in HCN

In HCN , N is more electronegative than carbon,

each bond gives an oxidation number of – 1 to N. There are

three covalent bonds, the oxidation number of N in

HCN is taken as – 3

Now HCN

+1 + x – 3 = 0  x = + 2

Peroxide linkage

Peroxide linkage

Peroxide linkage

Peroxide linkage

Type III. In a molecule containing two or more atoms

of same or different elements in different oxidation states.

(i) Oxidation number of S in Na 2 S 2 O 3

By usual method

2 2 3

NaSO

 2 × (+1) + 2 × x + 3 (–2) = 0 or 2 + 2 x – 6 = 0

or x = 2.

But this is unacceptable as the two sulphur atoms in

Na 2

S

2

O

3

cannot have the same oxidation number because on

treatment with dil. H 2 SO 4 , one sulphur atom is precipitated

while the other is oxidised to SO 2.

Na SO HSO NaSO SO S HO

2 2 3 2 4 2 4 2 2

In this case, the oxidation number of sulphur is evaluated

from concepts of chemical bonding. The chemical structure of

Na 2 S 2 O 3 is

O

Na O S O Na

S

   

Due to the presence of a co-ordinate bond between two

sulphur atoms, the acceptor sulphur atom has oxidation number

of – 2 whereas the other S atom gets oxidation number of + 2.

2 × (+1) + 3 × (–2) + x × 1 + 1 × (– 2) = 0

(for Na ) (for O ) (for S ) (for coordinated S )

or + 2 – 6 + x – 2 = 0 or x = + 6

Thus two sulphur atoms in Na 2

S

2

O

3

have oxidation

number of – 2 and +6.

(ii) Oxidation number of chlorine in CaOCl 2

(bleaching powder)

In bleaching powder, Ca ( OCl ) Cl , the two Cl atoms

are in different oxidation states i.e. , one Cl

-

having

oxidation number of – 1 and the other as OCl

having

oxidation number of +1.

(iii) Oxidation number of N in NH 4 NO 3

By usual method N 2 H 4 O 3 ; 2 x + 4 × (+1) + 3 × (–1) = 0

2 x + 4 – 3 = 0 or 2 x = + 1 (wrong)

No doubt NH 4

NO

3

has two nitrogen atoms but one N has

negative oxidation number (attached to H ) and the other has

positive oxidation number (attached to O ). Hence the evaluation

should be made separately for

4

NH and

3

NO

4

NH x + 4 × (+1) = +1 or x = – 3

3

NO x + 3 (– 2) = – 1 or x = + 5.

(iv) Oxidation number of Fe in Fe 3

O

4

In Fe 3 O 4 , Fe atoms are in two different oxidation states.

Fe 3

O

4

can be considered as an equimolar mixture of FeO [iron

(II) oxide] and Fe 2 O 3 [iron (III) oxide]. Thus in one molecule of

Fe 3

O

4

, two Fe atoms are in + 3 oxidation state and one Fe atom

is in + 2 oxidation state.

(v) Oxidation number of S in sodium

tetrathionate ( Na 2

S

4

O

6

Its structure can be represented as follows,

O O

Na O S S S S ONa

O O

 

The two S - atoms which are linked to each other have

oxidation number zero. The oxidation number of other S - atoms

can be calculated as follows

Let oxidation number of S = x.

 2 × x + 2 × 0 + 6 × ( – 2) = – 2

(for S ) (for S–S ) (for O )

x = + 5.

Balancing of oxidation-reduction reactions

Though there are a number of methods for balancing

oxidation – reduction reactions, two methods are very

important. These are,

(1) Oxidation number method

(2) Ion – electron method

(1) Oxidation number method : The method for

balancing redox reactions by oxidation number change method

was developed by Johnson****. In a balanced redox reaction, total

increase in oxidation number must be equal to the total decrease

in oxidation number. This equivalence provides the basis for

balancing redox reactions. This method is applicable to both

molecular and ionic equations. The general procedure involves

the following steps,

(i) Write the skeleton equation (if not given, frame it)

representing the chemical change.

(ii) Assign oxidation numbers to the atoms in the

equation and find out which atoms are undergoing oxidation

and reduction. Write separate equations for the atoms

undergoing oxidation and reduction.

(iii) Find the change in oxidation number in each

equation. Make the change equal in both the equations by

multiplying with suitable integers. Add both the equations.

(iv) Complete the balancing by inspection. First balance

those substances which have undergone change in oxidation

number and then other atoms except hydrogen and oxygen.

Finally balance hydrogen and oxygen by putting H 2 O molecules

wherever needed.

The final balanced equation should be checked to ensure

that there are as many atoms of each element on the right as

there are on the left.

(v) In ionic equations the net charges on both sides of the

equation must be exactly the same. Use H

ion/ions in acidic

reactions and OH

ion/ions in basic reactions to balance the

charge and number of hydrogen and oxygen atoms.

The following example illustrate the above rules,

Step : I Cu HNO CuNO NO HO

3 32 2 2

( Skeleton equation )

to lower state of oxidation. Such a reaction, in which a substance

undergoes simultaneous oxidation and reduction is called

disproportionation and the substance is said to

disproportionate****.

Following are the some examples of disproportionation ,

2

0

2

2

1

2 2 2

H O  HO  HO  O

1

4

7

3

5

  

KClOKClOKCl

3

3

2

1

2 2

0

 

PNaOHHONaH POPH

(4) Cl NaOH NaCl NaClO HO

hot

conc

2

5

3

1

( .)

0

2

 

 If an element is in its highest possible oxidation state in a

compound, it can act as an oxidising agent. for example,

KMnO 4 , K 2 Cr 2 O 7 , HNO 3 , H 2 SO 4 , HClO 4 etc.

 If an element is in its lowest oxidation state in a compound,

it can act as a reducing agent. For example, H 2 S , H 2 C 2 O 4 ,

FeSO 4

, Na 2

S

2

O

3

, SO

2

, SnCl 2

, many metals etc.

 The strength of oxyacids of chlorine decrease in the

order. HClO 4 > HClO 3 > HClO 2 > HClO

 If highly electronegative element is in its highest

oxidation state in a compound that compound can act as

powerful oxidant. For example, KClO 4 , KClO 3 , KBrO 3 ,

KIO

3

etc.

 If an element is in intermediate oxidation state in a

compound, it can act as both oxidising & reducing agent.

For example, H 2 O 2 , H 2 SO 3 , HNO 3 , SO 2 etc.

Oxidation, Reduction

2 2

HO reduces

4

MnO ion to [KCET (Med.) 2000]

(a)

Mn (b)

2 

Mn

(c)

3 

Mn (d)

Mn

2. When a sulphur atom becomes a sulphide ion

[AMU 1999]

(a) There is no change in the composition of atom

(b) It gains two electrons

(c) The mass number changes

(d) None of these

3. The ultimate products of oxidation of most of hydrogen

and carbon in food stuffs are [DCE 2001]

(a) HO

2

alone (b)

2

CO alone

(c) HO

2

and

2

CO (d) None of these

4. When P reacts with caustic soda, the products are

3

PH

and.

2 2

NaHPO This reaction is an example of

[IIT 1980; Kurukshetra CEE 1993; CPMT 1997]

(a) Oxidation

(b) Reduction

(c) Oxidation and reduction (Redox)

(d) Neutralization

5. Which one of the following does not get oxidised by

bromine water [MP PET/PMT 1988]

(a)

 2

Fe to

 3

Fe (b)

Cu to

 2

Cu

(c)

 2

Mn to

4

MnO (d)

 2

Sn to

 4

Sn

6. In the reaction.

2 2 2

H S  NO  HO  NO  S HS

2

is

(a) Oxidised (b) Reduced

(c) Precipitated (d) None of these

7. The conversion of

2

PbO to

3 2

Pb ( NO )is

(a) Oxidation

(b) Reduction

(c) Neither oxidation nor reduction

(d) Both oxidation and reduction

8. In the course of a chemical reaction an oxidant

[MP PMT 1986]

(a) Loses electrons

(b) Gains electrons

(c) Both loses and gains electron

(d) Electron change takes place

2

2 CuICuCuI , the reaction is [RPMT 1997]

(a) Redox (b) Neutralisation

(c) Oxidation (d) Reduction

decrease

increase

  • 1
    • 2

decrease

increase

decrease

increase

decrease

increase