Download 08 Redox Reactions

08 Redox Reactions
Chapter 08: Redox Reactions
8.0 Introduction
8.3 Oxidation number
8.1 Concept of oxidation and reduction
8.4 Change in oxidation number
8.2 Redox reactions
reactions)
8.5 Balancing redox reactions in terms of loss
and gain of electrons
(Oxidation
reduction
8.0 Introduction

Chemical reaction:
A reaction in which two substances chemically react with each other to give products is called chemical
reaction.
Chemical reactions
Precipitation
reactions

i.
ii.
iii.
iv.
v.
vi.
vii.
Oxidation-reduction
or
Redox reactions
Acid-base
Neutralization reactions
Importance of redox reactions:
The redox reactions are a very important group of reactions which occur with almost every element.
These reactions are involved in large number of processes in nature (biological and industrial).
They take place in burning of fuels such as gasoline, oil, natural gas, organic substances of carbon
and hydrogen and so on.
The redox reactions are involved in household bleaching.
The metallic elements are extracted from their ores by oxidation-reduction reactions.
The functioning of batteries is based on redox reactions.
The corrosion of metals involve redox reactions.
8.1 Concept of oxidation and reduction

i.
ii.
iii.
Oxidation:
According to the classical concept, “Oxidation is defined as the addition of oxygen or any other
electronegative element or removal of hydrogen or any other electropositive element.”
According to the modern concept, “Oxidation is a process which involves loss of electrons. It is also called
de-electronation.”
Oxidation increases the oxidation number of the element in the given substance and it involves,
a.
Addition of oxygen.
C + O2  CO2
b.
Addition of an electronegative element.
Zn + S  ZnS
c.
Removal of hydrogen.
H2S + Br2  2HBr + S
d.
Removal of an electropositive element/radical.
2KI + Cl2  2KCl + I2
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Chemistry Vol ‐ 1.2 (Med. and Engg.) 
i.
ii.
iii.

i.
Reduction:
According to the classical concept, “Reduction is defined as the addition of hydrogen or any other
electropositive element or removal of oxygen or any other electronegative element”
According to the modern concept, “Reduction is a process which involves gain of electrons. It is also
called electronation.”
Reduction decreases the oxidation number of the element in the given substance and it involves,
a.
Addition of hydrogen.
Cl2 + H2  2HCl
b.
Addition of an electropositive element.
2HgCl2 + SnCl2  Hg2Cl2 + SnCl4
c.
Removal of oxygen.
ZnO + C  Zn + CO
d.
Removal of an electronegative element/ radical.
2FeCl3 + H2  2FeCl2 + 2HCl
Oxidizing agent:
“The substance which gains one or more electrons (i.e.itself gets reduced) is called oxidizing agent or
oxidant.”
Oxidizing agent increases the oxidation number of an element in a given substance.
Eg. O2, Cl2, Br2, H2O2, HNO3, CO2, etc.
ii.

i.
Reducing agent:
“The substance which loses one or more electrons (i.e.,itself gets oxidized) is called reducing agent or
reductant.”
Reducing agent lowers the oxidation number of an element in a given substance.
Eg. H2, Fe, Cu, C, LiAIH4, Sodium amalgam etc.
ii.
8.2 Redox reactions (Oxidation reduction reactions)

i.
Redox reactions:
“The reactions in which both oxidation and reduction reactions occur simultaneously (together) are called
redox reactions.”
In these reactions, one substance acts as a reducing agent and itself gets oxidized while another substance
acts as an oxidizing agent and itself gets reduced.
Eg. Reaction between zinc and copper (II) salt occuring in a battery. In this reaction, zinc loses electrons
and gets oxidized, whereas Cu2+ ions gain electrons and get reduced.
ii.
loss of 2e : Oxidation
2
Zn(s) + Cu (aq)
2
+ Cu(s)
Zn (aq)
gain of 2e : Reduction
iii.
iv.
v.
vi.
2
In the direct redox reaction, the transferance of electrons is limited to very small distances and therefore,
no useful electrical work could be obtained. In these reactions, chemical energy appears as heat.
If the transferance of electrons from zinc to copper ions is allowed to occur through some metallic wires,
useful electrical work could be performed. Such redox reactions are called Indirect redox reactions and
electrical energy is produced during such spontaneous reactions instead of heat energy.
In this case, zinc acts as a reducing agent or reductant, while Cu2+ ions act as an oxidizing agent or oxidant.
Other examples of redox reactions are,
Zn + 2HCl  ZnCl2 + H2 
5Fe2+ + 8H+ + MnO 4  5Fe3+ + Mn2+ + 4H2O
Examples for substances which can act both as oxidising as well as reducing agents are SO2, H2O2, HNO2
etc.
Chapter 08: Redox Reactions

i.
ii.
Half reactions:
“Every redox reaction according to the electronic concept consists of two reactions known as
half-reactions.” These are:
Oxidation half reaction (loss of electrons).
Reduction half reaction (gain of electrons).
Eg. The reaction involving zinc and dilute HCl.
The two half reactions are:
(oxidation half reaction)
Zn(s)
Zn 2aq  + 2e
(reduction half reaction)
H2(g)
2H aq + 2e
 
Zn(s) + 2H aq 

(overall reaction)
Zn 2aq  + H2(g)
Types of redox reactions:
Redox reactions
Combination
reactions
Decomposition
reactions
Disproportionation
reactions
Displacement
reactions
Metal
displacement
Non-metal
displacement
i.
Combination reactions:
a.
“The chemical reactions in which two or more substances (elements or compounds) combine to form
a single substance are called combination reactions.”
b.
A combination reaction may be expressed as,
A + B  C
c.
For combination reaction to be a redox reaction one or both ‘A’ and ‘B’ must be in the elementary form.
0
0
+4 2
0
+2
3
Eg. 0
CO2(g) ; 3Mg(s) + N2(g)
C(s) + O2(g)
Mg3N2(s)
ii.
Decomposition reactions:
a.
“The chemical reactions in which a compound breaks up into two or more simple substances are
called decomposition reactions.”
b.
The decomposition reactions are the opposite of combination reactions.
c.
A decomposition reaction is the breakdown of a compound into two or more compounds atleast one
of which must be in the elemental state.
+1 +5 2
+1 2
0
0
+1 1
0
Eg.
Δ
Δ
2H2O(l) 
 2H2(g) + O2(g) ; 2KClO3(s) 
 2KCl(s) + 3O2(g)
Note:
All decomposition reactions are not redox reactions.
Eg. Decomposition of calcium carbonate is not a redox reaction.
+2 +4 2
+2 2
Δ
CaCO3(s) 
 CaO(s) +
iii.
+4 2
CO2(g)
Displacement reactions:
a.
“The reactions in which one ion (or atom) in a compound is replaced by an ion (or atom) of other
element are called displacement reactions.”
b.
Displacement reactions may be expressed as:
X + YZ  XZ + Y
3
Chemistry Vol ‐ 1.2 (Med. and Engg.) c.
1.
These reactions are of two types:
Metal displacement:
In these reactions, a metal in a compound can be displaced by another metal in the uncombined state.
0
0
+2 +6 2
Eg. +2 +6 2
CuSO4(aq) + Zn(s)  Cu(s) + ZnSO4(aq)
+5 2
2.
i.
ii.
iii.
iv.
iv.
0
0
+2 2

V2O5(s) + 5Ca(s) 
 2V(s) + 5CaO(s)
Non-metal displacement:
The non-metal displacement redox reactions include hydrogen displacement and a rarely occurring
reaction involving oxygen displacement.
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
Eg.
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g) 
Less active metals such as magnesium and iron react with steam to produce hydrogen gas.
0
+1 2
+2 2 +1
0
Eg.
Mg(s) + 2H2O(l)  Mg(OH)2(aq) + H2(g) 
Many metals, including those which do not react with cold water, are capable of displacing hydrogen
from acids.
0
+1 1
+2 1
0
Eg.
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g) 
Metals like cadmium and tin do not react with steam, but react with acids to displace hydrogen gas.
0
+1 1
+2 1
0
Eg.
Cd(s) + 2HCl(aq)  CdCl2(aq) + H2(g) 
Disproportionation reactions:
a.
In a disproportionation reaction, an element in one oxidation state is simultaneously oxidized and
reduced.
b.
Atleast one reacting substance in a disproportionation reaction always contains an element that has
both higher and lower oxidation states existing for that element.
Eg. The decomposition of hydrogen peroxide is an example of the reaction, where oxygen element
experiences disproportionation.
Reduced
+1 1
+1 2
2H2O(l)
2H2O2(aq)
0
O2(g)
+
Oxidized
In this reaction, oxidation number of ‘O’ decreases from  1 to  2 (in H2O) and increases from
 1 to 0 (in O2).
Note:
1.
All redox reactions are exothermic.
2.
Phosphorus, sulphur and chlorine undergo disproportionation in the alkaline medium as shown
below:
Reduced
0
P4(s) + 3OH

(aq)
3
+ 3H2O(l)
+1
PH3(g) + 3H 2 PO 2(aq)
Oxidized
In this reaction, oxidation number of ‘P’ increases from 0 to + 1 (in H 2 PO 2 ) and decreases
from 0 to  3 (in PH3).
4
Chapter 08: Redox Reactions
8.3 Oxidation number

i.
ii.
iii.
iv.
v.

i.
ii.
iii.
iv.
v.
vi.
vii.
Oxidation number:
“The charge which an atom appears to have, when all other atoms are removed from it as ions is known as
oxidation number.”
OR
“The oxidation number or oxidation state of an atom in a molecule or ion is defined as the number of
charges it would carry if electrons were completely transferred.”
Metals invariably have positive oxidation states, while non-metals may have positive or negative
oxidation states.
Transition metals usually display several oxidation states.
The highest positive oxidation state for s-block elements is equal to its group number but for p-block
elements it is equal to group number minus 10 (except for noble gases).
However, the highest negative oxidation state for p-block elements is equal to eight minus (8) the number
of electrons in the valence shell.
Eg.
In the third period, the highest positive oxidation state or number increases from + 1 to + 7 as shown below:
Group
Element
1
Na
2
Mg
13
Al
14
Si
15
P
16
S
17
Cl
Compounds
Oxidation
state of the
underlined
element in the
compound
NaCl
MgSO4
AlF3
SiCl4
IF5, P4O10
SF6, SO3
HClO4, Cl2O7
+1
+2
+3
+4
+5
+6
+7
Rules to assign oxidation number or oxidation states to an atom:
If there is a covalent bond between,
a.
two same atoms then, oxidation number of both the atoms will be zero.
Eg. In a chlorine molecule, the oxidation number of both the chlorine atoms is zero.
b.
two different atoms then, electrons are counted more towards electronegative atom and,
1.
more electronegative atom will be negative.
2.
less electronegative atom will be positive.
Eg. In a HCl molecule, the oxidation number of chlorine (more electronegative atom) is 1 and that
of hydrogen (lesser electronegative atom) is +1.
If there is a co - ordinate bond between two atoms then,
a.
Oxidation number of acceptor atom will be  2.
b.
Oxidation number of donor atom will be + 2.
The oxidation number of an element in the free or elementary state or in any of its allotropic forms is
always zero.
Eg. Oxidation numbers of helium in He, hydrogen in H2, Oxygen in O2 or O3 are zero.
The oxidation number of an element in a single (monoatomic) ion is same as the charge on the ion.
Eg. Oxidation number of K+ is + 1, and Ca2+ is + 2.
In binary compounds of metals and non-metals the oxidation number of metals is always positive while that
of non-metals is negative.
Eg. In NaCl, the oxidation number of sodium is + 1 and that of chlorine is  1.
In compounds formed by the combination of non-metallic atoms, the atom with higher electronegativity is
given negative oxidation number.
Eg. In HCl, the oxidation number of chlorine is  1 because of it’s high electronegativity.
In all compounds of hydrogen, the oxidation number of hydrogen is + 1 except in hydrides of active metals
such as LiH, NaH, KH, MgH2, etc., where hydrogen has the oxidation number of  1.
5
Chemistry Vol ‐ 1.2 (Med. and Engg.) viii. The oxidation number of oxygen is 2 in most of the compounds. However, there are two exceptions.
a.
The first exception is peroxides and superoxides in which oxygen atoms are directly linked to each
other.
Eg. In peroxides i.e. H2O2, each oxygen atom is assigned an oxidation number of  1 and in
superoxides i.e. KO2, RbO2 etc. each oxygen atom is assigned an oxidation number of  1/2.
b.
The second exception is found in compounds in which oxygen is bonded to fluorine.
Eg. In OF2 (oxygen difluoride) the oxidation no. of O is +2 and in O2F2 (dioxygen fluoride), the
oxidation no. of each O is +1.
ix. The most electronegative element, fluorine always has an oxidation number  1. For other halogens, the
oxidation number is generally  1, but there are exceptions, when these are bonded to a more
electronegative halogen atom or oxygen.
Eg. In HI, the oxidation number of ‘I’ is  1 but in IF5, it is + 5 and in IF7, it is + 7.
x.
For neutral molecule, the sum of the oxidation numbers of all the atoms is equal to zero.
Eg. In NH3, Nitrogen is in  3 oxidation state, whereas hydrogen is in + 1 oxidation state and there are
three hydrogen atoms present. Therefore the net charge is zero i.e.  3 and + 3 equals to zero.
8.4 Change in oxidation number
i.
ii.
iii.
iv.
The oxidation number is more oftenly termed as oxidation state.
In an oxidation process, the oxidation number of the element increases in the given substance whereas in a
reduction process, the oxidation number of the element decreases in the given substance.
Oxidising agent increases the oxidation number of an element in a given substance whereas Reducing
agent lowers the oxidation number of an element in a given substance.
Redox reactions involve the change in oxidation number of interacting species.
Eg. The reaction between zinc and hydrochloric acid.
The oxidation number of all the atoms are written above their respective symbols,
Oxidized
+1 1
0
Zn
+
2HCl
2+ 1
ZnCl2
0
+
H2
Reduced
In this reaction, the oxidation number (O. N.) of zinc increases from 0 to + 2 and that of hydrogen
decreases from + 1 to 0, while that of chlorine remains unchanged. Thus, zinc is oxidized while
hydrogen is reduced.
8.5 Balancing redox reactions in terms of loss and gain of electrons
Methods for balancing redox reactions
Oxidation number
method

Ion-electron method
(Half-reaction method)
Oxidation number method:
The oxidation number method for balancing the redox reactions follows the basic principle that the total
increase in oxidation number must be equal to total decrease in oxidation number.
Steps involved in balancing of redox reactions by the oxidation number method are given as follows:
Write the unbalanced net equation for the redox reaction. Balance the equation for all atoms except ‘H’ and
‘O’.
i.
6
Chapter 08: Redox Reactions
ii.
iii.
iv.
v.
vi.
vii.
Assign the oxidation numbers to all atoms in the reactants and the products, using the rules to assign
oxidation number. Identify the atoms undergoing change in oxidation numbers.
Show an increase in oxidation number per atom of the oxidized species and hence, the net increase in
oxidation number. Similarly, show a decrease in oxidation number per atom of the reduced species and the
net decrease in oxidation number.
Determine the factor that will make the total increase and decrease in oxidation numbers equal. Insert these
coefficients into the equation.
Balance oxygen atoms by adding H2O to the side containing less O atoms, one H2O for one O atom.
Balance H atoms by adding H+ ions, to the side with less H atoms.
If the reaction occurs in basic medium, then add OH ions, equal to the number of H+ on both sides of the
equation. The H+and OH ions appearing on the same side of the reaction are combined to give H2O
molecules.
Check the balanced equation to make sure that the reaction is balanced with respect to both the number of
atoms of each element and the charges.
Note: For a reaction in alkaline medium all the steps are applicable. However, if the reaction occurs in acidic
medium, point (vi) is omitted.
a.
Reaction occurring in acidic medium:
2
3+
H2O2(aq) + Cr2O 7(aq)
 O2(g) + Cr(aq)
1.
Balance Cr atom and assign oxidation number to each atom.
2
3+
H2O2(aq) + Cr2 O7(aq)
 O2(g) + 2Cr(aq)
 
2.

+1 1

+6 2

0
+3
Identify the atoms undergoing change in oxidation number. The oxidation number of Cr reduces from
+ 6 to + 3. i.e., undergoes reduction by gain of three electrons and is an oxidizing agent. The
oxidation number of ‘O’ increases from  1 to 0 by loss of two electrons. i.e., it acts as a reducing
agent and itself gets oxidized.
2
3+
H2O2(aq) + Cr2 O7(aq)
 O2(g) + 2Cr(aq)

1
3.
4.

+6

0

+3
Gain of e
Loss of e
Find the total increase in oxidation number for the oxidized atom and total decrease in oxidation
number for the reduced atom.
i.
Increase in oxidation number : 2O ( 1)  2O (0).
Increase per atom = + 1
Net increase = + 2.
ii.
Decrease in oxidation number : 2Cr (+ 6)  2Cr (+ 3)
Decrease per atom =  3
Net decrease =  6.
Choose the factors that will make net increase and decrease equal. The net increase must be
multiplied by 3. Hence, the coefficient 3 is needed for the O species on both sides.
2
3+
 3O2(g) + 2Cr(aq)
3H2O2(aq) + Cr2 O7(aq)
5.
Balance the equation for O atoms by adding H2O to the side with less O atoms. There are 13 (O)
atoms on the left side and 6 (O) atoms on the right side. Hence, add 7 H2O molecule to the right side.
2
3+
3H2O2(aq) + Cr2 O7(aq)
 3O2(g) + 2Cr(aq)
+ 7H2O(l)
6.
Balance the equation for H atoms by adding H+ ions on the side with less H atoms. There are 6 (H)
atoms on left side whereas 14 (H) atoms on right side. Hence add 8H+ to the left side.
2
+
3+
+ 8H (aq)
 3O2(g) + 2Cr(aq)
+ 7H2O(l)
3H2O2(aq) + Cr2 O7(aq)
Thus, the equation is balanced for both charge as well as atoms.
7
Chemistry Vol ‐ 1.2 (Med. and Engg.) b.
Reaction occurring in basic medium:
2
Zn(s) + NO 3(aq)  NH3(g) + Zn(OH)6(aq)
1.
Assign oxidation number to all atoms.

Zn(s) + NO3(aq)
 NH3(g) + Zn(OH)62(aq)


0
2.
+5 2

3 +1
 
+4
–2 +1
Identify the atoms undergoing changes in oxidation numbers. The oxidation number of ‘N’ reduces from +
5 to  3 and that of ‘Zn’ increases from 0 to + 4. Hence, N is an oxidizing agent and is itself reduced by the
gain of electrons. ‘Zn’ acts as a reducing agent and is itself oxidized by loss of electrons.

2
Zn(s) + NO3(aq)
 NH3(g) + Zn(OH)6(aq)

0


+5
3

+4
Gain of e
Loss of e
3.
4.
Find the net increase in oxidation number for the oxidized atom and net decrease in oxidation number
for the reduced atom.
i.
Increase in oxidation number : Zn (0)  Zn (4)
Increase per atom = Net increase = +4
ii.
Decrease in oxidation number : N (+5)  N (3)
Decrease per atom = Net decrease = 8.
Choose the factors that will make the net increase equal to the net decrease. The net increase is
multiplied by 2. Hence, a coefficient 2 is needed for both species of Zn.

2
2Zn(s) + NO3(aq)
 NH3(g) + 2Zn(OH)6(aq)
5.
Balance the ‘O’ atoms by adding H2O molecule to the side with less ‘O’ atoms. The left side has 3
oxygen atoms and right side has 12 oxygen atoms. Therefore, balance the equation for ‘O’ atoms by
adding 9H2O to the left side.

2
2Zn(s) + NO3(aq)
+ 9H2O(l) NH3(g) + 2Zn(OH)6(aq)
6.
‘H’ atoms are balanced by adding H+ ions to the side with less ‘H’ atoms. There are 18 H atoms on
the left side and 15 on the right side. Add 3H+ ions to the right side to balance ‘H’ atoms.

2
+
+ 9H2O(l)  NH3(g) + 2Zn(OH)6(aq)
+ 3H (aq)
2Zn(s) + NO3(aq)
7.
Since the reaction occurs in basic medium, add OH ions equal to the number of H+ ions, on both
sides of the equation. Hence add 3OH ions on both sides.


2
+
2Zn(s) + NO3(aq)
+ 9H2O(l) + 3OH (aq)
 NH3(g) + 2Zn(OH)6(aq)
+ 3H (aq)
+ 3OH(aq)
8.
3H+ and 3OH on the right side will be neutralized to form 3H2O.


2
+ 9H2O(l) + 3OH (aq)
 NH3(g) + 2Zn(OH)6(aq)
+ 3H2O(l)
2Zn(s) + NO3(aq)
9.
Cancel 3H2O molecules on both sides of equation.
The final equation is

2
 NH3(g) + 2Zn(OH)6(aq)
2Zn(s) + NO 3(aq) + 6H2O(l) + 3OH (aq)
The equation is balanced for both atoms and charges.
8
Chapter 08: Redox Reactions

Ion-electron method (Half reaction method):
The Ion-electron method for balancing the redox reactions follows the principle of balancing the equations
using half reactions.
Steps involved in balancing of redox reaction by the ion-electron method are given as follows:
i.
Write the unbalanced equation for the redox reaction and assign the oxidation number to all the atoms in the
reactants and the products.
ii.
Divide the equation into two half equations. One half equation describes oxidation, which involves
increase in oxidation number of an oxidized species and the other half equation describes reduction in
which the oxidation number of the reduced species decreases.
iii. Balance the atoms except ‘O’ and ‘H’ in each half equation. Balance oxygen atoms by adding H2O to the
side with less ‘O’ atoms.
iv. Balance ‘H’ atoms by adding ‘H+’ ions to the side with less ‘H’ atoms.
v.
Balance the charges by adding appropriate number of electrons to the right side of oxidation half equation
and to the left of reduction half equation.
vi. Multiply half equations by suitable factors to equalize the number of electrons in the two half equations.
Then add two half equations and cancel the number of electrons on both sides of the equation.
vii. If the reaction occurs in basic medium, then add OH ions, equal to the number of H+ ions, on both sides of the
equation. H+ and OH ions appearing on the same side of the equation are combined to give H2O molecules.
viii. Check that the equation is balanced for both, the atoms and the charges.
Note: For the reaction in an alkaline medium, all the steps are applicable. However, if the reaction occurs in acidic
medium, point (vii) is omitted.
a.
Reaction occurring in acidic medium:
3+
2+
2
SO2(g) + Fe(aq)
 Fe(aq)
+ SO 4(aq)
(acidic)
1.
Write the unbalanced equation and assign oxidation number to all atoms.

2

SO2(g) + Fe3(aq)
 Fe(aq)
+ SO 24(aq)
+4 2
+3
+2
+6 2

Gain of e
Loss of e
2.
Divide the equation into two half equations, an oxidation half equation and a reduction half equation.
The oxidation number of ‘S’ increases from + 4 to + 6 and that of Fe decreases from + 3 to + 2.
Hence, S is oxidized and Fe is reduced.

Oxidation half equation: SO2(g)  SO 24(aq)

2
 Fe(aq)
Reduction half equation: Fe3(aq)
3.
Balance the half equations for ‘O’ atoms by adding H2O to the side with less ‘O’ atoms. Hence, add
2H2O to the left side of oxidation half equation whereas no addition of H2O to reduction half equation.

Oxidation: SO2(g) + 2H2O(l)  SO 24(aq)

2
 Fe(aq)
Reduction: Fe3(aq)
4.
Balance ‘H’ atoms by adding ‘H+’ ions to the side with less ‘H’ atoms. Hence, add ‘4H+’ to the right
side of oxidation half equation and no addition to reduction half equation.


+ 4H (aq)
Oxidation: SO2(g) + 2H2O(l)  SO 24(aq)

2
Reduction: Fe3(aq)
 Fe(aq)
5.
Add 2e to the right side of oxidation half equation and le to the left side of reduction half equation
to balance the charges.


+ 4H (aq)
+2e
Oxidation: SO2(g) + 2H2O(l)  SO 24(aq)

2
Reduction: Fe3(aq)
+ e  Fe(aq)
9
Chemistry Vol ‐ 1.2 (Med. and Engg.) 6.
Multiply reduction half equation by ‘2’ to equalize the number of electrons in the two half equations
and then add the two half equations.


SO2(g) + 2H2O(l)  SO 24(aq)
+ 4H (aq)
+ 2e

2
2Fe3(aq)
+ 2e  2Fe(aq)


2

SO2(g) + 2Fe3(aq)
+ 2H2O(l)  SO 24(aq)
+ 2Fe(aq)
+ 4H (aq)
b.
1.
Hence, the equation is balanced in terms of number of atoms and charges.
Reaction occurring in basic medium:
2+ I2(aq)  S 4O 6(aq
(basic)
S 2 O 23(aq)
) + I (aq)
Balance ‘S’ and ‘I’ atoms and assign oxidation number to all atoms.
22+ I2(aq)  S4 O6(aq.)
+ 2 I-(aq)
2S2 O3(aq)

+2 2

0

+2.5 2

1
Loss of e
Gain of e
2.
Divide the equation into two half equations, an oxidation half equation and a reduction half equation.
The oxidation number of ‘S’ increases from + 2 to + 2.5 and that of ‘I’ decreases from 0 to  1.
Therefore ‘S’ is oxidized and ‘I’ is reduced.
22i.
Oxidation half equation: 2S2 O3(aq)
 S4 O 6(aq)
ii.
3.
4.
Reduction half equation: I2(aq)  2I-(aq)
Number of oxygen atoms is already balanced for the oxidation step. In the reduction step, there is no
oxygen atom involved.
Balancing for the charge:
Add 2 electrons on the R.H.S. of the oxidation half equation and 2 electrons on the L.H.S. of the
reduction half equation.
22Oxidation: 2S2 O3(aq)
 S4 O6(aq)
+ 2e
Reduction: I2(aq) + 2e  2I-(aq)
Adding two half equations we get,
22
2S2 O3(aq)
+ I2(aq)  S4 O6(aq)
+ 2I (aq)
Thus, the equation is balanced for all atoms and charges.
Note: Stock Notation:
i.
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.
ii.
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.
iii. Thus, Aurous chloride and Auric chloride are written as Au(I)Cl and Au(III)Cl3. Similarly, Stannous
chloride and Stannic chloride are written as Sn(II)Cl2 and Sn(IV)Cl4. This change in oxidation number
implies change in oxidation state, which helps to identify whether the species is present in oxidized form or
reduced form.
No. Compounds Oxidation state of metal ion Representation or stock notation
i.
HAuCl4
+3
HAu(III)Cl4
ii.
FeO
+2
Fe(II)O
iii.
Fe2O3
+3
Fe2(III)O3
iv.
CuI
+1
Cu(I)I
v.
CuO
+2
Cu(II)O
vi.
MnO
+2
Mn(II)O
vii
MnO2
+4
Mn(IV)O2
10
Chapter 08: Redox Reactions

Difference between valency and oxidation number:
No.
Valency
Oxidation number (O.N.)
i. It is the combining capacity of the element. No O.N. is the charge (real or imaginary) which an
atom has or appears to have when all the atoms
(+) or () sign is attached to it.
are removed as ions. It may have (+) or () sign.
ii. Valency of an element is usually fixed.
O.N. of an element may have different values. It
depends on the nature of compound in which it
is present.
iii. Valency is always a whole number.
O.N. of the element may be a whole number or
fractional.
iv. Valency of the element cannot be zero except O.N. of the element may be zero.
noble gases.
Note: Electrode potential:
i.
The electrical potential difference set up between the metal and its solution is known as electrode potential.
ii.
The electrode potential is a measure of tendency of an electrode in a half cell to gain or lose electrons.
iii. If the concentration of the ions is 1 mol L1 solution, the electrode potential is called standard electrode
potential (E)
iv. Higher the standard electrode potential (E), stronger is the oxidizing agent.
v.
All standard potentials are taken as reduction potentials.

i.
Applications of redox reactions:
Metallurgy: In metallurgy, metals are extracted from their ores and they are then purified. The extraction
and purification of metals use redox reactions in different steps.
Eg. a.
Sulphide minerals are converted to oxides by roasting.
heat
2ZnS(s) + 3O2(g) ¾¾
 2ZnO(s) + 2SO2(g)




+2 2
0
+2 2
+4 2
The oxidation number of ‘S’ increases from  2 to + 4 and that of ‘O’ decreases from 0 to  2.
b.
ZnO is reduced by coke to prepare Zn.
heat
ZnO(s) + C(s) ¾¾
 Zn(s) + CO(g)


ii.

0
+2

0
+2
The oxidation number of ‘Zn’ decreases from + 2 to 0 and that of ‘C’ increases from 0 to + 2.
Batteries: The electricity produced in batteries or galvanic cells is due to redox reactions occurring in them.
Eg. Daniel cell involves the transfer of electrons from Zn to Cu through a wire connecting the two
electrodes which are dipped in the solutions of their own ions. Zn is oxidized by transferring two electrons
to Cu2+ ions which are reduced. Thus,
2+
2+
Zn(s) + Cu (aq)
 Zn (aq)
+ Cu(s)

0
iii.
iv.


+2

0
+2
The oxidation number of Zn increases from 0 to + 2 and that of Cu decreases from + 2 to 0.
Bleaching: Decolourization or lightening of coloured material uses redox reaction and is called bleaching.
Eg. a.
NaOCl is used as an oxidizing agent in bleaching of clothes to remove stains.
b.
Chlorine is used as an oxidizing agent to bleach wood pulp into white paper.
c.
H2O2 is used as an oxidizing agent to bleach dark hair by a redox reaction.
Combustion: Burning of a substance with oxygen in air is called combustion.
Eg. CH4 burns by oxidation with oxygen.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)

4 +1

0

+4 2

+1 2
11
Chemistry Vol ‐ 1.2 (Med. and Engg.) v.
The oxidation number of ‘C’ increases from  4 to + 4 and that of ‘O’ decreases from 0 to  2. Hence, it is a
redox reaction. CH4 is a reducing agent and O2 is an oxidizing agent.
Corrosion: Corrosion is the destruction of metals by oxidation.
Eg. Rusting of iron is its oxidation by oxygen of air in presence of moisture.
Moisture
4Fe(s) + 3O2(g) 
 2Fe2O3.H2O(s)
H2O
Corroded Fe
vi.
The method employed in preventing corrosion of iron is by zinc coating. This method is called
Galvanization.
Respiration:
a.
The process of breathing and using oxygen for various biological redox reactions is called respiration.
b.
The biological redox reactions provide energy required for living organisms. The overall effect of
respiration is similar to combustion.
Eg. Glucose reacts with oxygen to give CO2 and H2O releasing energy.
C6H12O6 + 6O2  6CO2  + 6H2O + energy
0
0
+4 –2
–2
The oxidation number of ‘C’ increases from 0 to + 4 and that of ‘O’ decreases from 0 to  2.
Hence, it is a redox reaction.
Quick Review

Oxidation:
L
Loss
of

Reduction:
G
Gain
of

Oxidizing agent:
(Oxidant)
O
Oxidizing
agent

Reducing agent:
(Reductant)
R
E
Electrons
is
E
O
Eg.

Na(s)  Na (aq)
+ e
Eg.

Cl2(g) + 2e  2Cl(aq)
Eg.

F2(g) + 2e  2F(aq)
Oxidation
R
Electrons
is
Reduction
E
A
(F2 is an oxidizing agent)
Electron
E
Acceptor
D
Eg.
Reducing
agent
12
2

Fe(aq)
 Fe3(aq)
+ e
(Fe2+ is a reducing agent)
Electron
Donor
Chapter 08: Redox Reactions

Oxidation numbers of some of the elements in their compounds:
Oxidation
Element
Number
Element
Oxidation
Number
Oxidation
Element
Number
H
+1, 1
Co
+3, +2
Fe
+3, +2
Li
+1
Ni
+2
Sn
+4, +2
Be
+2
Cu
+2, +1
Sb
+5, +3, 3
B
+3
Zn
+2
Te
+6, +4, 2
C
+4, +2, 4
Ga
+3
I
Ge
+4, 4
Xe
+6, +4, +2
N
+5, +4, +3
+2, +1, 3, 2, 1
+7, +5, +1, 1
O
+1, +2, 1/2, 1, 2
As
+5, +3, 3
Cs
+1
F
1
Se
+6, +4, 2
Ba
+2
Na
+1
Br
+5, +3, +1, 1
La
+3
Mg
+2
Kr
+4, +2
Hf
+4
Al
+3
Rb
+1
Ta
+5
Si
+4, 4
Sr
+2
W
+6, +4
P
+5, +3, 3
Y
+3
Re
+7, +6, +4
S
+6, +4, +2, 2
Zr
+4
Os
+8, +4
Nb
+5, +4
Ir
+4, +3
Cl
+7, +6, +5, +4, +3,
+1, 1
K
+1
Mo
+6, +4, +3
Pt
+4, +2
Ca
+2
Tc
+7, +6, +4
Au
+3, +1
Sc
+2 ,+3
Ru
+8, +6, +4, +3
Hg
+2, +1
Ti
+4, +3, +2
Pd
+4, +2
Tl
+3, +1
V
+5, +4, +3, +2
Ag
+1
Pb
+4, +2
Cr
+6, +5, +4, +3
Cd
+2
Bi
+5, +3
Po
+2
In
+3
At
1
+7, +6, +4,
Mn
+3, +2
13
Chemistry Vol ‐ 1.2 (Med. and Engg.) Multiple Choice Questions
8.
Change of hydrogen into proton is _______.
(A) Oxidation of hydrogen
(B) Acid-base reaction
(C) Reduction of hydrogen
(D) Displacement reaction
9.
In the reaction, H2S+NO2  H2O + NO + S,
H2S is _______.
(A) Oxidized
(B) Reduced
(C) Precipitated
(D) None of these
10.
In a reaction between zinc and iodine, in
which zinc iodide is formed, what is being
oxidized?
[NCERT 1975]
(A) Zinc ions
(B) Iodide ions
(C) Zinc atom
(D) Iodine atom
11.
When a piece of copper wire is dipped in
AgNO3 solution, the colour of the solution
turns blue due to _______.
(A) Formation of soluble complex
(B) Oxidation of copper
(C) Oxidation of silver
(D) Reduction of copper
12.
Which one of the following does NOT get
oxidized by bromine water?
(B) Cu+ to Cu2+
(A) Fe2+ to Fe3+
(C) Mn2+ to MnO 4 (D) Sn2+ to Sn4+
13.
In which of the following reactions, the
underlined substance has been oxidized?
(A) Br2 + H2S  2HBr + S
(B) 2HgCl2 + SnCl2  Hg2Cl2+ SnCl4
(C) Cl2 + 2KI  2KCl + I2
(D) 2Cu2+ + 4I  Cu2I2 + I2
14.
Reduction is a process which involves
_______.
(A) gain of electrons
(B) loss of electronegative element
(C) decrease in the oxidation number of one
of the atoms
(D) all of these
15.
Reduction is defined as _______.
(A) Increase in positive valency
(B) Gain of electrons
(C) Loss of protons
(D) Decrease in negative valency
16.
H2O2 reduces MnO 4 ion to _______.
[KCET (Med.) 2000]
(A) Mn+
(B) Mn2+
(C) Mn3+
(D) Mn
8.0 Introduction
1.
The metallic elements are extracted from their
ores by _______.
(A) precipitation reactions
(B) acid-base neutralization
(C) redox reactions
(D) complexometric titrations
8.1 Concept of oxidation and reduction
2.
3.
4.
5.
6.
7.
14
According to classical concept, oxidation
involves which of the following?
(A) Addition of oxygen.
(B) Addition of electronegative radical.
(C) Removal of either hydrogen or some
electropositive radical.
(D) All of these.
According to modern concept, oxidation is
which of the following?
(A) Electronation
(B) De-electronation
(C) Addition of oxygen
(D) Addition of electronegative element
Oxidation involves _______.
[NCERT 1971, 81; CPMT 1980, 82, 83;
MP PMT 1983]
(A) Loss of electrons
(B) Gain of electrons
(C) Increase in the valency of negative part
(D) Decrease in the valency of positive part
Oxidation is a process which involves _______.
(A) gain of electrons
(B) loss of an electronegative radical
(C) gain of an electropositive radical
(D) increase in the oxidation number of one
of the atoms
When Sn2+ changes to Sn4+ in a reaction,
_______.
[CPMT 1981]
(A) it loses two electrons
(B) it gains two electrons
(C) it loses two protons
(D) it gains two protons
Consider the reaction, Zn+Cu2+Zn2+ + Cu
With reference to the above, which one of the
following is the CORRECT statement?
(A) Cu is oxidized to Cu2+.
(B) Zn is oxidized to Zn2+.
(C) Zn2+ is oxidized to Zn.
(D) Cu2+ is oxidized to Cu.
Chapter 08: Redox Reactions
17.
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
25.
An oxidizing agent is a substance which
_______.
(A) gains electrons
(B) loses an electronegative radical
(C) undergoes decrease in the oxidation
number of one of its atoms
(D) undergoes any one of the above changes
18.
The reaction of KMnO4 and
_______.
(A) Oxidation of Mn in
production of Cl2
(B) Reduction of Mn in
production of H2
(C) Oxidation of Mn in
production of H2
(D) Reduction of Mn in
production of Cl2
26.
In the course of a chemical reaction, an
oxidant _______.
[MP PMT 1986]
(A) Loses electrons
(B) Gains electrons
(C) Loses as well as gains electron
(D) Gains an electronegative radical.
27.
Which of the following is an oxidizing
substance?
[CPMT 1997]
(B) CO
(A) C2H2O2
(D) CO2
(C) H2S
28.
In a conjugate pair of reductant and oxidant,
the oxidant has _______.
(A) Higher oxidation number
(B) Lower oxidation number
(C) Same oxidation number
(D) Either of these.
29.
In which of the following reactions, hydrogen
acts as an oxidizing agent?
(A) With iodine to give hydrogen iodide
(B) With lithium to give lithium hydride
(C) With nitrogen to give ammonia
(D) With sulphur to give hydrogen sulphide
30.
In the reaction,
2FeCl3 + H2S  2FeCl2 + 2HCl + S
(A) FeCl3 acts as an oxidizing agent
(B) Both H2S and FeCl3 are oxidized
(C) FeCl3 is oxidized while H2S is reduced
(D) H2S acts as an oxidizing agent
31.
Of the four oxyacids of chlorine the strongest
oxidizing agent in dilute aqueous solution is
_______.
(B) HClO3
(A) HClO4
(C) HClO2
(D) HClO1
Which of the following CANNOT act as an
oxidizing agent?
[CPMT 1996]
(B) KMnO4
(A) O2
(D) K2Cr2O7
(C) I2
HCl results in
KMnO4 and
KMnO4 and
KMnO4 and
KMnO4 and
19.
The conversion of PbS to Pb is _______.
(A) dissociation
(B) reduction
(C) oxidation
(D) electrolysis
20.
In which of the following reactions, the
underlined substance has been reduced?
(A) CO + CuO  CO2 + Cu
(B) CuO + 2HCl  CuCl2 + H2O
(C) 4H2O (g) + 3Fe  4H2 (g) + Fe3O4
(D) C + 4HNO3  CO2 + 2H2O + 4NO2
21.
In the reaction,
4Zn+ NO3 +7H2O 4Zn2+ + NH 4 +10OH,
the substance reduced is _______.
(A) Zn
(B) H2O

(C) NO3
(D) NH 4
22.
23.
24.
Which of the following is an example of
reduction?
(A) CuO  Cu2O
(B) [Fe(CN)6]4–  [Fe(CN)6]3–
(C) KI  I2
(D) H2S  S
In the chemical reaction,
Ag2O + H2O + 2e  2Ag + 2OH
(A) water is oxidized.
(B) electrons are reduced.
(C) silver is oxidized.
(D) silver is reduced.
In which one of the following reactions,
nitrogen is NOT reduced?
(A) NO2  NO 2 (B) NO3  NO
(C)
NO3  NH 4 (D)
NH 4  N2
32.
33.
Reducing agent is a substance which can
_______.
[CPMT 1971, 74, 76, 78, 80;
NCERT 1976]
(A) accept electrons (B) accept protons
(C) donate electrons (D) donate protons
15
Chemistry Vol ‐ 1.2 (Med. and Engg.) 34.
Sodium amalgam is useful as which of the
following?
(A) Oxidizing agent (B) Reducing agent
(C) Catalyst
(D) Bleaching agent
35.
Oxidation number of chlorine increases from
1 to 0. Hence HCl is _______ agent.
(A) redox
(B) oxidizing
(C) reducing
(D) complex
36.
In a conjugate pair of reductant and oxidant,
the reductant has _______.
(A) Lower oxidation number
(B) Higher oxidation number
(C) Same oxidation number
(D) Either of these
37.
38.
39.
Nitric oxide acts as a reducing agent in which
of the following reaction?
(A) 4NH3 + 5O2  4NO + 6H2O
(B) 2NO + 3I2 + 4H2O  2 NO3
+ 6I + 8H+
(C) 2NO + H2SO3  N2O + H2SO4
(D) 2NO + H2S  N2O + S + H2O
Strongest reducing agent is _______.
[CPMT 1977; BHU 1984, 96; MP PET 1990;
AMU 1999]
(B) Cl
(A) F
(C) Br
(D) I
Which substance is serving as a reducing
agent in the following reaction?
14H+ + Cr2 O72 + 3Ni2Cr3+ +7H2O+ 3Ni2+
[CBSE PMT 1994; AFMC 2000;
DPMT 2001]
(B) Ni
(A) H2O
(D) Cr2 O72
(C) H+
40.
In the reaction between ozone and hydrogen
peroxide, H2O2 acts as _______.
[RPET 2000]
(A) Oxidizing agent
(B) Reducing agent
(C) Bleaching agent
(D) Both oxidizing and bleaching agent
41.
In the reaction,
Ag2O + H2O2  2Ag + H2O + O2, H2O2
acts as _______.
[BHU 2004]
(A) Reducing agent
(B) Oxidizing agent
(C) Bleaching agent
(D) None of the above
16
42.
In C + H2O  CO + H2, H2O acts as :
[AFMC 1988]
(A) Oxidizing agent
(B) Reducing agent
(C) Both oxidizing agent and reducing
agent.
(D) Neither oxidizing agent nor reducing
agent.
8.2 Redox reactions
43.
A chemical reaction in which oxidation and
reduction processes take place simultaneously
is known as ________ reaction.
(A) redox
(B) precipitation
(C) complexometric (D) titration
44.
Which of the following behaves as both
oxidising and reducing agent? [AFMC 1995]
(B) SO2
(A) H2SO4
(C) H2S
(D) HNO3
45.
The compound that can work both as
oxidising and reducing agent is
[CPMT 1986; MP PET 2000]
(A) KMnO4
(B) H2O2
(D) K2Cr2O7
(C) BaO2
46.
A redox reaction is _______.
(A) exothermic
(B) endothermic
(C) can neither be exothermic nor
endothermic i.e. H = 0
(D) can be either exothermic or endothermic
47.
Which of the following statements is
CORRECT?
(A) Oxidation of a substance is followed by
reduction of another.
(B) Reduction of a substance is followed by
oxidation of another.
(C) Oxidation
and
reduction
are
complementary reactions.
(D) It is necessary that both oxidation and
reduction should take place in the same
reaction.
48.
Which of the following is NOT an example of
redox reaction?
[NCERT Exemplar]
(A) CuO + H2  Cu + H2O
(B) Fe2O3 + 3CO  2Fe + 3CO2
(C) 2K + F2  2KF
(D) BaCl2 + H2SO4  BaSO4 + 2HCl
Chapter 08: Redox Reactions
49.
50.
In the reaction,
P + NaOH  PH3 + NaH2PO2
[MP PET 2004]
(A) P is oxidised only
(B) P is reduced only
(C) P is oxidised as well as reduced
(D) Na is reduced
In the following reaction,
4P + 3KOH + 3H2O  3KH2PO2 + PH3,
Which of the following is TRUE?
[Pb. PMT 2002]
(A) P is oxidised as well as reduced
(B) P is reduced only
(C) P is oxidised only
(D) None of these
51.
2CuI  Cu + CuI2, the reaction is _______.
[RPMT 1997]
(A) Redox
(B) Neutralisation
(C) Oxidation
(D) Reduction
52.
2
2
Co(s) + Cu (aq)
 Co(aq)
+ Cu(s). The reaction
is an example for which of the following?
(A) Oxidation reaction
(B) Reduction reaction
(C) Redox reaction
(D) None of these
53.
54.
55.
Which is the best description of the behaviour
of bromine in the reaction given below?
H2O + Br2  HOBr + HBr
[CBSE PMT 2004]
(A) It is oxidised only
(B) It is reduced only
(C) It acts as a proton acceptor only
(D) It gets both oxidised and reduced
Which of the following represents a redox
reaction?
(A) NaOH + HCl  NaCl + H2O
(B) BaCl2 + H2SO4  BaSO4 + 2HCl
(C) CuSO4 + 2H2O  Cu(OH)2 + H2SO4
(D) Zn + 2HCl  ZnCl2 + H2
Which of the following is a redox reaction?
[CBSE PMT 1997]
(A) H2SO4 with NaOH.
(B) In atmosphere, O3 from O2 by lightning.
(C) Nitrogen oxides from nitrogen and
oxygen by lightning.
(D) Evaporation of H2O.
56.
Of the following reactions, only one is a redox
reaction. Identify it
(A) Ca(OH)2 + 2HCl  CaCl2 + 2H2O
(B) BaCl2 + MgSO4  BaSO4 + MgCl2
(C) 2S2 O72  + 2H2O  4SO 24  + 4H+
(D)
Cu2S + 2FeO  2Cu + 2Fe + SO2
57.
Which of the following is a redox-reaction?
[AIEEE 2002]
(A) 2Na[Ag(CN)2] + Zn  Na2[Zn(CN)4]
+ Ag
(B) BaO2 + H2SO4  BaSO4 + H2O2
(C) N2O5 + H2O  2HNO3
(D) AgNO3 + K Ag2+ + KNO3
58.
Which of the following is a redox reaction?
[MP PMT 2003]
(A) P2O5 + 2H2O  H4P2O7
(B) 2AgNO3 + BaCl2  2AgCl + Ba(NO3)2
(C) BaCl2 + H2SO4  BaSO4 + 2HCl
(D) Cu + 2AgNO3  2Ag + Cu(NO3)2
59.
What is ‘A’ in the following reaction

2
2
2Fe3(aq)
 Sn (aq)
 2Fe(aq)
A?
(A)
Sn
3
(aq)
(B)
[MP PET 2003]
4
Sn (aq)
(C)
2
Sn (aq)
(D)
Sn
60.
Which of the following reactions involves
oxidation-reduction?
[NCERT 1972; AFMC 2000;
Pb. CET 2004; CPMT 2004]
(A) NaBr + HCl  NaCl + HBr
(B) HBr + AgNO3  AgBr + HNO3
(C) H2 + Br2  2HBr
(D) 2NaOH + H2SO4  Na2SO4 + 2H2O
61.
Which of the following is NOT a redox
reaction?
(A) 2Na + Cl2  2NaCl
(B) C + O2  CO2
(C) AgNO3 + NaCl  AgCl + NaNO3
(D) Zn + H2SO4  ZnSO4 + H2
62.
Which of the following is a redox reaction?
(A) NaCl + KNO3  NaNO3 + KCl
(B) CaC2O4+ 2NH4Cl  CaCl2 + H2C2O4
(C) Mg(OH)2+2NH4ClMgCl2+2NH4OH
(D) Zn + 2AgCN  2Ag + Zn(CN)2
17
Chemistry Vol ‐ 1.2 (Med. and Engg.) 63.
64.
65.
The violent reaction between sodium and
water is an example of _______.
(A) reduction
(B) oxidation
(C) redox reaction
(D) neutralisation reaction
The reaction
2H2O2(aq) 2H2O(l) + O2(g) is an example for
which of the following?
(A) oxidation
(B) reduction
(C) disproportionation
(D) neither oxidation nor reduction
In the reaction,
3Cl2 + 6NaOH  NaClO3 + 5NaCl + 3H2O
the element which loses as well as gains
electrons is _______.
(A) Na
(B) O
(C) Cl
(D) None of these
66.
Which of the following elements does NOT
show disproportionation tendency?
[NCERT Exemplar]
(A) Cl
(B) Br
(C) F
(D) I
67.
Identify disproportionation reaction.
[NCERT Exemplar]
(A) CH4 + 2O2  CO2 + 2H2O
(B) CH4 + 4Cl2  CCl4 + 4HCl
(C) 2F2 + 2OH  2F– + OF2 + H2O
(D) 2NO2 + 2OH  NO 2  NO3  H 2 O
68.
Which of the following change represents a
disproportionation reaction ?
(A) Cl2 + 2OH–  ClO– + Cl– + H2O
(B) Cu2O + 2H+  Cu + Cu2+ + H2O
3
(C)
(D)
71.
72.


P4(s)+ 3OH (aq)
+3H2O(l) P H3 g  + 3H 2 P O 2(aq)
(D)
All of the above
Which of the following halogens always show
only one oxidation state ?
(A) Cl
(B) F
(C) Br
(D) I
74.
Fluorine does not show positive oxidation
state due to which of the following reasons?
(A) Absence of s-orbitals
(B) Absence of p-orbitals
(C) Absence of d-orbitals
(D) Highest electronegativity
75.
The characteristic oxidation
in free metals is _______.
(A) Minus one
(B)
(C) One
(D)
76.
The oxidation number of hydrogen in LiH is
_______.
(A) + 1
(B) – 1
(C) 2
(D) 0
77.
Hydrogen can have oxidation number/s of
_______.
(A)  1 only
(B) + 1 only
(C) 0 only
(D)  1, 0, + 1
78.
O.N. of hydrogen in KH, MgH2 and NaH
respectively would be _______.
(A) –1, – 1 and –1
(B) +1, + 1, and + 1
(C) +2, +1 and –2
(D) –2, –3 and –1
79.
In which of the following oxidation number of
chlorine is + 5?
(A) Cl
(B) ClO
(C) ClO 2
(D) ClO 3
80.
In which of the following compounds, an
element exhibits two different oxidation
states?
[NCERT Exemplar]
(A) NH2OH
(B) NH4NO3
(D) N3H
(C) N2H4
8.3 Oxidation number
69.
70.
18
Oxidation number of metals is always _______.
(A) positive
(B) negative
(C) zero
(D) neutral
The oxidation number of an element in a
compound is evaluated on the basis of certain
rules. Which of the following rules is NOT
correct in this respect? [NCERT Exemplar]
(A) The oxidation number of hydrogen is
always +1.
(B) The algebraic sum of all the oxidation
numbers in a compound is zero.
The oxidation number of oxygen in peroxide
is _______.
(A) 2
(B) 1
(C) +1
(D) +2
The element with atomic number 9 can exhibit
oxidation state of _______.
(A) + 1
(B) + 3
(C)  1
(D) + 5
73.
1
(C)
An element in the free or the
uncombined state bears oxidation
number zero.
In all its compounds, the oxidation
number of fluorine is – 1.
number of atoms
[NCERT 1975]
Two
Zero
Chapter 08: Redox Reactions
81.
In which of the following compounds, iron has
the lowest oxidation number?
[MNR 1984]
(A) Fe(CO)5
(B) Fe2O3
(C) K4[Fe(CN)6]
(D) FeSO4.(NH4)2SO4.6H2O
82.
A metal ion M3+ loses 3 electrons, its oxidation
number will become _______. [CPMT 2002]
(A) + 3
(B) + 6
(C) 0
(D)  3
83.
The oxidation number of P in HP2 O7 ion is
_______.
(A) + 5
(B) + 6
(C) + 7
(D) + 3
84.
The oxidation number of Fe in [Fe(CN)6]3 ion
is _______.
(A) + 2
(B) + 3
(C)  2
(D)  3
85.
86.
87.
88.
The oxidation number of Mn is + 7 in
_______.
(A) manganese dioxide
(B) manganese chloride
(C) manganese sulphate
(D) potassium permanganate
In the conversion of K2Cr2O7 to K2CrO4, the
oxidation number of chromium _______.
(A) remains same
(B) increases
(C) decreases
(D) none of these
The oxidation number of C in sucrose
(C12H22O11) is _______.
(A) + 4
(B) + 3
(C) + 2
(D) zero
The change in oxidation number of Pb and Cl
in the reaction,

PbO2 +ClClO+Pb  OH 3 respectively are
_______.
(A) + 2, + 2
(C) + 2,  1
(B)
(D)
 2, + 2
 2, + 1
89.
The oxidation state of S in S2 O82 is _______.
(A) + 2
(B) + 4
(C) + 6
(D) + 7
90.
Oxidation number of iodine varies from
_______.
[CPMT 1982]
(A) – 1 to + 1
(B) – 1 to + 7
(C) + 3 to + 5
(D) – 1 to + 5
91.
Match List I with List II and select the correct
answer using the codes given below the lists:
List I
List II
(Compound)
(Oxidation state of N)
(A) NO2
i. + 5
(B) HNO
ii. – 3
(C) NH3
iii. + 4
(D) N2O5
iv. + 1
Codes :
(A) A B C D
ii. iii. iv. i.
(B) A B C D
iii. i. ii. iv.
(C) A B C D
iii. iv. ii. i.
(D) A B C D
ii. iii. i. iv.
92.
When SO2 is passed through acidic solution of
potassium dichromate, then chromium
sulphate is formed. Change in valency of
chromium is _______.
[CPMT 1979]
(A) + 4 to + 2
(B) + 5 to + 3
(C) + 6 to + 3
(D) + 7 to + 2
93.
The oxidation number of sulphur in S8, S2F2,
H2S respectively are _______.
[IIT 1999]
(A) 0, + 1 and  2
(B) + 2, + 1 and  2
(C) 0, + 1 and + 2
(D)  2, + 1 and  2
94.
Among the following, identify the species
with an atom in + 6 oxidation state.
[IIT Screening 2000]

(A) MnO 4
(B) Cr(CN)36
(C)
NiF62 
(D)
CrO2Cl2
95.
The oxidation number of Cr in CrO5 is _______.
(A) +3
(B) +5
(C) +6
(D) 0
96.
Calculate the oxidation number of the
underlined element Ba2XeO6.
(A) + 4
(B) + 8
(C) + 5
(D) + 6
97.
Oxidation number of C in CH3OH, CH2O,
HCOOH and C2H2 is respectively _______.
(B) +2, 0, +2, 2
(A) 2, 0, +2, 1
(C) 2, 0, +2, 0
(D) 2, 4, +2, 2
98.
The oxidation state of oxygen in H2O is
_______.
(A) 1
(B) +2
(C) 2
(D) +1
19
Chemistry Vol ‐ 1.2 (Med. and Engg.) 99.
Which of the following arrangements
represent increasing oxidation number of the
central atom?
[NCERT Exemplar]


2
(A) CrO 2 , ClO3 , CrO 4 , MnO 4
(B)
ClO3 , CrO 24  , MnO 4 , CrO 2
(C)
CrO 2 , ClO3 , MnO 4 , CrO 42 
(D)
CrO 24 , MnO 4 , CrO 2 , ClO3
100. The compound in which oxidation state of
metal is zero is _______.
(A) Fe2(CO)9
(B) Ni(CO)4
(C) Fe3(CO)9
(D) All of the above
101. The oxidation state of phosphorus is + 3 in
which of the following?
(A) Orthophosphorous acid
(B) Orthophosphoric acid
(C) Pyrophosphoric acid
(D) Metaphosphoric acid
102. In the compounds KMnO4 and K2Cr2O7, the
highest oxidation state is of which of the
following element?
(A) Potassium
(B) Manganese
(C) Chromium
(D) Oxygen
103. Oxidation number of nitrogen in NH3 is :
[CPMT 1979; Pb CET 2004]
(A) – 3
(B) + 3
(C) 0
(D) + 5
104. The oxidation number of chlorine in HOCl is
_______.
(A) – 1
(B) 0
(C) + 1
(D) + 2
105. Sulphur has lowest oxidation number in which
of the following?
[EAMCET 1993]
(A) H2SO3
(B) SO2
(C) H2SO4
(D) H2S
106. Oxidation number of oxygen in ozone (O3) is
_______. [MP PET 2000; MP PMT 2001]
(A) + 3
(B) – 3
(C) – 2
(D) 0
109. Oxidation number of Pt in [Pt(C2H4)Cl3]– is
_______.
[MLNR 1993]
(A) +1
(B) +2
(C) +3
(D) +4
110. The sum of the oxidation numbers of all the
carbons in C6H5CHO is _______.
[EAMCET 1986]
(A) + 2
(B) 0
(C) + 4
(D) – 4
111. Calculate the oxidation number of the
underlined element K4P2O7
(A) +3
(B) 5
(C) +5
(D) +6
112. In which compound, oxidation state of
nitrogen is +1?
[MP PMT 1989]
(A) NO
(B) N2O
(D) N2H4
(C) NH2OH
113. Carbon is in the lowest oxidation state in
_______.
[NCERT 1979; MH CET 1999]
(A) CH4
(B) CCl4
(D) CO2
(C) CF4
114. The oxidation states of sulphur in the anions
SO32  ,S2 O 42  and S2 O62 follows the order:
[CBSE PMT 2003]
2
2
2
(A) S2 O 6  S2 O 4  SO3
(B)
S2 O 24   SO32   S2 O62
(C)
SO32   S2 O 42   S2 O62 
(D)
S2 O 24   S2 O 62   SO32
115. H2S acts only as a reducing agent while SO2
can act both as a reducing and oxidising agent
because _______.
[AMU 1999]
(A) S in H2S has – 2 oxidation state and S in
SO2 has oxidation state + 4
(B) Hydrogen in H2S is more +ve than
oxygen
(C) Oxygen is more – ve in SO2
(D) None of the above
107. Oxidation number of cobalt in K[Co(CO)4] is
[KCET 1996]
_______.
(A) + 1
(B) + 3
(C) – 1
(D) – 3
116. In the conversion Br2  BrO3 , the
oxidation state of bromine changes from
[EAMCET 1990; AMU 1999;
_______.
RPMT 2002]
(A) – 1 to 0
(B) 0 to – 1
(C) 0 to + 5
(D) 0 to – 5
108. The charge on cobalt in [Co(CN)6]3– is
[CPMT 1985, 93]
_______.
(A) – 6
(B) – 3
(C) + 3
(D) + 6
117. Oxidation state of chlorine in perchloric acid
is _______.
[EAMCET 1989]
(A) – 1
(B) 0
(C) – 7
(D) + 7
20
Chapter 08: Redox Reactions
118. Sn++ loses two electrons in a reaction. What
will be the oxidation number of tin after the
reaction?
(A) + 2
(B) Zero
(C) + 4
(D) – 2
119. Oxidation number of carbon in H2C2O4 is
_______.
[CPMT 1982]
(A) + 4
(B) + 3
(C) + 2
(D) – 2
120. The oxidation state of Cr in [Cr(NH3)4Cl2]+
[AIEEE 2005]
is _______.
(A) +3
(B) +2
(C) +1
(D) 0
121. The oxidation number of Fe and S in iron
pyrites are _______.
[RPMT 1997]
(A) 4, – 2
(B) 2, – 1
(C) 3, – 1.5
(D) 3, – 1
+
122. The oxidation number of N in N2H5 is
[Pb. PMT 2001]
_______.
(A) – 3
(B) – 2
(C) – 1
(D) + 2

6
123. The oxidation state of I in H 4 IO is _______.
[CBSE PMT 1994]
(A) + 7
(B) + 5
(C) + 1
(D) – 1
124. The brown ring complex compound is
formulated as [Fe(H2O)5NO]SO4. The
oxidation state of iron is _______.
[EAMCET 1987; IIT 1987; MP PMT 1994;
AIIMS 1997; DCE 2000]
(A) 1
(B) 2
(C) 3
(D) 0
125. Oxidation state of oxygen in F2O is _______.
[BHU 1982; UPSEAT 2001;
MH CET 2002]
(A) + 1
(B) + 2
(C) –1
(D) –2
126. Oxidation number of P in KH2PO2 is which of
the following?
[CPMT 1987; MH CET 1999]
(A) + 1
(B) + 3
(C) + 5
(D) – 4
127. The oxidation number of phosphorus in
Ba(H2PO2)2 is _______.
[Kurukshetra CEE 1998; DCE 2004]
(A) – 1
(B) + 1
(C) + 2
(D) + 3
128. Oxidation number of Xenon in XeOF2 is
_______.
(A) 0
(B) 2
(C) 4
(D) 3
129. In XeO3 and XeF6 the oxidation state of Xe is
_______.
[MP PET 2003]
(A) + 4
(B) + 6
(C) + 1
(D) + 3
130. Oxidation numbers of the two nitrogen atoms
present in ammonium nitrate are respectively ?
(A) +3 and +3
(B) 0 and 0
(C) –3 and +5
(D) –1 and –1
131. Oxidation numbers of two Cl atoms in
bleaching powder, CaOCl2 are _______.
(A) – 1, – 1
(B) + 1, – 1
(C) + 1, + 1
(D) 0, – 1
132. In which compound chlorine has highest
oxidation number?
(A) KCl
(B) HClO
(C) HClO2
(D) HClO4
133. What is the oxidation number of sulphur in
Na2S4O6?
[AIIMS 1998; DCE 1999]
2
3
(B)
(A)
3
2
3
5
(D)
(C)
5
2
134. When CuSO4 reacts with KI, the oxidation
number of Cu changes by _______.
[BHU 1997]
(A) 0
(B) – 1
(C) 1
(D) 2
135. What is the oxidation number of Co in
[Co(NH3)4ClNO2] _______.
[BHU 1999]
(A) + 2
(B) + 3
(C) + 4
(D) + 5
136. The oxidation number of nickel in
K4[Ni(CN)4] is _______.
[JIPMER 1999]
(A) – 2
(B) – 1
(C) + 2
(D) 0
137. Oxidation number of nitrogen in NaNO2 is
_______.
[Pb. CET 2000]
(A) + 2
(B) + 3
(C) + 4
(D) – 3
138. In which of the following compounds, the
oxidation number of carbon is not zero?
(A) C12H22O11
(B) HCHO
(D) CH3COOH
(C) CH3CHO
21
Chemistry Vol ‐ 1.2 (Med. and Engg.) 139. In the substance Mg(HXO3), the oxidation
number of X is _______.
(A) 0
(B) +2
(C) +3
(D) +5
140. What will be the oxidation state of copper in
YBa2Cu3O7, if oxidation state of (Y) is +3 ?
(A) 7/3
(B) 7
(C) 3 and 5
(D) 3/7
141. The oxidation number of P in
SO 24
and that of
respectively _______.
(A) 3, +6 and +6
(B) +5, +6 and +6
(C) +3, +6 and +5
(D) +5, +3 and +6
PO34 ,
of S in
Cr in Cr2 O72 are
[Assam CEE 2015]
142. The oxidation state of chromium in the final
product formed by the reaction between KI
and acidified potassium dichromate solution is
which of the following?
[AIEEE 2005]
(A) +4
(B) +6
(C) +2
(D) +3
143. Oxidation number of S in H2S2O7 is _______.
(A) +4
(B) –6
(C) –5
(D) +6
144. The oxidation number of sulphur in H2S is
_______.
(A) –2
(B) +3
(C) +2
(D) –3
145. The oxidation state of S in H2SO4 is _______.
(A) +8
(B) +6
(C) +5
(D) +4
146. The oxidation state of sulphur in SO 24 is
_______.
[Bihar MEE 1996]
(A) +4
(B) 2
(C) +6
(D) –6
147. 3CuO + 2NH3  3Cu + N2 + 3H2O
In the above conversion, the oxidation number
of nitrogen is changing from _______.
(A) +5 to 0
(B) 0 to +2
(C) +2 to 0
(D) –3 to –5
148. Oxidation number of
_______.
(A) From + 5 to – 3
(B) From – 5 to – 3
(C) From – 5 to + 3
(D) From+10 to + 6
22
nitrogen
can
be
149. Oxidation state of nitrogen is CORRECTLY
given for _______.
(A)
(B)
(C)
(D)
Compound
HNO3
NH2OH
(N2H5)2SO4
Mg3N2
Oxidation state
–5
+1
+2
3
150. In which of the compounds does manganese
exhibit highest oxidation number?
(A) MnO2
(B) Mn3O4
(D) MnSO4
(C) K2MnO4
151. In which of the following compounds, the
oxidation number of iodine is fractional?
(B) I3
(A) IF7
(D) IF3
(C) IF5
152. The element, which shows minimum
oxidation number in its compound is _______.
[RPET 1992]
(A) Fe
(B) Mn
(C) Ca
(D) K
153. Oxidation number of N in HNO3 is _______.
[BHU 1997]
(A) –3.5
(B) +3.5
(C) –3, +5
(D) +5
154. Oxidation number of P in Mg2P2O7 is
_______.
[CPMT 1989; MP PMT 1995]
(A) +3
(B) +2
(C) +5
(D) –3
155. The oxidation number of hydrogen in MgH2 is
_______.
[CPMT 1976]
(A) +1
(B) –1
(C) +2
(D) –2
156. The oxidation number of carbon in CH2O is
_______.
[IIT 1982; EAMCET 1985; MNR 1990;
UPSEAT 2001; CPMT 1997]
(A) –2
(B) +2
(C) 0
(D) +4
157. The oxidation number of carbon in CH2Cl2 is
_______.
[CPMT 1976; Pb. PET 1999; AFMC 2004]
(A) 0
(B) + 2
(C) –2
(D) + 4
158. Which one of the following
oxidation number of iodine?
(A) KI3
(B)
(D)
(C) IF5
has the highest
[CPMT 1982]
KI
KIO4
Chapter 08: Redox Reactions
159. The oxidation state of Mo in Mo 2 Cl84  ion is
[EAMCET 1994]
_______.
(A) –4
(B) –2
(C) +6
(D) +2
160. The oxidation number of Cr in K2Cr2O7 is
_______.
[CPMT 1981, 85, 90, 93, 99; KCET 1992;
BHU 1988, 98; AFMC 1991, 99;
EAMCET 1986; MP PMT 1996, 99, 2002;
MP PET/PMT 1998; Bihar CEE 1995;
RPET 2000]
(A) +6
(B) –7
(C) +2
(D) –2
161. The valency of Cr in the complex
[Cr(H2O)4Cl2]+ is _______. [MP PMT 2000]
(A) 1
(B) 3
(C) 5
(D) 6
162. The oxidation number of Ba in barium
[Pb. PMT 2002]
peroxide is _______.
(A) +6
(B) +2
(C) 1
(D) +4
163. If HNO3 changes into N2O, the oxidation
number is changed by _______.
[BHU 1997; AFMC 2001]
(A) + 2
(B) –1
(C) 0
(D) +4
164. The oxidation states of the most
electronegative element in the products of the
reaction of BaO2 with dilute H2SO4 are
_______.
[IIT 1991; CBSE PMT 1992; BHU 2000]
(A) 0 and –1
(B) –1 and –2
(C) –2 and 0
(D) –2 and +1
165. Thiosulphate reacts differently with iodine and
bromine in the reactions given below:
2S2 O32   I 2  S4 O62   2I 
S2 O32   2Br2  5H 2 O  2SO 24   2Br   10H 
Which of the following statements justifies the
above dual behaviour of thiosulphate?
[NCERT Exemplar]
(A) Bromine is a stronger oxidant than
iodine.
(B) Bromine is a weaker oxidant than
iodine.
(C) Thiosulphate undergoes oxidation by
bromine and reduction by iodine in
these reactions.
(D) Bromine undergoes oxidation and iodine
undergoes reduction in these reactions.
166. The most common oxidation state of an
element is –2. The number of electrons present
in its outermost shell is _______.
[BHU 1983; NCERT 1974; CPMT 1977]
(A) 4
(B) 2
(C) 6
(D) 8
167. When KMnO4 is reduced with oxalic acid in
acidic solution, the oxidation number of Mn
changes from _______.
[MNR 1987; MP PET 2000;
CBSE PMT 2000; UPSEAT 2000, 02;
BHU 2003; AMU 2002]
(A) 7 to 4
(B) 6 to 4
(C) 7 to 2
(D) 4 to 2
168. In the chemical reaction
Cl2 + H2S  2HCl + S, the oxidation
number of sulphur changes from _______.
[MP PMT 1999]
(A) 0 to 2
(B) 2 to 0
(C) – 2 to 0
(D) – 2 to – 1
169. When KMnO4 acts as an oxidising agent and
ultimately forms [MnO4]–2, MnO2, Mn2O3,
Mn+2 then the number of electrons transferred
in each case respectively are _______.
[AIEEE 2002]
(A) 4, 3, 1, 5
(B) 1, 5, 3, 7
(C) 1, 3, 4, 5
(D) 3, 5, 7, 1
8.5 Balancing redox reactions in
terms of loss and gain of electrons
170. In acidic medium, reaction : MnO 4  Mn2+
[MP PMT 1989]
is an example of :
(A) Oxidation by three electrons
(B) Reduction by three electrons
(C) Oxidation by five electrons
(D) Reduction by five electrons
171. Which of the following equation is a balanced
[EAMCET 1980]
one ?
(A) 5BiO3 + 22H+ + Mn2+  5Bi3+
+ 7H2O + MnO 4
(B)
5BiO3 + 14H+ + 2Mn2+  5Bi3+
+ 7H2O + 2 MnO 4
(C)
2BiO3 + 4H+ + Mn2+  2Bi3+
+ 2H2O + MnO 4
(D)
6BiO3 + 12H+ + 3Mn2+  6Bi3+
+ 6H2O + 3 MnO 4
23
Chemistry Vol ‐ 1.2 (Med. and Engg.) 172. Which of the following equation is a balanced
equation?
(A) FeS2 + 4O2  8SO2 + 2Fe2O3
(B) 4FeS2 + 6O2  2Fe2O3 + 8SO2
(C) 4FeS2 + 11O2  2Fe2O3 + 8SO2
(D) 2FeS2 + 11O2  Fe2O3 + 6SO2
173. In a balanced equation,
H2SO4 + x HI H2S + y I2 + zH2O,
the values of x, y, z are _______.
[EAMCET 2003]
(A) x = 3, y = 5, z = 2
(B) x = 4, y = 8, z = 5
(C) x = 8, y = 4, z = 4
(D) x = 5, y = 3, z = 4
174. In the balanced equation,
5H2O2 + XClO2 + 2OH–  XCl– + YO2
+ 6H2O
The reaction is balanced if _______.
[CPMT–2000]
(A) X = 5, Y = 2
(B) X = 2, Y = 5
(C) X = 4, Y = 10
(D) X = 5, Y = 5
175. In the equation,
4M + 8CN– + 2H2O + O2  4[M(CN)2]–
+ 4OH–
Identify the metal M.
[AFMC 1998]
(A) Cobalt
(B) Iron
(C) Gold
(D) Zinc
176. The number of moles of KMnO4 reduced by
one mole of KI in alkaline medium is:
[CBSE PMT 2005]
(A) One fifth
(B) five
(C) One
(D) Two
177. The product of oxidation of I– with MnO 4 in
alkaline medium is _______.
[IIT-JEE Screening 2004]

(A) IO3
(B) I2
(C)
IO–
(D)
IO 4
178. In the following equation
ClO3 + 6H+ + X  Cl- + 3H2O, X
is _______.
(A) O
(B) 6e–
(D) 5e–
(C) O2
24
179. The number of electrons involved in the
reduction of Cr2 O72 in acidic solution to Cr3+
[EAMCET 1983]
is _______.
(A) 0
(B) 2
(C) 3
(D) 5
180. C2H6 (g) + nO2  CO2 (g) + H2O(l)
In this equation, the ratio of the coefficients of
CO2 and H2O is _______.
[KCET 1992]
(A) 1 : 1
(B) 2 : 3
(C) 3 : 2
(D) 1 : 3
181. In the following reaction,
Cr2 O72 +14H+ + 6I  2Cr3+ + 3H2O + 3I2.
Which element is reduced?
[CPMT 1976]
(A) Cr
(B) H
(C) O
(D) I
182. 2MnO 4 +5H2O2+ 6H+  2Z + 5O2 + 8H2O.
In this reaction Z is _______.
[Rajasthan PMT 2002]
(A) Mn+2
(B) Mn+4
(D) Mn
(C) MnO2
183. For the redox reaction,
MnO 4 + C2 O 42 + H+  Mn2+ + CO2 + H2O
the CORRECT coefficients of the reactants for
the balanced reaction are _______.
[IIT 1988, 92; BHU 1995; CPMT 1997;
RPMT 1999; DCE 2000; MP PET 2003]
(A)
(B)
(C)
(D)
MnO 4
C2O 42
H+
2
16
5
2
5
5
16
16
16
2
2
5
184. In which reaction there is a change in valency?
[NCERT 1971; CPMT 1971]
(A) 2NO2  N2O4
(B) 2NO2 + H2O  HNO2 + HNO3
(C) NH4OH  NH4+ + OH
(D) CaCO3  CaO + CO2
185. In which of the following reactions there is no
change in valency?
[NCERT 1974; CPMT 1978]
(A) 4KClO3  3KClO4 + KCl
(B) SO2 + 2H2S  2H2O + 3S
(C) BaO2 + H2SO4  BaSO4 + H2O2
(D) 2BaO + O2  2BaO2
Chapter 08: Redox Reactions
186. The more positive the value of E, the greater
is the tendency of the species to get reduced.
Using the standard electrode potential of redox
couples given below, find out which of the
following is the strongest oxidising agent.
E values:
Fe3+/Fe2+ = + 0.77; I2(s)/I– = + 0.54;
Cu2+/Cu = + 0.34; Ag+/Ag = + 0.80 V
[NCERT Exemplar]
3+
(B) I2(s)
(A) Fe
(C) Cu2+
(D) Ag+
Applications of redox reactions:
187. Which of the following involves the redox
reaction?
(A) Corrosion of metals
(B) Extraction of metals from their ores
(C) Respiration
(D) All of these
188. Sulphide minerals are converted to oxides by
_______.
(A) cracking
(B) coke
(C) roasting
(D) bleaching
189. Bleaching is _______.
(A) decolourization of coloured material
(B) extraction and purification of metals
(C) destruction of metals by oxidation
(D) using oxygen for biological redox
reactions
190. The reaction involved in household bleaching
is _______.
(A) acid-base neutralisation
(B) redox
(C) precipitation reaction
(D) complex reaction
191. ______ is used as an oxidizing agent to bleach
dark hair by a redox reaction.
(B) HO2
(A) H2O
(D) HO
(C) H2O2
192. To remove stains from clothes ______ is used.
(A) NaCl
(B) NaOH
(C) Na2O2
(D) NaOCl
193. The burning of methane in air by oxidation is
an example of _______.
(A) bleaching
(B) metallurgy
(C) combustion
(D) respiration
194. ________ is the destruction of metals by
oxidation.
(A) Combustion
(B) Bleaching
(C) Roasting
(D) Corrosion
195. In the rusting of iron, iron has been _______.
(A) Oxidized
(B) Reduced
(C) Vapourised
(D) Decomposed
196. Corroded iron is _______.
(A) FeO3.H2O
(B)
(C) 2Fe2O3.H2O
(D)
2FeO3. H2O
Fe2O3.2H2O
197. The metal used in galvanizing of iron is
_______.
[MP PET 1985, 96]
(A) Pb
(B) Zn
(C) Al
(D) Sn
198. Prevention of corrosion of iron by zinc coating
is called _______.
[MP PMT 1993; CPMT 2002]
(A) Galvanization
(B) Cathodic protection
(C) Electrolysis
(D) Photo–electrolysis
199. Respiration is _______.
(A) Oxidation
(B)
(C) Both (A) and (B) (D)
Reduction
None of these
200. The overall effect of respiration is similar to
which of the following?
(A) Photosynthesis (B) Corrosion
(C) Combustion
(D) Bleaching
Miscellaneous
201. The process in which oxidation number
increases is known as _______. [CPMT 1976]
(A) Oxidation
(B) Reduction
(C) Auto-oxidation
(D) None of the above
202. Following reaction describes the rusting of
iron 4Fe + 3O2  4Fe3+ + 6O2.
Which one of the following statement is
INCORRECT ?
[NCERT 1981; MNR 1991; AIIMS 1998]
(A) This is an example of a redox reaction.
(B) Metallic iron is reduced to Fe3+.
(C) Fe3+ is an oxidising agent.
(D) Metallic iron is a reducing agent.
25
Chemistry Vol ‐ 1.2 (Med. and Engg.) 203. The process of conversion of SnCl4 to SnCl2 is
_______.
(A) oxidation
(B) reduction
(C) neither reduction nor oxidation
(D) both oxidation and reduction
204. When iron or zinc is added to CuSO4
solution, copper is precipitated. It is due to
which of the following?
[CPMT 1974, 79]
+2
(A) Oxidation of Cu
(B) Reduction of Cu+2
(C) Hydrolysis of CuSO4
(D) Ionisation of CuSO4
205. SnCl2 gives a precipitate with a solution of
HgCl2.In this process, HgCl2 is _______.
[CPMT 1983]
(A) Reduced
(B) Oxidised
(C) Converted into a complex compound
containing both Sn and Hg
(D) Converted into a chloro complex of Hg
206. Hot concentrated sulphuric acid is a
moderately strong oxidizing agent. Which of
the following reactions does NOT show
[NEET P-II 2016]
oxidizing behaviour?
(A) CaF2 + H2SO4  CaSO4 + 2HF
(B) Cu + 2H2SO4  CuSO4 + SO2 +2H2O
(C) 2S + 2H2SO4  2SO2 + 2H2O
(D) C + 2H2SO4  CO2 + 2SO2 + 2H2O
207. Which of the following acts both as an
oxidizing as well as reducing agent?
(A) HNO3
(B) HNO2
(C) HI
(D) H2SO4
208. In which of the following reactions oxidation
and reduction is occurring?
(A) AgNO3 + HCl  AgCl + HNO3
(B) H2 + Cl2  2HCl
(C) BaCl2 + H2SO4  BaSO4 + 2HCl
(D) KI + HCl  KCl + HI
209. Which one of the following reactions does
NOT involve either oxidation or reduction?
[EAMCET 1982]

(A) VO2  V2O3
(B)
26
Na  Na+
(C)
CrO24  Cr2 O72
(D)
Zn2+  Zn
210. The atomic number of an element which
shows the oxidation state of +3 is:
[CPMT 1989, 94]
(A) 13
(B) 32
(C) 33
(D) 17
211. Most common oxidation number of an
element is 1. The number of electrons in its
outermost shell is _______.
(A) 7
(B) 1
(C) 8
(D) 4
212. Calculate the oxidation number of sulphur in
S2 O72 .
(A) +6
(B) 6
(C) +7
(D) +4
213. Oxidation number of sulphur in S2 O32 is
_______.
(A) – 2
(C) + 6
(B)
(D)
[CPMT 1979]
+2
0
214. Oxidation state of Fe in Fe3O8 is ________.
[CBSE 1999]
(A) + 3/2
(B) + 4/5
(C) + 15/4
(D) + 16/3
215. In ferrous ammonium sulphate oxidation
number of Fe is _______.
[CPMT 1988]
(A) +3
(B) +2
(C) +1
(D) –2
216. Oxidation number of carbon in carbon
suboxide (C3O2) is _______.
(A) +2/3
(B) +4/3
(C) +4
(D) 4/3
217. E values of some redox couples are given
below. On the basis of these values, choose
the CORRECT option.
E values:
Br2/Br– = + 1.09; Ag+ /Ag(s) = + 0.80
Cu2+/Cu(s) = + 0.34; I2(s)/I– = + 0.54
[NCERT Exemplar]
(A) Cu will reduce Br–
(B) Cu will reduce Ag
(C) Cu will reduce I–
(D) Cu will reduce Br2
Chapter 08: Redox Reactions
218. In the following reaction, which is the species
being oxidized?
3

2
2Fe(aq)
+ 2I(aq)
 I2(aq) + 2Fe(aq)
(A)
(C)
Fe3+
I2
(B)
(D)
I
Fe2+
219. Oxidation number of carbon in diamond is
_______.
(A) – 4
(B) + 4
(C) 0
(D) + 2
220. The largest oxidation number exhibited by an
element depends on its outer electronic
configuration. With which of the following
outer electronic configurations the element
will exhibit largest oxidation number?
[NCERT Exemplar]
1
2
(B) 3d3 4s2
(A) 3d 4s
5
1
(C) 3d 4s
(D) 3d5 4s2
221. In which one of the following compounds the
oxidation number of carbon is zero?
(B) C6H6
(A) C2H6
(C) CH3Cl
(D) C12H22O11
222. Which one of the following nitrates will leave
behind a metal on strong heating?
[AIEEE 2003]
(A) Ferric nitrate
(B) Copper nitrate
(C) Manganese nitrate
(D) Silver nitrate
223. In which one of the following changes there
are transfer of five electrons?
[NCERT 1982]
(A) MnO 4 Mn2+
(B)
CrO 2-4  Cr3+
(C)
MnO 4 MnO2
(D)
Cr2 O72 2Cr3+
224. How many moles of K2Cr2O7 are reduced by
1 mole of formic acid ?
(A) 1/3Mole
(B) 1Mole
(C) 2/3Mole
(D) 2Mole
225. Which halide is not oxidised by MnO2?
[MNR 1985; JIPMER 2000]
(A) F
(B) Cl
(C) Br
(D) I
226. In the reaction,
8Al + 3Fe3O4  4Al2O3 + 9Fe, the number
of electrons transferred from reductant to
oxidant is _______.
(A) 8
(B) 4
(C) 16
(D) 24
227. Oxidation number of S in S2Cl2 is _______.
[CPMT 1979]
(A) +1
(B) – 1
(C) + 6
(D) 0
228. To prevent rancidification of food material,
which of the following is added?
[CPMT 1996]
(A) Reducing agent
(B) Anti-oxidant
(C) Oxidizing agent
(D) None of these
229. The ultimate products of oxidation of most of
hydrogen and carbon in food stuffs are
_______.
[DCE 2001]
(A) H2O alone
(B) CO2 alone
(C) H2O and CO2
(D) None of these
230. A particular gas bleaches the colour of flowers
by reduction while the other by oxidation.
Name the gases.
[EAMCET 1980]
(A) CO and Cl2
(B) SO2 and Cl2
(C) H2S and Br2
(D) NH3 and SO2
231. The oxidation number and covalency of
sulphur in the sulphur molecule (S8) are
respectively _______.
[NCERT 1977]
(A) 0 and 2
(B) 6 and 8
(C) 0 and 8
(D) 6 and 2
232. Using the standard electrode potential, find out
the pair between which redox reaction is NOT
feasible.
E values:
Fe3+/Fe2+ = + 0.77; I2/I– = + 0.54;
Cu2+/Cu = + 0.34; Ag+/Ag = + 0.80 V
[NCERT Exemplar]
3+
–
(A) Fe and I
(B) Ag+ and Cu
(C) Fe3+ and Cu
(D) Ag and Fe3+
27
Chemistry Vol ‐ 1.2 (Med. and Engg.) Answers to MCQs
1. (C)
11. (B)
21. (C)
31. (D)
41. (A)
51. (A)
61. (C)
71. (B)
81. (A)
91. (C)
101. (A)
111. (C)
121. (A)
131. (B)
141. (B)
151. (B)
161. (B)
171. (B)
181. (A)
191. (C)
201. (A)
211. (A)
221. (D)
231. (A)
2. (D)
12. (C)
22. (A)
32. (C)
42. (A)
52. (C)
62. (D)
72. (C)
82. (B)
92. (C)
102. (B)
112. (B)
122. (B)
132. (D)
142. (D)
152. (D)
162. (B)
172. (C)
182. (A)
192. (D)
202. (B)
212. (A)
222. (D)
232. (D)
3. (B)
13. (C)
23. (D)
33. (C)
43. (A)
53. (D)
63. (C)
73. (B)
83. (B)
93. (A)
103. (A)
113. (A)
123. (A)
133. (D)
143. (D)
153. (D)
163. (D)
173. (C)
183. (A)
193. (C)
203. (B)
213. (B)
223. (A)
4. (A)
14. (D)
24. (D)
34. (B)
44. (B)
54. (D)
64. (C)
74. (D)
84. (B)
94. (D)
104. (C)
114. (B)
124. (B)
134. (C)
144. (A)
154. (C)
164. (B)
174. (B)
184. (B)
194. (D)
204. (B)
214. (D)
224. (A)
5. (D)
15. (B)
25. (D)
35. (C)
45. (B)
55. (C)
65. (C)
75. (D)
85. (D)
95. (C)
105. (D)
115. (A)
125. (B)
135. (A)
145. (B)
155. (D)
165. (A)
175. (C)
185. (C)
195. (A)
205. (A)
215. (B)
225. (A)
6. (A)
16 (B)
26. (B)
36. (A)
46. (A)
56. (D)
66. (C)
76. (B)
86. (A)
96. (B)
106. (D)
116. (C)
126. (A)
136. (D)
146. (C)
156. (C)
166. (C)
176. (D)
186. (D)
196. (C)
206. (A)
216. (B)
226. (D)
7. (B)
17. (B)
27. (D)
37. (B)
47. (C)
57. (A)
67. (D)
77. (D)
87. (D)
97. (A)
107. (C)
117. (D)
127. (B)
137. (B)
147. (C)
157. (A)
167. (C)
177. (A)
187. (D)
197. (B)
207. (B)
217. (D)
227. (A)
8. (A)
18. (D)
28. (A)
38. (D)
48. (D)
58. (D)
68. (D)
78. (A)
88. (B)
98. (C)
108. (C)
118. (C)
128. (C)
138. (C)
148. (A)
158. (D)
168. (C)
178. (B)
188. (C)
198. (A)
208. (B)
218. (B)
228 (B)
9. (A)
19. (B)
29. (B)
39. (B)
49. (C)
59. (B)
69. (A)
79. (D)
89. (D)
99. (A)
109. (B)
119. (B)
129. (B)
139. (C)
149. (D)
159. (C)
169. (C)
179. (C)
189. (A)
199. (C)
209. (C)
219. (C)
229. (C)
10. (C)
20. (C)
30. (A)
40. (B)
50. (A)
60. (C)
70. (A)
80. (B)
90. (B)
100. (D)
110. (D)
120. (A)
130. (C)
140. (A)
150. (C)
160. (A)
170. (D)
180. (B)
190. (B)
200. (C)
210. (A)
220. (D)
230. (B)
Hints to MCQs
4.
It is the process in which electrons are lost
(de-electronation).
6.
Sn2+  Sn4+ + 2e. This is an oxidation
reaction.
7.
Zn + Cu2+  Zn2+ + Cu
Here Zn is oxidized to Zn2+.
8.
The change of hydrogen into proton involves
the loss of a hydrogen atom. Hence it is an
oxidation reaction.
10.
16.
2MnO4 +5H2O2+6H+ Mn2++5O2+8H2O
17.
S + 2e  S2
20.
(A)
(B)
(C)
(D)
Oxidation
0
0
Zn + I2  ZnI2
In this reaction, Zn atom gets oxidised to
Zn2+ ion and iodine gets reduced to I.
11.
2Ag+ + Cu  Cu++ + 2Ag.
12.
O.N. of Mn is already oxidized with change in
oxidation number from +2 to +7, whereas all
other ions can get oxidized.
28
CO  CO2
+2
+2
+1
0
CuO  CuCl2
H2O  H2
0
+4
C  CO2
Only in option (C), ON. of hydrogen
decreases from + 1 to 0 and hence H2O gets
reduced to H2.
+2 –1
Reduction
+4
+2
21.
+5
3
NO 3  NH 4
Here, O.N. of N decreases from +5 in NO 3 to
3 in NH 4 and hence NO 3 gets reduced to NH 4 .
23.
Ag2O + H2O + 2e  2Ag + 2OH
Here the O.N. of Ag decreases from +1 in
Ag2O to 0 in Ag. Therefore, silver is reduced.
Chapter 08: Redox Reactions
24.
(A)
(B)
+4
+3
45.
Hydrogen peroxide (H2O2) acts as an oxidising
as well as reducing agent.
+5
+2
46.
All redox reactions are exothermic.
48.
BaCl2 + H2SO4  BaSO4 + 2HCl
This is an example of double displacement
reaction. Here, the oxidation number of all the
interacting species remains the same after the
reaction. Hence, it is not a redox reaction.
NO2  NO2 , N is reduced.
NO3  NO, N is reduced.
+5
(C)
(D)
29.
0
–3

3

4
NO  NH , N is reduced.
–3
0
0
+1 1
NH 4  N2 , N is oxidised.
2Li + H2  2Li H
Here, H2 acts as an oxidizing agent since the
O.N. of Li increases from 0 to + 1 and that of
H2 decreases from 0 to 1.
+3
2
0
+2
Reduction
49.
3
0
1
P +NaOH 
 P H 3  NaH 2 P O 2
30.
2FeCl3 + H2S 2 FeCl2 + 2 HCl + S
Here, FeCl3 acts as an oxidizing agent and
oxidizes H2S to S. Hence option (A) is correct.
50.
The oxidation number of P changes from 0 to
+1 (in KH2PO2) and 0 to –3 (in PH3). Hence P
is oxidised as well as reduced.
31.
HClO1 is the strongest oxidizing agent. The
correct order of oxidizing power is
53.
H2O + Br2  HOBr + HBr
+1
+3
0
+7
+5
HClO1 > HClO2 > HClO3 > HClO4
32.
Because I2 is a reducing agent.
33.
Reducing agent donates electrons.
37.
2NO + 3I2 + 4 H2O  2 NO3 + 6 I + 8 H+
+5
+2
Here, NO acts as a reducing agent and reduces
I2 to I. Since the O.N. of nitrogen changes
from + 2 in NO to + 5 in NO 3 , it undergoes
oxidation whereas in reaction (C) and (D), the
oxidation number of nitrogen changes from +2
in NO to +1 in N2O, thereby undergoing
reduction. Hence NO acts as a reducing agent
in reaction (B).
38.
54.
The oxidation number of Ni changes from 0 to
+2. Hence, Ni acts as a reducing agent in the
given equation.
40.
H2O2 acts as a reducing agent in the reaction
between O3 and H2O2.
41.
Reduction (oxidizing agent)
Ag2O + H2O2  2Ag + H2O + O2
Oxidation (reducing agent)
–1
Only in reaction (D), the O.N. of element
changes.
0
+1
0
+2
Zn + 2HCl  ZnCl2 + H2
55.
(A)
H2SO4 + 2NaOH  Na2SO4 + 2H2O
is a neutralization reaction.
0
(B)
Iodine (I) is the least electronegative among
the given set of ions. Therefore, electrons are
more easily given away thereby resulting in Ibeing the strongest reducing agent.
39.
+1
In the above reaction, the oxidation number of
Br2 increases from zero (in Br2) to +1 (in
HOBr) and decreases from zero (in Br2) to –1
(in HBr). Thus Br2 is oxidised as well as
reduced and hence it is a redox reaction.
0
2O3  3O2
O.N. of O in both O3 and O2 is zero,
therefore, it is not a redox reaction but is
actually a decomposition reaction.
0
0
+2
2
2N2 + O2  2N O
Here O.N. of N increases from 0 in N2
to + 2 in NO and that of O decreases
from 0 in O2 to  2 in NO therefore, it is
a redox reaction.
(D) H2O(l)  H2O(g)
Only state of matter changes and hence
there is no change in O.N.
Thus, option (C) is correct.
(C)
56.
Only in reaction (D), the O.N. of the elements
undergo a change
+1 2
+2
0
0
+4
Cu2S + 2 FeO  2 Cu + 2 Fe + SO2
29
Chemistry Vol ‐ 1.2 (Med. and Engg.) i.e., O.N. of Cu decreases from + 1 to 0 and
that of Fe decreases from + 2 to 0 while that of
S increases from  2 to + 4. Therefore, it is a
redox reaction.
57.
58.
67.
–4 +1
O. N. of Ag changes from +2 to 0 (undergoing
reduction) whereas that of Zn changes from 0
to +4 (undergoing oxidation).
Oxidation
0
+2
0
H2
0
None of the element in reaction (C) undergoes
a change in O.N., therefore, (C) is not a redox
reaction.
+1 1
+1 +5 2
+1 1
0
+1
+1
0
2 Na + 2H2O  2NaOH + H2
It is a redox reaction since here Na is oxidised
to NaOH while H2O is reduced to H2.
0
65.
66.
30
+5
Disproportionation reactions are those
reactions in which the same substance gets
oxidised as well as reduced. Fluorine is the
most electronegative element of the entire
periodic table. In its compounds, it always has
an oxidation number of –1. Hence, F can only
be reduced and so, it does not show
disproportionation tendency.
+3 – 2
+2 –1
+1 –2
+5 – 2
+1 – 2
The element with atomic number 9 is Fluorine
and the oxidation number of fluorine is always
1 in all of its compounds. Fluorine has the
electronic configuration 1s22s22p5. Hence, it
can exhibit oxidation state of –1.
77.
1 in metal hydrides, zero in H2 and + 1 in
halogen acids.
79.
(A)
(B)
(C)
–1
3Cl2+6NaOH NaClO3 + 5 NaCl + 3H2O.
Here, O.N. of Cl increases from 0 in Cl2 to + 5
in NaClO3 and decreases to  1 in NaCl,
therefore, Cl gets both oxidised and reduced.
–2 +1
–1
72.
+1 +5 2
Oxidation
Reduction
2+

+
Zn  Zn + 2e , Ag + e  Ag
–2 +1
+1 –1
Oxidation number of hydrogen is different in
different compounds.
Hydrogen exhibits + 1 oxidation number in
hydrogen halides, –1 in hydrides and zero in
H2 molecule.
In fact, it is a double decomposition reaction.
63.
2F2 + 2 OH–  2F– + OF2 + H2O
+4 –1
70.
AgNO3 + NaCl  AgCl + NaNO3
62.
(C)
0
In the reaction (A) Cl2, reaction (B) Cu and
reaction (C) P, undergo disproportionation
reaction.
reduction
61.
CH4 + 4Cl2  CCl4 + 4HCl
68.
+1 –1
0
+ Br2  2HBr
–2
(B)
+4 –2
Oxidation
+1
(D) 2NO2 +2OH–  N O 2 + NO3– + H2O.
In option (D) nitrogen is both oxidized i.e.
(O.N. increases from +4 to +5) as well as
reduced i.e., (O.N. decreases from + 4 to + 3).
Hence, option (D) is an example of
disproportionation reaction.
2Fe3+ + Sn2+  2Fe2+ + Sn4+
60.
+4 –2
CH4 + 2O2  CO2 + 2H2O.
0
Cu + 2AgNO3  Cu(NO3)2 + 2Ag
This is a redox reaction.
59.
0
(A)
–4 +1
Reduction
+1
Disproportionation reactions are those
reactions in which the same substance gets
oxidised as well as reduced.
(D)
80.
O.N. of Cl in Cl =  1
O.N. of Cl in ClO = x  2 =  1
or x = +1
O.N. of Cl in ClO 2 = x + 2  ( 2) =  1
or x = + 3
O.N. of Cl in ClO 3 = x + 3  ( 2) =  1
or x = + 5
Nitrogen exhibits two different oxidation
states in NH4NO3.
NH4NO3  NH 4 + NO3
i.
Let the oxidation number of nitrogen be
‘x’ in NH 4
x + 4 = +1
x = –3

the oxidation state of nitrogen = –3 in
NH 4
Chapter 08: Redox Reactions
ii.
Let the oxidation number of nitrogen be
‘y’ in NO3
y + (–2  3) = –1
y = +5
the oxidation state of nitrogen = +5 in
NO3

88.
82.
In metal carbonyls, the O.N. of metal is
always zero.
(A) O.N. of Fe in Fe (CO)5 is x + 5  0 = 0
or x = 0
(B) O.N. of Fe in Fe2O3 is 2x + 3  ( 2) = 0
or x = + 3
(C) O.N. of Fe in K4[Fe(CN)6] is
4  (+ 1) + x + 6  ( 1) = 0 or x = + 2
(D) O.N.
of
Fe
in
FeSO4
of
FeSO4.(NH4)2SO4.6H2O
= x + 6 + 4 ( 2) = 0
x2=0
x=+2
Thus, option (A) is correct.
 1 x
H P
2


(+ 1) + (2x) + ( 2  7) =  1 or x = + 6.
84.
1 
 x


 Fe (CN)6 

x + (1  6) =  3
or
x=3+6
x=+3
+1
x
K Mn O4
1 + x + (2  4) = 0 or x = +7.
86.
O.N. of Cr in K2Cr2O7 is
2  (+ 1) + 2 x + (2  7) = 0 or x = + 6
Similarly, O.N. of Cr in K2CrO4 is
2  (+1) + x + (2  4) = 0 or x = + 6.
87.

x
1
2
C12
H 22
O11
12x + 22  (+1) + 11  ( 2) = 0 or x = 0.
1
92.
The oxidation no. of chromium in K2Cr2O7 is
+6 and in Cr2(SO4)3 is +3.
0
+1 1 +1 2
93.
S8, S2F2, H2S
Here O.N. of S in S8, S2F2 and H2S
respectively are 0, + 1 and  2.
94.
O.N. of Cr in CrO2Cl2 is
x + 2  ( 2) + 2  ( 1) = 0
or x  4  2 = 0 or x = + 6
95.
The structure of CrO5 is
O
O
Cr
O
O
O
i.e., it has four peroxide bonds each having an
O.N. of 1 and one double bond in which
O.N. of O is  2. Therefore, the O.N. of Cr in
CrO5 is
x + 4  (1) + 1 (2) = 0 or x = + 6
96.
Ba2XeO6 = (2  2) + x + (–2  6)
= 4 + x – 12 = 0,
x=8
97.
CH3OH: x + 3 + (–2) + 1 = 0
x + 3 –2 + 1 = 0
x+2=0
x = –2
2
85.

+2
1
O.N of S = x
2x + (–2  8) = –2
2x – 16 = –2
2x = –2 + 16 = 14
14
x=
= +7
2
2 
O 7 
3
+12
1
89.
M3+  M6+ + 3e. Thus, the O.N. of
metal = + 6.
83.
1
PbO2 + (Cl)  (ClO) + Pb (OH) 3
Change in O.N. of Pb = + 2  4 =  2
Change in O.N. of Cl = + 1  ( 1) = + 2
Hence option (B) is correct.
Thus, nitrogen exhibits two different
oxidation states in NH4NO3.
81.
+4 2
CH2O: x + 2 + (–2) = 0
x = 0.
HCOOH: 1 + x + (–2) + (–2) + 1 = 0
x–2=0
x=2

C2H2: 2x + 2 = 0
2x = –2
x = –1
Oxidation number of C in CH3OH, CH2O,
HCOOH and C2H2 is –2, 0, 2, and –1
respectively.
31
Chemistry Vol ‐ 1.2 (Med. and Engg.) 99.
The oxidation states of each central atom in
the given species are as follows:
+3
+5
+6
+7
CrO2–, ClO3–, Cr O 24  , Mn O 4
The arrangement represented in option (A) is
the correct answer.
101. The molecular formula of orthophosphorous
acid is H3PO3.
Oxidation state of P is 3(1) + x + 3(2) = 0.
therefore, x = +3
102. In KMnO4 : 1 + x + (–2  4) = 0
1+x–8=0
x=7

Oxidation no. of Mn = +7.


In K2Cr2O7 : +2 + 2x + (–2  7) = 0
2x – 12 = 0
x = 6.
Oxidation no. of Cr = +6
Manganese has the highest oxidation state.


105. H 2SO3  4 ; SO2  4


H 2SO4  6 ; H 2S  2 .
108. In [Co(CN)6]3– complex, Co shows +3
oxidation state.
2 +1
1
109. [Pt(C2H4) Cl3]–
x + 2(– 2) + 4(+ 1) + 3(– 1) = –1
x – 4 + 4 – 3 = – 1 or x = – 1 + 3, x = + 2
110. In C6H5CHO :
6x + 5 + x + 1 + (–2) = 0
7x + 4 = 0
4
x= 
7
Sum of the oxidation numbers of all the
carbons in
 4   4 
C6H5CHO = 6x + x = 6  
+ 

 7   7 
 28
=
=–4
7
111. K4P2O7
4(+1) + 2x + 7(–2) = 0
2x – 10 = 0
2x = +10

x = +5
112. The oxidation state of N in NO, N2O, NH2OH
and N2H4 are +2, +1, –1, –2 respectively.
32
114. S2 O 24   SO 32   S2 O 62 
Oxi. state of sulphur in S2 O 24  = + 3
Oxi. state of sulphur in SO 32  = + 4
Oxi. state of sulphur in S2 O 62  = + 5
115. In H2S, sulphur shows –2 oxidation state and
in SO2 it shows +4 oxidation state. Hence, SO2
shows both oxidising and reducing properties.
5
0
 BrO3 , in this reaction oxidation
116. Br2 
state changes from 0 to + 5.

117. HClO4
1 + x  2  4 = 0; 1 + x  8 = 0
x = 8  1 = +7 oxidation state.
118. Sn2+  Sn4+ + 2e–

119. H 2 C2 O 4
2 + 2x – 2  4 = 0; 2x = 8 – 2 = 6
6
x=
= + 3.
2
120. [Cr(NH3)4Cl2]+
x + 4  (0) 2 = 1
 x + 0  2 = 1.
 x = 1 + 2 = + 3.
121. The molecular formula of iron pyrites is FeS2

FeS2
x4=0
x = +4

FeS2
4 + 2x = 0
2x = 4
4
x =
= 2.
2
124. In this complex, iron is a central metal atom
showing + 2 oxidation state.
2 2 1
128. XeOF2
x + (– 2) + 2(– 1) = 0 or x – 2 – 2 = 0 or x = +4
129. The oxidation state of Xe in both XeO3 and
XeF6 is + 6


XeO3
XeF6
x–23=0
x = +6
x–6=0
x = +6
Chapter 08: Redox Reactions
130. NH4(NO3)2 comprises of NH +4 and NO3 ions.

In NO3 , x + 3  (–2) = –1
x – 6 = –1
x=5
In NH4(NO3)2, x + 4 + 5 – 6 = 0
x = –3
The oxidation numbers of the two nitrogen
atoms present in NH4(NO3)2 are –3 and +5
respectively.
131. Two Cl atoms show +1 and –1 oxidation state.

10
5
= .
4
2
+2
 K2SO4 + CuI2
134. CuSO4 + 2KI 
+1
2CuI2 
 Cu 2 I 2 +I2

135. [Co(NH3 ) 4 ClNO 2 ]
x + 4(0) + 1(–1) + 1(–1) = 0
x+0–1–1=0
x – 2 = 0; x = +2

136. K 4 [Ni(CN)4 ]
= 4  (+1) + x + 4  (–1) = 0
= +4 + x – 4 = 0  x = 0
137. Let the oxidation number of N in NaNO2 be x.
1 + x + (–2)  2 = 0
1 + x – 4 = 0; x = +3
138. (A) C12H22O11 : 12(x) + 22(+1) + 11(– 2) = 0
12x + 22 – 22 = 0 or x = 0
(B) HCHO : x + 2(+1) + (– 2) = 0
x + 2 – 2 = 0 or x = 0
(C) CH3CHO : 2x + 4(+ 1) – 2 = 0
2x + 4 – 2 = 0
2x + 2 = 0 or x = 1
(D) CH3COOH : 2x + 4(+1) + 2(– 2) = 0
2x + 4 – 4 = 0 or 2x = 0
or x = 0
Hence option (C) is correct.
+2

+3 + 4 + 3x – 14 = 0
3x – 7 = 0
x = 7/3
141.
PO34
SO24 
Cr2 O72 
x – 8 = 3 x – 8 = 2 2x – 14 = 2
x = 3 + 8 x = 2 + 8 2x = 2 + 14
x = +5
x = +6
2x = 12
x = +6
142. K2Cr2O7 + 6KI + 7H2SO4  4K2SO4
+ Cr2(SO4)3 + 7H2O + 3I2
133. Na 2 S4 O6
2 + 4x –12 = 0
4x = 10  x =
140. YBa2Cu3O7
+1 2
139. Mg(HXO3)
(+ 2) + (+ 1) + x + 3(– 2) = 0 or +3 + x – 6 = 0
x = + 3.
*
+3
Cr2 (SO4 )3  2Cr +3SO2-4
144. H2S [O.N. of H = +1]
(+1)  2 + x = 0
2 + x = 0; x = – 2
+1 2
145. H2SO4  2(+ 1) + x + 4(– 2) = 0
 + 2 + x – 8 = 0 or x = + 6
146. SO24
x + 4(–2) = – 2
x–8=–2
x = – 2 + 8 or x = + 6
147. In CuO, x –2 = 0, x = 2
In Cu, the oxidation no. is 0

Change in the oxidation no is +2 to 0.
149. (A)
(B)
(C)
O.N. of N in HNO3 = +1 + x+ 3 (2) = 0
or x = + 5
O.N. of N in NH2OH= x +2 1  2+1= 0
or x = 1
O.N. of N in  N 2 H5 2 =2(2x + 5)= + 2
2
or x = 2
O.N. of N in Mg3N2 = 3  (+ 2) + 2x = 0
or x = 3
Thus, option (D) is correct.
(D)
150. Highest O.N. of Mn in K2MnO4 is 2+ x  8= 0
or x = + 6
while in all other compounds O.N. of Mn is
lower than 6, i.e.,
+4
+8/3
+2
MnO2, Mn3O4, MnSO4
151. In I3 , the O.N. of I = 3x =  1
or x = 1/3.
33
Chemistry Vol ‐ 1.2 (Med. and Engg.) x +1 –2 –1
152. O. N. of Fe : + 2, + 3
Mn : + 3
Ca : + 2
K : + 1.
Hence option (D) is correct.
161. [Cr(H2O)4Cl2]+
x + 8(+ 1) + 4 (– 2) + 2(– 1) = + 1
x+8–8–2=+1
x=+3
+1 x –2
153. H N O3
1 + (x) +3(– 2) = 0 or x – 5 = 0 or x = + 5
+2
x –2
154. Mg2P2O7
2(+ 2) + 2x + 7(– 2) = 0
+ 4 + 2x – 14 = 0
+2x = + 10
x=+5
–1
165. 2S+2 O2 + 0I +2.5
S4 O62 + 2I–
2 3
2
x +1 –2
156. CH2O
x + 2(+ 1) – 2 = 0
x+2–2=0
or x = 0
+2
x +1 1
CH2Cl2
x + 2(+1) + 2(1) = 0 or x = 0
+1 x
(A)
KI3 = (+ 1) + (3x) = 0
3x = – 1; x =
1
3
+1 x
(B)
KI = (+ 1) + (x) = 0
x=–1
x –1
(C)
IF5 = x + 5(– 1) = 0
x–5=0
x=+5
+1 x –2
(D)
x
KIO4 = (+ 1) + x + 4(– 2) = 0
x–8+1=0
x=+7
–2
159. Mo 2 Cl84
2x + 8(– 2) = – 4
2x – 16 = – 4
2x = + 12
x=+6
160. K2Cr2O7
2 + 2x – 2  7 = 0; 2x – 14 + 2 = 0
12
=+6
2x = 12; x =
2
34
N2O
2x – 2 = 0
2x = 2
2
x = = +1.
2
The change in oxidation number = 5  1 = +4.
164. BaO2 + H2SO4  H2O2 + BaSO4
In H2O2, oxygen shows –1 oxidation state and in
BaSO4, oxygen shows –2 oxidation state.
+2 x
158.

163. HNO3

1 + x – 6 = 0;
x = +5

155. MgH2
x + 2(+ 1) = 0
or x = – 2
157.
162. +2, it is a second group element.
0
+6
–1
S2 O32 + 2Br2 + 5H2O  2S O24 + 4Br–
+ 10H+
In the first reaction, the oxidation number of
sulphur changes from + 2 to + 2.5 when iodine
is used as an oxidising agent.
In the second reaction, the oxidation number
of sulphur changes from +2 to +6 when
bromine is used as an oxidising agent.
Hence, bromine is a stronger oxidant than
iodine.
166. Oxygen has 6 electrons in the outer most shell
and shows common oxidation state –2.
COOH
7
167. 5 |
+ 2KMnO 4 + 3H2SO4 
COOH
2
K2SO4 + 2MnSO 4 + 10CO2 + 8H2O
In this reaction, oxidation state of Mn changes
from +7 to +2.
169. In KMnO4, the oxidation number is given by :
1 + x – (2  4) = 0
1+x–8=0
x=7
In [MnO4]–2 : x + (–2  4) = –2
x=6
In MnO2 : x + (–2  2) = 0
x=4
In Mn2O3 : 2x + (–2  3) = 0
2x = 6, x = 3.
In Mn2+ : oxidation number = 2.
Chapter 08: Redox Reactions

The number of electrons transferred in the
case of [MnO4]–2, MnO2, Mn2O3, M2+ is
therefore 1, 3, 4, 5 respectively.
170. MnO 4 + 5e–  Mn2+
173. The values of x,y,z are 8, 4, 4 respectively
hence the reaction is
H2SO4 + 8HI  H2S + 4I2 + 4H2O
175. In the above given equation the change in the
oxidation no. is from 0 to +1 (i.e. in M, it is 0
and in [M(CN)2]–, it is +1) Among the given
options, Gold is the only metal existing in +1
oxidation state.
176. In alkaline medium,
2KMnO4 + KI + H2O  2MnO2 + 2KOH
+ KIO3
+6
207. In HNO2, the O.N. of N is + 3 which is less
than the maximum possible O.N. of + 5 and
more than the minimum O.N. of  3,
therefore, it can act both as an oxidising as
well as a reducing agent.
209. (A)
(B)
(C)
(D)
177. 6MnO4 + I– + 6OH–  6MnO24 + IO3
+ 3H2O
x –2

Reduction
In this reaction, three electrons are required
for the reduction of Cr2 O72 into Cr3+.
180. The balanced equation is
2C2H6 + 7O2  4CO2 +6H2O. Ratio of the
coefficients of CO2 and H2O is 4 : 6 or 2 : 3.
181. O. N. of Cr changes from + 6 to + 3,
undergoing reduction.
183. MnO4 + 8H+ + 5e  Mn2+ + 4H2O  2
C2 O

4
2
4

 2CO2 + 2e  5
2MnO + 5C2 O +16H  2Mn2+ +10CO2 + 8H2O
+
Thus the coefficient of MnO4 , C2 O24 and
H+ in the above balanced equation
respectively are 2, 5, 16.
186. The E values of the redox couple Ag+/Ag is
most positive amongst the four given redox
couples.

Ag+ has greater tendency to get reduced and is
the strongest oxidising agent.
202. The metallic iron is oxidised to Fe3+.
2
2
1
O. N. of ‘V’ changes from + 5 to + 3.
O. N. of ‘Na’ changes from 0 to 1.
O. N. of ‘Cr’ does not change (i.e. it is
same + 6)
O. N. of ‘Zn’ changes from + 2 to 0.
Hence option (C) is correct.
212. S2 O72
2

 Cr 3
179. Cr2 O7  3e 
2
4
+6
206. CaF2 + H2SO4  CaSO4 + 2HF
The oxidation number of S remains
unchanged. Hence, reaction given in option
(A) is the one that does not show oxidizing
behaviour of H2SO4.
1
 SnCl4  Hg 2 Cl2 In this
205. SnCl2  2HgCl2 
reaction, HgCl2 is reduced.

O. N. of S is
2(x) + 7(– 2) = –2
2x – 14 = –2
2x = + 12
x=+6
x –2
214. Fe3O8
3x + 8(– 2) = 0 or 3x – 16 = 0 or x =
16
3
215. O.N. of Fe in FeSO4.(NH4)2SO4.6H2O
= x + 6 + 4 ( 2) = 0
=x2=0

x=+2
216. O.N. of C in C3O2 = 3x + 2  ( 2) = 0
or x = + 4/3.
217. Copper will reduce Br2, if the E of the
following redox reaction is positive.
Cu + Br2  CuBr2
Cu  Cu2+ + 2e–; E = – 0.34 V
Br2 + 2e–  2Br–; E = + 1.09 V
Cu + Br2  CuBr2; E = + 0.75 V
E of the reaction is positive. Hence, Cu can
reduce Br2.
35
Chemistry Vol ‐ 1.2 (Med. and Engg.) 3+
1
218. 2Fe(aq) + 2I
0

(aq)
 I2(aq) + 2Fe(aq)
Here O.N. of I increases from 1 in I to 0 in
I2, therefore, I ions are oxidized to I2.

220. The highest oxidation number (O.N.) of any
transition element is given by:
O.N. = (n – 1)d electrons + ns electrons.
(A) 3d14s2: O.N. = 1 + 2 = 3
(B) 3d34s2: O.N. = 3 + 2 = 5
(C) 3d54s1 : O.N. = 5 + 1 = 6
(D) 3d54s2 : O.N. = 5 + 2 = 7
Hence, element having outer electronic
configuration 3d54s2 exhibits largest oxidation
number.
221. In C12H22O11, O.N of C is given by
12x + 22 – (2  11) = 0
12x + 22 – 22 = 0
12x = 0  x = 0.

222. 2AgNO3 
 2Ag + 2NO2 + O2
223. MnO4 + 5e  Mn2+
224. Equation is
Cr2 O72 + 8H+ + 3HCOOH  2Cr3+
+ 3CO2 + 7H2O
3 moles of formic acid reduces 1 mole
K2Cr2O7
1 mole of formic acid reduces 1/3 mole
K2Cr2O7.
225. Fluorine has highest E0 value and is more
reactive than MnO2.
228. To prevent rancidification of food material
anti-oxidants are added which are called
oxidation inhibitor.
230. SO2 bleaches by reduction while chlorine
bleaches colour of flowers by oxidation.
232. For a redox reaction to be feasible, Ecell
should be a positive value.
Now consider,
(A) 2Fe3+ + 2e–  2Fe2+; E = + 0.77 V
2I–  I2 + 2e–; E = – 0.54 V
2Fe3+ + 2I–  2Fe2+ + I2;
Ecell = + 0.23 V.

the given redox reaction is feasible.
36
Cu  Cu2+ + 2e– ; Ecell = – 0.34 V
2Ag+ + 2e–  2Ag; Ecell = +0.80 V
Cu + 2Ag+  Cu2+ + 2Ag;
Ecell = + 0.46 V.
the given redox reaction is feasible.
(B)
2+
2Fe3+ + 2e–  2Fe2+;
Ecell = + 0.77 V.
Cu  Cu2+ + 2e–; Ecell = – 0.34V.
2Fe3+ + Cu  2Fe2+ + Cu2+;
Ecell = + 0.43 V
the given redox reaction is feasible.
(C)

Ag  Ag+ + e–; Ecell = – 0.80 V
Fe3+ + e–  Fe2+; Ecell = + 0.77 V
Ag + Fe3+  Ag+ + Fe2+
Ecell = – 0.03 V
the given redox reaction is not feasible.
(D)

Chapter 08: Redox Reactions
Topic Test
1.
2.
3.
4.
5.
6.
7.
8.
The conversion of PbO2 to Pb(NO3)2 is
_______.
(A) Oxidation
(B) Reduction
(C) Neither oxidation nor reduction
(D) Both oxidation and reduction
Which of the following statement is NOT
CORRECT?
(A) Oxidant is a substance which increases
the oxidation number of other substance.
(B) In oxidation, there is increase in
oxidation number.
(C) The oxidation number of an oxidant
decreases.
(D) In oxidation, there is decrease in
oxidation number.
LiAlH4 is used as a/an _______.
(A) oxidizing agent (B) reducing agent
(C) bleaching agent (D) water softener
In the reaction, 2Na + Cl2  2NaCl, which
of the following is TRUE?
(A) Na has been oxidized while Cl2 has been
reduced.
(B) Na has been reduced while Cl2 has been
oxidized.
(C) No substance has been oxidized or
reduced as it is not a redox reaction.
(D) NaCl cannot be obtained by this
reaction.
Which of the following is NOT a redox
reaction?
(A) 2Rb + 2H2O  2RbOH + H2
(B) 2CuI2  2CuI + I2
(C) 2H2O2  2H2O + O2
(D) 4KCN + Fe(CN)2  K4Fe(CN)6
Oxidation state of hydrogen in CaH2 is
_______.
(A) +1
(B) –1
(C) + 2
(D) 0
In which one of the following compounds, the
oxidation number of oxygen is positive?
(B) Na2O2
(A) H2O2
(D) H2O
(C) OF2
Which of the following arrangements
represent increasing oxidation number of S?
(A) H2SO3, H2S, H2SO4, H2S2O3
(B) H2S2O3, H2SO3, H2S, H2SO4
(C)
(D)
H2S, H2SO3, H2SO4, H2S2O3
H2S, H2S2O3, H2SO3, H2SO4
9.
The value of x in the partial redox equation
 Mn2+ + 4H2O is
MnO 4 + 8H+ + xe– 
_______.
(A) 5
(B) 3
(C) 1
(D) 0
10.
The oxidation number of an element in its
elementary state is _______.
(A) zero
(B) one
(C) half its atomic number
(D) always negative
11.
In which of the following compounds,
oxidation state of the metal atom is +3?
(A) AuCl3
(B) Mg3N2
(D) KClO3
(C) V2O5
12.
Identify the reaction in which the oxidation
number of zinc decreases from +2 to 0?
(A) Zn + 2HCl  ZnCl2 + H2

(B) 2ZnS + 3O2 
 2ZnO + 2SO2
(C) Zn + CuSO4  ZnSO4 + Cu

(D) ZnO + C 
 Zn + CO
13.
During respiration, the oxidation number of
carbon _______.
(A) decreases from 0 to –2
(B) increases from 0 to +2
(C) decreases from +6 to +4
(D) increases from 0 to +4
14.
An element (Z = 20) loses electrons to achieve
noble gas electronic configuration. The outer
electronic configuration of the ion formed is
3s2 3p6. The number of electrons (x) lost and
the ion formed is _______ respectively.
(A) 1, Ca+
(B) 2, Ca2+
3+
(D) 4, Ti4+
(C) 3, Sc
15.
The oxidation number of iron in Fe3O4 is
_______.
(A) + 2
(B) + 3
(C) 8/3
(D) 2/3
Answers to Topic Test
1.
5.
9.
13.
(B)
(D)
(A)
(D)
2.
6.
10.
14.
(D)
(B)
(A)
(B)
3.
7.
11.
15.
(B)
(C)
(A)
(C)
4. (A)
8. (D)
12. (D)
37