NCERT Solutions for Class 11 Chemistry Chapter 7 The Redox Reactions

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Chemistry
Chapter : 8
Chapter Name : Redox Reactions

Q8.1 Assign oxidation number to the underlined elements in each of the following species:
(a)NaH2 P O (b)NaHS O (c)H4 P O7 (d)K2 Mn O
–
– 4 4 –2
– –––– 4
–
–

(e)CaO (f )NaBH (g)H2 S O7 (h) KAl(S O ) ⋅ 12H2 O
–– 2 –
– 4
–2
– –
– 4 2

Answer. (a) Let the oxidation number of P be x.
We know that,
Oxidation number of Na = +1
Oxidation number of H = +1
Oxidation number of O = –2
NaH2 PO4

Then, we have
1(+1) + 2(+1) + 1(x) + 4(−2) = 0

⇒ 1 + 2 + x − 8 = 0

⇒ x = +5

Hence, the oxidation number of P is +5.

+1

(b) N a H SO 4

Then, we have
1(+1) + 1(+1) + 1(x) + 4(−2) = 0

⇒ 1 + 1 + x − 8 = 0

⇒ x = +6

Hence, the oxidation number of S is + 6.

(c) H P O
4 2 7

Then, we have
4(+1) + 2(x) + 7(−2) = 0

⇒ 4 + 2x − 14 = 0

⇒ 2x = +10

⇒ x = +5

Hence, the oxidation number of P is + 5.

(d) K MnO
2 4

Then, we have
2(+1) + x + 4(−2) = 0

⇒ 2 + x − 8 = 0

⇒ x = +6

Hence, the oxidation number of Mn is + 6.

+2 x
(e)
CaO2

Then, we have
(+2) + 2(x) = 0

⇒ 2 + 2x = 0

⇒ x = −1

(f) 9NaBH 4

Then, we have
1(+1) + 1(x) + 4(−1) = 0

⇒ 1 + x − 4 = 0

⇒ x = +3

Hence, the oxidation number of B is + 3.

(g) H S O
2 2 7

Then, we have
2(+1) + 2(x) + 7(−2) = 0

⇒ 2 + 2x − 14 = 0

⇒ 2x = 12

⇒ x = +6

Hence, the oxidation number of S is + 6.

(h ) KAl(SO ) ⋅ 12H2 O
––4
–– 2

Then, we have
1(+1) + 1(+3) + 2(x) + 8(−2) + 24(+1) + 12(−2) = 0

⇒ 1 + 3 + 2x − 16 + 24 − 24 = 0

⇒ 2x = 12

⇒ x = +6

Or,
We can ignore the water molecule as it is a neutral molecule. Then, the sum of the oxidation
numbers of all atoms of the water molecule may be taken as zero. Therefore, after ignoring the
water molecule, we have
1(+1) + 1(+3) + 2(x) + 8(−2) = 0

⇒ 1 + 3 + 2x − 16 = 0

⇒ 2x = 12

⇒ x = +6

Hence, the oxidation number of S is + 6.
Page : 280 , Block Name : Exercise

Q8.2 What are the oxidation numbers of the underlined elements in each of the following and how
do you rationalise your results?
Kl3

H2 S 4 O 6
–
––

(c) Fe O
3 4

(d) CH CH OH
3 2

(e) CH COOH
3

Answer. (a) Kl 3

In Kl , the oxidation number (O.N.) of K is +1. Hence, the average oxidation number of I is − .
1
3
3

However, O.N. cannot be fractional. Therefore, we will have to consider the structure of KI3to nd
the oxidation states.
In a Kl molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.
3

Hence, in a Kl molecule, the O.N. of the two I atoms forming the l2> molecule is 0, whereas the
3

O.N. of the I atom forming the coordinate bond is as 1.
(b)
H2 S 4 O 6
–
––
Now, 2(+1) + 4(x) + 6(−2) = 0

⇒ 2 + 4x − 12 = 0

⇒ 4x = 10

1
⇒ x = +2
2

However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the
molecule.

The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0.

(c) Fe O
3 4

On taking the O.N. of O as 2, the O.N. of Fe is found to be +2 . However, O.N. cannot be
2

3

fractional.
Here, one of the three Fe atoms exhibits the O.N. of +2 and the other two Fe atoms exhibit the
O.N. of +3.

(d) CH CH OH
3 2

2(x) + 6(+1) + 1(−2) = 0

or, 2x + 4 = 0

or, x = −2

u
^ϵ 2
Hence, the O.N. of C is a

(e) CH COOH
3

2(x) + 4(+1) + 2(−2) = 0

or, 2x = 0

or, x = 0

However, 0 is average O.N. of C. The two carbon atoms present in this molecule are present in
different environments. Hence, they cannot have the same oxidation number. Thus, C exhibits the
oxidation states of +2 and –2 in CH COOH. 3

Page : 280 , Block Name : Exercise

Q8.3 Justify that the following reactions are redox reactions:
(a) CuO(s) + H2 (g) → Cu(s) + H2 O(g)

(b) Fe2 O3 (s) + 3cO(g) → 2Fe(s) + 3CO2 (g)

(c) 4BCl3 (g) + 3LiAlH4 (s) → 2B2 H6 (g) + 3LiCl(s) + 3AlCl3 (s)

(d) 2K(s) + F2 (g) → 2K + F − (s)

(e) 4NH3 (g) + 5O2 (g) → 4NO(g) + 6H2 O(g)

Answer. (a) CuO (s)
+ H2(g) longrightarrow Cu (s)
+ H2 O(g)

Let us write the oxidation number of each element involved in the given reaction as:
+2 θ

Cu O(s) + H2(g) -> Cu (s)
+ H2 O(g)

Here, the oxidation number of Cu decreases from +2 in CuO to 0 in Cu i.e., CuO is reduced to Cu.
Also, the oxidation number of H increases from 0 in H to +1 in H O i.e., H is oxidized to H O.
2 2 2 2

Hence, this reaction is a redox reaction.

(b) Fe O
2 3(s)
+ 3CO(g) ⟶ 2Fe(s) + 3CO2(g)

Let us write the oxidation number of each element in the given reaction as:

Here, the oxidation number of Fe decreases from +3 in Fe O to 0 in Fe i.e., Fe O is reduced to
2 3 2 3

Fe. On the other hand, the oxidation number of C increases from +2 in CO to +4 in CO i.e., CO is 2

oxidized to CO . Hence, the given reaction is a redox reaction.
2

(c) 4BCl (g) + 3LiAlH (s) → 2B H (g) + 3LiCl(s) + 3AlCl (s)
3 4 2 6 3

The oxidation number of each element in the given reaction can be represented as:

In this reaction, the oxidation number of B decreases from +3 in BCl to –3 in B H . i.e., BCl is
3 2 6 3

reduced toB H . Also, the oxidation number of H increases from –1 in LiAlH4 to +1 in B H i.e.,
2 6 2 6

LiAlH4 is oxidized to B H . Hence, the given reaction is a redox reaction.
2 6

(d) 2K(s) + F (g) → 2K + F−
2

The oxidation number of each element in the given reaction can be represented as:

(e) 4NH (g) + 5O (g) → 4NO(g) + 6H O(g)
3 2 2

The oxidation number of each element in the given reaction can be represented as:

Here, the oxidation number of N
increases from –3 in NH to +2 in NO. On the other hand, the oxidation number of O decreases
3 2

from 0 in O to –2 in NO and H O i.e., O is reduced. Hence, the given reaction is a redox
2 2 2

reaction.

Page : 280 , Block Name : Exercise

Q8.4 Fluorine reacts with ice and results in the change:
H2 O(s) + F2 (g) → HF(g) + HOF(g)

Justify that this reaction is a redox reaction:

Answer. Let us write the oxidation number of each atom involved in the given reaction above its
symbol as:

Here, we have observed that the oxidation number of F increases from 0 in F to +1 in HOF. Also,
2

the oxidation number decreases from 0 in F to –1 in HF. Thus, in the above reaction, F is both
2

oxidized and reduced. Hence, the given reaction is a redox reaction.

Page : 280 , Block Name : Exercise

Q8.5 Calculate the oxidation number of sulphur, chromium and nitrogen in
Suggest structure of these compounds. Count for the fallacy.
2− −
H SO , Cr O
2 5 2 and NO
7 3

(i) H2 SO5

2(+1) + 1(x) + 5(−2) = 0
Answer.
⇒ 2 + x − 10 = 0

⇒ x = +8

However, the O.N. of S cannot be +8. S has six valence electrons. Therefore, the O.N. of S cannot be
more than +6.
The structure of H SO is shown as follows:
2 5

Now, 2(+1) + 1(x) + 3(−2) + 2(−1) = 0

⇒ 2 + x − 6 − 2 = 0

⇒ x = +6

Therefore, the O.N. of S is +6.
(ii)Cr O
2−
2 7

2(x) + 7(−2) = −2

⇒ 2x − 14 = −2

⇒ x = +6

Here, there is no fallacy about the O.N. of Cr inCr O
2−
2 7

The structure of Cr O is shown as follows:
2−
2 7

Here, each of the two Cr atoms exhibits the O.N. of +6.
(iii) NO
−

3

1(x) + 3(−2) = −1

⇒ x − 6 = −1

⇒ x = +5

Here, there is no fallacy about the O.N. of N in NO
−

3

The structure of NO is shown as follows:
−

3

The N atom exhibits the O.N. of +5.

Page : 280 , Block Name : Exercise

Q8.6 Write the formula for the following compounds:
(a) Mercury(II) chloride
(b) Nickel(II) sulphate
(c) Tin(IV) oxide
(d) Thallium(I) sulphate
(e) Iron(III) sulphate
(f) Chromium(III) oxide
Answer. (a) Mercury (II) chloride: HgCl 2

(b) Nickel (II) sulphate: NiSO 4

(c) Tin (IV) oxide: SnO2

(d) Thallium (I) sulphate: TI SO
2 4

(e) Iron (III) sulphate: Fe (SO )
2 4 3

(f) Chromium (III) oxide: Cr O 2 3

Page : 280 , Block Name : Exercise

Q8.7 Suggest a list of the substances where carbon can exhibit oxidation states from –4 to +4 and
nitrogen from –3 to +5.

Answer. The substances where carbon can exhibit oxidation states from –4 to +4 are listed in the
following table.

The substances where nitrogen can exhibit oxidation states from –3 to +5 are listed in the
following table.

Page : 280 , Block Name : Exercise

Q8.8 While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents
in their reactions, ozone and nitric acid act only as oxidants. Why?

Answer. In sulphur dioxide (SO2), the oxidation number (O.N.) of S is +4 and the range of the O.N.
that S can have is from +6 to –2.Therefore, SO2 can act as an oxidising as well as a reducing agent.
In hydrogen peroxide (H O ) the O.N. of O is –1 and the range of the O.N. that O can have is from
2 2

0 to –2. O can sometimes also attain the oxidation numbers +1 and +2. Hence, H O can act as an
2 2

oxidising as well as a reducing agent.
In ozone , the O.N. of O is zero and the range of the O.N. that O can have is from 0 to –2.
Therefore, the O.N. of O can only decrease in this case. Hence, O3 acts only as an oxidant.
In nitric acid (HNO ), the O.N. of N is +5 and the range of the O.N. that N can have is from +5 to –
3

3. Therefore, the O.N. of N can only decrease in this case. Hence, HNO acts only as an oxidant.
3

Page : 280 , Block Name : Exercise

Q8.9 Consider the reactions:

(a) 6CO2 (g) + 6H2 O(l) → C6 H12 O6 (aq) + 6O2 (g)

(b) O3 (g) + H2 O2 (l) → H2 O(l) + 2O2 (g)

Why it is more appropriate to write these reactions as:
(a) 6CO2 (g) + 12H2 O(l) → C6 H12 O6 (aq) + 6H2 O(l) + 6O2 (g)

(b) O3 (g) + H2 O2 (I) → H2 O(l) + O2 (g) + O2 (g)

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

Answer. (a)The process of photosynthesis involves two steps.
Step 1:
H O decomposes to give H and O .
2 2 2

2H2 O(l) ⟶ 2H2(s) + O2(s)

Step 2: TheH produced in step 1 reduces CO , thereby producing glucose (C H
2 2 6 12 O6 ) and H O.
2

6CO2(g) + 12Hz(g) ⟶ C6 H12 O6(s) + 6H2 O(l)

Now, the net reaction of the process is given as:

It is more appropriate to write the reaction as given above because water molecules are also
produced in the process of photosynthesis.
The path of this reaction can be investigated by using radioactive H O in place of H O.
2
18
2

(b) O is produced from each of the two reactants O and H O . For this reason, O is written
2 3 2 2 2

twice.
The given reaction involves two steps. First, O decomposes to form O and O. In the second step,
3 2

H O reacts with the O produced in the rst step, thereby producing H O and O .
2 2 2 2

The path of this reaction can be investigated by using H O or O .
18 18
2 2 3

Page : 280 , Block Name : Exercise

Q8.10 The compound AgF is an unstable compound. However, if formed, the compound acts as a
2

very strong oxidizing agent. Why?

Answer. The oxidation state of Ag in AgF is +2. But, +2 is an unstable oxidation state of Ag.
2

Therefore, whenever AgF is formed, silver readily accepts an electron to form Ag . This helps to
+
2

bring the oxidation state of Ag down from +2 to a more stable state of +1. As a result, AgF acts as 2

a very strong oxidizing agent.

Page : 281 , Block Name : Exercise

Q8.11 Whenever a reaction between an oxidising agent and a reducing agent is carried out, a
compound of lower oxidation state is formed if the reducing agent is in excess and a compound of
higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving
three illustrations.

Answer. Whenever a reaction between an oxidising agent and a reducing agent is carried out, a
compound of lower oxidation state is formed if the reducing agent is in excess and a compound of
higher oxidation state is formed if the oxidising agent is in excess. This can be illustrated as
follows:
(i) P and F are reducing and oxidising agents respectively.
4 2

If an excess of P is treated with F , then PF will be produced, wherein the oxidation number
4 2 3

(O.N.) of P is +3.
P4 ( excess ) + F2 ⟶ PF3

However, if P4 is treated with an excess of F2, then PF5 will be produced, wherein the O.N. of P is
+5.
P4 ( excess ) + F2 ⟶ PF5

(ii)K acts as a reducing agent, whereas O2 is an oxidising agent.
If an excess of K reacts with O2, then K2O will be formed, wherein the O.N. of O is –2.
4K( excess ) + O2 ⟶ 2K2 O

However, if K reacts with an excess of O2, then K2O2 will be formed, wherein the O.N. of O is –1.
2K + O2 ( excess ) ⟶ K2 O2

(iii)C is a reducing agent, while O2 acts as an oxidising agent.
If an excess of C is burnt in the presence of insuf cient amount of O2, then CO will be produced,
wherein the O.N. of C is +2.
C( excess ) + O2 ⟶ CO

On the other hand, if C is burnt in an excess of O2, then CO2 will be produced, wherein the O.N. of
C is +4.
C( excess ) + O2 ⟶ CO

Page : 281 , Block Name : Exercise

Q8.12 How do you count for the following observations?
(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as
oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium
permanganate as an oxidant. Why? Write a balanced redox equation for the reaction.
(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get
colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour
of bromine. Why?

Answer. (a) In the manufacture of benzoic acid from toluene, alcoholic potassium permanganate is
used as an oxidant because of the following reasons.
(i) In a neutral medium, OH ions are produced in the reaction itself. As a result, the cost of
−

adding an acid or a base can be reduced.

(ii) KMnO and alcohol are homogeneous to each other since both are polar. Toluene and alcohol
4

are also homogeneous to each other because both are organic compounds. Reactions can proceed
at a faster rate in a homogeneous medium than in a heterogeneous medium. Hence, in alcohol,
KMnO and toluene can react at a faster rate.
4

The balanced redox equation for the reaction in a neutral medium is give as below:

(b) When conc.H SO is added to an inorganic mixture containing bromide, initially HBr is
2 4

produced. HBr, being a strong reducing agent reduces H SO to SO with the evolution of red
2 4 2

vapour of bromine.
2NaBr + 2H2 SO4 ⟶ 2NaHSO4 + 2HBr

2HBr + H2 SO4 ⟶ Br2 + SO2 + 2H2 O

But, when conc. H SO is added to an inorganic mixture containing chloride, a pungent smelling
2 4

gas (HCl) is evolved. HCl, being a weak reducing agent, cannot reduce H SO to SO . 2 4 2

2NaCl + 2H SO ⟶ 2NaHSO + 2HCl .
2 4 4

Page : 281 , Block Name : Exercise

Q8.13 Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the
following reactions:
(a) 2AgBr(s) + C6 H6 O2 (aq) → 2Ag(s) + 2HBr(aq)+

C6 H4 O2 (aq)
+ −
(b) HCHO(l) + 2[Ag(NH3 ) ] (aq) + 3OH (aq) → 2Ag(s)+
2

−
HCOO (aq) + 4NH3 (aq) + 2H2 O(l)
2+ − −
(c) HCHO (1) + 2Cu (aq) + 5OH (aq) → Cu2 O(s) + H COO

(aq) + 3H2 O(l)

(d) N H (l) + 2H O (l) → N (g) + 4H O(l)
2 4 2 2 2 2

(e) Pb(s) + PbO (s) + 2H SO (aq) → 2PbSO (s) + 2H O(l)
2 2 4 4 2

Answer. (a) Oxidised substance →C H O 6 6 2

Reduced substance → AgBr
Oxidising agent → AgBr
Reducing agent → C H O 6 6 2

(b)Oxidised substance → HCHO
Reduced substance →
+
[Ag (NH3 ) ]
2

Oxidising agent →
+
[Ag (NH3 ) ]
2

Reducing agent → HCHO
(c) Oxidised substance → HCHO
Reduced substance → Cu
2+

Oxidising agent → Cu
2+

Reducing agent → HCHO
(d) Oxidised substance →N H 2 4

Reduced substance → H O 2 2

Oxidising agent → H O 2 2

Reducing agent → N H 2 4

(e) Oxidised substance → Pb
Reduced substance → PbO 2

Oxidising agent → PbO 2

Reducing agent → Pb

Page : 281 , Block Name : Exercise

Q8.14 Consider the reactions:
2− 2− −
2S2 O (aq) + I2 (s) → S4 O (aq) + 2I (aq)
3 6
2−
2− 2SO4 − +
S2 O aq) + 2Br2 (l) + 5H2 O(l) → (aq) + 4Br (aq) + 10H
3

Answer. The average oxidation number (O.N.) of S in S O is +2. Being a stronger oxidising agent
2−
2 3

than I2, Br2 oxidises S O to SO , in which the O.N. of S is +6. However, I2 is a weak oxidising
2− 2−
2 3 4

agent. Therefore, it oxidises S O to
2−
2 3

, in which the average O.N. of S is only +2.5. As a result, S O reacts differently with
2− 2−
S4 O 2
6 3

iodine and bromine.

Page : 281 , Block Name : Exercise

Q8.15 Justify giving reactions that among halogens, uorine is the best oxidant and among
hydrohalic compounds, hydroiodic acid is the best reductant.

Answer. F can oxidize Cl to Cl , Br to Br and I to I
− − −
2 2 2, 2

On the other hand, Cl , Br , and I cannot oxidize F to F . The oxidizing power of halogens
−
2 2 2 2

increases in the order ofI < Br < Cl < F . Hence, uorine is the best oxidant among
2 2 2 2

halogens.
HI and HBr can reduce H SO to SO , but HCl and HF cannot. Therefore, HI and HBr are stronger
2 4 2

reductants than HCl and HF.
2HI + H2 SO4 ⟶ I2 + SO2 + 2H2 O

2HBr + H2 SO4 ⟶ Br2 + SO2 + 2H2 O
− 2+ + −
1 can reduce Cu to Cu , but Br
− 2+
4I + 2Cu ⟶ Cu2 I2(s) + I2(aq)
(aq) (aq)

Hence, hydroiodic acid is the best reductant among hydrohalic compounds.
Thus, the reducing power of hydrohalic acids increases in the order of HF < HCl < HBr < HI.

Page : 281 , Block Name : Exercise

Q8.16 Why does the following reaction occur?
4− − +
XeO (aq) + 2F (aq) + 6H (aq) → XeO3 (g) + F2 (g) + 3H2 O(l)
6

What conclusion about the compound Na XeO (of which XeO is a part) can be drawn from the
4−
4 6 6

reaction.

Answer. The given reaction occurs because XeO oxidises F and F reduces XeO .
4− − − 4−

6 6

0

-> X
+ −1 +
XeO + 2F + 6H 3(g) + F2(x) + 3H2 O(l)
6(aq) (aq) (aq)

In this reaction, the oxidation number (O.N.) of Xe decreases from +8 in
to +6 in XeO3 and the O.N. of F increases from –1 in F to O in F Hence, we can conclude
4− −
XeO 2
6

that Na XeO is a stronger oxidising agent than F .
−
4 6

Page : 281 , Block Name : Exercise

Q8.17 consider the reactions:
(a) H3 PO2 (aq) + 4AgNO (aq) + 2H2 O(l) → H3 PO4 (aq)+
3

4Ag(s) + 4HNO3 (aq)

(b) H3 PO2 (aq) + 2CuSO4 (aq) + 2H2 O(I) → H3 PO4 (aq) + 2Cu(s)

+H2 SO4 (aq)
+ − −
(c) C6 H5 CHO(I) + 2[Ag(NH3 ) ] (aq) + 3OH (aq) → C6 H5 COO
2

(aq) + 2Ag(s) + 4NH3 (aq) + 2H2 O(l)

2+(aq) −
(d) C6 H5 CHO(l) + 2Cu + 5OH (aq) → No change

observed.

Answer. Cu and Cu act as oxidising agents in reactions (a) and (b)respectively.
2+ 2+

In reaction (c), Cu oxidises C H CHO to C H COO , but in reaction (d), Cu cannot
2+ − 2+
6 5 6 5

oxidise C H CHO. 6 5

Hence, we can say that Cu is a stronger oxidising agent than
2+

.
2+
Cu

Page : 281 , Block Name : Exercise

Q8.18 Balance the following redox reactions by ion-electron method:
− −
MnO (aq) + I (aq) → MnO2 (s) + I2 (s)( in basic medium)
4
− 2+ −
MnO (aq) + SO2 (g) → Mn (aq) + HSO (aq)
4 4
2+ 3+
(c) H2 O2 (aq) + Fe (aq) → Fe (aq) + H2 O(l) (in acidic

solution)
2− 3+ 2−
Cr2 O + SO2 (g) → Cr (aq) + SO (aq)( in acidic solution )
7 4

Answer. (a) Step 1: The two half reactions involved in the given reaction are:
−1
Oxidation half reaction:
I(aq) ⟶ I2(s)

+7 +4

Reduction half reaction: M nO
−
⟶M nO2(aq)
4(aq)

Step 2: Balancing I in the oxidation half reaction, we have:
−
2I ⟶ I2(s)
(aq)

Now, to balance the charge, we add 2 e– to the RHS of the reaction.
− −
2I ⟶ I2(x) + 2e
(aq)

Step 3: In the reduction half reaction, the oxidation state of Mn has reduced from +7 to +4. Thus, 3
electrons are added to the LHS of the reaction.
− −
MnO + 3e ⟶ MnO2(aq)
4(aq)

Now, to balance the charge, we add 4 OH– ions to the RHS of the reaction as the reaction is taking
place in a basic medium.
− − −
MnO + 3e ⟶ MnO2(cq) + 4OH
4(oq)

Step 4: In this equation, there are 6 O atoms on the RHS and 4 O atoms on the LHS. Therefore, two
water molecules are added to the LHS.
− − −
MnO + 2H2 O + 3e ⟶ MnO2(ca) + 4OH
4(cq)

Step 5: Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the
reduction half reaction by 2, we have:
− −
6I ⟶ 3I2(s) + 6e
(aq)

− − −
2MnO + 4H2 O + 6e ⟶ 2MnO2(s) + 8OH
4(aq) (aq)

Step 6: Adding the two half reactions, we have the net balanced redox reaction as:
-> 3I
− − −
6I + 2MnO + 4H O + 2MnO 2 + 8OH
(l) 2(s) 2(s)
(eq) 4(cy) (aq)

(b)Following the steps as in part (a), we have the oxidation half reaction as:
SO2(x) + 2H2 O(l)

And the reduction half reaction as:
-> Mn
− + − 2+
MnO + 8H + 5e + 4H2 O(l)
4(αq) (aq) (aq)

Multiplying the oxidation half reaction by 5 and the reduction half reaction by 2, and then by
adding them, we have the net balanced redox reaction as:
+ 2H O + H -> 2Mn
− + 2+ −
2MnO + 5SO 2(g) 2 + 5HSO
(l)
4(aq) (aq) (aq) 4(aq)

(c) Following the steps as in part (a), we have the oxidation half reaction as:
2+ 3+ −
Fe ⟶ Fe + e
(αq) (aq)

And the reduction half reaction as:
+ −
H2 O2(aq) + 2H + 2e ⟶ 2H2 O(l)
(aq)

Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we
have the net balanced redox reaction as:
H O
2 + 2Fe
2(aq)
+ 2H -> 2Fe
2+

(aq)
+ 2H O
+

(aq)
3+

(aq)
2 (l)

(d) Following the steps as in part (a), we have the oxidation half reaction as:
-> SO
2− + −
SO2(g) + 2H2 Ol) + 4H + 2e
4 (aq)

And the reduction half reaction as:
-> 2Cr
2− + − 3+
Cr2 O + 14H + 6e + 7H2 O(l)
7 (aq) (aq)

Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we
have the net balanced redox reaction as:
-> 2Cr
2− + 3+ 2−
Cr2 O + 3SO2(g) + 2H + 3SO + H2 O(l)
7 (aq) (aq) 4

Page : 282 , Block Name : Exercise

Q8.19 Balance the following equations in basic medium by ion-electron method and oxidation
number methods and identify the oxidising agent and the reducing agent.
-> PH
−
P4(s)
+ OH −(oq)
+ HPO 3(g) 2

-> NO
− −
N2 H4(l) + ClOaq (g) + Cl
(g)

-> ClO +H
− +
Cl2 O7(g) + H2 O2(aq) + O2(g)
2(aq) (aq)

Answer. (a)

The O.N. (oxidation number) of P decreases from 0 in P to –3 in PH3 and increases from 0 inP
4 4

to + 2 inHPO . Hence, P4acts both as an oxidizing agent and a reducing agent in this reaction.
−

2

Ion–electron method:
The oxidation half equation is:
P4s → HPO2 − (aq)

The P atom is balanced as:
P →4HPO2- (aq)
4

The O atom is balanced by adding 8 H2O molecules:
P +8H O →4(aq)
4 2

The H atom is balanced by adding 12 H+ ions:
P +8H O→4HPO (aq) + 12H +
−
4 2 2

The charge is balanced by adding e– as:
P +8H O →4HPO (aq) + 12H + +8e− …(i)
−
4 2 2

The reduction half equation is:
P4(s) ⟶ P H3(g)

The P atom is balanced as:
P →4PH
4 3(g)

The H is balanced by adding 12 H+ as:
P +12H+→4PH
4 3(g)

The charge is balanced by adding 12e– as:
P +12H++12e-→4PH
4 …(ii) 3(g)

By multiplying equation (i) with 3 and (ii) with 2 and then adding them, the balanced chemical
equation can be obtained as:
5P (s) +24H O→12HPO
−
4 2 2

– +8PH 3(g) + 12H +

As, the medium is basic, add 12OH– both sides as:
5P (s) +12H O +12OH- →12HPO +8PH
−
4 2 2 3(g)

This is the required balanced equation.

Oxidation number method:
Let, total no of P reduced = x
∴Total no of P oxidised = 4– x
P (s) +OH- →xPH + 4-xHHPO … (i)
−
4 3(g) 2

Total decrease in oxidation number of P = x × 3 = 3x
Total increase in oxidation number of P = (4 – x) × 2 = 8 – 2x
∵ 3x = 8 – 2x x = 8/5 From (i),
5P4 (s) +5OH- →8PH ) +12HPO
−
3(g) 2

Since, reaction occurs in basic medium, the charge is balanced by adding 7OH– on LHS as:
5P (s) +12OH- →8PH +12HPO
−
4 3(g) 2

The O atoms are balanced by adding 12H O as: 2

5P (s) + 12H O + 12OH-→ +12HPO + 8PH
−
4 2 2 3(g)

This is the required balanced equation.

(b)

The oxidation number of N increases from – 2 in N H to + 2 in NO and the oxidation number of
2 4

Cl decreases from + 5 in ClO to – 1 in Cl . Hence, in this reaction, N H is the reducing agent
− −

3 2 4

and ClO is the oxidizing agent.
−

3

Ion–electron method:

(c)

The oxidation number of Cl decreases from + 7 in Cl O to + 3 in ClO and the oxidation number
−
2 7 2

of O increases from – 1 in H O to zero in O . Hence, in this reaction, Cl O is the oxidizing
2 2 2 2 7

agent and H O is the reducing agent.
2 2

Ion–electron method:

Page : 282 , Block Name : Exercise

Q8.20 What sorts of informations can you draw from the following reaction ?
(CN) + 2OH
2(g)
-> CN + CNO
−

(aq)
−

(g)
+ H O
−

(aq) 2 (l)

Answer. The oxidation numbers of carbon in (CN) , CN are +3, +2 and +4
2 − −
and CNO

respectively. These are obtained as shown below:
Let the oxidation number of C be x.
(CN)2

2(x – 3) = 0
∴x = 3
−
CN

x – 3 = –1
∴x = 2
CNO–
x – 3 – 2 = –1
∴x = 4
The oxidation number of carbon in the various species is:

It can be easily observed that the same compound is being reduced and oxidised simultaneously in
the given equation. Reactions in which the same compound is reduced and oxidised is known as
disproportionation reactions. Thus, it can be said that the alkaline decomposition of cyanogen is
an example of disproportionation reaction.

Page : 282 , Block Name : Exercise

Q8.21 The Mn ion is unstable in solution and undergoes disproportionation to give Mn ,
3+ 2+

MnO and H ion. Write a balanced ionic equation for the reaction.
+
2

Answer. The given reaction can be represented as:
3+ 2+ +
Mn ⟶ Mn + MnO2(s) + H
(αq) (aq) (aq)

The oxidation half equation is:
+4
3+
Mn ⟶M nO2(s)
(aq)

The oxidation number is balanced by adding one electron as:
3+ −
Mn ⟶ MnO2(s) + e
(aq)

The charge is balanced by adding 4H+ ions as:
3+ + −
Mn ⟶ MnO2(s) + 4H + e
(α) (aq)

The O atoms and H+ ions are balanced by adding 2H O molecules as: 2

The reduction half equation is:

3+ 2+
Mn ⟶ Mn
(aq) (aq)

The oxidation number is balanced by adding one electron as:

The balanced chemical equation can be obtained by adding equation (i) and (ii) as:

Page : 282 , Block Name : Exercise

Q8.22 Consider the elements:
Cs, Ne, I and F
(a) Identify the element that exhibits only negative oxidation state.
(b) Identify the element that exhibits only positive oxidation state.
(c) Identify the element that exhibits both positive and negative oxidation states.
(d) Identify the element which exhibits neither the negative nor does the positive oxidation state.

Answer. (a) F exhibits only negative oxidation state of –1.
(b) Cs exhibits positive oxidation state of +1.
(c) I exhibits both positive and negative oxidation states. It exhibits oxidation states of – 1, + 1, +
3, + 5, and + 7.
(d) The oxidation state of Ne is zero. It exhibits neither negative nor positive oxidation states.

Page : 282 , Block Name : Exercise

Q8.23 Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of
chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox
change taking place in water.

Answer. The given redox reaction can be represented as:
-> Cl
− 2−
Cl2(s) + SO2(αq) + H2 O(l) + SO
(αq) 4

The oxidation half reaction is.

Page : 282 , Block Name : Exercise

Q8.24 Refer to the periodic table given in your book and now answer the following questions:
(a) Select the possible non metals that can show disproportionation reaction.
(b) Select three metals that can show disproportionation reaction.

Answer. In disproportionation reactions, one of the reacting substances always contains an
element that can exist in at least three oxidation states,

(a) P, Cl, and S can show disproportionation reactions as these elements can exist in three or more
oxidation states,
(b) Mn, Cu, and Ga can show disproportionation reactions as these elements can exist in three or
more oxidation states.

Page : 282 , Block Name : Exercise

Q8.25 In Ostwald’s process for the manufacture of nitric acid, the rst step involves the oxidation
of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of
nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen?

Answer. The balanced chemical equation for the given reaction is given as:

Thus, 68 g of NH reacts with 160 g of O
3 2

Therefore, 10g of NH3 reacts with g of O , or 23.53 g of O .
160×10
2 2
68

But the available amount of O is 20 g. 2

Therefore, O is the limiting reagent (we have considered the amount of O to calculate the
2 2

weight of nitric oxide obtained in the reaction).
Now, 160 g of O gives 120g of NO.
2

Therefore, 20 g of O gives g of N, or 15 g of NO.
120×20
2
160

Hence, a maximum of 15 g of nitric oxide can be obtained.

Page : 282 , Block Name : Exercise

Q8.26 Using the standard electrode potentials given in the Table 8.1, predict if the reaction
between the following is feasible:
3+ −
(a) Fe (aq) and I (aq)

+
(b) Ag (aq) and Cu(s)

3+
(c) Fe (aq) and Cu(s)

3+
(d) Ag(s) and Fe (aq)

2+
(e) Br2 (aq) and Fe (aq)

Answer. (a) The possible reaction between Fe is given by
3+ −
+ I
(aq) (aq)

E° for the overall reaction is positive. Thus, the reaction between Fe and I is feasible.
3+ −

(aq) (aq)

(b) The possible reaction between Ag is given by,
+
+ Cu(s)
(aq)

E° positive for the overall reaction is positive. Hence, the reaction between Ag is
+
+ Cu(s)
(aq)

feasible.
(c) The possible reaction between Fe and Cu is given by,
3+

(aq) (s)

E° positive for the overall reaction is positive. Hence, the reaction between Fe and Cu is
3+

(aq) (s)

feasible.
(d) The possible reaction between Fe and Ag is given by,
3+

(aq) (s)

Here, E° for the overall reaction is negative. Hence, the reaction between Fe 3+

(aq)
and Cu (s)
is not
feasible.

(e) The possible reaction between Br and Fe is given by,
2+
2(aq)
(aq)

Here, E° for the overall reaction is positive. Hence, the reaction between Br and Fe is
2+
2(aq) (aq)

feasible.

Page : 282 , Block Name : Exercise

Q8.27 Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO with silver electrodes 3

(ii) An aqueous solution AgNO with platinum electrodes 3

(iii) A dilute solution of H SO with platinum electrodes
2 4

(iv) An aqueous solution of CuCl with platinum electrodes. 2

Answer. (i) AgNO ionizes in aqueous solutions to form Ag and NO ions.On electrolysis, either
+ −

3 3

ions or H O molecules can be reduced at the cathode. But the reduction potential of Ag
+ +
Ag 2

ions is higher than that of H O. 2

Hence, Ag ions are reduced at the cathode. Similarly, Ag metal or H O molecules can be
+
2

oxidized at the anode. But the oxidation potential of Ag is higher than that of H O molecules. 2

Therefore, Ag metal gets oxidized at the anode.
(ii) Pt cannot be oxidized easily. Hence, at the anode, oxidation of water occurs to liberate O . At 2

the cathode, Ag ions are reduced and get deposited.
+

(iii) H SO ionizes in aqueous solutions to give H+ and SO ions.
2−
2 4 4
+ 2−
H2 SO4(aq) ⟶ 2H + SO
(aq) 4

On electrolysis, either of H ions or H O molecules can get reduced at the cathode. But the
+
2

reduction potential of H ions is higher than that of H Omolecules.
+
2

Hence, at the cathode, H ions are reduced to liberate H gas.
+
2

On the other hand, at the anode, either of SO ions or H O molecules can get oxidized. But the
2−

4 2

oxidation of SO involves breaking of more bonds than that of H O molecules. Hence, SO
2− 2−

4 2 4

ions have a lower oxidation potential thanH O. Thus, H O is oxidized at the anode to liberate O
2 2 2

molecules.
(iv) In aqueous solutions, CuCl ionizes to give Cu and Cl ions as:
2+ −
2

2+ −
CuCl2(aq) ⟶ Cu + 2Cl
(aq) (aq)

On electrolysis, either of Cu ions or H O molecules can get reduced at the cathode. But the
2+
2

reduction potential of Cu is more than that of H O molecules.
2+
2

Hence,Cu ions are reduced at the cathode and get deposited.
2+

Similarly, at the anode, either of Cl or H O is oxidized. The oxidation potential of H O is higher
−
2 2

than that of Cl .
−

But oxidation of H2O molecules occurs at a lower electrode potential than that of Cl ions
−

because of over-voltage (extra voltage required to liberate gas). As a result, Cl .
−

ions are oxidized at the anode to liberate Cl gas. 2

Similarly, at the anode, either of Cl– or H2O is oxidized. The oxidation potential of H2O is higher
than that of Cl .
−

Page : 283 , Block Name : Exercise

Q8.28 Arrange the following metals in the order in which they displace each other from the
solution of their salts.
Al, Cu, Fe, Mg and Zn.

Answer. A metal of stronger reducing power displaces another metal of weaker reducing power
from its solution of salt.
The order of the increasing reducing power of the given metals is Cu < Fe < Zn < Al < Mg.
Hence, we can say that Mg can displace Al from its salt solution, but Al cannot displace Mg.
Thus, the order in which the given metals displace each other from the solution of their salts is
given below:
Mg>Al> Zn> Fe,>Cu

Page : 283 , Block Name : Exercise

Q8.29 Given the standard electrode potentials,

+ +
K /K = −2.93V, Ag /Ag = 0.80V

2+
Hg /Hg = 0.79V

2+ 3+
Mg /Mg = −2.37V ⋅ Cr /Cr = −0.74V

Arrange these metals in their increasing order of reducing power.

Answer. The lower the electrode potential, the stronger is the reducing agent. Therefore, the
increasing order of the reducing power of the given metals is Ag < Hg < Cr < Mg < K.

Page : 283 , Block Name : Exercise

Q8.30 Depict the galvanic cell in which the reaction Zn(s) + 2Ag (aq)^
+ ′ 2+
at Zn (aq) + 2Ag(s)

takes place, further show:
(i) which of the electrode is negatively charged,
(ii) the carriers of the current in the cell, and
(iii) individual reaction at each electrode.

Answer. The galvanic cell corresponding to the given redox reaction can be represented as:
2+
∣
Zn Zn ∥Ag+ |Ag
∣ (aq) ∥ (aq)

(i) Zn electrode is negatively charged because at this electrode, Zn oxidizes to Zn2+ and the
leaving electrons accumulate on this electrode.
(ii) Ions are the carriers of current in the cell.
(iii) The reaction taking place at Zn electrode can be represented as:
2+ −
Zn(s) ⟶ Zn + 2e
(aq)

And the reaction taking place at Ag electrode can be represented as:
+ −
Ag + e ⟶ Ag
(aq) (s)

Page : 283 , Block Name : Exercise