Ncert Exercises & Solutions

8.19 Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.

(a) P4(s) + OH–(aq) → PH3(g) + HPO2– (aq)

(b) N2H4(l) + ClO3 –(aq) → NO(g) + Cl–(g)

(a) The O.N. (oxidation number) of P decreases from 0 in

P_{4} o - 3 in {PH}_{3}

and increases from 0 in

P_{4} to + 2 in {HPO}_{2}^{-}

. Hence,

P_{4}

acts both as an oxidizing agent and a reducing agent in this reaction.

Ion-electron method:

The oxidation half equation is:

P_{4 (s)} \to {HPO}_{2 (aq)}^{-}

The P atom is balanced as:

{\overset{0}{P}}_{4 (s)} \to 4 H \overset{2 +}{P} O_{2}^{-} (aq)

The O.N. is balanced by adding 8 electrons as:

P_{4 (s)} \to 4 {HPO}_{2 (aq)}^{-} + 8 e^{-}

The charge is balanced by adding

12 {OH}^{-}

as:

P_{4 (s)} + 12 {OH}_{(aq)}^{-} \to 4 {HPO}_{2 (aq)}^{-} + 8 e^{-}

The H and O atoms are balanced by adding

4 H_{2} O

as:

P_{4 (s)} + 12 {OH}_{(aq)}^{-} \to 4 {HPO}_{2}^{-}_{(aq)} + 4 H_{2} O_{(l)} + 8 e^{-} . . . . . . . (i)

The reduction half equation is:

P_{4 (s)} \to {PH}_{3 (g)}

The P atom is balanced as

\overset{0}{P_{4 (s)}} \to 4 \overset{- 3}{P} H_{3 (g)}

The O.N. is balanced by adding 12 electrons as:

P_{4 (s)} + 12 e^{-} \to 4 {PH}_{3 (g)}

The charge is balanced by adding

12 {OH}^{-}

as:

P_{4 (s)} + 12 e^{-} \to 4 {PH}_{3 (g)} + 12 {OH}^{-}_{(aq)}

The O and H atoms are balanced by adding

12 H_{2} O

as:

P_{4 (s)} + 12 H_{2} O_{(l)} + 12 e^{-} \to 4 {PH}_{3 (g)} + 12 {HO}^{-}_{(aq)} (ii)

By muliplying equation (i) and 3 and (ii) with 2 and then adding them, the balanced chemical equation can be obtained as:

5 P_{4 (s)} + 12 H_{2} O_{(l)} + 12 {HO}^{-}_{(aq)} \to 8 {PH}_{3 (g)} + 12 {HPO}_{2}^{-}_{(aq)}

(b)

The oxidaion number of N increases from -2 in

N_{2} H_{4}

to +2 in NO and the oxidation number of Cl decreases from +5 in

{ClO}_{3}^{-}

to -1 in

{Cl}^{-}

. Hence, in this reaction,

N_{2} H_{4}

is the reducing agent and

{ClO}_{3}^{-}

is the oxidizing agent.

Ion-electron method:

The oxidation half equation is:

{\overset{- 2}{N}}_{2} H_{4 (l)} \to \overset{+ 2}{N} O_{(g)}

The N atoms are balanced as:

N_{2} H_{4 (l)} \to 2 {NO}_{(g)}

The oxidation number is balanced by adding 8 electrons as :

N_{2} H_{4 (l)} \to 2 {NO}_{(g)} + 8 e^{-}

The charge is balanced by adding 8

{OH}^{-}

ions as:

N_{2} H_{4 (l)} + 8 {OH}^{-}_{(aq)} \to 2 {NO}_{(g)} + 8 e^{-}

The O atoms are balanced by adding

6 H_{2} O

as:

N_{2} H_{4 (l)} + 8 {OH}^{-}_{(aq)} \to 2 {NO}_{(g)} + 6 H_{2} O_{(l)} + 8 e^{-} . . . . . (i)

The reduction half equation is:

\overset{+ 5}{C} {lO}^{-}_{3 (aq)} \to \overset{- 1}{C} {l^{-}}_{(aq)}

The oxidation number is balanced by adding 6 electrons as:

{ClO}_{3 (aq)}^{-} + 6 e^{-} \to {Cl}_{(aq)}^{-}

The charge is balanced by adding

6 {OH}^{-}

ions as:

{ClO}_{3 (aq)}^{-} + 6 e^{-} \to {Cl}_{(aq)}^{-} + 6 {OH}_{(aq)}^{-}

The O atoms are balanced by adding

3 H_{2} O

as:

{ClO}_{3 (aq)}^{-} + 3 H_{2} O_{(l)} + 6 e^{-} \to {Cl}_{(aq)}^{-} + 6 {OH}_{(aq)}^{-} . . . . . . . (ii)

The balanced equation can be obtained by multiplying equation (i) with 3 and equation (ii) with 4 and then adding them as:

3 N_{2} H_{4 (l)} + 4 {ClO}_{3 (aq)}^{-} \to 6 {NO}_{(g)} + 4 {Cl}_{(aq)}^{-} + 6 H_{2} O_{(l)}

Oxidation number method:

Total decrease in oxidation number of N = 2 x 4 = 8

Total increase in oxidation number of Cl = 1 x 6 = 6

On multiplying

N_{2} H_{4}

with 3 and

{ClO}_{3}^{-}

with 4 to balance the increase and decrease in O.N., we get:

3 N_{2} H_{4 (l)} + 4 {ClO}_{3 (aq)}^{-} \to {NO}_{(g)} + {Cl}_{(aq)}^{-}

The N and Cl atoms are balanced as:

3 N_{2} H_{4 (l)} + 4 {ClO}_{3 (aq)}^{-} \to 6 {NO}_{(g)} + 4 {Cl}_{(aq)}^{-}

The O atoms are balanced by adding

6 H_{2} O

as:

3 N_{2} H_{4 (l)} + 4 {ClO}_{3 (aq)}^{-} \to 6 {NO}_{(g)} + 4 {Cl}_{(aq)}^{-} + 6 H_{2} O_{(l)}

This is the required balanced equation.

(c)

The oxidation number of Cl decreases from +7 in

{Cl}_{2} O_{7}

to +3 in

{ClO}_{2}^{-}

and the oxidation number of O increases from -1 in

H_{2} O_{2}

to zero in

O_{2}

. Hence, in this reaction,

{Cl}_{2} O_{7}

is the oxidizing agent and

H_{2} O_{2}

is the reducing agent.

Ion-electron method:

The oxidation half equation is:

H_{2} {\overset{- 1}{O}}_{2 (aq)} \to {\overset{0}{O}}_{2 (g)}

The oxidation number is balanced by adding 2 electrons as:

H_{2} O_{2 (aq)} \to O_{2 (g)} + 2 e^{-}

The charge is balanced by adding

2 {OH}^{-}

ions as:

H_{2} O_{2 (aq)} + 2 {OH}_{(aq)}^{-} \to O_{2 (g)} + 2 e^{-}

The oxygen atoms are balanced by adding

2 H_{2} O

as:

H_{2} O_{2 (aq)} + 2 {OH}_{(aq)}^{-} \to O_{2 (g)} + 2 H_{2} O_{(l)} + 2 e^{-} (i)

The reduction half equation is:

\overset{+ 7}{C} l_{2} O_{7 (g)} \to \overset{+ 3}{C} {lO}_{2 (aq)}^{-}

The Cl atoms are balanced as:

{Cl}_{2} O_{7 (g)} \to 2 {ClO}_{2 (aq)}^{-}

The oxidation number is balanced by adding 8 electrons as:

{Cl}_{2} O_{7 (g)} + 8 e^{-} \to 2 {ClO}_{2 (aq)}^{-}

The charge is balanced by adding

6 {OH}^{-}

as:

{Cl}_{2} O_{7 (g)} + 8 e^{-} \to 2 {ClO}_{2 (aq)}^{-} + 6 {OH}_{(aq)}^{-}

The oxygen atoms are balanced by adding

3 H_{2} O

as:

{Cl}_{2} O_{7 (g)} + 3 H_{2} O_{(l)} 8 e^{-} \to 2 {ClO}_{2 (aq)}^{-} + 6 {OH}_{(aq)}^{-} (ii)

The balanced equation can be obtained by multiplying the equation (i) with 4 and adding equation (ii) to its as:

{Cl}_{2} O_{7 (g)} + 6 H_{2} O_{2 (aq)} + 2 {OH}_{(aq)}^{-} \to 2 {ClO}_{(aq)}^{-} + 4 O_{2 (g)} + 5 H_{2} O_{(l)}

Oxidation number method:

Total decrease in oxidation number of

{Cl}_{2} O_{7} = 4 \times 2 = 8

Total increase in oxidation number of

H_{2} O_{2} = 2 \times 1 = 2

By multiplying

H_{2} O_{2} and O_{2}

with 4 to balance the increase and decrease in the oxidation number, we get:

{Cl}_{2} O_{7 (g)} + 4 H_{2} O_{2 (aq)} \to {ClO}_{2 (aq)}^{-} + 4 O_{2 (g)}

The Cl atoms are balanced as:

{Cl}_{2} O_{7 (g)} + 4 H_{2} O_{2 (aq)} \to 2 {ClO}_{2 (aq)}^{-} + 4 {OH}_{2 (g)}

The O atoms are balanced by adding

3 H_{2} O

as:

{Cl}_{2} O_{7 (g)} + 4 H_{2} O_{2 (aq)} \to 2 {ClO}_{2 (aq)}^{-} + 4 O_{2 (g)} + 3 H_{2} O_{(l)}

The H atoms are balanced by adding

2 {OH}^{-} and 2 H_{2} O

as:

{Cl}_{2} O_{7 (g)} + 4 H_{2} O_{2 (aq)} + 2 {OH}_{(aq)}^{-} \to 2 {ClO}_{2 (aq)}^{-} + 4 O_{2 (g)} + 5 H_{2} O_{(l)}

This is the required balanced equation.

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