Ncert Exercises & Solutions

22. Calculate the oxidation number of phosphorus in the following species.

(a) ${HPO}_{3}^{2 -}$

(b) ${PO}_{4}^{3 -}$

(a) Ion electron method Write the skeleton equation for the given reaction.

{MnO}_{4}^{-} (aq) + {SO}_{2} (g) \to {Mn}^{2 +} (aq) + {HSO}_{4}^{-} (aq)

Find out the elements which undergo change in O.N.

Divide the given skeleton into two half equations.

Reduction half equation :

{MnO}_{4}^{-} (aq) \to {Mn}^{2 +} (aq)

Oxidation half equation : ${SO}_{2} (g) \to {HSO}_{4}^{-} (aq)$

To balance reduction half equation

In acidic medium, balance H and O-atoms

{MnO}_{4}^{-} (aq) + 8 H^{+} + 5 e^{-} \to {Mn}^{2 +} (aq) + H_{2} O (l)

To balance the complete reaction

2 {MnO}_{4}^{-} (aq) + 16 H^{+} + 10 e^{-} \to 2 {Mn}^{2 +} (aq) + 8 H_{2} O (l)

5 {SO}_{2} (g) + 10 H_{2} O (l) \to 5 {HSO}_{4}^{-} (aq) + 15 H^{+} (aq) + 10 e^{-} 2 {MnO}_{4}^{-} (aq) + 5 {SO}_{2} (g) + 2 H_{2} O (l) + H^{+} (aq) \to 2 {Mn}^{2 +} (aq) + 5 {HSO}_{4}^{-} (aq)

(b) Oxidation number method Write the skeleton equation for the given reaction.

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

O.N. increases by 4 per N-atom

Multiply NO by 2 because in N₂H₄ there are 2N atoms

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

Total increase in O.N. of

N = 2 \times 4 = 8 (8 e^{-} lost)

Total decrease in O.N. of

Cl = 1 \times 6 = 6 (6 e^{-} gain)

Therefore, to balance increase or decrease in O.N. multiply

N_{2} H_{4}

by 3, 2NO by 3 and

{ClO}_{3}^{-}, {Cl}^{-}

by 4

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

Balance O and H-atoms by adding

6 H_{2} O

to RHS

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

(c) Ion electron method Write the skeleton equation for the given reaction.

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

Find out the elements which undergo a change in O.N.

Divide the given skeleton equation into two half equations.

Reduction half equation :

{Cl}_{2} O_{7} \to {ClO}_{2}^{-}

Oxidation half equation :

H_{2} O_{2} \to O_{2}

To balance the reduction half equation

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

To balance the oxidation half equation

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

To balance the complete reaction

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

4 H_{2} O_{2} (aq) \to 4 O_{2} (g) + 8 H^{+} + 8 e^{-} {Cl}_{2} O_{7} (g) + 4 H_{2} O_{2} (aq) \to 2 {ClO}_{2}^{-} (aq) + 3 H_{2} O (l) + 4 O_{2} (g) + 2 H^{+} + (aq)

This represents the balanced redox reaction.