Ncert Exercises & Solutions

26. Balance the following ionic equations.

(a) ${Cr}_{2} O_{7}^{2 -} + H^{+} + I^{-} \to {Cr}^{3 +} + I_{2} + H_{2} O$

(b) ${Cr}_{2} O_{7}^{2 -} + {Fe}^{2 +} + H^{+} \to {Cr}^{3 +} + {Fe}^{3 +} + H_{2} O$

(d) ${MnO}_{4}^{-} + H^{+} + {Br}^{-} \to {Mn}^{2 +} + {Br}_{2} + H_{2} O$

(a) Write the O.N. of all atoms above their respective symbols.

O.N. decreases by, 3 per Cr-atom

Divide the given equation into two half reactions

Reduction half reaction : ${Cr}_{2} O_{7} \to {Cr}^{3 +}$

Oxidation half reaction : $I^{-} \to I_{2}$

To balance reduction half reaction.

${Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 {Cr}^{3 +} + 7 H_{2} O$

To balance oxidation half reaction.

$2 I^{-} \to I_{2} + 2 e^{-}$

To balance reaction by electrons gained and lost

${Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 {Cr}^{3 +} + 7 H_{2} O$

$6 I^{-} \to 3 I_{2} + 6 e^{-} {Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 I^{-} \to 2 {Cr}^{3 +} + 3 I_{2} + 7 H_{2} O$

This gives the final balanced ionic equations.

(b) Write the skeletal equation of the given reaction

${Cr}_{2} O_{7}^{2 -} (aq) + {Fe}^{2 +} (aq) \to {Cr}^{3 +} (aq) + {Fe}^{3 +} (aq)$

Write the O.N. of all the elements above their respective symbols.

Divide the given equation into two half reactions

Oxidation half reaction : ${Fe}^{2 +} (aq) \to {Fe}^{3 +} (aq)$

Reduction half reaction : ${Cr}_{2} O_{7}^{2 -} (aq) \to {Cr}^{3 +} (aq)$

To balance oxidation half reaction

${Fe}^{2 +} (aq) \to {Fe}^{3 +} (aq) + e^{-}$

To balance reduction half reaction

${Cr}_{2} O_{7}^{2 -} (aq) + 6 e^{-} \to 2 {Cr}^{3 +} (aq)$

Balance charge by adding H⁺ ions.

${Cr}_{2} O_{7}^{2 -} (aq) + 14 H^{+} + 6 e^{-} \to 2 {Cr}^{3 +} (aq)$

Balance O atoms by adding H₂O molecules

${Cr}_{2} O_{7}^{2 -} (aq) + 14 H^{+} + 6 e^{-} \to 2 {Cr}^{3 +} (aq) + 7 H_{2} O (l)$

To balance the reaction

$6 {Fe}^{2 +} (aq) \to 6 {Fe}^{3 +} (aq) + 6 e^{-}$

${Cr}_{2} O_{7}^{2 -} (aq) + 14 H^{+} (aq) + 6 e^{-} \to 2 {Cr}^{3 +} (aq) + 7 H_{2} O (l) {Cr}_{2} O_{7}^{2 -} (aq) + 6 {Fe}^{2 +} (aq) + 14 H^{+} (aq) \to 2 {Cr}^{3 +} (aq) + 7 H_{2} O (l) + 6 {Fe}^{3 +} (aq)$

(c) Write the O.N. of all atoms above their respective symbols.

Divide the skeleton equation into two half-reactions.

Reduction half reaction :

${MnO}_{4}^{-} \to {Mn}^{2 +}$

Oxidation half reaction :

${SO}_{3}^{2 -} \to {SO}_{4}^{2 -}$

To balance reduction half reaction

${MnO}_{4}^{-} + 8 H^{+} + 5 e^{-} \to {Mn}^{2 +} + 4 H_{2} O$

To balance oxidation half reaction

${SO}_{3}^{2 -} \to {SO}_{4}^{2 -} + 2 e^{-}$

Balance charge by adding H⁺ ions.

${SO}_{3}^{2 -} \to {SO}_{4}^{2 -} + 2 H^{+} + 2 e^{-}$

Balance O-atoms by adding H₂O

${SO}_{3}^{2 -} + H_{2} O \to {SO}_{4}^{2 -} + 2 H^{+} + 2 e^{-}$

To balance the reaction

$2 {MnO}_{4}^{-} + 16 H^{+} + 10 e^{-} \to 2 {Mn}^{2 +} + 8 H_{2} O$

$5 {SO}_{3}^{2 -} + 5 H_{2} O \to 5 {SO}_{4}^{2 -} + 10 H^{+} + 10 e^{-} 2 {MnO}_{4}^{-} + 5 {SO}_{3}^{2 -} + 6 H^{+} \to 2 {Mn}^{2 +} + 5 {SO}_{4}^{2 -} + 3 H_{2} O$

This represents the correct balance redox equation.

(d) Write the O.N. of all the atoms above their respective symbols.

Divide skeleton equation into two half reactions

Reduction half reaction : ${MnO}_{4}^{-} \to {Mn}^{2 +}$

Oxidation half reaction :

${Br}^{-} + {Br}_{2}$

To balance half reaction

${MnO}_{4}^{-} + 8 H^{+} + 5 e^{-} \to {Mn}^{2 +} + 4 H_{2} O$

To balance oxidation half reaction

$2 {Br}^{-} \to {Br}_{2} + 2 e^{-}$

To balance the reaction

$2 {MnO}_{4}^{-} + 16 H^{+} + 10 e^{-} \to 2 {Mn}^{2 +} + 8 H_{2} O$

$10 {Br}^{-} \to 5 {Br}_{2} + 10 e^{-} 2 {MnO}_{4}^{-} + 10 {Br}^{-} + 16 H^{+} \to 2 {Mn}^{2 +} + 5 {Br}_{2} + 8 H_{2} O$

This represents the correct balance ionic equation.

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions