Question

Balance the following equation :

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

Accepted Answer

The skeleton equation along with the oxidation number of each atom is :

$\overset{+ 7 - 2}{{MnO}_{4}^{-}} + \overset{+ 1}{H^{+}} + \overset{+ 1 - 1}{H_{2} O_{2}} \to \overset{2 +}{{Mn}^{2 +}} + \overset{0}{O_{2}} + \overset{+ 1 - 2}{H_{2} O}$

In the above reaction, oxygen undergoes oxidation as its oxidation number increases from $- 1$ in $H_{2} O_{2}$ to $0$ in $O_{2}$ (Oxidation number increases by 1 per atom). $Mn$ undergoes reduction as its oxidation number decreases from $+ 7$ in ${MnO}_{4}^{-}$ to $+ 2$ in ${Mn}^{2 +}$ ion. (Oxidation number decreases by 5 per atom).

In order to balance the increase and decrease of oxidation numbers, multiply $H_{2} O_{2}$ by $2$ and $O_{2}$ by 2. Then we get,

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

To balance oxygen atoms, add $3$ $H_{2} O$ molecules on the product side. Then we get,

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

To balance hydrogen atoms, add $3 H^{+}$ ions on the reactant side. Then the balanced equation is,

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

Odisha Board Solutions for Chapter: Redox Reaction, Exercise 2: QUESTIONS

Odisha Board Chemistry Solutions for Exercise - Odisha Board Solutions for Chapter: Redox Reaction, Exercise 2: QUESTIONS

Questions from Odisha Board Solutions for Chapter: Redox Reaction, Exercise 2: QUESTIONS with Hints & Solutions

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