Balance the following redox reaction in basic medium:

${M n}^{2 +} + H_{2} O_{2} ⟶ M n O_{2}$

{M n}^{2 +} + H_{2} O_{2} + 2 {O H}^{-} ⟶ M n O_{2} + 2 H_{2} O

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{2 M n}^{2 +} + H_{2} O_{2} + 2 {O H}^{-} ⟶ M n O_{2} + 3 H_{2} O

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{3 M n}^{2 +} + H_{2} O_{2} + 2 {O H}^{-} ⟶ M n O_{2} + 3 H_{2} O

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None of these

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Solution

The correct option is C ${M n}^{2 +} + H_{2} O_{2} + 2 {O H}^{-} ⟶ M n O_{2} + 2 H_{2} O$
The unbalanced redox equation is as follows:
${M n}^{2 +} + H_{2} O_{2} ⟶ M n O_{2}$
All atoms other than H and O are balanced.
The oxidation number of Mn changes from 2 to 4. The change in the oxidation number is 2.
The oxidation number of O changes fro -1 to -2. The change in he oxidation number per O atom is 1. The total change in oxidation number for 2 O atoms is 2.
The increase in the oxidation number is balanced with decrease in the oxidation number.
The O atoms are balanced.
To balance H atoms, add two water molecules on RHS and two hydroxide ions to LHS.
${M n}^{2 +} + H_{2} O_{2} + 2 {O H}^{-} ⟶ M n O_{2} + 2 H_{2} O$
This is the balanced chemical equation.

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