Balance the following redox reactions in basic medium:

$H_{2} O_{2} + M n O_{4}^{-} ⟶ O_{2} + M n O_{2}$

3 H_{2} O_{2} + 3 M n O_{4}^{-} ⟶ {3 O}_{2} + 2 M n O_{2} + 2 {O H}^{-} + 2 H_{2} O

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3 H_{2} O_{2} + 2 M n O_{4}^{-} ⟶ {3 O}_{2} + 2 M n O_{2} + 2 {O H}^{-} + 2 H_{2} O

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3 H_{2} O_{2} + 5 M n O_{4}^{-} ⟶ {3 O}_{2} + 2 M n O_{2} + 2 {O H}^{-} + 2 H_{2} O

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None of these

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Solution

The correct option is B $3 H_{2} O_{2} + 2 M n O_{4}^{-} ⟶ {3 O}_{2} + 2 M n O_{2} + 2 {O H}^{-} + 2 H_{2} O$
The unbalanced redox equation is as follows:
$H_{2} O_{2} + M n O_{4}^{-} ⟶ O_{2} + M n O_{2} (b a s i c)$
All atoms other than H and O are balanced.
The oxidation number of Mn changes from +7 to +4. The change in the oxidation number is 3.
The oxidation number of O changes from -1 to 0. The change in the oxidation number per O atom is 1. The total change in oxidation number for 2 O atoms is 2.
To balance the increase in the oxidation number with decrease in the oxidation number multiply $M n O_{4}^{-}$ and $M n O_{2}$ with 2 and multiply $H_{2} O_{2}$ and $O_{2}$ with 3..
$3 H_{2} O_{2} + 2 M n O_{4}^{-} ⟶ 3 O_{2} + 2 M n O_{2} (b a s i c)$
To balance O atoms, add four water molecules on RHS.
$3 H_{2} O_{2} + 2 M n O_{4}^{-} ⟶ 3 O_{2} + 2 M n O_{2} + 4 H_{2} O (b a s i c)$
To balance H atoms, add 2 $H^{+}$ on LHS.
$3 H_{2} O_{2} + 2 M n O_{4}^{-} + 2 H^{+} ⟶ 3 O_{2} + 2 M n O_{2} + 4 H_{2} O (b a s i c)$
Since the reaction occurs in basic medium, add 2 $O H^{-}$ ions on either side.
$3 H_{2} O_{2} + 2 M n O_{4}^{-} + 2 H^{+} + 2 O H^{-} ⟶ 3 O_{2} + 2 M n O_{2} + 4 H_{2} O + 2 O H^{-} (b a s i c)$
On LHS, 2 $H^{+}$ ions combine with 2 $O H^{-}$ ions to form 2 water molecules which cancels with two water molecules on RHS.
$3 H_{2} O_{2} + 2 M n O_{4}^{-} ⟶ {3 O}_{2} + 2 M n O_{2} + 2 {O H}^{-} + 2 H_{2} O$
This is the balanced chemical equation.

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2. H_{2} O_{2} + H O C l \to C l^{-} + H_{3} O^{+} + O_{2}

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