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Potassium Permanganate

Balance the f...

Question

Balance the following reaction.
$M n O_{4}^{-} + H_{2} O_{2} \to M n O_{2} + O_{2}$

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Solution

To balance ${M n O_{4}}^{-} + H_{2} O_{2} ⟶ M n O_{2} + O_{2}$
(1) Oxidation $H_{2} {- 1 O}_{2} ⟶ {0 O}_{2} + 2 e^{-}$
Reduction $+ 7 M n O_{4}^{-} + 3 e^{-} ⟶ + 4 M n O_{2}$

(2) Balancing the atoms & charge
$H_{2} O_{2} + 2 O H^{-} ⟶ O_{2} + 2 e^{-}$
$M n O_{4}^{-} + 3 e^{-} ⟶ M n O_{2} + 4 O H^{-}$
Balance $O$ -atoms,
$H_{2} O_{2} + 2 O H^{-} ⟶ O_{2} + 2 e^{-} + 2 H_{2} O ⟶ (1)$
$M n O_{4}^{-} + 3 e^{-} + 2 H_{2} O ⟶ M n O_{2} + 4 O H^{-} ⟶ (2)$

(3) Electron gain equivalent to electron lost
$(1) \times 3$ & $(2) \times 2$
$\Rightarrow 3 H_{2} O_{2} + 6 O H^{-} ⟶ 3 O_{2} + 6 e^{-} + 6 H_{2} O ⟶ (3)$
$2 M n O_{4} + 6 e^{-} + 4 H_{2} O ⟶ 2 M n O_{2} + 8 O H^{-} ⟶ (4)$
Adding & simplifying $(3)$ & $(4)$ we get,
$3 H_{2} O_{2} + 2 M n O_{4}^{-} ⇌ 2 M n O_{2} + 3 O_{2} + 2 O H^{-} + 2 H_{2} O$

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