Complete and balance the following chemical equation:
$M n O^{-} + H_{2} O + I^{-} \to$

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Solution

Here, $+ 7 M n O_{4}^{-} ⟶ + 4 M n O_{2} I^{-} ⟶ I_{2}^{\circ}$
Reduction: $2 M 7 + n + 6 e^{-} ⟶ 2 M 4 + n$
Oxidation: $6 I^{-} - 6 e^{-} ⟶ 6 I^{\circ}$
Balancing $O$ atoms in reduction reaction,
$2 M n O_{4}^{-} + 4 H_{2} O ⟶ 2 M n O_{2} + 8 O H^{-}$

Overall reaction is:
$2 M n O_{4}^{-} + 6 I^{-} + 4 H_{2} O ⟶ 2 M n O_{2} + 3 I_{2} + 8 O H^{-}$

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