Balance the following equations by ion electron method.
${M n O_{4}}^{-} + F e^{2 +} + H^{+} \to M n^{2 +} + F e^{3 +} + H_{2} O$ .

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Solution

$M n O_{4}^{-} + F e^{2 +} + H^{+} \to M n^{2 +} + F e^{3 +} + H_{2} O$
$1$ st step: Splitting into two half reactions,
$M n O_{4}^{-} + H^{+} \to M n^{2 +} + H_{2} O; (R e d u c t i o n h a l f r e a c t i o n)$ $F e^{2 +} \to F e^{3 +} (O x i d a t i o n h a l f r e a c t i o n)$
$2$ nd step: Adding hydrogen ions to the side deficient in hydrogen,
$M n O_{4}^{-} + 8 H^{+} \to M n^{2 +} + 4 H_{2} O$
$3$ rd step: Adding electrons to the sides deficient in electrons,
$M n O_{4}^{-} + 8 H^{+} + 5 e \to M n^{2 +} + 4 H_{2} O$
$F e^{2 +} \to F e^{3 +} + e$
$4$ th step: Balancing electrons in both half reactions,
$M n O_{4}^{-} + 8 H^{+} + 5 e \to M n^{2 +} + 4 H_{2} O$
$5 F e 2 + \to 5 F e^{3 +} + 5 e$
$5$ th step: Adding both the half reactions,
$M n O_{4}^{-} + 8 H^{+} + 5 F e^{2 +} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$ .

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