Balance the following equations by ion electron method.
$M n O_{2} + H C l \to M n C l_{2} + H_{2} O + C l_{2}$ .

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Solution

$M n O_{2} + H C l \to M n C l_{2} + H_{2} O + C l_{2}$
Ionic equation,
$M n O_{2} + H^{+} + C l^{-} \to M n^{2 +} + H_{2} O + C l_{2}$
$1$ st step: Splitting into two half reactions,
$C l^{-} \to C l_{2} (O x i d a t i o n h a l f r e a c t i o n)$ ; $M n O_{2} + H^{+} \to M n^{2 +} + H_{2} O (R e d u c t i o n h a l f r e a c t i o n)$
$2$ nd step: Adding $H^{+}$ ions to the side deficient in hydrogen,
$C l^{-} \to C l_{2}; M n O_{2} + 4 H^{+} \to M n^{2 +} + 2 H_{2} O$
$3$ rd step: Making atoms equal on both sides,
$2 C l^{-} \to C l_{2}; M n O_{2} + 4 H^{+} \to M n^{2 +} + 2 H_{2} O$
$4$ th step: Adding electrons to the side deficient in electrons,
$2 C l^{-} \to C l_{2} + 2 e$
$M n O_{2} + 4 H^{+} + 2 e \to M n^{2 +} + 2 H_{2} O$
$5$ th step: Adding both the half reactions,
$2 C l^{-} + M n O_{2} + 4 H^{+} \to C l_{2} + M n^{2 +} + 2 H_{2} O$
Converting it into molecular form,
$M n O_{2} + 2 C l^{-} + 4 H^{+} + 2 C l^{-} \to C l_{2} + M n^{2 +} + 2 C l^{-} + 2 H_{2} O$
or
$M n O_{2} + 4 H C l \to M n C l_{2} + C l_{2} + 2 H_{2} O$

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