Balance the following equations by ion electron method:
$C l_{2} + {I O_{3}}^{-} + O H^{-} \to {I O_{4}}^{-} + C l^{-} + H_{2} O$ .

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Solution

$C l_{2} + I O_{3}^{-} + O H^{-} \to I O_{4}^{-} + C l^{-} + H_{2} O$

$1$ st step : Splitting into two half reactions,

$I O_{3}^{-} + O H^{-} \to I O_{4}^{-} + H_{2} O (O x i d a t i o n h a l f r e a c t i o n)$ ; $C l_{2} \to C l^{-} (R e d u c t i o n h a l f r e a c t i o n)$

$2$ nd step : Adding $O H^{-}$ ions,

$I O_{3}^{-} + 2 O H^{-} \to I O_{4}^{-} + H_{2} O$

$3$ rd step : Adding electrons to the sides deficient in electrons,

$I O_{3}^{-} + 2 O H^{-} \to I O_{4}^{-} + H_{2} O + 2 e$

$C l_{2} + 2 e \to 2 C l^{-}$

$4$ th step: Adding both the half reactions,

$I O_{3}^{-} + 2 O H^{-} + C l_{2} \to I O_{4}^{-} + 2 C l^{-} + H_{2} O$ .

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