Balance the following equations by ion electron method.
${C r_{2} O_{7}}^{2 -} + I^{-} + H^{+} \to C r^{3 +} + I_{2} + H_{2} O$ .

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Solution

$C r_{2} O_{7}^{2 -} + I^{-} + H^{+} \to C r^{3 +} + I_{2} + H_{2} O$
$1$ st step: Splitting into two half reactions,
$C r_{2} O_{7}^{2 -} + H^{+} \to C r^{3 +} + H_{2} O (R e d u c t i o n h a l f r e a c t i o n)$ ; $I^{-} \to I_{2} (O x i d a t i o n h a l f r e a c t i o n)$
$2$ nd step: Adding hydrogen ions to the side deficient in hydrogen,
$C r_{2} O_{7}^{2 -} + 14 H^{+} \to C r^{3 +} + 7 H_{2} O$
$3$ rd side: Making atoms equal on both sides,
$C r_{2} O_{7}^{2 -} + 14 H^{+} \to 2 C r^{3 +} + 7 H_{2} O; 2 I^{-} \to I_{2}$
$4$ th step: Adding electrons to the sides deficient in electrons,
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e \to 2 C r^{3 +} + 7 H_{2} O$
$2 I^{-} \to I_{2} + 2 e$
$5$ th step: Balancing electrons,
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 I^{-} \to 2 C r^{3 +} + 3 I_{2} + 7 H_{2} O$ .

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