Balance the following equation by ion electron method.
${C r_{2} O_{7}}^{2 -} + H^{+} + {C_{2} O_{4}}^{2 -} \to C r^{3 +} + C O_{2} + H_{2} O$

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Solution

${C r_{2} O_{7}}^{2 -} + H^{+} + {C_{2} O_{4}}^{2 -} \to C r^{3 +} + C O_{2} + H_{2} O$
$6 e^{-} + C r_{2}^{+ 12} O_{7}^{2 -} = 2 C r^{3 +}$ $r e d u c t i o n$
$C_{2}^{+ 6} O_{4}^{2 -} ⟶ 2 C^{+ 8} O_{2} + 2 e$ $\times 3$ $o x i d a t i o n$
Balancing the charges and then adding both equations,
${C r_{2} O_{7}}^{2 -} + 3 {C_{2} O_{4}}^{2 -} \to 2 C r^{3 +} + 6 C O_{2}$
Balancing oxygen,
${C r_{2} O_{7}}^{2 -} + 3 {C_{2} O_{4}}^{2 -} \to 2 C r^{3 +} + 6 C O_{2} + 7 H_{2} O$
Balancing hydrogen,
$14 H^{+} + {C r_{2} O_{7}}^{2 -} + 3 {C_{2} O_{4}}^{2 -} \to 2 C r^{3 +} + 6 C O_{2} + 7 H_{2} O$
Since it is an alkaline medium. Balancing $14 O H^{-}$ on both sides to neutralize $14 H^{+}$ .
Such that $14 H^{+} + 14 O H^{-} ⟶ 14 H_{2} O$
$14 H_{2} O + {C r_{2} O_{7}}^{2 -} + 3 {C_{2} O_{4}}^{2 -} \to 2 C r^{3 +} + 6 C O_{2} + 7 H_{2} O$
$A n s :$ $14 H^{+} + {C r_{2} O_{7}}^{2 -} + 3 {C_{2} O_{4}}^{2 -} \to 2 C r^{3 +} + 6 C O_{2} + 7 H_{2} O$

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