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

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Solution

$M n O_{4}^{-} + C_{2} O_{4}^{2 -} + H^{+} \to M n^{2 +} + C O_{2} + H_{2} O$
$1$ st step: Splitting into two half reactions,
$M n O_{4}^{-} + H^{+} \to M n^{2 +} + H_{2} O; C_{2} O_{4}^{2 -} \to 2 C O_{2}$
$2$ nd step: Balancing electrons to the side deficient in electrons,
$M n O_{4}^{-} + 8 H^{+} + 5 e \to M n^{2 +} + 4 H_{2} O$
$C_{2} O_{4}^{2 -} \to 2 C O_{2} + 2 e$
$3$ rd step: Balancing electrons in both half reactions,
$2 [M n O_{4}^{-} + 8 H^{+} + 5 e \to M n^{2 +} + 4 H_{2} O]$ ;
$5 [C_{2} O_{4}^{2 -} \to 2 C O_{2} + 2 e]$
$4$ th step: Adding both the half reactions,
$2 M n O_{4}^{-} + 16 H^{+} + 5 C_{2} O_{4}^{2 -} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O$ .

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Similar questions

{M n O_{4}}^{-} + H^{+} + H_{2} O_{2} \to M n^{2 +} + H_{2} O + O_{2}

M n O_{4}^{2 -}, C_{2} O_{4}^{2 -}

H^{+}

M n O_{4}^{-} + C_{2} O_{4}^{2 -} + H^{+} \to M n^{2 +} + C O_{2} + H_{2} O

C o^{2 +} + N O_{2}^{-} + H^{+} \to C o^{3 +} + N O + H_{2} O

In the reaction
$C_{2} O_{4}^{2 -} + M n O_{4}^{-} + H^{+} ⟶ M n^{2 +} + C O_{2} + H_{2} O$ the reductant is

M n O_{4} - + C_{2} O_{42^{-}} + H + \to C O_{2} + M n + 2 + H_{2} O

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