What is the coefficient of oxalate ion in the balanced form of following reaction?
$M n O_{4}^{-} + C_{2} O_{4}^{2 -} + H^{+} ⟶ M n^{2 +} + C O_{2} + H_{2} O$

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Solution

The correct option is D 5
$[M n O_{4}^{-} + 8 H^{+} + 5 e^{-} ⟶ M n^{2 +} + 4 H_{2} O] \times 2..... (i)$
$[C_{2} O_{4}^{2 -} ⟶ 2 C O_{2} + 2 e^{-}] \times 5..... (i i)$
Adding equation (i) and (ii) we get
$2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} ⟶ 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O$
Thus, the coefficient of oxalate ion, $C_{2} O_{4}^{2 -}$ is 5.

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