For the redox reaction,
${MnO}_{4}^{-} + C_{2} O_{4}^{2 -} + H^{+} \to {Mn}^{2 +} + {CO}_{2} + H_{2} O$
The correct coefficients of the reactants for the balanced equation are

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$16 {MnO}_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 2 H^{+}$

$2 {MnO}_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+}$

$2 {MnO}_{4}^{-} + 16 C_{2} O_{4}^{2 -} + 5 H^{+}$

$5 {MnO}_{4}^{-} + 16 C_{2} O_{4}^{2 -} + 2 H^{+}$

answer is B.

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Detailed Solution

(i) First balance the main element in the reaction
Reduction: ${M n O}_{4}^{-} \to {M n}^{+ 2}$
Oxidation: $C_{2} O_{4}^{2 -} \to 2 {C O}_{2}$
(ii) Balance oxygen atom on both sides by adding H2O to oxygen-defi cient side.
Reduction: ${MnO}_{4}^{-} \to {Mn}^{+ 2}$
Oxidation: $C_{2} O_{4}^{2 -} \to 2 {CO}_{2}$
(iii) Balance hydrogen atom on both sides by adding H+ to hydrogen defi cient side.
Reduction: ${MnO}_{4}^{-} + 8 H^{+} \to {Mn}^{+ 2} + 4 H_{2} O$
Oxidation: $C_{2} O_{4}^{2 -} \to 2 {CO}_{2}$
(iv)Balance the charge by adding e– s on the side that has more number of charges.
Reduction: ${MnO}_{4}^{-} + 8 H^{+} + 5 e^{-} \to {Mn}^{+ 2} + 4 H_{2} O$
Oxidation: $C_{2} O_{4}^{2 -} \to 2 {CO}_{2} + 2 \bar{e}$
In order to balance the electrons, we multiply the oxidation reaction by ‘5‘ and reduction reaction by ‘2‘.
Add both equations, we get the balanced equation. The balanced chemical equation in acidic medium will be
$5 C_{2} O_{4}^{2 -} + 2 {MnO}_{4}^{-} + 16 H^{+} \to 2 {Mn}^{+ 2} + 8 H_{2} O + 10 {CO}_{2}$