For the reaction of $M n O_{2} + C_{2} O_{4}^{2 -} + H^{+} \to M n^{2 +} + C O_{2} + H_{2} O$ the correct whole number stoichiometric coefficients of $M n O_{4}$ , $C_{2}$ $O_{4}$ and $H^{+}$ are respectively :

2 , 5 , 16

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16 , 5 , 2

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5 , 16 , 2

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2 , 16 , 5

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Solution

The correct option is D 2 , 16 , 5
${M n}^{+ 7} O_{2} + C_{2}^{+ 3} {O_{4}}^{2 -} + H^{+} ⟶ {M n}^{+ 2} + C^{+ 4} O_{2} + H_{2} O \cdot$

$\begin{matrix} Reduction & = 8 H^{+} + M n O_{4}^{+ 7} \to {M n}^{2 +} + 4 H_{2} O \cdot (i) HaIf Reaction & Now Balance the electrons \end{matrix}$

$\begin{matrix} Oxidation Half Reaction C_{2} O_{4}^{2 -} ⟶ 2 C O_{2} + 2 e^{-} - (2) \end{matrix}$

$Now Add both Reaction (1) & (2)$
$2 {M n O_{4}}^{-} + 16 H^{+} + {5 C}_{2} O 4^{2 -} \to 2 {M n}^{2 +} + 10 C O_{2} + 8 H_{2} O$

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Similar questions

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

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

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

M n O_{4}^{-}, 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

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

H^{+}

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

M n O_{4}^{-}, C_{2} O_{4}^{2 -} a n d H^{+}

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