$M n O_{4}^{-} + C_{2} O_{4}^{2 -} + H^{+} \to C O_{2} + H_{2} O + M n^{2 +}$
The correct coefficients of $M n O_{4}^{-}, C_{2} O_{4}^{2 -}$ 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 A $2, 5, 16$
To balance the atoms other than $H$ and $O$ , coefficient $2$ is added in front of $C O_{2}$ .
C is oxidized. Its oxidation number increases from $+ 3$ to $+ 4$ . The net increase in the oxidation number is $+ 2$ .
Mn is reduced. The oxidation number decreases from $+ 7$ to $+ 2$ . The net decrease in the oxidation number is $+ 5$ .
Hence, coefficient of C is multiplied with $5$ and the coefficient of Mn is multiplied with $2$ .
To balance oxygen atoms, $7$ water molecules are added to right. To balance H atoms, $15$ protons are added to left.
Hence, the balanced reaction is $2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 10 C O_{2} + 8 H_{2} O + 2 M n^{2 +}$ .

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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^{+}

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

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

For the redox reaction
MnO−4+C2O−24+H+⟶Mn2++CO2+H2O
the correct coefficients of the reactants MnO−4, C2O2−4 and H+ for the balanced reaction are

2, 5, 16

16, 5, 2

5, 16, 2

2, 16, 5

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