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

C_{2} O_{4}^{2 -}

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M_{n} O_{4}^{-}

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M n^{2 +}

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H^{+}

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Solution

The correct option is A $C_{2} O_{4}^{2 -}$
In the reaction, $C_{2} O_{4}^{2 -}$ acts as reducing agent since Carbon gets oxidised from +3 to +4

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Similar questions

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

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

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

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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