In the reaction between acidified $K M n O_{4}$ and hot oxalic acid, the species that gains electrons is:

K^{+}

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M n O_{4}^{-}

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

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C O_{2}

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Solution

The correct option is C $M n O_{4}^{-}$
The reaction between potassium permanganate and oxalic acid is a redox reaction.

During the reaction, potassium permanganate (which is a strong oxidizing agent) is reduced and oxalic acid is oxidized.

Thus, $M n O_{4}^{-}$ ion gains electrons.

Option B is correct.

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A salt which gives $C O_{2}$ with hot $H_{2} S O_{4}$ and also decolourises acidified $K M n O_{4}$ on warming is:

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

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