The oxidising agent in the following reaction is:

$2 M n O_{4}^{-} + 16 H^{+} + 5 C_{2} O_{4}^{- 2} \to 2 M n^{2 +} + 8 H_{2} O + 10 C O_{2}$

M n O_{4}^{-}

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H^{+}

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

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both

M n O_{4}^{-}

and

H^{+}

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Solution

The correct option is A $M n O_{4}^{-}$
Oxidation state of $M n$ decreases from $+ 7$ to $+ 2$ while that of carbon increases from $+ 3$ to $+ 4$ and hence, is oxidised.
Therefore, $M n$ itself gets reduced and acts as an oxidizing agent.

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Similar questions

2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} ⟶ 10 {C O}_{2} + 2 {M n}^{2 +} + 8 H_{2} O

M n O_{4}^{-}

2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O

20 m L

0.1 M M n O_{4}^{-}

2 M n O_{4}^{-} + 5 H_{2} C_{2} O_{4} + 6 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O

K M n O_{4} (m . w . = 158)

C O_{2}

5 {C_{2} O_{4}}^{2 -} + 2 {M n O_{4}}^{-} + 16 H^{+} \to 10 C O_{2} + 2 M n^{2 +} + 8 H_{2} O

K M n O_{4}

M n O_{4}^{-}

4.56 \times 10^{- 3} M s^{- 1}

2 M n O_{4}^{-} + 10 I^{-} + 16 H^{+} \to 2 M n^{2 +} + 5 I_{2} + 8 H_{2} O

I_{2}

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