What is the change in the oxidation state of Mn, in the reaction of $M n O_{4}^{-}$ with $H_{2} O_{2}$ in acidic medium?

7 \to 4

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6 \to 4

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7 \to 2

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6 \to 2

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Solution

The correct option is B $7 \to 2$
In acidic medium:

$3 H_{2} O_{2} + 2 K M n O_{4} ⟶ 2 M n O_{2} + 2 K O H + 3 O_{2} + 2 H_{2} O$

$7 + M n O_{4}^{-} \to M n^{2 +}$

The oxidation state of $M n$ changes from +7 to +2.

Hence, option C is correct.

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Similar questions

K M n O_{4}

M n^{7 +}

_{4}

+ 7

The Oxidation state of Mn in $K_{2} M n O_{4}$ is

{F e}^{2 +}

M n O_{4}^{-}

{C r}_{2} O_{7}^{2 -}

M n

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

4

+ 6

+ 7

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