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Oxides of Sulphur

The charge re...

Question

The charge required for the reduction of $1 m o l$ of ${M n O_{4}}^{-}$ to $M n O_{2}$ is:

1 F

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3 F

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5 F

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6 F

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Solution

The correct option is B $3 F$
$M n O_{4}^{-} + 4 H^{+} + 3 e^{-} \to M n O_{2} + 2 H_{2} O$

Oxidation number of Mn in ${M n O_{4}}^{-}$ = +7
Oxidation number of Mn in $M n O_{2}$ = +4

So, $3 F$ of charge will be required to reduce $1 m o l e$ of $M n O_{4}^{-}$ .

Hence, option B is correct.

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