When $0.1 m o l M n O_{4}^{2 -}$ is oxidised, the quanitity of electricity required to completely oxidise $M n O_{4}^{2 -}$ to $M n O_{4}^{-}$ is : -

96500 C

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9650 C

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2 \times 96500 C

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96.50 C

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Solution

The correct option is B $9650 C$
$\begin{matrix} + 6 M n O_{4}^{2 -} & \to & + 7 M n O_{4}^{-} + e^{-} 0.1 m o l e & 0.1 m o l e \end{matrix}$
Charge required $= 0.1 F$
$= 0.1 \times 96500$
$= 9650 C$

Hence, option (c) is correct.

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