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Faraday's Second Law

When 0.1 mol ...

Question

When 0.1 mol $M n O_{4}^{2 -}$ is oxidised, quantity of electricity required to completely oxidise $M n O_{4}^{2 -}$ to $M n O_{4}^{-}$ is

96500 C

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\times

96500 C

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9650 C

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96.50 C

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Solution

The correct option is C 9650 C
The oxidation reaction is
$+ 6 M n O_{4}^{2 -} 0.1 m o l \to + 7 M n O_{4}^{-} +^{-}$
$Q = 0.1 \times F = 0.1 \times 96500 C = 9650 C$

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