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Question

In order to completely oxidize 0.1 mol of $M n O_{4}^{2 -}$ to permanganate ion. The quantity of electricity required is:

96500 C

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2 \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 $9650 C$
$M n O_{4}^{2 -} \to M n O_{4}^{⊖} + e^{-}$
1 F = 1 mol of $M n O_{4}^{⊖}$
0.1 F = 0.1 mol of $M n O_{4}^{⊖}$
0.1 F = $0.1 \times 96500 = 9650 C$

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