Faraday's Laws

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Question

When 0.1 mol $M n O_{4}^{2 -}$ is oxidised, the quantity of electricity required to completely oxidise $M n O_{4}^{2 -}$ to $M n O_{4}^{1 -}$ is

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Solution

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

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