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Question
How much charge is required for the following reduction:
1 mol of \[\ce{MnO^-_4}\] to Mn2+?
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Solution
\[\ce{\underset{(1 mol)}{MnO^-_4} + 8H^+ + \underset{(5 mol)}{5e^-} -> Mn^{2+} + 4H2O}\]
∴ Required charge = 5 Faradays
= 5 × 96500 C
= 4.825 × 105 C
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