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Question
How much charge is required for the following reduction:
1 mol of \[\ce{Cu^{2+}}\] to \[\ce{Cu}\]?
Solution
\[\ce{\underset{(1 mol)}{Cu^{2+}} + \underset{(2 mol)}{2e^-} -> Cu}\]
∴ Required charge = 2 Faradays
= 2 × 96500 C
= 1.93 × 105 C
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