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Question
How much charge is required for the following reduction?
1 mol of Cu2+ to Cu.
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Solution
The given reaction:
\[\ce{\underset{(1 mol)}{Cu^{2+}} + \underset{(2 mol)}{2e-} -> Cu}\]
∴ 2 moles of electrons are needed for the reduction of 1 mole of Cu2+ to Cu.
∴ 2 mole electrons = 2 Faradays
= 2 × 96500 C
= 1.93 × 105 C
= 193000 C
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