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Question
How much charge is required for the following reduction:
1 mol of \[\ce{Al^{3+}}\] to \[\ce{Al}\]?
Solution
\[\ce{\underset{(1 mol)}{Al^{3+}} + \underset{(3 mol)}{3e^-} -> Al}\]
∴ Required charge = 3 Faradays
= 3 × 96500 C
= 2.895 × 105 C
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