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Question
How much charge is required for the following reduction?
1 mol of Al3+ to Al.
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Solution
The given reaction is:
\[\ce{\underset{(1 mol)}{Al^{3+}} + \underset{(3 mol)}{3e-} -> Al}\]
∴ 3 moles of electrons are needed for the reduction of 1 mole of Al3+ to Al.
3 mole electrons = 3 Faradays
= 3 × 96500 C
= 2.895 × 105 C
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