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How much charge is required for the following reductions : (i) 1 mol of Al^(3+) to Al? (ii) 1 mol of Cu^(2+) to Cu ? (iii) 1 mol of MnO_(4)^(-)to Mn^(2+)? |
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Answer» Solution :(i) The electrode reaction is REPRESENTED as: `Al^(3+) + 3e^(-) to Al` `therefore` Quantity fo charge required for reduction of 1 mol of `Al^(3+)` =`3 F = 3 xx 96500 C = 289500 C` (ii) The electrode reaction is represented by `Cu^(2+) + 2E^(-) to Cu` `therefore` Quantity of charge required for reduction of 1 mol of `Cu^(2+)` `=2F= 2 xx 96500 C = 193000 C` (iii) The electrode reaction is represented as: `=5F = 5 xx 96500 C = 482500 C` |
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