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10g fairly concentrated solution of `CuSO_(4)` is electrolyzed using `0.01F` of electricity. Calculate: (a) The weight of resulting solution (b) Equivalents of acid or alkali in the solution. |
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Answer» Electrode process during electrolysis of aqueous `CuSO_(4)` may be given as: `Cu^(2+) + 2e^(-) rarr Cu` (Cathode) `2OH^(-) rarr H_(2)O + 1//2 O_(2) + 2e^(-)` (Anode) Mass of copper deposited at cathode by 0.01 faraday charge `= 0.01 xx 31.75` `= 0.3175g` (Here, 31.75 is the equivalent mass of `Cu^(2+)`) Mass of oxygen evolved by 0.01 faraday charge `= 0.01 xx 8 = 0.08 g` Total weight loss from solution `= 0.3175 + 0.08 = 0.3975 g` Mass of resulting solution `= 10 - 0.3975` `= 9.6025 g` After deposition of `Cu^(2+) and OH^(-)` ions at the respective electrodes, `H_(2)SO_(4)` will prevail in the solution. 0.01 faraday of electricity will result in 0.01 equivalent of acid. |
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