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.

Explanation

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.