Ten grams of a fairly concentrated solution of cupric sulphate is electrolysed using 0.01 faraday of electricity. Calculate (i) the weight of the resulting solution and (ii) the number of equivalent of acid or alkali in the solution. (Cu = 63.5, F = 96500 coulombs)

Ten grams of a fairly concentrated solution of cupric sulphate is elec

1.	Ten grams of a fairly concentrated solution of cupric sulphate is electrolysed using 0.01 faraday of electricity. Calculate (i) the weight of the resulting solution and (ii) the number of equivalent of acid or alkali in the solution. (Cu = 63.5, F = 96500 coulombs)
Answer» Solution :(i) Since mole of electric charge used = 0.01 F `therefore` eq/ of Cu deposited = 0.01 `therefore` WT. of Cu deposited `= 0.01 xx 31.75 = 0.3175g`. `("eq. wt. of Cu " = (63.5)/(2) = 31.75)` Eq. of oxygen discharged = 0.01 `therefore` wt. of oxygen discharged `= 0.01 xx 8 = 0.08g`. `("eq. wt. of Cu " = (16)/(2) = 8)` Total loss in weight due to electrolysis = 0.3175 + 0.08 = 0.3975 g `therefore` the weight of the resulting solution = 10 - 0.3975 = 9.6025 g. (ii) Electrolysis of `CuSO_(4)` solutions follows through `CuSO_(4) = Cu^(2+) + SO_(4)^(2-)` At cathode : `Cu^(2+) + 2E^(-) RARR Cu` At anode : `SO_(4)^(2-) + H_(2)O rarr H_(2)SO_(4) + (1)/(2)O_(2) uarr + 2e^(-)` Thus `H_(2)SO_(4)` is produced during electrolysis. `therefore` eq. of `H_(2)SO_(4)` produced = 0.01 (due to passage of 0.01 faraday)