Question

Ten gram of a fairly concentrated solution of cupric sulphate is electrolysed using $0.01$ faraday of electricity.
Calculate:
(i) The mass of the resulting solution.
(ii) The number of equivalents of acid or alkali in the solution.

Answer 1

Electrode process during electrolysis of aqueous $CuS{O}_4$ may be given as:
$C{u}^{2+}+2e^{-}\to Cu\left(Cathode\right)$
$2O{H}^{-}\to H_{2}O+\frac{1}{2}{O}_2+2e^{-}\left(Anode\right)$
Mass of copper deposited at cathode by $0.01$ faraday charge
$=0.01\times 31.75$
$=0.3175g$
(Here, $31.75$ is the equivalent mass of $C{u}^{2+}$)
Mass of oxygen evolved by $0.01$ faraday charge
$=0.01\times 8=0.08g$
Total weight loss from solution $=0.3175+0.08=0.3975g$
Mass of resulting solution $=10-0.3975$
$=9.6025g$
After deposition of $C{u}^{2+}$ and $O{H}^{-}$ ions at the respective electrodes, $H_{2}S{O}_4$ will prevail in the solution. $0.01$ faraday of electricity will result in $0.01$ equivalent of acid.