A solution of CuSO₄ is electrolysed for 10 minutes with a current of 1.5 ampers. What is the mass of copper deposited at the cathode ?

t = 600 s charge = current × time
= 1.5 A × 600 s = 900 C
According to the reaction:
$\begin{array}{l}{\mathrm{Cu}}_{\left(\mathrm{aq}\right)}^{2+}+2{\mathrm{e}}^{-}\to {\mathrm{Cu}}_{\left(\mathrm{s}\right)}\\ {\mathrm{E}}_{\mathrm{Cu}\left(\mathrm{II}\right)}=\frac{63.5}{2}=31.75\end{array}$
$\mathrm{W}=\frac{\mathrm{EIt}}{96500}$
$\mathrm{w}=\frac{31.75\times 1.5\times 10\times 60}{96500}=0.296\mathrm{gr}$