CuSO 4 solution is electrolysed for 15 minutes to deposit 0 . 4725   g of copper at the cathode. The…

The reaction at cathode takes place as below:

${\mathrm{Cu}}^{2+ }+ 2{\mathrm{e}}^{-} \to \mathrm{Cu}\left(\mathrm{s}\right)$

$$\begin{gathered} \mathrm{For} 63 \mathrm{g} \mathrm{of} \mathrm{Cu} = 2\mathrm{F} \mathrm{electricity} \mathrm{required} \\ \mathrm{So}, \mathrm{for} 0.475 \mathrm{g}, \mathrm{electricity} \mathrm{required}= \end{gathered}$$

$$\begin{gathered} \frac{2\mathrm{F}}{63}\times 0.4725=\frac{2\times 96500\times 0.4725}{63} \\ =1447.5 C \end{gathered}$$

According to Faradays's law:

$$\begin{gathered} \mathrm{Q}=\mathrm{It}\to \mathrm{I}=\frac{\mathrm{Q}}{\mathrm{t}} \\ \mathrm{I}=\frac{1447.5}{15\times 60}=1.608 \mathrm{C} \end{gathered}$$