Chemistry - Electrochemistry Question with Solution | TestHub

In the electrolysis of cupric sulphate, the reaction that occurs at cathode is $${\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}⟶\mathrm{Cu}$$ Thus $2\mathrm{F}$ or $2\times 96500\mathrm{C}$ of electricity is required to deposit $=1\mathrm{mol}$ of $\mathrm{Cu}=63.5\mathrm{g}$ of $\mathrm{Cu}$ It means that to deposit $63.5\mathrm{g}$ of $\mathrm{Cu}$, the amount of electricity required $=2\times 96500\mathrm{C}$ So, to deposit $0.634\mathrm{g}$ of $\mathrm{Cu}$, the amount of $$\begin{array}{rl}& \text{ electricity required }=\frac{2\times 96500}{63.5}\times 0.634\\ & \approx 1930\mathrm{C}\end{array}$$