A solution of copper sulphate is electrolysed for $10$ minutes with a current of $1.5$ amperes. The mass of copper deposited at the cathode is: (Given: Molar mass of $\mathrm{Cu}$ $=63$ g mol${}^{-1}$; $1F=96487$ C mol${}^{-1}$)

Solution & Explanation

\textbf{Answer:} (B) 0.2938 g \textbf{Solution:} At the cathode: $\mathrm{Cu}{}^{2+}+2\mathrm{e}{}^{-}\to \mathrm{Cu}$ ($n=2$). Charge passed: $Q=It=1.5\times (10\times 60)=1.5\times 600=900\text{C}$. Using $w=\frac{M\cdot Q}{n\cdot F}$, $$w=\frac{63\times 900}{2\times 96487}=\frac{56700}{192974}\approx 0.2938\text{g}.$$