Chemistry - Electrochemistry Question with Solution | TestHub

$\begin{array}{rl}& {\mathrm{Cu}}_{(\mathrm{s})}^{2+}+2{\mathrm{e}}^{-}⟶{\mathrm{Cu}}_{(\mathrm{s})}\\ & \text{ Mole ratio }=\frac{1\mathrm{mol}}{2{\mathrm{mole}}^{-}}\\ & \mathrm{W}=\frac{\mathrm{I}(\mathrm{A})\times \mathrm{t}(\mathrm{s})}{96500\left(\mathrm{C}/\mathrm{mol}{\mathrm{e}}^{-}\right)}\times \text{mole ratio }\times \text{ molar mass }\\ & 4.8\mathrm{g}=\frac{\mathrm{I}(\mathrm{A})\times 30\times 60}{96500\left(\mathrm{C}/\mathrm{mol}{\mathrm{e}}^{-}\right)}\times \frac{1\mathrm{mol}}{2\mathrm{mol}{\mathrm{e}}^{-}}\times 63.5\mathrm{g}{\mathrm{mol}}^{-1}\\ & \mathrm{I}(\mathrm{A})=\frac{4.8\times 96500\times 2}{63.5\times 30\times 60}=8.1\mathrm{A}\end{array}$