Calculate current in ampere required to deposit \(4.8 \mathrm{~g} \mathrm{Cu}\) from it's salt solution in 30 minutes. \(\left[\right.\) Molar mass of \(\left.\mathrm{Cu}=63.5 \mathrm{~g} \mathrm{~mol}^{-1}\right]\)

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