Calculate the time required in second to deposit \(6.35 \mathrm{~g}\) copper from its salt solution by passing 5 ampere current.
[Molar mass of \(\mathrm{Cu}=63.5 \mathrm{~g} \mathrm{~mol}^{-1}\) ]

\(\mathrm{Cu}_{(\mathrm{s})}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}_{(\mathrm{s})} \)
\( \text {Mole ratio}=\frac{1 \mathrm{~mol}}{2 \mathrm{~mole}^{-}} \)
\( \mathrm{W}=\frac{\mathrm{I}(\mathrm{A}) \times \mathrm{t}(\mathrm{s})}{96500\left(\mathrm{C} / \mathrm{mol} \mathrm{e}^{-}\right)} \times \text {mole ratio }\) \(\times \text { molar mass}\)
\( 6.35 \mathrm{~g}=\frac{5 \times \mathrm{t}}{96500\left(\mathrm{C} / \mathrm{mol} \mathrm{e}^{-}\right)} \times \frac{1 \mathrm{~mol}}{2 \mathrm{~mol} \mathrm{e}}~ \times\) \(63.5 \mathrm{~g} \mathrm{~mol}^{-1} \)
\( \mathrm{t}=\frac{6.35 \times 96500 \times 2}{5 \times 63.5}=3860 \text { seconds}\)