10 F of electricity is passed through the solutions of silver nitrate, copper sulphate and ferric chloride. The amount of metal deposited at cathode in each case respectively are

$$\begin{gathered} {\mathrm{Ag}}^{+}+\underset{1\mathrm{mol}}{1\mathrm{F}}⟶\mathrm{Ag};{\mathrm{Cu}}^{2+}+\underset{2\mathrm{mol}}{2\mathrm{F}}\stackrel{-}{⟶}\mathrm{Cu} \\ 1\mathrm{F} 2\mathrm{F} \end{gathered}$$
${\mathrm{Fe}}^{2+}+\underset{\begin{array}{c}3\mathrm{mol}\\ 3\mathrm{F}\end{array}}{⟶}\mathrm{Fe}$
Gram of Ag liberated $=\frac{108\times 10}{1}=1080\mathrm{g}$
Gram of Cu liberated $\text{=}\frac{63.5\times 10}{2}=317.5\mathrm{g}$
Gram of Fe liberated $=\frac{56\times 10}{3}=186.7\mathrm{g}$