If three faradays of electricity are passed through the solutions of ${\mathrm{AgNO}}_3,{\mathrm{CuSO}}_4\text{ and }{\mathrm{AuCl}}_3$, the molar ratio of the cations deposited at the cathodes will be

$$\begin{gathered} {\mathrm{Ag}}^{+}+\hspace{0.17em}{\mathrm{e}}^{-}\to \mathrm{Ag} \\ 1 \mathrm{F}\hspace{0.17em}= 1 \mathrm{mole} \mathrm{Ag} \\ 3\mathrm{F}\hspace{0.17em}= 3 \mathrm{moles}\hspace{0.17em}\mathrm{Ag} \end{gathered}$$

$$\begin{gathered} {\mathrm{Cu}}^{2+}+\hspace{0.17em}2{\mathrm{e}}^{-}\to \mathrm{Cu} \\ 2\mathrm{F}= 1 \mathrm{mole} \mathrm{Cu} \\ 3 \mathrm{F}\hspace{0.17em}= \frac{3}{2} \mathrm{mole} \mathrm{Cu} \end{gathered}$$

$$\begin{gathered} {\mathrm{Au}}^{3+}+\hspace{0.17em}3{\mathrm{e}}^{-}\to \mathrm{Au} \\ 3\mathrm{F}\hspace{0.17em}= 1 \mathrm{mole} \mathrm{Au} \end{gathered}$$

$\mathrm{Ratio} \mathrm{of} \mathrm{Ag}:\hspace{0.17em}\mathrm{Cu} :\hspace{0.17em}\mathrm{Au} ::\hspace{0.17em}3: \frac{3}{2}:1 = 6:3:2$