If three faradays of electricity is passed through the solutions of AgNO₃ , CuSO₄ and AlCl₃ the mole ratio of cations deposited at the cathodes will be

${\mathrm{AgNO}}_3\to {\mathrm{Ag}}^{+}+{\mathrm{NO}}_3^{-};{\mathrm{Ag}}^{+}+{\mathrm{e}}^{-}\to \mathrm{Ag}$
1 F of electricity deposits 1 mole Ag
$\therefore$3 F gives 3 moles
${\mathrm{CuSO}}_4\to {\mathrm{Cu}}^{+2}+{\mathrm{SO}}_4^{-2};{\mathrm{Cu}}^{+2}+2{\mathrm{e}}^{-}\to \mathrm{Cu}$
2 F of electricity deposits 1 mole of Cu
$\therefore$3 F gives 1.5 moles
${\mathrm{AuCl}}_3\to {\mathrm{Au}}^{+3}+3{\mathrm{Cl}}^{-};{\mathrm{Au}}^{+3}+3{\mathrm{e}}^{-}\to \mathrm{Au}$
3 F electricity deposits 1 mole of Au
$\therefore$ ratio of No. of moles of Ag, Cu & Au deposited = 3 : 1.5 : 1 (or) 6 : 3 : 2