One Faraday of electricity is passed through molten Al₂O₃, aqueous solution of CuSO₄ and molten NaCl taken in three different electrolytic cells connected in series. The mole ratio of Al, Cu and Na deposited at the respective cathode

The charge carried by 1 mole of electron is one faraday. Thus for a reaction  $M^{n+}+n{e}^{-}\to M$

nF = 1 mole of M

$\underset{\underset{1F}{3F}}{A{l}^{3+}}+3e^{-}\to \underset{1\underset{1/3mol}{mol}}{Al}$

$\underset{\underset{1F}{2F}}{C{u}^{2+}}+2e^{-}\to \underset{1\underset{1/2mol}{mol}}{Cu}$

$\underset{1F}{N{a}^{+}}\to \underset{1mol}{Na}$

The mole ratio of Al, Cu and Na deposited the respective cathode is $\frac{1}{3}:\frac{1}{2}:1 or 2:3:6$