One faraday of electricity deposits one mol of Na from the molten salt but `(1)/(3)` mol of Al from an aluminium salt. Why?

Explanation

The reactions at cathode for the deposit of Na and Al are `Na^(+)+e^(-)toNa and Al^(3+)+3e^(-)toAl`
Thus, 1 faraday deposits 1 mol of Na whereas 3 faradays are required for depositing one mol of Al.