What is the amount of electricity required to deposit one mole of aluminium from a solution of `AlCl_(3)` ?

Explanation

`CaCl_(2)hArr Ca^(2+)+2Cl^(-)`
`Ca^(2+)+2e^(-) to Ca`
Charge on 2 mol of electrons
`=(2mol)xx(96500C" "mol^(-1))=193,000C`
Thus to deposit 1 mol of Ca, electricity required
=193000C