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

Explanation

`AlCl_(3)hArrAl^(3+)+3Cl^(-)`
`Al^(3+)+underset(3mol)(3e^(-)) to underset(1mol)(Al)`
3 mol of electrons are needed to depoint one mole of aluminium.
Thus amount of electricity required.
=Charge an 3 mol of electrons
`=(3mol)xx96500"C "mol^(-1)`
`=3xx96500C=2,89,500C`