If 1 mol electron is passed through each of `AgNO_(3)` , `CuSO_(4)` and `AiCl^(3)` , solutions , calculate the ratio of number of moles of Ag , Cu and Al formed at cathode .

Explanation

`Ag^(+)+erarrAg" "Cu^(2+)+2e rarrCu" "Al^(3+)+3erarrAl`
`"Charge of 1 mol electron is 1 F"`
`"So, 1 F "-="1 mol Ag" -=(1)/(2)" mol Cu"-=(1)/(3)" mol Al"`
`Ag:Cu:Al=1:(1)/(2):(1)/(3)=6:3:2`.