Correct Answer - C
`Al^(3+) + 3e^(-) to Al`
`Cu^(2+) + 2 e^(-) to Cu`
`Na^(+) + e^(-) to Na`
Thus, 1 F will deposit `(1)/(3)` mol Al , `(1)/(2)` mol Cu and 1 mol Na , i.e., moles deposited are in the ratio `(1)/(3) : (1)/(2): 1 ` , i.e.,`2 : 3 : 6` or `1 : 1.5 : 3` .