Three faradays of electricity are passed through molten `Al_2O_3` aqueous solution of `CuSO_4` and molten `NaCl` taken in deffernt electrolytic cells. The amout of `Al,Cu` and Na deposited at the cathodes will be in the ration of .

Options

A. 1 mol `:` 2 mol `:` 3 mol
B. 3 mol `:` 2 mol `:` 1 mol
C. 1 mol `:` 1.5 mol `:` 3 mol
D. 1.5 mol `:` 2 mol `:` 3 mol

Answer

C

Explanation

`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` .