1 mol of electrons pass through each of the solution of `AgNO_(3), CuSO_(4) and AlCl_(3) ` when Ag, Cu and Al are deposited. Molar ratio of their deposition will be:

Options

A. 1 : 1 : 1
B. 6 : 3 : 2
C. 6 : 3 : 1
D. 1 : 3 : 6

Answer

B

Explanation

1mole of electrons `equiv` 1F electricity
`AgNO_(3) : Ag^(+) + e^(-) rightarrow Ag`, `Cu^(2+) + 2e^(-) rightarrow Cu, AlCl_(3) : Al^(3+) + 3e^(-) rightarrow Al`
`Rightarrow 1F` eletricity will deposit : 1mole `Ag, (1)/(2)` mole `Ag,(1)/(2)` mole Cu and `(1)/(3)` mole Al
`Rightarrow` Molar ratio will be : 6:3:2