If 3 faradays of electricity is passed through the solutions of `AgNO_3` , `CuSO_4` and `AuCl_3` , the molar ratio of the cations deposited at the cathodes will be

Options

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

Answer

D

Explanation

`AgNO_3 : Ag^(+)+underset(1 F)(e^(-))tounderset(1 mol)(Ag)`
3 Faraday of electricity deposit 3 moles of Ag.
`CuSO_4: Cu^(2+)+underset(2 F)(2e^(-))tounderset(1 mol)(Cu)`
3 Faraday of electricity deposit `3/2` moles of Cu.
3 Faraday of electricity deposit 1 moles of Au.
Thus, Ag : Cu : Au= 3 : 3/2 :1 i.e. 6 : 3 : 2.