If three 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. a)` 1:1:1`
B. `b)1:2:3`
C. c)`3:2:1`
D. d)`6:3:2`

Answer

D

Explanation

The increasing order of depostion of cations at the cathode is
`Cu ^(2+) lt Ag ^(+) lt Au ^(3+)`
`E prop Z`
`Ag ^(+) + e ^(- ) to Ag`
`Cu ^(2+) + 2e ^(-) to Cu`
`Au ^(3+) + 3e ^(-) to Au`
3 Faradays liberate 1 mole of Au, 3 moles of Ag and 3/2 moles of Cu. Thus molar ratio of Ag: Cu : Au is `3: 3//2:1 or 6:3:2.`