If three faradyas 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

`Ag^(+)+e^(-)rarrAg`
3F of electricity will produce 3 moles of Ag.
`Cu^(2+)+2e^(-)rarrCu`
3F of electricity will produce 3/2 moles of Cu
`Au^(3+)+3e^(-)rarrAu`
3F of electricity will produce 1 mole of Au `therefore` Molar ratio of
`Ag:Cu:Au`
or `6:3:2`