Passage of three faraday of charge through aqueous solution of `AgNO_(3), CuSO_(4), Al(NO_(3))_(3)` and `NaCl` respectively will deposit themetals in the ratio (molar):

Options

A. `1 : 2 : 3 : 1`
B. `6 : 3 : 2 : 6`
C. `6 : 3 : 0 : 0`
D. `3 : 2 : 1 : 6`

Answer

C

Explanation

`3` eq. of `Ag` and `Cu`, zero equivalent of `Na` and `Al` will be deposited.
Note electrolysis of `NaCl_(aq.)` and `AlCl_(3_(aq.))` does not give `Na` and `Al` metal at cathode.
Thus molar ratio is `(3)/(1) : (3 )/(2)` for `Ag : Cu`
or `3 : (3)/(2)` or `6 : 3 : 0 : 0` for `Ag : Cu : Al : Na`