Passage of three faraday of charge through aqueous solution of `AgNO_(3), CuSO_(4), AlCl_(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. `3:2:1:0`

Explanation

(B) `Zn+ Fe^(2+) rarrFe + Zn^(2+)`
`E_(cell)=E_(cell)^0+0.059/2log(([Fe^(2+)])/([Zn^(2+)]))` ,
0.2905 = `E_(cell)=E_(cell)^0+0.059/2log(0.001/0.01)`
`E_(cell)` = 0.2905+0.0295=0.32V,
Now `E_(cell)=0.059/2log_10K_c` , 0.32 =`0.059/2log_10K_c`
`:.` `K_c`= `10^(0.32//0.0295)`