On passing C ampere of current for time t sec through 1 litre of 2(M) `CuSO_(4)` solution (atomic weight of Cu=63.5), the amount of m of Cu (in g) deposited on cathode will be:-

Options

A. `m= (Ct)/(63.5 xx 96500)`
B. `m = (Ct)/(31.25 xx 96500)`
C. `m = (C xx 96500)/(31.25 xx n)`
D. `m = (31.75 xx C xx t)/96500`

Answer

D

Explanation

According to Faraday’s law of electrolysis
`m prop Ct` or m = ZCt
where C = current, t = time
`Z = ("Equivalent weight of metal")/96500`
Eq. wt of Cu `=63.5/2 (therefore Cu^(2+) to Cu)`
`Z = 63.5/(2 xx 96500)`
`therefore m = (63.5 xx C xx t)/(2 xx 96500) = (31.75 xx C xx t)/96500`