A `5A` current in passed through a solution of zinc sulphate for `40 min`. These amount of zinc deposited at the cathode is

Options

A. `40.65 g`
B. `0.4065 g`
C. `4.065 g`
D. `65.04 g`

Answer

C

Explanation

The electrode reaction is
`Zn^(2+)+2e^(-) rarr Zn(s)`
Thus, equivalent mass of `E = ("Atomic mass")/("Change in oxidation number")`
`= 65.39 u//2`
According to Faradays laws of electrolysis
`(m)/(E) = (Q)/(F) = (It)/(F)`
`m = (ItE)/(F)`
`((5 A)(30 min) ((60 sec)/(min))((65.39 g eq^(-1))/(2)))/((96,500 C eq^(-1)))`
`= 4,065 g`
Alternatively, according to the electrode reaction
`2F (or 2 xx 96500 C)` deposit `65.39 g` of `Zn`,
Thus, `1C` will deposite `(65.39)/(2 xx 96500)g` of `Zn` and
`12,000 C` charge will deposite `(5.39 xx 12, 0000)/(2 xx 96,500) = 4.065 g Zn`
`(Q = It = 5 A) (40 min) ((60 sec)/(min)) = 12,000 C)`