`CuSO_(4)` solution is electrolysed for 15 minutes to deposity 0 . 4725 g of copper at the cathode. The current in amperes required is
(Faraday = 96500 C `mol^(-1)` , atomic weight of copper = 63)

Options

A. 0.804
B. 1.608
C. 1.206
D. 0.402

Answer

B

Explanation

Given,
Time (t) = 15 minutes ` = 15 xx 60 s`
Deposited weight of copper at cathode (w) = 0 . 4725 g
Atomic weight of copper = 63
` 1 F = 96500 C mol^(-1)`
As we know that, according of Faraday.s first law of electrolysis
w = zit
or ` w = ("Equivalent weight" xx i xx t)/(96500)`
`:. 0.4725 = (63 xx i xx 15 xx 60)/( 2 xx 96500) " "` (`:. Eq. wt. "Of copper" = (63)/(2)`)
`:. i = (0.4725xx 2 xx 96500)/(63 xx 15 xx 60)`
`i = 1.608 A `