Question

$CuS{O}_4$ solution is electrolysed for $15$ minutes to deposit $0.4725g$ of copper at the cathode. The current in amperes requited is
(Faraday = $96.500Cmo{l}^{-1}$. atomic weight of copper = $63$)

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

Answer 1

Answer: (B)

Given,

Time $(t)=15$ minutes $=15\times 60s$

Deposited weight of copper at cathode $(w)=0.4725g$

Atomic weight of copper $=63$

$1F=96500Cmo{l}^{-1}$

As we know that, according to Faraday's first law of electrolysis,

$w=Zit$

or $w=\frac{\text{ Equivalent weight }\times i\times t}{96500}$

$\therefore 0.4725=\frac{63\times i\times 15\times 60}{2\times 96500}$

$(\because$ Eq. wt. of copper $=\frac{63}{2})$

$\therefore i=\frac{0.4725\times 2\times 96500}{63\times 15\times 60}$

$\therefore i=1.608A$