Question

$CuSO_4$ solution is electrolysed for $15$ minutes to deposit $0.4725\, g$ of copper at the cathode. The current in amperes requited is
(Faraday = $96.500 \,C\,mol^{-1}$. 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 $=15 \times 60 \,s$

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 to Faraday's first law of electrolysis,

$w=Z i t$

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

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

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

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

$\therefore i=1.608 \,A$