Question

A current of $0.25A$ is passed through $CuS{O}_4$ solution placed in voltmeter for $45minutes$. The amount of $Cu$ deposited on cathode is (At.weight of $Cu=63.6$)

• A. $0.20g$
• B. $0.22g$
• C. $0.25g$
• D. $0.30g$

Answer 1

The correct option is A $0.20g$

Given :-

Time =45 min =45×60sec = 2700sec

Current=0.25A

$C{u}^{2+}+2e^{-}\to Cu$

At.weight of Cu=63.6

$\text{Electricity passed =45×60×0.25=675C electricity deposit copper = x × 2 × 96500 C of electricity will deposit copper.}$

$M=Z\times I\times T$

$x=0.20g$