Question

$2Faraday$ of electricity is passed through a solution of $CuS{O}_4$ . The mass of copper deposited at the cathode is:

• A. $0g$
• B. $63.5g$
• C. $2.0g$
• D. $127.0g$

Answer 1

The correct option is B $63.5g$

Given, $charge\left(Q\right)=2\times F$
Given: Atomic mass of $Cu=63.5u$ and Valency of the metal $Z=2$
We have, $CuS{O}_4\to C{u}^{2+}+{S{O}_4}^{2-}$
$C{u}^{2+}$ + $2e-$ $\to$ $Cu$
$1mol$ $2mol\left(2F\right)$ $1mol=63.5g$
Alternatively.
$W=\frac{Z}{Q}=\frac{E}{F}\times 2F=2E$
= $\frac{2\times 63.5}{2}=63.5g$