Question

Two Faraday of electricity is passed through a solution of $CuS{O}_4$. The mass of copper deposited at the cathode is (at. mass of Cu = 63.5 u)

• A. 0 g
• B. 63.5 g
• C. 2 g
• D. 127 g

Answer 1

The correct option is B 63.5 g

Given Q = 2F
Atomic mass of Cu = 63.5u
Valcency 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 1mol = 63.5g
2F
Alternatively,
$W=ZQ=\frac{E}{F}.2F=2E=\frac{2\times 63.5}{2}=63.5g$