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 amu)

• 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 $Cu=63.5amu$

Valency of metal $=2$

$CuS{O}_4\to C{u}^{2+}+S{O}_4^{2-}$

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

1 mol 2 mol 1 mol $=63.5g$

2 mole of electron means 2 Faraday.

$\therefore$ Option B is correct