Question

Two Faraday of electricity is passed through a solution of CuSO₄. The mass of copper deposited at the cathode is ______. (at. mass of Cu = 63.5 amu)

Options

• 0 g

• 63.5 g

• 2 g

• 127 g

Answer 1

Two Faraday of electricity is passed through a solution of CuSO₄. The mass of copper deposited at the cathode is 63.5 g.

Explanation:

Given: For copper deposition:

\[\ce{Cu^{2+} + 2e- -> Cu_{(s)}}\]

This means 2 Faradays of electricity are required to deposit 1 mole of copper (63.5 g).

By using Faraday’s first law:

W = `(E * Q)/F`    ...(i)

Where:

W = mass deposited

E = `M/z = 63.5/2` (equivalent weight of Cu)

Q = 2 F (charge passed)

F = 1 Faraday

By substituting values in equation (i), we get

W = `(63.5/2 xx 2F)/F`

= 63.5 g

∴ The mass of copper deposited is 63.5 g.