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Question: 3 Faradays of electricity is passed through solution of CuSO₄. The number of moles of copper deposit...

3 Faradays of electricity is passed through solution of CuSO₄. The number of moles of copper deposited on the cathode will be.

A

1.5

B

2

C

2.5

D

1.25

Answer

1.5

Explanation

Solution

The half-reaction is:

Cu2++2e−⟶Cu\mathrm{Cu^{2+} + 2e^- \longrightarrow Cu}

Here, 2 moles of electrons (2 Faradays) deposit 1 mole of copper. Therefore, with 3 Faradays:

Moles of Cu=32=1.5 moles\text{Moles of Cu} = \frac{3}{2} = 1.5\, \text{moles}