Two Faraday of electricity is passed through solution of `CuSO_(4)`. The mass of copper deposited at the cathode is : (Atomic mass of Cu = 63.5 amu)

Options

A. 0g
B. (63.5g)
C. 2g
D. 127g

Explanation

To find the mass of copper deposited at the cathode when two Faraday of electricity is passed through a solution of CuSO₄, we can follow these steps:

### Step 1: Understand the Reaction
When electricity is passed through a CuSO₄ solution, copper ions (Cu²⁺) are reduced to copper metal (Cu) at the cathode. The half-reaction for this process is:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]

### Step 2: Determine the Number of Moles of Electrons
According to Faraday's law of electrolysis, 1 Faraday corresponds to the charge of 1 mole of electrons (approximately 96500 coulombs). Since we are passing 2 Faraday of electricity, the number of moles of electrons (n) can be calculated as:
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