`1L` of `1M CuSO_(4)` solution is electrolyzed using `Pt` cathode and `Cu` anode. After passing `2F` of electricity, the `[Cu^(2+)]` will be

Options

A. 0
B. `M`
C. `M//2`
D. `M//4`

Explanation

To solve the problem, we need to analyze the electrolysis of a 1M CuSO₄ solution using a platinum cathode and a copper anode. We will determine the concentration of Cu²⁺ ions after passing 2 Faraday of electricity.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- We have 1 liter of 1M CuSO₄ solution.
- This means we initially have 1 mole of Cu²⁺ ions in the solution.

2. **Understand the Electrolysis Process:**
- During electrolysis, Cu²⁺ ions are reduced at the cathode to form copper metal (Cu) according to the half-reaction:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu (s)} \]
- At the anode, copper metal (Cu) is oxidized to Cu²⁺ ions, but since we are using a copper anode, the copper will dissolve into the solution, maintaining the Cu²⁺ concentration.

3. **Calculate the Amount of Copper Deposited:**
- The amount of electricity passed is 2 Faraday (F).
- 1 Faraday corresponds to the charge required to deposit 1 mole of Cu from Cu²⁺ ions.
- Therefore, 2 Faraday will deposit 1 mole of Cu:
\[ 2F \rightarrow 1 \text{ mole of Cu} \]

4. **Determine the Change in Cu²⁺ Concentration:**
- Initially, we have 1 mole of Cu²⁺ ions.
- Since 1 mole of Cu²⁺ ions is reduced to form 1 mole of Cu, and the copper anode will dissolve an equivalent amount of copper back into the solution, the net change in Cu²⁺ concentration is as follows:
- 1 mole of Cu²⁺ is consumed (reduced to Cu).
- 1 mole of Cu is produced from the anode (oxidized back to Cu²⁺).
- Therefore, after passing 2 Faraday, the concentration of Cu²⁺ ions will be:
\[ \text{Final moles of Cu}^{2+} = 1 \text{ mole (initial)} - 1 \text{ mole (deposited)} + 1 \text{ mole (dissolved)} = 1 \text{ mole} \]

5. **Calculate the Final Concentration of Cu²⁺:**
- Since the volume of the solution remains 1 liter, the final molarity of Cu²⁺ ions is:
\[ \text{Molarity} = \frac{\text{Number of moles}}{\text{Volume in liters}} = \frac{1 \text{ mole}}{1 \text{ L}} = 1M \]

### Final Answer:
The concentration of Cu²⁺ ions after passing 2 Faraday of electricity will be **1M**.