The passage of current liberates `H_2` at cathode and `Cl_2` at anode. The solution is-

Options

A. Copper chloride in water
B. NaCl in water
C. `H_(2)SO_(4)`
D. Water

Explanation

To solve the question regarding the liberation of \( H_2 \) at the cathode and \( Cl_2 \) at the anode, we need to identify the solution in which these reactions occur. The key points to consider are the electrolysis of different solutions and the products formed at the electrodes.

### Step-by-Step Solution:

1. **Understanding Electrolysis**: When an electric current passes through an electrolyte, it causes the decomposition of the electrolyte into its constituent ions. The ions migrate towards the electrodes, where reduction and oxidation reactions occur.

2. **Identifying the Electrolyte**: We need to find a solution that can produce \( H_2 \) at the cathode and \( Cl_2 \) at the anode. The key candidates mentioned are:
- Copper(II) chloride in water
- Sodium chloride in water
- Sulfuric acid
- Water

3. **Analyzing Each Solution**:
- **Copper(II) Chloride (CuCl2)**: Electrolysis of this solution would produce \( Cu \) at the cathode and \( Cl_2 \) at the anode, not \( H_2 \).
- **Sodium Chloride (NaCl)**: When NaCl is dissolved in water, it dissociates into \( Na^+ \) and \( Cl^- \). During electrolysis:
- At the cathode, \( 2H_2O + 2e^- \rightarrow H_2 + 2OH^- \) occurs, liberating \( H_2 \).
- At the anode, \( 2Cl^- \rightarrow Cl_2 + 2e^- \) occurs, liberating \( Cl_2 \).
- **Sulfuric Acid (H2SO4)**: This would produce \( H_2 \) at the cathode, but \( O_2 \) at the anode, not \( Cl_2 \).
- **Water**: Pure water does not produce \( H_2 \) and \( Cl_2 \) since it lacks chloride ions.

4. **Conclusion**: The only solution that liberates \( H_2 \) at the cathode and \( Cl_2 \) at the anode is sodium chloride in water.

### Final Answer:
The solution is Sodium Chloride (NaCl) in water.