In which of the following aqueous solutions `H_(2)` and `O_(2)` are not liberated at cathode and anode respectively on electrolysis using inert electrodes?

Options

A. `H_(2)SO_(4)` solutions
B. NaOH solution
C. `Na_(2)SO_(4)` solution
D. `AgNO_(3)` solution

Explanation

To solve the question, we need to analyze the electrolysis of different aqueous solutions and determine in which case hydrogen (H₂) is not liberated at the cathode and oxygen (O₂) is not liberated at the anode when using inert electrodes.

### Step-by-Step Solution:

1. **Understanding Electrolysis**:
- During electrolysis, reduction occurs at the cathode (where electrons are gained) and oxidation occurs at the anode (where electrons are lost).
- In aqueous solutions, the products formed depend on the ions present and their respective reduction and oxidation potentials.

2. **Identifying the Products**:
- At the cathode, we typically expect hydrogen gas to be produced when water is reduced. However, if a more easily reducible ion is present, it will be reduced instead.
- At the anode, we usually expect oxygen gas to be produced from the oxidation of water. However, if a more easily oxidizable ion is present, it will be oxidized instead.

3. **Analyzing Each Solution**:
- **Sulfuric Acid (H₂SO₄)**:
- At the cathode: H⁺ ions are reduced to produce H₂ gas.
- At the anode: SO₄²⁻ ions oxidize to produce O₂ gas.
- **Sodium Sulfate (Na₂SO₄)**:
- At the cathode: H⁺ ions are reduced to produce H₂ gas.
- At the anode: SO₄²⁻ ions oxidize to produce O₂ gas.
- **Silver Nitrate (AgNO₃)**:
- At the cathode: Ag⁺ ions are reduced to produce Ag (solid), not H₂ gas.
- At the anode: NO₃⁻ ions can oxidize, producing O₂ gas.
- **Copper(II) Sulfate (CuSO₄)**:
- At the cathode: Cu²⁺ ions are reduced to produce Cu (solid), not H₂ gas.
- At the anode: SO₄²⁻ ions oxidize to produce O₂ gas.

4. **Conclusion**:
- From the analysis, we see that in the case of **Silver Nitrate (AgNO₃)**, hydrogen gas is not liberated at the cathode (instead, silver is deposited), and oxygen gas is still liberated at the anode. Therefore, the correct answer is **Silver Nitrate (AgNO₃)**.

### Final Answer:
**Silver Nitrate (AgNO₃)** is the solution in which H₂ is not liberated at the cathode and O₂ is liberated at the anode during electrolysis using inert electrodes.