Which pair of electrolytes could not be distinguished by the products of electrolysis using inert electrodes?

Options

A. 1 M `CuSO_(4)` solution, 1 M `CuCl_(2)` solution
B. 1 M KCl solution, 1 M KI solution
C. 1 M `AgNO_(3)` solution, 1 M `Cu(NO_(3))_(2)` solution
D. 1 M KCl solution, 1 M NaCl solution

Explanation

To determine which pair of electrolytes could not be distinguished by the products of electrolysis using inert electrodes, we will analyze the electrolysis of the given pairs of electrolytes. The key is to identify if the products formed at the anode and cathode are the same for both electrolytes.

### Step-by-Step Solution:

1. **Identify the Electrolytes**: The question provides four pairs of electrolytes. We will analyze each pair:
- Pair 1: CuSO4 and CuCl2
- Pair 2: KCl and KI
- Pair 3: AgNO3 and Cu(NO3)2
- Pair 4: KCl and NaCl

2. **Electrolysis of CuSO4**:
- **At the Cathode**: Cu²⁺ ions are reduced to Cu (s).
- **At the Anode**: Water is oxidized to O₂ gas.
- **Products**: Cu (s) at cathode, O₂ gas at anode.

3. **Electrolysis of CuCl2**:
- **At the Cathode**: Cu²⁺ ions are also reduced to Cu (s).
- **At the Anode**: Cl⁻ ions are oxidized to Cl₂ gas.
- **Products**: Cu (s) at cathode, Cl₂ gas at anode.
- **Conclusion**: Different products, can be distinguished.

4. **Electrolysis of KCl**:
- **At the Cathode**: Water is reduced to H₂ gas.
- **At the Anode**: Cl⁻ ions are oxidized to Cl₂ gas.
- **Products**: H₂ gas at cathode, Cl₂ gas at anode.

5. **Electrolysis of KI**:
- **At the Cathode**: Water is reduced to H₂ gas.
- **At the Anode**: I⁻ ions are oxidized to I₂.
- **Products**: H₂ gas at cathode, I₂ at anode.
- **Conclusion**: Different products, can be distinguished.

6. **Electrolysis of AgNO3**:
- **At the Cathode**: Ag⁺ ions are reduced to Ag (s).
- **At the Anode**: Water is oxidized to O₂ gas.
- **Products**: Ag (s) at cathode, O₂ gas at anode.

7. **Electrolysis of Cu(NO3)2**:
- **At the Cathode**: Cu²⁺ ions are reduced to Cu (s).
- **At the Anode**: Water is oxidized to O₂ gas.
- **Products**: Cu (s) at cathode, O₂ gas at anode.
- **Conclusion**: Different products, can be distinguished.

8. **Electrolysis of KCl and NaCl**:
- **At the Cathode**: Water is reduced to H₂ gas.
- **At the Anode**: Cl⁻ ions are oxidized to Cl₂ gas.
- **Products**: H₂ gas at cathode, Cl₂ gas at anode.
- **Conclusion**: Same products for both electrolytes.

### Final Answer:
The pair of electrolytes that could not be distinguished by the products of electrolysis using inert electrodes is **KCl and NaCl** (Option 4).