Two aqueous solutions A and B containing solute `CuSO_(4)` and `NaBr` respectively were electrolysed using platinum electrodes. The `pH` of the resultins will show `a//an :`

Options

A. Increase in both the solutions
B. Decrease in both the solutions
C. Increases in A and decrease in B
D. Decrease in A and increase in B

Explanation

To solve the question regarding the pH changes in the electrolysis of two aqueous solutions, CuSO4 and NaBr, we can break down the process into clear steps.

### Step-by-Step Solution:

1. **Identify the Electrolysis Reactions**:
- For **CuSO4**:
- At the **cathode**: Cu²⁺ ions are reduced to copper metal (Cu).
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad (\text{Reduction}) \]
- At the **anode**: Water is oxidized to produce oxygen gas and hydrogen ions (H⁺).
\[ 2\text{H}_2\text{O} \rightarrow 4\text{H}^+ + \text{O}_2 + 4e^- \quad (\text{Oxidation}) \]
- **Result**: The concentration of H⁺ ions increases, leading to a decrease in pH.

2. **Analyze the Electrolysis of NaBr**:
- For **NaBr**:
- At the **cathode**: Water is reduced to produce hydrogen gas and hydroxide ions (OH⁻).
\[ 2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^- \quad (\text{Reduction}) \]
- At the **anode**: Bromide ions (Br⁻) are oxidized to produce bromine gas (Br₂).
\[ 2\text{Br}^- \rightarrow \text{Br}_2 + 2e^- \quad (\text{Oxidation}) \]
- **Result**: The concentration of OH⁻ ions increases, leading to an increase in pH.

3. **Determine pH Changes**:
- For **CuSO4**:
- The increase in H⁺ ions results in a **decrease in pH**.
- For **NaBr**:
- The increase in OH⁻ ions results in an **increase in pH**.

4. **Conclusion**:
- The pH of solution A (CuSO4) decreases, while the pH of solution B (NaBr) increases.

### Final Answer:
The pH of the resulting solutions will show a decrease in A (CuSO4) and an increase in B (NaBr).