What will be PH of aqueous solution of electrolyte in electrolytic cell during electrolysis of `CuSO_(4) (Aq)` between graphite electrodes ?

Options

A. pH = 14.0
B. pHgt 7.0
C. pHlt 7.0
D. pH = 7.0

Explanation

To determine the pH of the aqueous solution of copper sulfate (CuSO₄) during electrolysis between graphite electrodes, we can follow these steps:

### Step 1: Identify the Electrolytic Cell Components
In the electrolytic cell, we have:
- Electrolyte: Aqueous solution of CuSO₄
- Electrodes: Graphite electrodes (anode and cathode)

### Step 2: Understand the Reactions at the Electrodes
During electrolysis:
- At the **cathode** (negative electrode), reduction occurs. Copper ions (Cu²⁺) gain electrons to form solid copper (Cu):
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu (s)} \]
- At the **anode** (positive electrode), oxidation occurs. Water is oxidized to produce oxygen gas (O₂), protons (H⁺), and electrons:
\[ 2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^- \]

### Step 3: Analyze the Effect of Reactions on pH
- The oxidation of water at the anode produces protons (H⁺). The increase in H⁺ concentration will lead to a decrease in pH.
- The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
- As the concentration of H⁺ increases due to the oxidation of water, the pH will drop below 7, indicating an acidic solution.

### Step 4: Conclusion
Since the production of H⁺ ions at the anode will result in an acidic environment, the pH of the solution during the electrolysis of CuSO₄ will be less than 7.

### Final Answer
The pH of the aqueous solution of CuSO₄ during electrolysis will be **less than 7**.

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