Which substance is obtained in the solution on electrolysis of aqueous `CuSO_4` solution using graphite electrodes?

Options

A. `H_(2)O`
B. `H_(2)SO_(4)`
C. `Na_(2)SO_(4)`
D. `CU(OH)_(2)`

Explanation

To determine which substance is obtained in the solution during the electrolysis of aqueous CuSO₄ using graphite electrodes, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Electrolyte**: The electrolyte in this case is aqueous copper sulfate (CuSO₄). When dissolved in water, it dissociates into Cu²⁺ ions and SO₄²⁻ ions.

2. **Set Up the Electrolysis Process**: When electrolysis is performed, two electrodes are used: a cathode (negative electrode) and an anode (positive electrode). Graphite electrodes are commonly used due to their conductivity.

3. **Reactions at the Cathode**: At the cathode, reduction occurs. The Cu²⁺ ions in the solution gain electrons and are reduced to form solid copper (Cu). The half-reaction can be written as:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu (s)} \]

4. **Reactions at the Anode**: At the anode, oxidation occurs. Water (H₂O) is oxidized to produce oxygen gas (O₂), protons (H⁺), and electrons. The half-reaction can be written as:
\[ 2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^- \]

5. **Effect on the Solution**: As the electrolysis proceeds, Cu²⁺ ions are reduced and deposited as solid copper on the cathode. The oxidation of water at the anode produces H⁺ ions, which increase the acidity of the solution. The remaining sulfate ions (SO₄²⁻) from the dissociation of CuSO₄ combine with the H⁺ ions to form sulfuric acid (H₂SO₄).

6. **Conclusion**: Therefore, the substance obtained in the solution after the electrolysis of aqueous CuSO₄ using graphite electrodes is sulfuric acid (H₂SO₄).

### Final Answer:
The substance obtained in the solution is **H₂SO₄**.