Two platinum electrodes were immersed in a solution of `CuSO_(4)` and electric current was passed through the solution. After some time, it was found that colour of `CuSO_(4)` disappeared with evolution of gas at the electrode. The colourless solution contains.

Explanation

To solve the problem, we need to analyze the electrolysis of a copper sulfate (CuSO₄) solution when electric current is passed through it. Here’s a step-by-step breakdown of the process:

### Step 1: Understanding Electrolysis
When electric current is passed through an electrolyte solution, it undergoes electrolysis. In this case, the electrolyte is copper sulfate (CuSO₄) dissolved in water.

**Hint:** Electrolysis involves the breakdown of a compound into its ions when an electric current is applied.

### Step 2: Ionization of Copper Sulfate
Copper sulfate dissociates in solution into copper ions (Cu²⁺) and sulfate ions (SO₄²⁻):
\[ \text{CuSO}_4 \rightarrow \text{Cu}^{2+} + \text{SO}_4^{2-} \]

**Hint:** Remember that ionic compounds dissociate into their constituent ions in solution.

### Step 3: Reactions at the Electrodes
- **At the Cathode (Negative Electrode):** Reduction occurs. The positively charged 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. Here, we need to consider the competition between water and sulfate ions. The oxidation potential of water is higher than that of sulfate ions, so water will be oxidized:
\[ 2\text{H}_2\text{O} \rightarrow 4\text{H}^+ + \text{O}_2 + 4e^- \]

**Hint:** Identify which species is oxidized and which is reduced at each electrode based on their oxidation potentials.

### Step 4: Resulting Solution Composition
After the electrolysis, the copper ions are reduced to solid copper and precipitate out of the solution, causing the blue color of the copper sulfate to disappear. The water is oxidized to produce hydrogen ions (H⁺) and oxygen gas (O₂). The remaining ions in the solution are sulfate ions (SO₄²⁻) and hydrogen ions (H⁺), which combine to form sulfuric acid (H₂SO₄):
\[ \text{H}^+ + \text{SO}_4^{2-} \rightarrow \text{H}_2\text{SO}_4 \]

**Hint:** Consider what happens to the ions left in the solution after the electrolysis process.

### Conclusion
The colorless solution after electrolysis contains sulfuric acid (H₂SO₄).

**Final Answer:** The colorless solution contains H₂SO₄ (sulfuric acid).