In electrolysis of dilute `H_2SO_4` using platinum electrodes .

Options

A. `Cl_2 ` is obtained at cathode
B. ` NH_3` is produced at anode
C. `h_2` is evolved at cathode
D. `O_2` is produced

Explanation

To solve the question regarding the electrolysis of dilute H₂SO₄ using platinum electrodes, we will follow these steps:

### Step 1: Identify the Electrolyte
The electrolyte in this case is dilute sulfuric acid (H₂SO₄). When dissolved in water, it dissociates into ions:
- H₂SO₄ → 2H⁺ + SO₄²⁻

### Step 2: Identify the Electrodes
In electrolysis, we have two electrodes:
- **Cathode**: The electrode where reduction occurs (gains electrons).
- **Anode**: The electrode where oxidation occurs (loses electrons).

### Step 3: Determine the Reactions at the Electrodes
1. **At the Cathode**:
- The cathode is negatively charged, attracting positive ions (H⁺).
- The reduction reaction occurs here:
\[ 2H^+ + 2e^- \rightarrow H_2(g) \]
- This means hydrogen gas (H₂) is produced at the cathode.

2. **At the Anode**:
- The anode is positively charged, attracting negative ions (OH⁻ and SO₄²⁻).
- The oxidation reaction occurs here:
\[ 4OH^- \rightarrow O_2(g) + 2H_2O + 4e^- \]
- This means oxygen gas (O₂) is produced at the anode.

### Step 4: Conclusion
From the above reactions:
- **At the Cathode**: Hydrogen gas (H₂) is produced.
- **At the Anode**: Oxygen gas (O₂) is produced.

### Final Answer
- **Cathode Product**: H₂ gas
- **Anode Product**: O₂ gas

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