Electrolysis of dil. `H_(2)SO_(4)` liberates gases at anode and cathode respectively

Options

A. `O_(2), SO_(2)`
B. `SO_(2), O_(2)`
C. `O_(2), H_(2)`
D. `H_(2), SO_(2)`

Answer

To solve the question regarding the electrolysis of dilute \( H_2SO_4 \) and the gases liberated at the anode and cathode, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Electrolysis**:
- Electrolysis is a process that uses electrical energy to drive a non-spontaneous chemical reaction. In this case, we are considering the electrolysis of dilute sulfuric acid (\( H_2SO_4 \)).

2. **Components of Dilute \( H_2SO_4 \)**:
- Dilute \( H_2SO_4 \) consists mainly of water and sulfuric acid. The presence of water is significant because it will participate in the electrolysis process.

3. **Identifying the Electrodes**:
- During electrolysis, there are two electrodes: the anode (positive electrode) and the cathode (negative electrode).

4. **Reactions at the Anode**:
- At the anode, oxidation occurs. In dilute \( H_2SO_4 \), water is present in excess. The oxidation of water produces oxygen gas:
\[ 2 H_2O \rightarrow O_2 + 4 H^+ + 4 e^- \]
- Therefore, oxygen gas is liberated at the anode.

5. **Reactions at the Cathode**:
- At the cathode, reduction occurs. The \( H^+ \) ions from the dissociation of \( H_2SO_4 \) gain electrons to form hydrogen gas:
\[ 2 H^+ + 2 e^- \rightarrow H_2 \]
- Thus, hydrogen gas is liberated at the cathode.

6. **Conclusion**:
- In summary, during the electrolysis of dilute \( H_2SO_4 \):
- **At the Anode**: Oxygen gas is produced.
- **At the Cathode**: Hydrogen gas is produced.

### Final Answer:
The gases liberated at the anode and cathode respectively are **Oxygen** and **Hydrogen**.