Which of the following reaction is preferred at anodes during the electrolysis of `H_(2)SO_(4)` at higher concentration ?

Options

A. `2 H_(2) O(l) rarr O_(2)(g)+4H^(+)(aq)+4e^(-)`
B. `2SO_(4)^(2-)(aq)rarrS_(2)O_(8)^(2-)(aq)+2e^(-)`
C. Both a and b
D. None of the above

Explanation

To determine which reaction is preferred at the anode during the electrolysis of concentrated \( H_2SO_4 \), we need to analyze the possible reactions and their conditions.

### Step-by-Step Solution:

1. **Understanding Electrolysis of \( H_2SO_4 \)**:
- When \( H_2SO_4 \) is electrolyzed, it dissociates into ions: \( H^+ \) and \( SO_4^{2-} \).
- Water is also present in the solution, which can participate in the electrolysis.

2. **Identifying Possible Oxidation Reactions**:
- At the anode, oxidation occurs. The two main candidates for oxidation are:
- Oxidation of water:
\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]
- Oxidation of sulfate ions (\( SO_4^{2-} \)):
\[ 2SO_4^{2-} \rightarrow S_2O_8^{2-} + 2e^- \]

3. **Comparing Oxidation Potentials**:
- The oxidation potential of water is relatively high, making it easier to oxidize water to oxygen gas.
- The oxidation potential of sulfate ions is lower, meaning it requires more energy to oxidize them compared to water.

4. **Effect of Concentration**:
- In concentrated \( H_2SO_4 \), the concentration of water is reduced. This affects the availability of water for oxidation.
- As the concentration of \( H_2SO_4 \) increases, the sulfate ions become more favorable for oxidation because there is less water available.

5. **Conclusion**:
- In concentrated \( H_2SO_4 \), the oxidation of sulfate ions to peroxydisulfate (\( S_2O_8^{2-} \)) becomes the preferred reaction at the anode due to the decreased availability of water.
- Therefore, the preferred reaction at the anode during the electrolysis of concentrated \( H_2SO_4 \) is:
\[ 2SO_4^{2-} \rightarrow S_2O_8^{2-} + 2e^- \]