Volume of gases evolved when dill. `H_(2)SO_(4)` is electrolysed using 2F at STP

Options

A. 22.4L
B. 11.2L
C. 33.6L
D. 44.8L

Answer

To solve the problem of calculating the volume of gases evolved when dilute \( H_2SO_4 \) is electrolyzed using 2F at STP, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Electrolysis of \( H_2SO_4 \)**:
- When dilute \( H_2SO_4 \) is electrolyzed, it dissociates into \( H^+ \) ions and \( SO_4^{2-} \) ions. Water also contributes \( H^+ \) and \( OH^- \) ions during the electrolysis.

2. **Identifying Reactions at Electrodes**:
- **At the Cathode**: \( H^+ \) ions are reduced to form hydrogen gas (\( H_2 \)):
\[ 2H^+ + 2e^- \rightarrow H_2(g) \]
- **At the Anode**: Water is oxidized to form oxygen gas (\( O_2 \)):
\[ 4OH^- \rightarrow O_2(g) + 4H^+ + 4e^- \]

3. **Moles of Gases Produced**:
- From the cathode reaction, 1 mole of \( H_2 \) is produced for every 2 moles of electrons.
- From the anode reaction, 1 mole of \( O_2 \) is produced for every 4 moles of electrons.
- Since 2 Faradays (2F) of charge are used, we can calculate the total moles of gases produced.

4. **Calculating Moles of Gases**:
- 2 Faradays correspond to 2 moles of electrons.
- For the cathode reaction:
- 2 moles of electrons produce 1 mole of \( H_2 \).
- For the anode reaction:
- 2 moles of electrons can produce \( \frac{2}{4} = 0.5 \) moles of \( O_2 \).
- Total moles of gases produced:
\[ 1 \text{ mole of } H_2 + 0.5 \text{ moles of } O_2 = 1.5 \text{ moles of gas} \]

5. **Volume of Gases at STP**:
- At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 liters.
- Therefore, the total volume of gases produced:
\[ \text{Volume} = 1.5 \text{ moles} \times 22.4 \text{ L/mole} = 33.6 \text{ liters} \]

### Final Answer:
The volume of gases evolved when dilute \( H_2SO_4 \) is electrolyzed using 2F at STP is **33.6 liters**.

---