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 determining the volume of gases evolved when dilute \( H_2SO_4 \) is electrolyzed using 2 Faraday at STP, we can follow these steps:

### Step 1: Understand the Electrolysis Reaction
When dilute \( H_2SO_4 \) is electrolyzed, it dissociates into ions. The electrolysis of \( H_2SO_4 \) can be represented by the following half-reactions:
- At the cathode (reduction):
\[ 2H^+ + 2e^- \rightarrow H_2(g) \]
- At the anode (oxidation):
\[ 2H_2O \rightarrow O_2(g) + 4H^+ + 4e^- \]

### Step 2: Calculate Moles of Gases Produced
From the half-reactions, we see that:
- 2 moles of \( H^+ \) produce 1 mole of \( H_2 \) gas.
- 2 moles of \( H_2O \) produce 1 mole of \( O_2 \) gas.

Thus, for every 2 moles of electrons (2 Faraday), we produce:
- 1 mole of \( H_2 \)
- 1 mole of \( O_2 \)

This means a total of 2 moles of gas are produced (1 mole of \( H_2 \) and 1 mole of \( O_2 \)).

### Step 3: Use the Ideal Gas Law to Find Volume at STP
At Standard Temperature and Pressure (STP), 1 mole of an ideal gas occupies 22.4 liters. Therefore, the total volume of gas produced can be calculated as follows:
\[ \text{Total Volume} = \text{Total Moles of Gas} \times \text{Volume per Mole at STP} \]
\[ \text{Total Volume} = 2 \, \text{moles} \times 22.4 \, \text{liters/mole} = 44.8 \, \text{liters} \]

### Conclusion
The volume of gases evolved when dilute \( H_2SO_4 \) is electrolyzed using 2 Faraday at STP is **44.8 liters**.

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