Question

What volume of \(O_{2(g)}\) is produced when \(2F\) current is passed through a very dilute aqueous solution of \(H_2SO_4\) at STP?

• A. \(36 \text{ dm}^3\)
• B. \(48.6 \text{ dm}^3\)
• C. \(22.4 \text{ dm}^3\)
• D. \(11.2 \text{ dm}^3\)

Hint:

Remember the equivalent volumes of common gases at STP for \(1F\) charge:
\(H_2 \to 11.2 \text{ L} \text{ (needs } 2e^- \text{ per molecule)}\)
\(O_2 \to 5.6 \text{ L} \text{ (needs } 4e^- \text{ per molecule)}\)
\(Cl_2 \to 11.2 \text{ L} \text{ (needs } 2e^- \text{ per molecule)}\)
Since \(1F\) produces \(5.6 \text{ L}\) of \(O_2\), \(2F\) will produce \(11.2 \text{ L}\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Electrolysis of dilute aqueous \(H_2SO_4\) involves the oxidation of water at the anode to produce oxygen gas. One Faraday (\(1F\)) of charge is the charge carried by one mole of electrons.

Step 2: Key Formula or Approach:
The half-reaction at the anode is:
\[ 2H_2O(l) \to O_2(g) + 4H^+(aq) + 4e^- \]
This shows that 4 moles of electrons (\(4F\) charge) are required to produce 1 mole of \(O_2\) gas.

Step 3: Detailed Explanation:
1. From the stoichiometry:
\(4F\) charge produces \(1 \text{ mole of } O_2\).
Therefore, \(2F\) charge will produce \((2/4) \times 1 = 0.5 \text{ mole of } O_2\).
2. Calculate the volume at STP (Standard Temperature and Pressure):
At STP, the molar volume of an ideal gas is \(22.4 \text{ dm}^3/\text{mol}\).
\[ \text{Volume} = \text{moles} \times \text{Molar Volume} \]
\[ \text{Volume} = 0.5 \text{ mol} \times 22.4 \text{ dm}^3/\text{mol} = 11.2 \text{ dm}^3 \]

Step 4: Final Answer:
The volume of \(O_2\) gas produced is \(11.2 \text{ dm}^3\).