0.04 \ F of charge will liberate how much of oxygen gas (at STP) from…

The correct answer is Option (3) → 224 mL

At the anode, water in acidified solution is oxidised to oxygen:

    \( \ce{2H2O -> O2 + 4H+ + 4e-} \)

So 4 F of charge (4 mol of electrons) liberate 1 mol of O₂, which occupies 22.4 L (22400 mL) at STP.

• Moles of \( \ce{O2} \):

      \( \displaystyle \frac{0.04\ \text{F}}{4\ \text{F mol}^{-1}} = 0.01\ \text{mol} \)

• Volume of \( \ce{O2} \) = 0.01 mol × 22400 mL mol⁻¹ = 224 mL