During electrolysis of acidified water, \(2.8 L\) of oxygen gas is released (at STP). The charge passed through the anode is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 48250 C
By Faraday’s first law, the amount of substance liberated at an electrode is proportional to the charge passed; one mole of electrons carries 1 F = 96500 C. At the anode, water is oxidised to oxygen:
\( \ce{2H2O (l) -> O2 (g) + 4H+ (aq) + 4e-} \)
So 4 mol of electrons (4 F) are needed for 1 mol of \( \ce{O2} \). At STP one mole of a gas occupies 22.4 L:
\( \displaystyle n(\ce{O2}) = \frac{2.8\ \text{L}}{22.4\ \text{L mol}^{-1}} = 0.125\ \text{mol} \)
\( Q = 0.125 \times 4 \times 96500 = 0.5 \times 96500 = \mathbf{48250\ C} \)
NCERT (2026-27 edition):
Electrolytic Cells and Electrolysis – …
