Passage of 96,500 coulomb of electricity liberates ________ L of O2 at 273°C and 2 atm during electr — Electrochemistry Chemistry Question

💡 Solution & Explanation

96500 C = 1 F. Oxidation of water: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$. 4 F produces 1 mol O2, so 1 F produces 0.25 mol. Using $PV=nRT$, $V = (0.25 \times 0.0821 \times 546) / 2 = 5.6$ L (using 273°C = 546 K). Therefore, correct answer is A.