How many liters of O2 at 298 K and .9869atm are produced in 3.25 hr in an electrolytic cell operating at a current of 0.0300 A?

Amount of electricity passed = I t = (0.0300 A) × (3.25 × 3600 s) = 351 C

1 mole of electrons carries 96500 C of electricity.
Number of moles of electrons passed = (351 C) / (96500 C/mol) = 0.003637 mol

4OH⁻(aq) → 2H₂O(l) + O₂(g) + 4e⁻
Mole ratio O₂ : electrons = 1 : 4
Number of moles of O₂ produced = (0.003637 mol) × (1/4) = 0.000909 mol

For the O₂ produced :
Pressure, P = 0.9869 atm
Volume, V = ? L
Number of moles, n = 0.000909 mol
Gas constant, R = 0.0821 L atm / (mol K)
Temperature, T = 298 K

Gas law : PV = nRT
Then, V = nRT/P

Volume of O₂ produced, V = 0.000909 × 0.0821 × 298 / 0.9869 L = 0.0225 L

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