Calculate the volume in cm³ of oxygen evolved at s.t.p. when a current of 5 A is passed through acidified water for 193 s. {F = 96500 C mol⁻¹, molar volume at s.t.p. = 22.4 dm³}

Correct answer: 0.056 dm³ (i.e., 56 cm³)

Explanation:

• Charge passed: Q = It = 5 A × 193 s = 965 C.
• Moles of electrons: n(e⁻) = Q/F = 965/96500 = 0.010 mol.
• At the anode: 2H₂O → O₂ + 4H⁺ + 4e⁻, so 4 e⁻ produce 1 mol O₂ → n(O₂) = 0.010/4 = 0.0025 mol.
• Volume at s.t.p.: V = n × 22.4 dm³ mol⁻¹ = 0.0025 × 22.4 = 0.056 dm³ = 56 cm³.

Thus, the correct option is 0.056 dm³ (which corresponds to 56 cm³ as asked).

• 224.000 dm³ — Far too large.
• 0.056 dm³ — Correct.
• 0.224 dm³ — Off by a factor of 4.
• 56.000 dm³ — Wrong units/magnitude; 56 cm³ is correct, not 56 dm³.