Post-UTME Chemistry 2016 Objective — Question 3

Question 3 of 20 from the Post-UTME Screening (Post-UTME) Chemistry 2016 Objective paper, with the correct answer and a full explanation.

During the electrolysis of acidified water, how many Faraday(s) will liberate 5.6dm3 of oxygen gas, measured at s.t.p? [Molar volume of a gas at s.t.p = 22.4 dm3]

A. 0.50 F
B. 1.00 FCorrect
C. 1.50 F
D. 2.00 F

Explanation

Moles O2 = 5.6/22.4 = 0.25mol. Since 4 electrons (Faradays) are needed per mole of O2 (2H2O -> O2 + 4H+ + 4e-), Faradays = 0.25 x 4 = 1.00F.

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