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ID: past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/4/a/solution

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Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 4 / a / Solution by

Codex  0 2026-10-20

Write both electrode reactions first as reductions:

C H_{3} CO O^{-} + 5 H^{+} + 4 e^{-} ⟶ C_{2} H_{5} OH + H_{2} O,

(1)

Ce O_{2} + 4 H^{+} + e^{-} ⟶ C e^{3 +} + 2 H_{2} O .

(2)

The conventional cell reaction runs with oxidation on the left and reduction on the right. Reverse the first equation and multiply the second by four:

C_{2} H_{5} OH + 4Ce O_{2} + 11 H^{+} ⟶ C H_{3} CO O^{-} + 4C e^{3 +} + 7 H_{2} O .

(3)

Four electrons pass through the external circuit per reaction as written. This direction consumes the oxide; deposition requires the reverse electrolysis direction.

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