 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/4/b/i/solution

Oxidize zinc at the anode and reduce nitrate to nitric oxide at the inert platinum cathode. Multiplying the oxidation half-reaction by three and the reduction half-reaction by two cancels six electrons:

3Zn (s) + 2N O_{3}^{-} (aq) + 8 H^{+} (aq) ⟶ 3Z n^{2 +} (aq) + 2NO (g) + 4 H_{2} O (l) .

The standard cell potential is the cathode reduction potential minus the anode reduction potential:

E^{\circ} = 0.957 - (- 0.762) = 1.719 V .

The written direction is spontaneous under standard conditions, since

Δ_{r} G^{\circ} = - 6 F E^{\circ} < 0

. Multiplying a half-reaction changes its extensive Gibbs energy, but does not multiply its intensive electrode potential.

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