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Zadanie 1e

4 pktobliczenioweśrednie (szac.)Paper 2, maj 2026, TZ1

Wzorowane na: Chemistry HL Paper 2, May 2026, TZ1, pytanie 1(e). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.

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Informacja do zadań 1a-1f.

Nitric acid, HNO3, is manufactured industrially from ammonia in a three-step process known as the Ostwald process.

Polecenie

Many rechargeable batteries use an alkaline electrolyte. In the nickel-cadmium cell, cadmium, Cd, is the anode.

Cd(OH)2(s) + 2 e− → Cd(s) + 2 OH−(aq) E⊖ = −0.81 V

NiO(OH)(s) + H2O(l) + e− → Ni(OH)2(s) + OH−(aq) E⊖ = +0.49 V

(i) Deduce the overall reaction for the cell.

(ii) Calculate the cell potential, Ecell⊖.

(iii) Calculate the standard change in Gibbs energy, ΔG⊖, in kJ mol−1, of the cell reaction. Use sections 1 and 2 of the data booklet.

If you did not obtain an answer for (ii) use 1.10 V, although this is not the correct answer.

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Odpowiedź z klucza

(i) Cd(s) + 2 NiO(OH)(s) + 2 H2O(l) → Cd(OH)2(s) + 2 Ni(OH)2(s) ✔

«reverse the Cd half-equation since Cd is the anode and is oxidized; double the Ni half-equation to balance electrons; add and cancel 2e- and 2OH-»

(ii) «Ecell⊖ = 0.49 − (−0.81) =» 1.30 «V» ✔

(iii) «ΔG⊖ = −nFEcell⊖» n = 2 AND ΔG⊖ = −2 × 96 500 × 1.30 = −250 900 «J mol−1» ✔

−251 «kJ mol−1» ✔

«ECF: if 1.10 V is used, as the question allows, ΔG⊖ = −2 × 96 500 × 1.10 = −212 300 J, i.e. −212 kJ mol−1»

Schemat punktowania

(i) [1] correct overall equation with correct formulas and balancing; electrons and any species common to both sides must be cancelled; ignore missing or incorrect state symbols. (ii) [1] 1.30 V; the anode potential is subtracted from the cathode potential. (iii) [2] M1: n = 2 used (two electrons are transferred in the cell reaction as written) with F = 96 500 C mol-1, giving ΔG in J mol-1; M2: correct value in kJ mol-1 with the negative sign. Award [2] for the correct final answer. For n = 1 award [1 max] for −125 kJ mol-1. Award [1 max] for (+)251 kJ mol-1 (sign error). Award [2] for −212 kJ mol-1 obtained from 1.10 V. ECF from (ii). Do not penalize units or significant figures.

Komentarz

The number of electrons transferred (n = 2) comes from the balanced overall equation, not from the Ni half-equation alone.

Umiejętności w zadaniu