Zadanie 1f

(i) Anode «oxidation»: H₂(g) + 2 OH⁻(aq) → 2 H₂O(l) + 2 e⁻ ✔

Cathode «reduction»: O₂(g) + 2 H₂O(l) + 4 e⁻ → 4 OH⁻(aq) ✔

(ii) electrons occupy discrete/quantized energy levels «each line is a transition between two fixed levels, not a continuous range of energies» ✔

the energy levels converge «get closer together» at higher energy ✔ «accept: the levels are unequally spaced, so the lines crowd together towards higher frequency»

(iii) «486.1 nm = 4.861 × 10⁻⁷ m»; «$f = c/\lambda = \dfrac{3.00 \times 10^{8}}{4.861 \times 10^{-7}} =$» 6.17 × 10¹⁴ «s⁻¹» ✔

«E = hf = 6.63 × 10⁻³⁴ × 6.17 × 10¹⁴ =» 4.09 × 10⁻¹⁹ «J» ✔

(i) [2] M1: correct anode half-equation; M2: correct cathode half-equation; both balanced for atoms and charge and written for an alkaline electrolyte; accept any correct multiples. Allow [1 max] for the acidic forms H2 -> 2H+ + 2e- AND O2 + 4H+ + 4e- -> 2H2O. Penalize the use of equilibrium arrows once only. (ii) [2] M1: electrons in discrete (separate) energy levels; M2: higher energy levels converge; accept other correct statements such as unequally spaced energy levels for M2. (iii) [2] M1: wavelength converted to metres and the frequency found, f = c/λ = 6.17 × 10^14 s-1; M2: E = hf = 4.09 × 10^-19 J. Award [2] for the correct final answer; accept E = hc/λ in one step. ECF from M1.