A-LevelChemistryChemical energetics9701/42 · Feb/Mar 2021 · Q213 Marks

Iron(II) compounds are generally only stable in neutral, non-oxidising conditions. It is difficult to determine the lattice energy of FeO experimentally.

(a)(i)Use data from the Data Booklet and this Born-Haber cycle to calculate the lattice energy, ΔΗlatt, of FeO(s) in kJ mol-1. [Figure 2.1] ΔΗlatt FeO(s) = .......................... kJ mol-1[2]
(a)(ii)Most naturally occurring samples of iron(II) oxide are found as the mineral wüstite. Wüstite has formula Fe20Ox. It contains both Fe2+ and Fe3+ ions. 90% of the iron is present as Fe2+ and 10% is present as Fe3+. Deduce the value of x. x = ..........................[1]
(a)(iii)State and explain how the lattice energy of FeO(s) compares to the lattice energy of CaO(s).[2]
(b)(i)Heating of FeO results in the formation of Fe3O4, as shown. reaction 1 4FeO → Fe + Fe3O4 Each formula unit of Fe3O4 contains one Fe2+ and two Fe3+ ions. Show how reaction 1 can be described as a disproportionation reaction.[1]
(b)(ii)Fe3O4(I) can be electrolysed using inert electrodes to form Fe. Write the half-equation for the reaction that occurs at the anode during the electrolysis of Fe3O4(1).[1]
(b)(iii)Calculate the maximum mass of iron metal formed when Fe3O4(I) is electrolysed for six hours using a current of 50A. Assume the one Fe2+ and two Fe3+ ions are discharged at the same rate. mass of iron = .......................... g[3]
(c)(i)LiFePO4 can be used in lithium-ion rechargeable batteries. When the cell is charging, lithium reacts with a graphite electrode to form LiC6. When the cell is discharging, the half-equations for the two processes that occur are as follows. anode half-equation LiC6 → 6C + Li⁺ + e¯ cathode half-equation Li⁺ + FePO4 + e¯ → LiFePO4 State one possible advantage of developing cells such as lithium-ion rechargeable batteries.[1]
(c)(ii)Use the cathode half-equation to determine the change, if any, in oxidation states of lithium and iron at the cathode during discharging. change in oxidation state during discharging metal from to lithium iron[1]
(c)(iii)Write the equation for the overall reaction that occurs when this cell is discharging.[1]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2021). The mark or mark type is shown in brackets.

[2] M1: Correct setup of Born-Haber cycle equation or identification of values (416, 248, 762, 1560, -141, 798); M2: Correct calculation of ΔH_lattice = -3915 (kJ mol-1) (ecf for arithmetic error if M1 is correct).
[1] 20 × [0.9(+2) + 0.1(+3)] −2x = 0; .. x = 21
[2] FeO has a more exothermic/more negative lattice energy (than CaO).
[2] Fe2+ has a smaller radius/higher charge density (than Ca2+).
[2] There is greater attraction/stronger ionic bonds (between Fe2+ and O2-).
[1] Fe2+ is reduced to Fe (oxidation state Fe +2 → 0).; Fe2+ is oxidised to Fe3+ (in Fe3O4) (oxidation state Fe +2 → +3). (BOTH bullets required)
[1] 2O2- → O2 + 4e-
[3] M1: Calculate total charge Q = It = 50 A × (6 hours × 3600 s/hour) = 1.08 × 10⁶ C. Calculate moles of electrons = Q / F = 1.08 × 10⁶ C / 96500 C/mol = 11.19 mol e⁻.
[3] M2: Determine the stoichiometry from Fe₃O₄. Fe₃O₄ contains 1 Fe²⁺ and 2 Fe³⁺. To produce 3 Fe atoms (from 1 Fe₃O₄ unit), 8 electrons are required. Therefore, moles of Fe produced = (3/8) × moles of electrons = (3/8) × 11.19 mol = 4.196 mol Fe (ecf from M1).
[3] M3: Calculate mass of Fe = moles × Ar = 4.196 mol × 55.8 g/mol = 234.2 g. (Allow 3 significant figures, e.g. 234 g) (ecf from M2)
[1] small size / compact, low mass, high voltage (OWTTE)
[1] Lithium: from +1 to +1; Iron: from +3 to +2
[1] LiC6 + FePO4 → LiFePO4 + 6C