Skip to main content
A-LevelChemistryChemical energetics9701/41 · Oct/Nov 2023 · Q317 Marks

Hydrogen peroxide is a liquid at 298K. It is moderately stable under room conditions but will decompose quickly if a catalyst is added. reaction 1 2H2O2(l) → 2H2O (l) + O2 (g)

  1. (a)(i)Define entropy.[1]
  2. (a)(ii)Predict the sign of the standard entropy change of reaction 1. Explain your answer. sign explanation[1]
  3. (b)Some bond energy data are shown in Table 3.1. Table 3.1 type of bond bond energy /kJmol⁻¹ O-O 150 O-H 460 O=O 496 Use the data in Table 3.1 to show that the enthalpy change of the following reaction is -196 kJ mol⁻¹. 2H2O2(g) → 2H2O(g) + O2(g)[1]
  4. (c)Some standard entropies, Sº, are shown in Table 3.2. Table 3.2 substance Sº /J K⁻¹ mol⁻¹ H2O2(l) +102 H2O(l) +70 The enthalpy change and Gibbs free energy change for the following reaction are shown. 2H2O2(l) → 2H2O(l) + O2(g) ΔHº = –196 kJ mol⁻¹ ΔGº = –238 kJ mol⁻¹ Use the data given to calculate the standard entropy of oxygen, Sº, O2(g). Sº, O2(g) =[3]
  5. (d)The decomposition of H2O2(aq) is catalysed by aqueous iron(III) chloride and by silver metal. Identify which of these two catalysts is acting as a homogeneous catalyst. Explain your answer. homogeneous catalyst explanation[1]
  6. (e)The E values for two electrode reactions are given. H2O2 + 2H+ + 2e⁻ ⇌ 2H2O E° = +1.77V Cr3+ + e⁻ ⇌ Cr2+ E° = -0.41V
  7. (e)(i)An electrochemical cell is constructed with the following half-cells (electrodes): • an acidified solution of H2O2, a platinum wire • Cr2+ mixed with Cr3+, a platinum wire. Identify the positive half-cell and calculate the standard cell potential, E°cell. positive half-cell E°cell =[1]
  8. (e)(ii)Calculate the value of ΔGº for the cell reaction that occurs, per mole of H2O2. ΔGº =[2]
  9. (f)The E values for two electrode reactions are given. H2O2 + 2H+ + 2e⁻ ⇌ 2H2O E° = +1.77V Co3+ + e⁻ ⇌ Co2+ Eº = +1.82V An electrochemical cell is constructed with the following half-cells. half-cell 1 an acidified solution of H2O2 under standard conditions, a platinum wire half-cell 2 a solution containing 0.020 moldm⁻³ Co³+ and 2.0 moldm⁻³ Co2+, a platinum wire
  10. (f)(i)Use the Nernst equation to calculate the value of E, the electrode potential of half-cell 2 under these conditions. E[2]
  11. (f)(ii)Write an equation for the cell reaction that occurs in this cell under these conditions.[1]
  12. (g)(i)Define enthalpy change of hydration, ΔHhyd.[1]
  13. (g)(ii)Aluminium fluoride, AlF3, is an ionic solid. Complete and label the energy cycle to show the relationship between: • the enthalpy change of solution of AlF3, ΔHsol • the lattice energy of AlF3, ΔHlatt • the enthalpy changes of hydration of Al3+ and F⁻, ΔHhyd Include state symbols for all substances and ions.[2]
  14. (g)(iii)Relevant data for this question are given. ΔHsol AlF3 = -209 kJ mol⁻¹ ΔHhyd Al3+ = -4690 kJ mol⁻¹ ΔHhyd F⁻ = -506 kJ mol⁻¹ Use these data and your energy cycle in (g)(ii) to calculate the ΔHlatt of AlF3. ΔHlatt of AlF3 =[1]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.

  1. [1] (number of possible) arrangements of particles / energy in a system OR measure / degree of disorder in / of a system
  2. [1] positive / + AND more gas molecules / particles in products OR more moles / molecules in products / RHS
  3. [1] ΔH = (2 × 150) – (1 × 496) (= −196) OR ΔH = (2 × 150) + (4 × 460) – (1 × 496) – (4 × 460) (= −196)
  4. [1] ΔG = ΔH -TΔS seen or used with correct signs
  5. -238 = -196 – 298ΔS
  6. ΔS = 42/298
  7. [1] ΔS = (+)0.141 / 0.1409 (kJ K-1 mol-1) OR (+)141 / 140.9 (J K–1 mol-1) ecf from ΔG = ΔH + TΔS
  8. [1] 141 = (2 × 70) + S(O2(g)) – (2 × 102) S, O2(g) = 205 / 204.94 (J K−¹ mol-1) ecf
  9. [1] iron(III) chloride / FeCl3 AND same state / phase as reactants / H2O2
  10. [1] hydrogen peroxide / H2O2 AND +2.18
  11. [1] ΔG = -nEcellF
  12. [1] ΔG = (-2 × 2.18 × 96500) = −420.7 (kJ mol-1) ecf
  13. [1] Nernst: (E =) E° + (0.059/z)log[ox] / [red] u/c OR (E =) E° + (RT/zF)ln[ox] / [red] OR (E =) 1.82 + (0.059/1)log(0.02/2)
  14. [1] E = 1.82 + (0.059 / 1)log(0.02 / 2) = (+)1.702 (V) min 2sf OR E = 1.82 + [(8.314 × 298 / 1) × (96 500) ] ln(0.02 /2) = (+)1.702
  15. [1] H2O2 + 2H+ + 2Co2+ → 2H2O + 2Co3+ ECF for reverse equation from (f)(i) if E > 1.77 V
  16. [1] enthalpy change when one mole of gaseous ions forms an aqueous solution / dissolves in water
  17. [1] Substances with state symbols: AlF3(s) AlF3(aq) OR Al3+(aq) + (3)F-(aq) Al3+(g) (3)F¯(g)
  18. [1] changes identified and three correct arrow directions (one given)
  19. [1] ΔHlatt = −4690 + (3 × – 506) – (−209) = −5999 (kJ mol-1)

Write your answer and get it marked

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.

Sign up free →

About this question

Topic
Chemical energetics, A-Level Chemistry (syllabus 9701). Structured question, 17 marks.
Source
A-Level Chemistry, Oct/Nov 2023 Paper 4 Variant 1, question 3.

Questions and mark schemes are from past examination papers and remain the copyright of the exam board. How Oracle Prep builds its content

Related Chemistry questions

© 2026 Oracle Prep