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A-LevelChemistryChemical energetics9701/42 · Feb/Mar 2025 · Q117 Marks

Silver, Ag, is a metal in the d-block of the Periodic Table.

  1. (a)Silver can form compounds containing either Ag+ or Ag2+ ions. Explain why silver is a transition element.[1]
  2. (b)(i)Table 1.1 gives data relevant to the Born–Haber cycle for silver(I) fluoride, AgF. Table 1.1 standard energy change value/kJ mol-1 first ionisation energy of silver +732 enthalpy change of atomisation of silver +289 enthalpy change of atomisation of fluorine +79 enthalpy change of formation of silver(I) fluoride -203 lattice energy of silver(I) fluoride -955 Write equations for the standard enthalpy changes described. Include state symbols. • standard enthalpy change of atomisation of silver • standard enthalpy change of formation of silver(I) fluoride[2]
  3. (b)(ii)Define lattice energy.[2]
  4. (b)(iii)Calculate the first electron affinity, EA₁, of fluorine, using data from Table 1.1. It may be helpful to draw a labelled energy cycle as part of the working for your answer. EA₁ = .............................................. kJ mol-1[2]
  5. (c)(i)Use the data in Table 1.2 to calculate the enthalpy change of solution, ΔHsol, of AgF(s). Table 1.2 energy change at 298 K value/kJ mol-1 lattice energy of AgF(s) -955 enthalpy change of hydration of Ag+(g) -464 enthalpy change of hydration of F¯(g) -506 ΔHsol of AgF(s) = ...................................... kJ mol-1[1]
  6. (c)(ii)Use your answer to (c)(i) to suggest whether AgF is soluble in water at 298K. Explain your answer.[1]
  7. (d)(i)Table 1.3 shows some data relevant to the silver(I) halides, AgCl to AgI. Table 1.3 silver(I) halide first electron affinity of halogen/kJ mol-1 lattice energy /kJ mol-1 AgCl -349 -905 AgBr -325 -890 AgI -295 -876 Explain the trend in the first electron affinities of the halogens, Cl to I.[2]
  8. (d)(ii)Explain the trend in the lattice energies of the silver(I) halides, AgCl to AgI.[1]
  9. (e)(i)An electrochemical cell is constructed using the electrodes shown in Table 1.4. Table 1.4 electrode half-equation E°/V 1 AgCl(s) + e¯ ⇌ Ag(s) + Cl-(aq) +0.222 2 Cu2+(aq) + 2e¯ ⇌ Cu(s) +0.342 Calculate the standard cell potential, E°cell. Construct an equation for the overall cell reaction. E°cell = ...................................... V equation .......................................................................................................................................................................[2]
  10. (e)(ii)In a different experiment, electrode 1 is set up using a saturated solution of KCl. Saturated KCl(aq) contains 36.0g of KCl per 100 cm³ of solution at 298K. The Nernst equation for electrode 1 is: E = E° + (0.059/Z) log (1/[Cl-(aq)]) Calculate the electrode potential, E, of electrode 1 under these conditions. E = ...................................... V[3]

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Mark scheme

Marking points from the official mark scheme (Feb/Mar 2025). The mark or mark type is shown in brackets.

  1. [1] (Ag(2+) is a) forms stable ion(s) with incomplete d orbitals / an incomplete d subshell
  2. [2] Ag(s) → Ag(g)
  3. Ag(s) + 1/2F2(g) → AgF(s)
  4. [2] ΔH/energy change when 1 mole of an ionic solid / lattice / crystal / compound is formed from gas phase ions / gaseous ion(s) (under standard conditions)
  5. [2] M1: 5 numbers and no multipliers -203 = +289 + 732 + 79 + EA1 + (−955)
  6. M2 correct sign and evaluation : EA1 = −348 (kJ mol-1) ecf min 3sf
  7. [1] enthalpy change of solution of AgF = −(−955) + (−464) + (−506) = -15 (kJ mol-1)
  8. [1] AgF is (slightly) soluble / yes AND (the enthalpy change of solution is slightly) exothermic / negative ecf from Q1(c)(i)
  9. [2] greater the distance between the nucleus and (the shells of the) electrons OR atomic radii increases / atomic size increases / more shells OR more shielding by inner shells
  10. AND the less attraction between nucleus / protons AND incoming electron / added electron
  11. [1] halide/X-/ions get larger (down the group) AND decreasing attraction between ions / weaker ionic bond
  12. [1] E°cell = +0.342 – (+0.222) = (+)0.120 (V) min 2sf
  13. [1] Cu2+ + 2Ag + 2Cl¯ → Cu + 2AgCl
  14. [3] M1: [Cl-(aq)] = 10 × 36.0 / 74.6 = 4.83 (4.8257) mol dm-3
  15. M2: E = 0.222 + (0.059/1) log (1/M1) ecf M1
  16. M3: E = (+)0.182 (V) min 2sf ecf M1

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About this question

Topic
Chemical energetics, A-Level Chemistry (syllabus 9701). Structured question, 17 marks.
Source
A-Level Chemistry, Feb/Mar 2025 Paper 4 Variant 2, question 1.

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