Skip to main content
A-LevelChemistryElectrochemistry9701/42 · May/June 2023 · Q98 Marks

An electrochemical cell is set up to measure E°cell of a cell consisting of an Fe3+ /Fe2+ half-cell and a Cl2/Cl- half-cell. Include all necessary substances. It is not necessary to state conditions used. [Important values, constants and standards] [Periodic Table]

  1. (a)Define standard cell potential, E°cell.[1]
  2. (b)Draw a labelled diagram of this electrochemical cell.[3]
  3. (c)The cell reaction for the electrochemical cell in (b) is shown. Cl2 + 2Fe2+ → 2Fe3+ + 2Cl- E°cell = +0.59V Calculate ΔG°, in kJ mol−1, for this cell reaction.[2]
  4. (d)(i)Another experiment is set up using the same electrochemical cell. In this experiment the Fe2+ concentration is 0.15moldm¯³. All other concentrations remain at their standard values. The Nernst equation is shown. E = E° + (0.059/z) log [oxidised species] / [reduced species] Use the Nernst equation to calculate the electrode potential, E, for the Fe3+ /Fe2+ half-cell in this experiment. [E° : Fe3+/Fe2+ = +0.77V] E =[1]
  5. (d)(ii)Use your answer to (d)(i) to calculate Ecell for this electrochemical cell. Ecell =[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 (May/June 2023). The mark or mark type is shown in brackets.

  1. [1] potential difference / voltage between the two half-cells / two electrodes (in a cell) under standard conditions
  2. [1] Correctly labelled salt bridge, voltmeter, and complete circuit [wires/salt bridge in/touching both solutions]
  3. [1] Correctly labelled Cl2, Cl-, and Pt electrode for the Cl2/Cl- half-cell, with good delivery system (no arrow required)
  4. [1] Correctly labelled Fe3+, Fe2+, and Pt electrode for the Fe3+/Fe2+ half-cell
  5. [1] M1 correct formula (∆G® ) = − nFEcell® OR correct substitution −2 × 96500 × 0.59
  6. [1] M2 correct final answer -114 kJ mol-1 (with conversion to kJ and 3sf) ecf
  7. [1] E = 0.77 + (0.059 / 1) log (1/0.15) use of z = 1; E = 0.82 V
  8. [1] Ecell = 0.59 + 0.77 – (answer to (d)(i)) = 0.54 V ecf

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
Electrochemistry, A-Level Chemistry (syllabus 9701). Structured question, 8 marks.
Source
A-Level Chemistry, May/June 2023 Paper 4 Variant 2, question 9.

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