Explain how electrolysis of molten lead(II) bromide produces lead and bromine, identifying which species is oxidised and which is reduced, and stating at which electrode each process occurs.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Model answer (4 marks)
Lead(II) bromide is melted so that Pb²⁺ and Br⁻ ions are free to move. The external DC supply forces electrons to flow from the anode to the cathode. At the cathode (negative electrode) Pb²⁺ ions accept two electrons (reduction) to give Pb(s). At the anode (positive electrode) Br⁻ ions lose two electrons (oxidation) to give Br₂(l). Thus Pb²⁺ is reduced at the cathode and Br⁻ is oxidised at the anode, producing lead metal and bromine gas.
Examiner tips
- State the electrode where each ion migrates and the direction of electron flow
- Use the words ‘reduction’ and ‘oxidation’ with the correct species
- Mention that the reaction is non‑spontaneous and requires an external power source
- Show the ion migration to the appropriate electrode
Common mistakes
- Confusing the anode and cathode roles
- Saying Pb is oxidised instead of reduced
- Omitting that the reaction needs a molten state or external power
Mark scheme (4 marks)
- Pb²⁺ ions migrate to the cathode (negative electrode) and are reduced by gaining electrons to form lead metal.
- Br⁻ ions migrate to the anode (positive electrode) and are oxidised by losing electrons to form bromine.
- The lead(II) bromide must be molten (or dissolved) so that the ions are free to move and carry the current.
- An external power source (direct current supply) drives the non-spontaneous reaction by forcing electron flow, distinguishing electrolysis from a spontaneous voltaic cell.
Key terms in this question
Related
- All IB DP Chemistry Standard Level (2023 syllabus) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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