A molten mixture of equal parts aluminium fluoride, AlF₃, and sodium chloride, NaCl, undergoes electrolysis - VCE - SSCE Chemistry - Question 7 - 2019 - Paper 1

To determine which product is formed during the electrolysis of the molten mixture, we need to consider the processes happening at both the cathode and anode.

1. At the Cathode: In electrolysis, the cations are attracted to the cathode. In this case, we have Al³⁺ ions from AlF₃ and Na⁺ ions from NaCl. Aluminium ions will preferentially be reduced because aluminium's standard reduction potential is more negative than sodium's. Thus, aluminium metal is produced at the cathode:

   2Al^{3+} + 6e^- → 2Al (s)

2. At the Anode: The anions, F⁻ from AlF₃ and Cl⁻ from NaCl, are attracted to the anode. Chlorine gas forms preferentially due to the higher oxidation potential of Cl⁻ compared to F⁻. The reaction at the anode for chlorine is:

   2Cl^- → Cl₂ (g) + 2e^-

Based on these reactions, the correct statement is C: Aluminium metal will be produced at the cathode and chlorine gas will be produced at the anode.