When molten aluminum chloride (AlCl3) is electrolyzed, the following half-reactions occur at the anode and cathode:
At the anode (oxidation): 2Cl^-(l) -> Cl2(g) + 2e^-
Chloride ions (Cl^-) are oxidized to chlorine gas (Cl2) by losing two electrons.
At the cathode (reduction): Al^3+(l) + 3e^- -> Al(l)
Aluminum ions (Al^3+) are reduced by gaining three electrons to form liquid aluminum (Al).
Overall, the balanced equation for the electrolysis of molten aluminum chloride can be represented as:
2AlCl3(l) -> 2Al(l) + 3Cl2(g)
At the anode, chloride ions are oxidized to chlorine gas, while at the cathode, aluminum ions are reduced to liquid aluminum.