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The half reactions are Al+3 + 3 e- -> Al (at the cathode) and 2 Cl-1 -> Cl2 + 2 e- at the anode. Applying the proper integers by which to multiply the half reactions so that the number of electrons received at the cathode equals the number of electrons given up at the anode gives an overall reaction of 2 Al+3 + 6 Cl-1 -> 2 Al + 3 Cl2.

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Q: What are the chemical reactions that take place during the electrolysis of molten aluminum chloride?
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