In this reaction, aluminum is being oxidized and copper II ions are being reduced. The balanced half-reactions for this process are:Oxidation half-reaction: Al Al + 3eReduction half-reaction: Cu + 2e CuSince we have 2 moles of aluminum being oxidized, the total number of electrons released in the oxidation half-reaction is:2 moles Al 3 moles e / 1 mole Al = 6 moles eNow, we need to find the number of electrons required for the reduction of 3 moles of copper II ions:3 moles Cu 2 moles e / 1 mole Cu = 6 moles eIn this reaction, the number of electrons released by the oxidation of aluminum is equal to the number of electrons required for the reduction of copper II ions. Therefore, the total number of electrons transferred in this reaction is 6 moles of electrons.