Half reaction of

2 Al2O3 + 3 C → 4 Al + 3 CO2

SO3^2- + Cu^2+ --> SO4 ^2- + Cu+

To determine the half-reaction for the given equation, we need to break down the reaction into two half-reactions: one for the oxidation and one for the reduction process.

First, let's identify the elements that are being oxidized and reduced in the reaction.

In this reaction, aluminum (Al) is being reduced (gaining electrons) from the +3 oxidation state to zero oxidation state, while carbon (C) is being oxidized (losing electrons) from the zero oxidation state to the +4 oxidation state.

Now let's set up the half-reactions:

1. Reduction Half-Reaction:
Al3+ + 3e- → Al

2. Oxidation Half-Reaction:
C → CO2 + 4e-

The reduction half-reaction represents the reduction of Al3+ ions to elemental aluminum (Al) with the gain of 3 electrons. The 3e- term is added on the reactant side to balance the charge.

The oxidation half-reaction represents the oxidation of carbon (C) in the reactant side to carbon dioxide (CO2) with the loss of 4 electrons. The 4e- term is added on the product side to balance the charge.

It is important to note that when balancing the half-reactions, the number of electrons in both the reduction and oxidation half-reactions should be equal.

Overall, the balanced equation for the given reaction can be obtained by multiplying the reduction half-reaction by 2 and the oxidation half-reaction by 3 to balance the number of electrons gained and lost:

2Al3+ + 6e- → 2Al (reduction)
3C → 3CO2 + 12e- (oxidation)

Adding the two half-reactions together cancels out the electrons:

2Al3+ + 6e- + 3C + 12e- → 2Al + 3CO2

Simplifying this equation gives:

2Al2O3 + 3C → 4Al + 3CO2

Therefore, the half-reactions for the given equation are:
Reduction half-reaction: Al3+ + 3e- → Al
Oxidation half-reaction: C → CO2 + 4e-