Use half-cells to balance the following redox reaction, and label the oxidation and reduction half-reactions.

Also, note that bromide is a spectator ion and can be eliminated.
FeBr3(aq) +Mg(s) = Fe(s) + MgBr2(aq)
Fe^+2 -2e +Br^- e+ + Mg^+2 -2e =Fe^+2 -2e +MgBr2(aq) -2e e+

I gave you the two half cells with a note to add them. What more do you want? If you don't understand something I wrote let me know in detail what it is.

Oh my gosh I apologize my brother wrote that for me I told him that I already did it but he posted it for me trying to help thank you

No sweat.

To balance the redox reaction using half-cells, we need to identify the oxidation and reduction half-reactions.

First, let's write the unbalanced half-reactions for the redox reaction:

Oxidation Half-Reaction: FeBr3(aq) → Fe(s)
Reduction Half-Reaction: Mg(s) → MgBr2(aq)

Now, let's balance each half-reaction separately.

Oxidation Half-Reaction:
1. Determine the oxidation states of each element.
FeBr3: Fe = +3, Br = -1
Fe: Fe = 0
2. Identify the changes in oxidation states.
The oxidation state of Fe changes from +3 to 0.
3. Balance the number of atoms of each element other than H and O.
Since there is one Fe on each side, the number of atoms of Fe is already balanced.
4. Balance the charges by adding electrons (e-).
The overall charge of FeBr3 is 0, while the overall charge of Fe is also 0. Hence, there is no need to add electrons to balance the charges.
5. Verify the balance in terms of mass and charge.
The oxidation half-reaction is as follows:
FeBr3(aq) → Fe(s)

Reduction Half-Reaction:
1. Determine the oxidation states of each element.
Mg: Mg = 0
MgBr2: Mg = +2, Br = -1
2. Identify the changes in oxidation states.
The oxidation state of Mg changes from 0 to +2.
3. Balance the number of atoms of each element other than H and O.
Since there is one Mg on each side, the number of atoms of Mg is already balanced.
4. Balance the charges by adding electrons (e-).
The overall charge of Mg is 0, while the overall charge of MgBr2 is +2. Hence, we need to add 2 electrons to the left-hand side to balance the charges.
5. Verify the balance in terms of mass and charge.
The reduction half-reaction is as follows:
Mg(s) → MgBr2(aq) + 2e-

Now that we have balanced the oxidation and reduction half-reactions, we need to combine them to obtain the balanced redox reaction:

FeBr3(aq) + Mg(s) → Fe(s) + MgBr2(aq) + 2e-

Note: The bromide (Br-) ions are spectator ions and do not participate in the redox reaction. Hence, we can eliminate them from the balanced equation:

FeBr3(aq) + Mg(s) → Fe(s) + MgBr2(aq)

In summary, the oxidation half-reaction is FeBr3(aq) → Fe(s) and the reduction half-reaction is Mg(s) → MgBr2(aq).