Who can use half reactions to balance FeBr2 + Br2 = FeBr3?

Oxidation half equation :
Fe²⁺(aq) → Fe³⁺(aq) + e⁻
Then, 2Fe²⁺(aq) → 2Fe³⁺(aq) + 2e⁻ ...... (*)

Reduction half equation :
Br₂(aq) + 2e⁻ → 2Br⁻(aq) ...... (#)

(*) + (#), and cancel 2e⁻ on the both sides :
2Fe²⁺(aq) + Br₂(aq) → 2Fe³⁺(aq) + 2Br⁻(aq)

Add 4Br⁻(aq) to the both sides of the above equation :
2Fe²⁺(aq) + 4Br⁻(aq) + Br₂(aq) → 2Fe³⁺(aq) + 6Br⁻(aq)

Group the cations and anions into compounds. The overall full equation is :
2FeBr₂(aq) + Br₂(aq) → 2FeBr₃(aq)

First you have to determine what's being oxidized and what's being reduced. For that you have to assign oxidation numbers to each of the elements on both sides of the equation. I assume you're familiar with the rules for assigning oxidation numbers.

(+2)(-1)...(0)....(+3)(-1)
FeBr2 + Br2 = FeBr3

Iron is being oxidized from +2 to +3, so the oxidation half-reaction should focus on iron:

Fe(+2) --> Fe(+3)

(I'm writing the oxidation numbers in parentheses because I cannot easily make superscripts.)

The mass is already balanced because there is exactly one iron ion on each side. Now we have to balance the charge. Add 1 electron to the product side, and you get:

Fe(+2) --> Fe(+3) + e(-)

There's your oxidation half-reaction. The molecular bromine, Br2, is being reduced to bromide ions, Br(-)

Br2 --> Br(-)

Start by balancing the mass.

Br2 --> 2Br(-)

Now balance the charge. Each Br2 molecule has to absorb 2 electrons to be reduced to two bromide ions:

Br2 + 2e(-) --> 2Br(-)

Now you have your half-reactions:

Fe(+2) --> Fe(+3) + e(-)
Br2 + 2e(-) --> 2Br(-)

Now you have to balance the number of electrons between them. The oxidation HR releases one electron, but the reduction HR accepts two, so you have to multiply the oxidation HR by 2.

2Fe(+2) --> 2Fe(+3) + 2e(-)
Br2 + 2e(-) --> 2Br(-)

Now you can recombine them:

2Fe(+2) + Br2 + 2e(-) --> 2Fe(+3) + 2e(-) + 2Br(-)

Eliminate the electrons, since they are the same on both sides:

2Fe(+2) + Br2 --> 2Fe(+3) + 2Br(-)

And if you are required to do so, recombine ions to form neutral compounds:

2FeBr2 + Br2 --> 2FeBr3

(Now you've probably noticed that before we recombined the ions, we had 2 bromide ions on the product side; now we have six. Well, remember, there were already four bromide ions present from the 2 units of FeBr2, and those were spectators, meaning they did not change during the reaction.)

I hope that helps. Good luck!

strange question... from one of our top contributors to the chemistry section..
.. "who can use half reactions to balance... "

but I'll give it a go... my answer..
.. MANY people can use half reactions to balance that particular equation including
.. (1) me, (2) YOU, (3) chemteam, (4) pisgachemist, (5) roger the mole, (6) Lucas C, etc

***********
proof you ask?

let's apply these 7 seven steps (which you undoubtedly have memorized by now).

steps to balancing a redox rxn
.. (1) identify the oxidation state of each and every atom
.. (2) determine what is oxidized and what is reduced
.. (3) write half reactions.. include electrons
.. (4) balance electrons in half reactions
.. (5) combine half reactions and cancel electrons
.. (6) rearrange as necessary and add counter ions
.. (7) add an remaining chemical species, THEN balance them

like this

*** 1 ***
Fe in FeBr2 is in the +2 oxidation state.. why? Br has the higher electronegativity (EN)
.. ... .absconds with the electrons it needs to get a full outer octet.. 1 each
Br in FeBr2 is in the -1 oxidation state.. see above reasoning

Br in Br2 is in the 0 oxidation state... WHY? each Br has the same EN
.. . .. .the overall Br2 has no charge (see any +1 or -2 or whatever after Br2?)
.. .. .. so 2 x Br = 0 ----> Br = 0

Fe in FeBr3 is in the +3 oxidation state.. again Br is -1 because of higher EN
Br in FeBr3 is -1

*** 2 ***
since reduction = reduction in charge (technically reduction in oxidation state)
since oxidation is the opposite of reduction.. in INCREASE in charge
.. ALL the Fe went from +2 to +3 and was oxidized
.. SOME Br went from 0 to -1 and was reduced... the Br in Br2

*** 3 ***
half reactions
.. Br2(0) + 2 e's --> 2 Br(-)... . reduction half
.. Fe(2+) ----> Fe(3+) + 1 e.. . .oxidation half

note the Br2... I did not write Br(0) + 1 e- ---> Br(-)... why not?
because Br2 is diatomic.. you have to have 2 together. I have to deal with this
sooner or later, might as well be now.

*** 4 ***
balancing e's.. 1x the first rxn + 2x the 2nd gives
.. Br2(0) + 2 e's --> 2 Br(-)... . .. .. ..reduction half
.. 2 Fe(2+) ----> 2 Fe(3+) + 2 e's.. . .oxidation half

*** 5 ***
combining
.. Br2(0) + 2 e's + 2 Fe(2+)--> 2 Br(-) + 2 Fe(3+) + 2 e's

canceling e's
.. Br2(0) + 2 Fe(2+)--> 2 Br(-) + 2 Fe(3+)

*** 6 ***
rearranging
.. 2 Fe(2+) + Br2(0) ---> 2 Fe(3+) + 2 Br(-)

adding counter ions..to the left and then finishing the rest.. (because it's trivial)
.. 2 FeBr2 + Br2 ---> 2 FeBr3

*******
and as you can see. ... there are at least 2 of us on my list that can balance this equation via the half reaction method. That answer your question?

Br2 Febr3 Reaction