HIO3 + FeI2 + HCl --> FeCl3 + ICl

(in acidic solution)
Show and label the two balanced half-reactions (oxidation and reduction). Add H2Os, H+s or OH-s as needed.

I just wanted to know if the answer I got is correct

2FeI2 +6HCl +Cl +HIO3-->6FeCl3 +ICl +3H20

No, I don't think so.

First, the problem asks for the two balanced half equations. You have given the balanced equation.
Second, the balanced equation you have written is not balanced. For example, Fe = 2 on the left and 6 on the right.
H atoms don't balance.
Cl atoms don't balance.
I atoms don't balance.
and I quit counting.

I don't understand how I am supposed to give two separate half reactions.

To balance the given chemical equation in acidic solution and identify the half-reactions, you need to follow a few steps. Here's the correct balanced equation:

2FeI2 + 8HCl + HIO3 → 2FeCl3 + ICl + 4H2O

Now, let's separate the equation into two half-reactions: oxidation and reduction.

Oxidation Half-Reaction:
FeI2 → FeCl3

Since the iodide (I-) in FeI2 is oxidized to iodine (ICl), the oxidation half-reaction can be written as follows:

2I- → ICl

Reduction Half-Reaction:
HIO3 + 6H+ + 6e- → 4H2O

In the above half-reaction, iodine (I) in HIO3 is reduced to form water (H2O). To balance the charges, we add 6 electrons (e-) to the reactant side.

So, the overall balanced equation with the correct half-reactions is:

Oxidation: 2I- → ICl + 2e-
Reduction: HIO3 + 6H+ + 6e- → 4H2O

Your original balanced equation was close, but there were a few errors in the coefficients. The correct coefficients are shown above.