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.

Jillian--It won't do much good for us to balance these for you. You need to learn to do them yourself. It's a matter of following the rules. Tell us what your hang up is and let us help you through them.

I don't know which are the half reactions.

To balance the given redox reaction in acidic solution, we need to identify the oxidation and reduction half-reactions.

First, let's determine the oxidation numbers of the elements in the compounds involved:

HIO3: In this compound, the oxidation number of iodine (I) is +5.
FeI2: In this compound, the oxidation number of iron (Fe) is +2.
HCl: In this compound, the oxidation number of chlorine (Cl) is -1.
FeCl3: In this compound, the oxidation number of iron (Fe) is +3.
ICl: In this compound, the oxidation number of iodine (I) is +1.

From the given reaction, we can observe that iodine is reduced from an oxidation state of +5 to +1, while iron is oxidized from +2 to +3. Therefore, the reduction half-reaction involves the reduction of iodine, and the oxidation half-reaction involves the oxidation of iron.

Half-Reactions:
Reduction Half-Reaction: I2 + 10e- + 12H+ → 2ICl + 6H2O
Oxidation Half-Reaction: Fe → Fe3+ + 3e-

Now, let's balance the number of atoms and charges on each side of the half-reactions:

Reduction Half-Reaction:
I2 + 10e- + 12H+ → 2ICl (balanced on atoms and charges)

Oxidation Half-Reaction:
Fe → Fe3+ + 3e- (balanced on atoms and charges)

Next, we need to balance the number of electrons transferred in each half-reaction, so that they cancel out when combined:

Reduction Half-Reaction:
I2 + 10e- + 12H+ → 2ICl + 6H2O (multiply by 6 to balance electrons)
6I2 + 60e- + 72H+ → 12ICl + 36H2O

Oxidation Half-Reaction:
Fe → Fe3+ + 3e-

Now, we can combine the two balanced half-reactions by multiplying them accordingly:

Overall Balanced Equation:
6I2 + 60e- + 72H+ + 6Fe → 12ICl + 36H2O + 6Fe3+

Remember to double-check the balanced equation and ensure that the number of atoms and charges are balanced on both sides.