Using the half- reaction method, balance the redox reaction shown below. (I-)+(Br2) ---> (IO3-)+(Br-)

Step 1. Break the reaction into the two half reactions.
Br2 ==> Br^-
I^- ==> IO3^-

Step 2. I will leave the first one for you to do. The second one is done as follows:
Determine the change in oxidation state of the element changing. In this case, I.
I on the left is -1. On the right is +5.

Step 3. Add electrons to the appropriate side to balance the change in oxidation state.
I^- ==> IO3^- + 6e

Step 4. Determine the charge on each side. The left side is -1; the right side is -7. Add H^+ in acid solution to balance the charge.
I^- ==> IO3^- + 6e + 6H^+

Step 5. Add water to the opposite side to balance the H^+.
I^- + 3H2O ==> IO3^- + 6e + 6H^+
This completes the balancing for this half reaction.

Post your work if you get stuck on finishing.

Iodine doesn't have an oxidation state of -1 in the molecule I-Br because it is a heteronuclear molecule. Br is more electronegative therefore it would have an oxidation state of +1.

Therefore Iodine would have an oxidation state of +1 not Br.

Sure! Continuing from Step 5, let's balance the first half-reaction, which is:

Br2 ==> 2Br^-

Let's follow the same steps as we did for the second half-reaction:

Step 2: Determine the change in oxidation state of the element changing. In this case, Br.
The oxidation state of Br on the left is 0, and on the right is -1.

Step 3: Add electrons to balance the change in oxidation state.
Br2 + 2e^- ==> 2Br^-

Step 4: Determine the charge on each side. The left side is 0, the right side is -2. Since this is in an acidic solution, we can add H^+ ions to balance the charge.
Br2 + 2e^- + 2H^+ ==> 2Br^- + 2H^+

Step 5: Add water to balance the H^+ ions.
Br2 + 2e^- + 2H^+ + 2H2O ==> 2Br^- + 2H^+ + 2H2O

Now we have both half-reactions balanced.

The balanced overall redox reaction is:
I^- + 3H2O + Br2 + 2e^- + 2H^+ ==> IO3^- + 6e + 6H^+ + 2Br^-

I hope this helps! Let me know if you have any further questions.

Step 6: Balance the first half-reaction, Br2 => Br^-

- Determine the change in oxidation state of bromine (Br) in the reaction:
Br2 on the left is 0. Br^- on the right is -1.
The oxidation state of bromine decreases from 0 to -1.

- Add electrons (e^-) to the appropriate side to balance the change in oxidation state:
Br2 => 2Br^-
(Note: The coefficient 2 is necessary to balance both bromine atoms)

Step 7: Check the charge on each side of the half-reaction.
On the left side: Br2, no charge.
On the right side: 2Br^-, -2 charge.
The charges are balanced.

The balanced half-reactions are:
Br2 => 2Br^- (oxidation half-reaction)
I^- + 3H2O => IO3^- + 6e^- + 6H^+ (reduction half-reaction)

Finally, add the two half-reactions together.
Multiply each half-reaction by the necessary coefficients to ensure the number of electrons transferred is the same.

6(I^- + 3H2O => IO3^- + 6e^- + 6H^+)
+ (Br2 => 2Br^-)
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6I^- + 6Br2 + 18H2O => 6IO3^- + 12e^- + 12H^+ + 12Br^-