Balance Redox using reactions by the 1/2 reaction method.
NaIO3 (aq) + KI (aq) -->„³ I2 (s)
IO3^- ==> I2
First balance the I so we are working with apples and apples and not apples and oranges.
I on left is +10; on right is 0
add 10 e to left side
2IO3^- + 10e ==> I2
charge on left is -12; right is zero.
Add H^+ to left and H2O to the right.
2IO3^- + 10e + 12H^+ ==> I2 + 6H2O
I^- ==> I2
balanced I (again for apples and apples)
2I^- ==>I2 + 2e
Now multiply 1st half equation by 1 and second by 5 and add them.
When finished you will find that you can divide all of the coefficients by 2.
Thank you so much!
To balance the given redox reaction using the half-reaction method, you need to split the reaction into two half-reactions: one for the oxidation half-reaction and one for the reduction half-reaction.
1. Identify the oxidation and reduction half-reactions:
The iodine (I) in NaIO3 is reduced from an oxidation state of +5 to 0 in I2. This means that the reduction half-reaction involves the iodine atoms. The iodine in KI is oxidized from -1 to 0 in I2. Therefore, the oxidation half-reaction involves the iodine atoms as well.
2. Balance the oxidation half-reaction:
Start by balancing the atoms other than oxygen or hydrogen first. In this case, the only element other than iodine is potassium (K). Since there is only one potassium atom on both sides of the equation, it is already balanced. Next, balance the oxygen atoms using water (H2O). Since there are 3 oxygen atoms on the left side and 0 on the right side, we need to add 3 H2O to the right side of the equation. This introduces 6 hydrogen atoms, so you need to balance the hydrogen atoms by adding H+ ions to the left side of the equation. Finally, balance the charge by adding electrons (e-) to the left side of the equation. The oxidation half-reaction is:
2I- (aq) -> I2 (s) + 2e-
3. Balance the reduction half-reaction:
Following the same process as in step 2, balance the atoms other than oxygen or hydrogen. In this case, we only have iodine to balance. Since there are two iodine atoms on the right side (I2), we need to balance it by multiplying the left side by a factor of 2. Next, balance the oxygen atoms by adding water (H2O). There are zero oxygen atoms on the left side, so add 3 H2O to the left side. This introduces 6 hydrogen atoms, so balance them by adding H+ ions to the right side. Finally, balance the charge by adding electrons (e-) to the right side of the equation. The reduction half-reaction is:
6H+ (aq) + 2IO3- (aq) + 6e- -> 3H2O (l) + I2 (s)
4. Balance the number of electrons transferred:
To ensure that the number of electrons transferred in the oxidation and reduction half-reactions are equal, we need to multiply both half-reactions by a factor such that the number of electrons in both reactions is the same. In this case, the oxidation half-reaction needs to be multiplied by a factor of 6 and the reduction half-reaction needs to be multiplied by a factor of 2. This will give us:
12I- (aq) -> 6I2 (s) + 12e-
12H+ (aq) + 4IO3- (aq) + 12e- -> 6H2O (l) + 2I2 (s)
5. Balance the overall reaction:
Now that the number of electrons transferred is balanced, you can combine the two half-reactions to form the balanced overall reaction. Add the oxidation half-reaction to the reduction half-reaction while canceling out any common species on both sides of the equation. The balanced overall reaction for NaIO3 (aq) + KI (aq) -> I2 (s) is:
12I- (aq) + 12H+ (aq) + 4IO3- (aq) + 6H2O (l) -> 6I2 (s) + 6H2O (l) + 2I2 (s)
Simplify the equation by removing the water molecules:
12I- (aq) + 12H+ (aq) + 4IO3- (aq) -> 6I2 (s) + 6H2O (l) + 2I2 (s)
And that's how you balance the given redox reaction using the half-reaction method.