To balance the redox equations, we will first assign oxidation numbers to all the elements in the equations. Then, we will balance the atoms and charges in each half-reaction separately.
For the equation IO3-(aq) -> I2(aq), let's assign the oxidation number for iodine (I) in IO3- as x. The oxidation number for oxygen (O) is -2.
Starting with IO3-(aq):
Oxidation number of I in IO3- = +x
Oxidation number of O in IO3- = -2
To balance the equation, we need to balance the number of atoms and the charge in each half-reaction.
Half-reaction for reduction:
I2(aq) -> I-(aq)
Balancing the atoms:
There is no need to balance any atoms in this half-reaction.
Balancing the charges:
Since I2(aq) has no charge, we only need to balance the charge on I-(aq). In order for the charge on I- to become balanced, we need two electrons (e-) on the right side:
I2(aq) + 2e- -> 2I-(aq)
Now let's balance the oxidation half-reaction.
Half-reaction for oxidation:
IO3-(aq) -> I2(aq)
Balancing the atoms:
There is one iodine atom on each side, so the atoms are balanced.
Balancing the charges:
Since IO3-(aq) has a charge of -1, and I2(aq) has no charge, we need to balance the charge by adding 3 electrons (e-) on the left side:
6IO3-(aq) + 6e- -> 3I2(aq)
To equalize the number of electrons, we need to multiply the reduction half-reaction by a factor of 3 and the oxidation half-reaction by a factor of 2:
2I2(aq) + 4e- -> 4I-(aq)
6IO3-(aq) + 6e- -> 3I2(aq)
Now the electrons are balanced:
2I2(aq) + 4e- -> 4I-(aq)
6IO3-(aq) + 6e- -> 3I2(aq)
To balance the overall equation, we can add the half-reactions together:
6IO3-(aq) + 6I-(aq) -> 3I2(aq) + 6IO3-(aq)
2I2(aq) + 4I-(aq) -> 4I-(aq) + 2I2(aq)
After canceling out the common species on both sides of the equation, the balanced redox equation is:
6IO3-(aq) + 6I-(aq) -> 5I2(aq)
Next, let's balance the equation S2O3(-2) -> S4O2(2-) and I3- -> I-.
For the equation S2O3(-2) -> S4O2(2-), let's assign the oxidation number for sulfur (S) as x.
Starting with S2O3(-2):
Oxidation number of S in S2O3(-2) = +x
Oxidation number of O in S2O3(-2) = -2
Balancing the atoms and charges for this equation:
Since the atoms and charges are already balanced, no additional steps are required.
Now, let's balance the equation I3- -> I-.
Starting with I3-:
Oxidation number of I in I3- = +3
Oxidation number of I in I- = -1
Balancing the atoms and charges for this equation:
To balance the charge, we need to add 2 electrons (e-) on the left side:
I3- + 2e- -> I-
Now, let's combine the two half-reactions:
S2O3(-2) -> S4O2(2-) + I3- + 2e- -> I-
To balance the number of electrons, we need to multiply the S2O3(-2) half-reaction by 2:
2S2O3(-2) -> 2S4O2(2-) + I3- + 2e- -> I-
Finally, we can add the half-reactions together:
2S2O3(-2) + I3- + 2e- -> 2S4O2(2-) + I-
After canceling out the common species on both sides of the equation, the balanced redox equation is:
2S2O3(-2) + I3- -> 2S4O2(2-) + I-