balance the redox equations:
IO3-(aq) -> I2(aq) and I-(aq) -> I2(aq)
balance the redox equation:
S2O3(-2) -> S4O2(2-) and I3- -> I-
IO3- + 6H+ + 5I- ==> 3I2 + 3H2O
S2O3^2- + I2 ==> S4O6^2- + 2I^-
(you made a typo; it should be S4O6 and not S4O2 and I balanced it with I2 and not I3 but I2 + I^- = I3^- and there is no oxidation-reduction for this part; i.e., I2 ==> 2I^- is the redox part.)
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-
To balance redox equations, you need to follow these steps:
Step 1: Assign oxidation numbers to all the elements.
- Oxidation numbers represent the charge an atom would have if electrons were transferred completely during the reaction.
Step 2: Identify which elements are being oxidized and which are being reduced.
- The element that increases its oxidation number is being oxidized (loses electrons).
- The element that decreases its oxidation number is being reduced (gains electrons).
Step 3: Write the half-reactions for the oxidation and reduction separately.
- Split the redox reaction into two half-reactions, one for oxidation and one for reduction.
Step 4: Balance the atoms involved in the half-reactions.
- Balance all the elements, except for hydrogen and oxygen, in each half-reaction.
Step 5: Balance the charge by adding electrons to one side of the equation.
- Add enough electrons to the side that has a higher positive charge (if it is the oxidation half-reaction) or to the side that has a higher negative charge (if it is the reduction half-reaction).
Step 6: Equalize the number of electrons in both half-reactions.
- Multiply each half-reaction by the appropriate factor to equalize the number of electrons gained and lost.
Step 7: Combine the half-reactions and cancel out electrons.
- Add the balanced half-reactions together to form the balanced overall redox equation.
- Make sure to cancel out electrons in the final equation.
Now let's apply these steps to the given equations:
Equation 1: IO3-(aq) -> I2(aq)
Step 1:
- The oxidation number of iodine (I) is changing from +5 in IO3- to 0 in I2.
Step 2:
- Iodine is being reduced (going from +5 to 0).
Step 3:
Oxidation half-reaction: IO3-(aq) -> I2
Reduction half-reaction: nothing to balance
Step 4:
Oxidation half-reaction: 6IO3-(aq) -> 5I2(aq)
Step 5:
Oxidation half-reaction: 6IO3-(aq) -> 5I2(aq) + 6e-
Step 6:
Oxidation half-reaction: 6IO3-(aq) + 6e- -> 5I2(aq)
Step 7:
Combine the half-reactions:
6IO3-(aq) + 6e- -> 5I2(aq)
Equation 2: I-(aq) -> I2(aq)
Step 1:
- The oxidation number of iodine (I) is changing from -1 in I- to 0 in I2.
Step 2:
- Iodine is being oxidized (going from -1 to 0).
Step 3:
Oxidation half-reaction: I-(aq) -> I2
Reduction half-reaction: Nothing to balance
Step 4:
Oxidation half-reaction: 2I-(aq) -> I2(aq)
Step 5:
Oxidation half-reaction: 2I-(aq) -> I2(aq) + 2e-
Step 6:
Oxidation half-reaction: 2I-(aq) + 2e- -> I2(aq)
Step 7:
Combine the half-reactions:
6IO3-(aq) + 6e- -> 5I2(aq)
2I-(aq) + 2e- -> I2(aq)
Final balanced equation:
6IO3-(aq) + 6I-(aq) -> 5I2(aq) + 6I-(aq)
Now you have balanced the redox equations.