Balance the following redox reaction using the half reaction method:
NaIO3+KI==> I2+H20+Na(+1)+K(+1)
I'll help you get started with oxidation states. I changes from +5 on the left in NaIO3 to 0 on the right in I2. I changes from -1 on the left in KI to 0 on the right in I2.
To balance the redox reaction using the half-reaction method, we need to split the reaction into two half-reactions: the oxidation half-reaction and the reduction half-reaction.
First, let's identify the oxidation and reduction half-reactions. In this case, the oxidation half-reaction involves the iodine atom going from the -1 oxidation state in I1- to the 0 oxidation state in I2. The reduction half-reaction involves the iodine atom going from the +5 oxidation state in IO3- to the -1 oxidation state in I1-.
Oxidation half-reaction: I1- → I2
Reduction half-reaction: IO3- → I1-
Now, let's balance each half-reaction step-by-step:
1. Balance the atoms other than hydrogen and oxygen:
Oxidation half-reaction: 2I1- → I2
Reduction half-reaction: IO3- → 6I1-
2. Balance the oxygen atoms:
Oxidation half-reaction: 2I1- → I2 + 2e-
Reduction half-reaction: IO3- + 6e- → 6I1-
3. Balance the charges by adding electrons:
Oxidation half-reaction: 2I1- → I2 + 2e-
Reduction half-reaction: IO3- + 6e- → 6I1- + 6e-
4. Balance the number of electrons:
Oxidation half-reaction: 2I1- → I2 + 2e-
Reduction half-reaction: 6IO3- + 12e- → 6I1- + 6e-
5. Multiply the half-reactions by appropriate coefficients to balance the number of electrons:
Oxidation half-reaction: 2I1- → I2 + 2e-
Reduction half-reaction: 3IO3- + 6e- → 3I1- + 3e-
Now, combine the two half-reactions:
2I1- + 3IO3- → I2 + 3I1-
Finally, balance the charges by adding appropriate coefficients:
6I1- + 9IO3- → 3I2 + 9I1-
The balanced redox reaction is:
6NaIO3 + 9KI → 3I2 + 9H2O + 6Na+ + 9K+
To balance a redox reaction using the half-reaction method, follow these steps:
Step 1: Assign oxidation numbers:
Start by assigning oxidation numbers to each element in the reaction. In this case:
NaIO3 + KI -> I2 + H2O + Na+ + K+
The oxidation number of Na is +1, the oxidation number of I in IO3 is +5 and in I2 is 0, the oxidation number of K is +1, and the oxidation number of O in H2O and IO3 is -2.
Step 2: Divide the reaction into two half-reactions:
Split the reaction into two half-reactions: oxidation and reduction.
The reduction half-reaction involves the gain of electrons, whereas the oxidation half-reaction involves the loss of electrons.
In this case, we have:
Oxidation half-reaction: NaIO3 -> Na+ + 3O2-
Reduction half-reaction: I- -> I2
Step 3: Balance the atoms and charges in each half-reaction:
Now, balance the number of atoms and charges in each half-reaction, excluding oxygen and hydrogen atoms for now.
Reduction half-reaction: I- -> I2
The iodine atoms are already balanced, but we need to balance the charge. Since I- has a charge of -1, and I2 has no charge, we need to add 2 electrons (e-) to the left side to balance the charges:
I- -> I2 + 2e-
Step 4: Balance the oxygen atoms:
Balance the number of oxygen atoms in each half-reaction by adding water molecules (H2O). In this case, there are no oxygen atoms in the reduction half-reaction, so we move on to the oxidation half-reaction.
Oxidation half-reaction: NaIO3 -> Na+ + 3O2-
We have 3 oxygen atoms on the right side, but only 1 oxygen atom on the left side. To balance this, we add 3 water molecules (H2O) to the left side:
NaIO3 + 3H2O -> Na+ + 3O2-
Step 5: Balance the hydrogen atoms:
Balance the number of hydrogen atoms in each half-reaction. In this case, there are no hydrogen atoms involved, so we move on to the next step.
Step 6: Balance the electrons:
Balance the number of electrons in each half-reaction by multiplying one or both of the half-reactions by integers.
In the oxidation half-reaction, there are no electrons involved. In the reduction half-reaction, we have 2 electrons (e-). To balance the electrons, we multiply the oxidation half-reaction by 2:
2NaIO3 + 6H2O -> 2Na+ + 6O2- + 2I2 + 4e-
Step 7: Add the two half-reactions together:
Now that the electrons are balanced, add the two half-reactions together. Make sure to cancel out any common species on both sides of the equation.
2NaIO3 + 6H2O + 2I- -> 2Na+ + 6O2- + 2I2 + 4e-
Cancel out the common species (6O2-):
2NaIO3 + 6H2O + 2I- -> 2Na+ + 2I2 + 4e-
Step 8: Combine the electrons:
Combine the species with electrons on one side of the equation. In this case, on the left side of the equation, we have 2I- and 4e-. We can add them together:
2NaIO3 + 6H2O + 2I- + 4e- -> 2Na+ + 2I2 + 4e-
Step 9: Simplify and finalize:
Since there are 4e- on both sides of the equation, we can cancel them out:
2NaIO3 + 6H2O + 2I- -> 2Na+ + 2I2
So, the balanced redox reaction using the half-reaction method is:
2NaIO3 + 6H2O + 2I- -> 2Na+ + 2I2