Question: Balance the following reactions by the oxidation number method.
a) I2 + HNO3 --> HIO3 + NO2 + H2O
so i have
N +5 to +4 net charge = -1
I 0 to +5 net charge = +5
i add coefficient where the oxidation number is changing.
in this case
I2 + 4HNO3 --> 2HIO3 + 4NO2 + H2O
i cant balance now.
it takes five I for every one N
So if you are using I2, you will change 10N
I2+10HNO3>> HIO3 + NO2 + H2O
start then with 2 in front of HIO3, 10 NO2, and .....
Read over my earlier post to you. I mentioned that most beginning students made a mistake in not using the same number of atoms; therefore, "you should add a 2 in front of the HIO3." That way the I2 goes from a total of zero on the left to 2HIO3 or 10 total on the right.
To balance the given reaction, follow these steps:
1. Assign oxidation numbers to each element in the reaction.
I2 + HNO3 --> HIO3 + NO2 + H2O
I: 0
H: +1
N: +5 to +4
O: -2 (unchanged)
2. Identify the element undergoing the oxidation and reduction.
In this case, iodine (I) is being oxidized from 0 to +5, and nitrogen (N) is being reduced from +5 to +4.
3. Balance the equations for oxidation and reduction separately.
For the oxidation half-reaction (I2 to HIO3), since iodine is increasing in oxidation number from 0 to +5, the coefficient of I2 should be 1. This means that the coefficient of HIO3 will also be 1. The overall balanced oxidation half-reaction is:
I2 --> 2HIO3
For the reduction half-reaction (+5 to +4), since nitrogen is decreasing in oxidation number, the coefficient of HNO3 should be 5. The coefficient of NO2 will also be 5. The overall balanced reduction half-reaction is:
5HNO3 --> 5NO2
4. Balance the number of atoms of each element on both sides of the half-reactions by adding H2O or H+ ions as needed.
Looking at the oxidation half-reaction, there are already 2 iodines and 2 hydrogens on the left side. On the right side, there are 2 iodines, 6 oxygens, and 2 hydrogens (from the HIO3 molecule). To balance the hydrogens, add 2 H2O on the left side:
I2 + 2H2O --> 2HIO3
5. Balance the charges on both sides of the half-reactions by adding electrons (e-) as needed.
The oxidation half-reaction is already balanced in terms of charges since there are no charges on either side.
6. Multiply the half-reactions by integers to make the number of electrons in each half-reaction equal.
Since there are no electrons involved in this example, we can skip this step.
7. Combine the half-reactions and cancel out common species.
By combining the previously balanced oxidation and reduction half-reactions, we have:
I2 + 2H2O + 5HNO3 --> 2HIO3 + 5NO2
This is the balanced equation for the given reaction:
I2 + 2H2O + 5HNO3 --> 2HIO3 + 5NO2
To balance this equation using the oxidation number method, follow these steps:
1. Write down the unbalanced equation:
I2 + HNO3 -> HIO3 + NO2 + H2O
2. Assign oxidation numbers to each element in the equation:
I2 + HNO3 -> HIO3 + NO2 + H2O
I: 0 -> +5 (increased)
N: +5 -> +4 (decreased)
H: +1 (unchanged)
O: -2 (unchanged)
3. Identify which elements' oxidation numbers have changed and write down the corresponding half-reactions:
Half-reaction for iodine:
I2 -> 2I^5+ (oxidation)
Half-reaction for nitrogen:
N^5+ -> N^4+ (reduction)
4. Balance the atoms in each half-reaction:
For the iodine half-reaction:
I2 -> 2I^5+
Since the number of iodine atoms is already balanced, there is nothing further to do.
For the nitrogen half-reaction:
N^5+ -> N^4+
Since the number of nitrogen atoms is already balanced, there is nothing further to do.
5. Balance the charge in each half-reaction by adding electrons (e-):
For the iodine half-reaction:
I2 -> 2I^5+ + 10e-
For the nitrogen half-reaction:
10e- + N^5+ -> N^4+
6. Multiply each half-reaction by the appropriate coefficient to equalize the number of electrons involved:
For the iodine half-reaction:
10I2 -> 20I^5+ + 100e-
For the nitrogen half-reaction:
100e- + N^5+ -> 100N^4+
7. Combine the half-reactions and simplify if necessary:
10I2 + HNO3 -> 10HIO3 + 100NO2 + H2O
Now the equation is balanced using the oxidation number method.