NH3(g) + 2O2 (g) = HNO3 (aq) + H2O (l)
How do I calculate the oxidation states of Nitrogen and figure out if it is reduced, oxidized or neither?
http://www.chemteam.info/Redox/Redox-Rules.html
Thank you for the link! Here is my answer below. I am not sure if I am figuring it right. Please advise.
NH3(g) + 2O2 (g) = HNO3 (aq) + H2O (l)
N is -3,H3 is +1, 2o2 is 0,
H is +1, N is X ??, O3 is -2, H2O is -2
Therefore X=-2
I am not sure how to figure N. Is it X or is N include NO3?
and it is
Oxidized
To calculate the oxidation states of nitrogen (N) in a compound, you can use the following guidelines:
1. The oxidation state of a neutral atom is 0.
2. The sum of the oxidation states in a molecule or ion must equal the overall charge of the species.
3. Oxygen (O) generally has an oxidation state of -2, unless it is in a peroxide where it is -1. Hydrogen (H) usually has an oxidation state of +1.
Let's apply these rules to the reaction NH3(g) + 2O2(g) = HNO3(aq) + H2O(l):
1. Start by assigning an oxidation state of 0 to H2O, since hydrogen and oxygen are in their usual oxidation states: H2O(l) --> 0
2. Now, consider NH3. Since hydrogen usually has an oxidation state of +1, and there are three hydrogens in NH3, the total oxidation state contributed by hydrogen is +3. This leaves N with an oxidation state of -3, as it balances out the charge: NH3(g) --> H3^+ + N^3- (-3 + 3 = 0)
3. Next, examine O2. Oxygen typically has an oxidation state of -2, and since there are two oxygens, the total oxidation state contributed by oxygen is -4. This leaves N with an oxidation state of +5, considering the overall charge: O2(g) --> 2O^- + 2e^- (-4 + 4 = 0)
4. Finally, look at HNO3. Since hydrogen usually has an oxidation state of +1, and there are three hydrogens in HNO3, the total oxidation state contributed by hydrogen is +3. Oxygen typically has an oxidation state of -2, and there are three oxygens in HNO3, making the total oxidation state contributed by oxygen -6. Considering the overall charge, we can determine the oxidation state of N: HNO3(aq) --> 3H^+ + NO3^- (+1 × 3 + (-6) = -2)
Now that we have the oxidation states of nitrogen in NH3 and HNO3, we can compare them:
In NH3, nitrogen has an oxidation state of -3. In HNO3, nitrogen has an oxidation state of +5. Nitrogen is therefore oxidized (its oxidation state has increased) from -3 to +5 during the reaction.
Please note that this is a basic method for calculating oxidation states, and in some cases, exceptions and special rules may apply.