NO2 ----> NO2^-1 + NO3^-1
which N do you use? I have N being +4 on the reactant side, then as a +3 on for the NO2^-1 and +5 for the NO3^-1. Do you add the N's together on the product side? Then what about the rest of the equation? You leave out all the O, . . . . I'm confused.
NO2 ==> NO2^-
NO2 ==> NO3^-
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Balance each half equation, multiply by appropriate numbers, add them, cancel anything common to both sides.
Ok, I tried that and this is what I am getting.
NO2 ----> NO2^-1
NO2 ----> NO3^-1
3(2e- + 2NO ----> 2NO^-1)
2(3e- + 2NO ----> 3NO^-1)
I am still getting the e- on the reactant side, so they won't cancel -- what am I doing wrong now?
For starters you are dropping one of the O atoms on NO2. Notice you wrote NO2 ==>NO2^-1 first, then followed that with 2e + 2NO ==> 2NO^-1.
Here is the way I balance these things. It isn't the easiest method around BUT it emphasizes oxidation state which I like to do.
Step 1. Write the half equation.
NO2 ==> NO2^-
Step 2. What is changing oxidation state?
N on the left is +4 and on the right is +3.
Step 3. Add electrons to make the oxidation change balance.
NO2 + e ==> NO2^-
Step 4. Count the charge on both sides and add H^+ (if the solution is acid) or OH^-(if the solution is basic) to balance the charge. The charge on the left is -1 (from the one electrons) and it is -1 on the right from the NO2^- ion; therefore, the charge balances and that half cell is done. But I'll finish the steps even though this equation is done.
Step 5. Add water molecules to the appropriate side to balance the H^+ or OH and the half equation should be finished.
Step 6. Remember to check you work.
a. atoms should balance.
b. oxidation charge should balance with electrons.
c. charge should balance.
For NO3^-. I'll just write in the steps and leave the narrative out but you can refer to the top to see what I have done for each step.
NO2 ==> NO3^-
N on left is +4, on the right is +5
NO2 ==> NO3^- + e
Charge on left is zero; on the right is -2 so add H^+.
NO2 ==> NO3^- + e + 2H^+
Charge on left is zero; on right is now zero which is what we want.
Add water to balance the H^+
NO2 + H2O ==> NO3^- + e + 2H^+
This step SHOULD balance the oxygen atoms and it does.
Check it.
a. atoms: 1 N on both sides, 2H on both sides, 3 O on both sides.
b. charge: zero on left; zero on right.
c. e and oxidation change: N changed fro +4 to +5. We added 1 electron to the +5 side so +5 + (-1) = _4 and that balances.
To determine the oxidation states of nitrogen in this equation, you need to follow a few steps:
Step 1: Identify the atoms and their states in the equation. In this case, we have NO2, NO2^-1, and NO3^-1.
Step 2: Assign an oxidation state to each element in the given compounds. Remember that oxidation states are hypothetical charges assigned to atoms in a compound, assuming that all bonding is purely ionic.
In NO2, we can let the oxidation state of oxygen (O) be -2 since it is usually -2 in compounds unless stated otherwise. Let's assume the oxidation state of nitrogen (N) is x.
Using the given equation, we can set up the following equation: x + 2(-2) = 0. Simplifying, we get: x - 4 = 0. Solving for x, we find that the oxidation state of nitrogen in NO2 is +4.
In NO2^-1, the oxidation state of oxygen remains -2. Let the oxidation state of nitrogen be y. Using the given equation, we can set up the equation: y + 2(-2) = -1. Simplifying, we get: y - 4 = -1. Solving for y, we find that the oxidation state of nitrogen in NO2^-1 is +3.
In NO3^-1, the oxidation state of oxygen remains -2. Let the oxidation state of nitrogen be z. Using the given equation, we can set up the equation: z + 3(-2) = -1. Simplifying, we get: z - 6 = -1. Solving for z, we find that the oxidation state of nitrogen in NO3^-1 is +5.
So, the oxidation states of nitrogen are:
- NO2: +4
- NO2^-1: +3
- NO3^-1: +5
To answer your specific question, when writing out the redox reaction, it is crucial to account for both the changes in oxidation states and the balanced transfer of electrons. In this equation, the nitrogen atoms are undergoing different oxidation state changes, which should be included in the balanced equation along with the changes for other elements, such as oxygen.