Posted by Amy on Monday, June 20, 2011 at 7:22am.
Hi this is a repost, and I have another 2 questions I posted yesterday. Please help me with these, I need help to verify my work, not confident if I'm doing these right.
Use oxidation numbers to identify if this reaction is REDOX reactions.
3NO2(g) + H2O(l) �� 2HNO3(aq) + NO(g)
3NO2 = 3 for N
= 2 for o2
H2O = +1 for H2
= 2 for O
>
2HNO3 = +1 for 2H
= 3 for N
= 2 for O3
NO = 3 for N
= 2 for O
Therefore Nitrogen is reduced (one 3 remains on right side)
hence REDOX has occured
Is this right? Also should I be multiplying oxidation numbers if there is a number in front ie. 2SO = 2 oxidation but should I multiply by the 2 infront so is now 4 oxidation for sulphur in 2SO? Please let me know because I haven't done this and if wrong need to correct for a lot of questions :(

Chemistry  DrBob222, Monday, June 20, 2011 at 3:40pm
What you have done confuses me; in a class I could ask a question or two and know how to answer.
The first equation is redox so you are correct; however, not for the reason you cite(3 N left over on one side versus the other).
The oxidation state for N in NO2 is +4 and not 3. How do you get that? Each O is 2 and there are two of them for a total of 4; therefore, for NO2 to be zero N must be +4. On the right side, each N is +5 in HNO3 and +2 in NO. This is a redox equation because you have something being oxidized and something being reduced (in this case it just happens to be N in NO2it goes both ways).
As to multiplying by the coefficient, it depends. To identify if redox has occurred, you want to see that EACH atom has changed and I don't multiply by the coefficient to determine that since it isn't needed. In fact, it can be confusing in some situations (such as this one).
So when do I multiply by the coefficient? When I am balancing equations, we want to know the TOTAL charge on side going to a TOTAL charge on the other. In the example you cite for 2SO, here are the two ways of doing it.
#1. Each O is 2, for SO to be zero then S must be +2.
#2. Each O is 2 and that x 2 = 4 for both O atoms. Therefore, both S atoms must be +4 or EACH must be +2. Doing it this way just increases the number or steps (the work) AND the chance of making a mistake.
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