4 Responses

1. Jen

   21|Mar|2008 1

   Hello.

   My answer would be D

   For A, B, C, D : initial oxidation state (O.S) of Nitrogen = +5
   irregardless of the coefficient of NO3^-1

   A: final O.S of Nitrogen = +3
   hence, change= +3-(+5) = -2

   B: final O.S of Nitrogen = +2
   hence, change = +2 – (+5) = -3

   C: final O.S of Nitrogen= +1
   hence, change = +1 – (+5) = -4

   D: final O.S of Nitrogen = -3
   hence, change = -3 + – (+5) = -8

   Reply

2. Gary

   21|Mar|2008 2

   My answer is D too.But i think the OS for N2 at the products in A should be +4 instead of +3

   As Jen said,oxidation state for Nitrogen at the reactants are all +5

   A: The OS for N2 at products is +4 so the change of OS is 1.
   B: The OS for N2 at products is +2 so the change of OS is 3.
   C: The OS for N2 at products is +1 so the change of OS is 4
   D:The OS for N2 at products is -3 so the change of OS is 8

   Reply

3. sean

   16|Apr|2008 3

   Hi Jennifer,

   Very good attempt indeed. The approach you use is correct.

   Just note that for (A), the O.S. for N should be +4 instead of +3. Check out the
   working again.

   Keep up the good work!

   To your Academic Success in Chemistry,
   Sean
   Founder & Master Trainer
   http://www.SimpleChemConcepts.com

   Reply

4. sean

   16|Apr|2008 4

   Hi Gary,

   Hope all is well with your revision for Mid-Year exams.

   Spot On! Answer is indeed D and your approach towards such Redox question is
   correct.

   Like what Jennifer has mentioned, the stoichiometry of each species does not
   affect the O.S. since the equation is balanced.

   Keep up the good work!

   To your Academic Success in Chemistry
   Sean
   Founder & Master Trainer
   http://www.SimpleChemConcepts.com

   Reply