Question

Balance the following redox equation and identify the element oxidized, the element reduced, the oxidizing agent, and the reducing agent. Show all of the work used to solve the problem.

NO2- + Al yields AlO2- + NH3

I need help

Answer 1

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Answer 2

First of all, let's start by defining oxidation and reduction:
-Oxidation is the loss of electrons from a species, so its charge, or what we refer to as its oxidation number, will become less negative and will increase.
-Reduction is the gain of electrons by a species, so its oxidation number will become more negative and will decrease.

-An oxidising agent in a redox reaction is in itself reduced (undergoes reduction).
-A reducing agent in a redox reaction is in itself oxidised (undergoes oxidation).

When we talk about species here, were referring to the oxidation numbers of individual atoms/ions, not the molecules or compounds as a whole.
http://www.mreisley.com/pdf/ion_reference.pdf
Here's a list of some of the common oxidation numbers atoms/ions...you can see that a lot of them are similar for those ions of elements in the same group in the periodic table. You don't need to remember them all, often you just need a few key ones and from that you can work out the rest.

Back to the question....we'll work out the oxidation number of each atom/ion present in all the reactants and products:

Reactants:
NO2- : This has an overall charge of -1. We know that the oxidation number/charge of the oxygen is 2-. But, we have 2 oxygens here, so there is an overall contribution is
2(2-) = 4-. This means that if NO2- is to have an overall charge of -1, the nitrogen atom must have an oxidation number of +3.

Al : The oxidation number of any individual atom (neutral - not an ion) on its own (not as part of any compound) will always be zero (in the same way, the oxidation number of any individual ion on its own will be the same as the charge on the ion). So, the oxidation number of Al will be 0.

Products:
AlO2- : Here, the Al is not on its own, so it could possibly have a charge/oxidation number which is not 0. Oncemore, we know that the oxidation number of O will be 2-, and, as there are two oxygens present in the molecule, the overall contribution will be 2(-2) = -4. We know that AlO2- has an overall charge, as a compound, of -1, so for this to be true the charge/oxidation number of Al must be +3, in this case.

NH3 : This compound is overall neutral (charge = 0). We know that each hydrogen will have an oxidation number of +1 in the compound, and with 3 present, will have an overall contribution of 3(+1) = +3. So, for ammonia to be neutral this means that the oxidation number of N, the single nitrogen present, must be -3.

So, lets look and see which species has had their oxidation numbers changed between the reactants and products:
- N is gone from +3 (in NO2-) to -3 (in NH3), so its oxidation number has DECREASED.
-Al has gone from 0 (on its own in Al) to +3 (in AlO2-), so its oxidation number has INCREASED.

From this, you should be able to work out which species are the oxidising and reducing agents for the reaction. Hope that helps! :)

Answer 3

wow thanks so much! That was a very detailed and thorough answer and I learned a lot!

Answer 4

Great job @Ciarán95 !

Answer 5

No problem @XxZeroxX :)

Answer 6

I believe that when balanced it would be OH- + NO2- + H2O + 2Al → 2AlO2- +NH3