Question

In the following chemical reaction, what is oxidized? What is reduced?

N2(g) + 2O2(g) --> 2NO2(g)

A) N_2
B) O_2
C) NO_2

Answer 1

You can determine this by comparing the charge of the element in the reactants with the charge of the element in the product.

An element is always neutral, meaning it has a charge of 0
As you see both reactants (the ones before the arrow) are both only elements. This means both reactants have a charge of 0.

The products overall charge is also 0 since it doesn't have any + or - on the right side of the formula. BUT, the elements in this product do have a charge.

Oxygen is almost always -2
we have 4 oxygen and 2 nitrogen
4x(-2) = -8
To make up for this, the nitrogen has to be +8.
We have 2 nitrogen so they both have a charge of +4

Now when an element increases in charge, it loses electrons. We call this oxidation.

When an element decreases in charge, it gains electrons. That's reduction.

This seems odd, but that's because electrons are negatively charged.

So oxygen decreases in charge (it goes from 0 to -8). That means it gains electrons, so oxygen is reduced.
Nitrogen goes from 0 to +8, it loses electrons and is therefore oxidized.


I hope this explains it :)

Answer 2

Thank you so much :) So just to be clear, N2 is oxidized? @thomaster