Worked example: Using oxidation numbers to identify oxidation and reduction | Khan Academy

What we have here is a reaction that involves iodine, manganese, oxygen, and hydrogen. What we want to do in this video is think about which of the elements are being oxidized in this reaction and which of the elements are being reduced in this reaction. Pause this video and see if you can figure that out before we work through it together.

All right, now let's work through it together. The way that I will tackle it—and you might have tackled it or I suggest you tackle it—is to figure out the oxidation numbers for each of the elements as we go into the reaction, as they are entering the reaction, and as they are exiting the reaction, or I guess you could say on either side of the reaction.

So first, let's look at this iodine right over here. Well, it has a—each iodine has a negative one charge. So it's quote hypothetical charge, which isn't so hypothetical in this case, which would be its oxidation number, is negative one. Now let's move over to this permanganate ion right over here. This one's a little bit more involved to figure out the oxidation numbers. But what we generally remember is that oxygen is quite electronegative. It is likely to hog two electrons.

When we think about hypothetical charge with oxidation numbers, oxygen is going to have a negative two oxidation number because it likes to hog those two extra electrons. So if each of these four oxygens has a hypothetical charge of negative 2, that would be negative 8 total. We see that this entire ion has a negative one charge. So that means that the manganese has to have a hypothetical charge—an oxidation number of plus seven.

So I just want to review that one again because this is a little bit involved. We said oxygen, we're going to go with a negative 2 because it likes to hog two electrons. We have four of them, so if you add all that together, you're at negative eight. The whole ion has a negative one. So what plus negative eight is going to be negative one? Well, positive seven. And so that's manganese's oxidation number as we enter into the reaction on this side of the reaction.

Then let's look at the water. Well, water—both the hydrogens and oxygens—these are ones you'll see a lot. This oxygen is going to have a negative two oxidation number, and each of those hydrogen atoms are going to have a plus one oxidation number because in that water molecule, if we know that the oxygen hogs the electrons, these are covalent bonds. But if we had to assign kind of a hypothetical charge where we said, all right, well let's just say the oxygen takes those two electrons, then each of those hydrogens will lose an electron and have a plus one oxidation number.

Now let's look at the right-hand side of this reaction. What's going on with these iodines here? Well, in this iodine molecule, they aren't gaining or losing electrons, so your oxidation number is zero. Then let's move on to the next compound. Each of these oxygens has an oxidation number of negative two. So what would be manganese's oxidation number? Well, the compound is neutral. Two oxygens at negative two is going to be negative four. So in order to be neutral, the manganese must be at plus four, an oxidation number of plus four.

And then last but not least, if we look at these hydroxide anions, each of the oxygen is going to have a negative two oxidation number, and then the hydrogen is going to have a plus one. We can confirm that that makes sense. Negative 2 plus 1 is going to be negative 1 for each of these ions.

So now let's just think about who's been oxidized and who's been reduced. Remember, oxidation is losing electrons (OIL RIG), reduction is gaining electrons, or reduction is a reduction in the oxidation number. So first, let's look at the iodine. We go from an oxidation number of negative 1 to zero. To go from an oxidation number of negative one to zero, you need to lose electrons, so it has been oxidized.

Let me write that down: the iodine has been oxidized. Now let's look at the manganese. We go from a plus seven to a plus four. Our oxidation number has gone down—it has been reduced. Now let's look at the oxygen. Well, everywhere the oxygen has an oxidation number of negative two, so nothing there. And then same thing for the hydrogens—plus one on both sides, so nothing there.

So the iodine has been oxidized, and the manganese has been reduced.