Worked example: Balancing a simple redox equation | Chemical reactions | AP Chemistry | Khan Academy

So what we have here is a redox reaction. Things are getting oxidized and reduced; that's the name, redox. But we want to balance this redox reaction, and when we talk about balancing a redox reaction, we want to make sure we conserve mass and charge on both sides of this reaction.

So how do we do that? Well, the first step is to assign oxidation numbers, or oxidation states, to each of the constituent elements on either side of the reaction. That will let us know who's getting oxidized and who's getting reduced, and then we can set up the half reactions, which we can then balance.

All right, so let's first look at this aluminum right over here. It has an oxidation state, or an oxidation number, of zero; it's just aluminum by itself. Then we can go to this hydrogen. This is really just a proton right over here; it has a plus one charge, and so its hypothetical charge, you could say, well, that would be plus one as well, which would be its oxidation number or its oxidation state.

Then, when we get on to the right side of this reaction, we see that the aluminum now has a positive three charge. So its oxidation number would also be positive three. We can see here that the hydrogen—actually, now we have two, so we're gonna have to deal with that later on—we only have one here, but now each of these hydrogens has an oxidation number of zero. They're not taking up electrons or giving away electrons; you just have two hydrogens bonded to each other.

All right, so now that we've assigned oxidation numbers, we can figure out who's getting oxidized and who is getting reduced. If you look at the aluminum, aluminum goes from an oxidation number of zero to plus three. So if your oxidation number is increasing, that means you're getting oxidized. You might also remember oil rig: oxidation is losing electrons. Because you're losing electrons, you're having a more positive charge, but your oxidation number is going up, so you're getting oxidized.

If you look at the hydrogen, we're going from a plus one to a zero. If your oxidation number is reducing, you are being reduced. So now let's set up both of the half reactions: the oxidized half reaction for aluminum and then the reduction for hydrogen.

So first for the aluminum, we have aluminum solid, and its half reaction is going to now form aluminum with a plus three charge in the aqueous state. So let's first balance this for just the number of aluminums we have; we have one aluminum on the left, one aluminum on the right, so that seems balanced. Now let's try to balance it for charge. So we have no charge here, so we should have a total of no charge on the right, but we have plus three here.

What we need to do is add some electrons. So let me add three electrons right over here, and there we have it; we have balanced it for charge. Now let's think about the hydrogen. If we take a proton on the left, let me just do it like this, and it's in an aqueous solution. On the right, we just have hydrogen molecules—neutral hydrogen molecules. So first, let us balance it for the number of hydrogens.

We have two on the right, one on the left, so we're going to have to put a two right over here. So we've balanced for the number of hydrogens, and now let's balance for charge. Let's see on the right-hand side we have no net charge. Well, on the left-hand side right over here, we have a positive 2 net charge.

So in order to balance this, I have to put two electrons on the left-hand side. So let me do that here—so two electrons I'm going to add over here—and now it looks like it is balanced for charge. Now the next thing we want to do is we want to balance the number of electrons that we have on the right-hand side and on the left-hand side in these half reactions.

So how can you do that? Once you have three here and you have two here, the least common multiple involving a little bit of your elementary school mathematics here of three and two is six. So we can get these both to six. How could we do that? Well, we could multiply this top half reaction by two, so let me do that.

If I multiply this top half reaction by two, and if I multiply this bottom half reaction by three—why does that work? Well, two times three electrons is going to give us six electrons and three times two electrons is going to give us six electrons. So let me now rewrite this. If I multiply by two, we're going to have two aluminums. So we have two aluminums in a solid state, and then on the right-hand side of this half reaction, I'm multiplying everything by two; so now I have two aluminums plus three in an aqueous solution, and then I'm multiplying these electrons by two as well, so plus six electrons.

Now let me do this down here. Three times two electrons—that is going to give us six electrons—and then I'm going to multiply this three times this two, so that's going to give us six hydrogen protons, I guess I could say. So plus six hydrogen protons that are in an aqueous solution, and then on the right-hand side, if I multiply by three, I have three hydrogen molecules, each of them with two hydrogen atoms.

And so now I have balanced the number of electrons on both sides. Now I'm going to add these two half reactions. If I add the two half reactions, what do I get? Let me do it down here. So on the left-hand side, I'm just going to add up all of this stuff right over here. So I'm going to get six electrons plus two aluminums plus six hydrogen protons plus six hydrogen protons.

Then on the right-hand side, I am going to add up all of this stuff. So I have those two aluminums now with a plus three charge in an aqueous solution. I'm going to have the three hydrogens—three hydrogen molecules, I should say. There's actually six hydrogens here, and then I have the six electrons plus six electrons.

I have six electrons on the left, six electrons on the right; I could cancel those out. And then what I have here is our balanced redox reaction—balanced for both mass and charge—and we are done.