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Video transcript

What we have depicted right over here is a combustion reaction. We have the hydrocarbon methane right over here. You take that. You take some molecular oxygen. You give them enough heat. And then, they are going to combust. And they're going to produce carbon dioxide, water, and then more energy than you put in. This is an exothermic reaction. More energy comes out than you put in. This is why fires keep spreading. This is why combustion is used to power things. But that's not what we're going to focus on in this video. In this video, we want to think about-- which of these components, of these molecules, what's being oxidized and what is being reduced? And to do that, let's first think about the oxidation states of the input atoms and the oxidation states of the outputs-- of the different constituents of these molecules. So I encourage you to now pause this video and try to figure that out on your own. So I'm assuming you've given a go at it. Now, let's work through this together. Let's first think about the methane. And I have a bunch of electronegativities here based on the Pauling scale listed out here, but let's just visualize methane. Methane is a carbon bonded to 4 hydrogens. In our oxidation state world-- even though, this, in reality, is a covalent bond-- we pretend like they're hypothetically ionic bonds. So we have to give the electron pair to one of the parties to the bond. If we look between carbon and hydrogen, carbon is more electronegative than hydrogen. So we will assume that carbon will be hogging the electrons and that hydrogen will be giving away the electrons. So carbon's going to take-- in this hypothetical world-- hypothetically take an electron from each of these hydrogens. It is going to have an oxidation state of negative 4, an electron from each of 4 hydrogens. Negative 4. And once again, we write the sign after the number probably so that we don't get these confused with exponents. Now, let's think about the hydrogens. Each of those hydrogens is having-- in this hypothetical world-- an electron taken away from it. We could say it has an oxidation state of plus 1, which we could write as 1 plus. Or we could just write a positive right over here-- a plus sign. Now, we have molecular oxygen-- oxygen bonded to oxygen. Well, all oxygens are created equal. Or we'll assume that these oxygens are created equal, that they're not different isotopes or anything like that. In this reality, there's no reason why one oxygen would hog any electrons from the other oxygen. In this world, the oxygen has an oxidation state-- when it's in this molecular oxygen form-- it has an oxidation state of 0 or an oxidation number of 0. Now, let's think about this side-- the products. Now, what's happening here with carbon dioxide? Carbon dioxide is a carbon double bonded to 2 different oxygens. We see oxygen is one of the most electronegative elements out there, definitely more electronegative than carbon. So in our hypothetical ionic bond world, we would say that oxygen would take 4 electrons. These bonds would all go to the different electrons. Each oxygen will take 2 electrons from the carbon. The carbon will lose 4 electrons. We could say the carbon loses 4 electrons. You lose 4 electrons. That gives you a hypothetical positive charge of positive 4. Each of these oxygens is gaining 2 electrons. So it gives them each a hypothetical charge of negative 2. And we see it nets out. Positive 4. 2 times negative 2 is negative 4. It all adds up. This is a neutral molecule. And we saw that over here-- for negative 4 plus 4 times positive 1. It all nets out to be neutral. And that makes sense because these are neutral molecules. And then, finally, you have water. And we've seen that multiple times already. Oxygen-- what, in reality, are covalent bonds with the hydrogens-- in our hypothetical ionic bond world, oxygen is a good bit more electronegative. So we're assuming it's going to take the electrons from the hydrogens. So each of the hydrogens loses an electron, giving it an oxidation number of 1. I could even write it like this, if you like. An oxidation number of positive 1. The oxygen has gained 2 electrons. So that gives it an oxidation number of negative 2. So now that we've done that, let's think about who is getting oxidized and who is being reduced. So let's first focus on the carbon. The carbon starts off at an oxidation number of negative 4. The reaction takes place. And then, carbon now has an oxidation number of positive 4. So how does something go from an oxidation number of negative 4 to positive 4? Well, the way to increase your charge or your hypothetical charge is to lose electrons. Every time you lose an electron, this becomes less negative. And eventually, it'll become positive. So you have to lose 8 electrons. So I'll write plus 8 electrons right over here. You take these electrons, give it to this carbon. You're going to get to this side of this reaction. And the way I'm writing right now, these are called "half reactions" where I'm independently focusing on each of the elemental components of these reactions. So here, you have carbon. In this reaction, carbon-- in our hypothetical oxidation number world-- has lost 8 electrons. What do we call it when you are losing these hypothetical electrons? Well, we can remind ourselves OIL RIG-- oxidation is losing electrons, reduction is gaining. Or LEO-- losing electrons is oxidation-- the lion says GER-- gaining electrons is reduction. So it's clear right over here that carbon is being oxidized. It is losing electrons. Oxidation is losing electrons. Carbon is oxidized. Let's think about the hydrogen. On the left-hand side, you have 4 hydrogens that each have an oxidation number of plus 1. On the right-hand side, you have 4 hydrogens. We're writing it a slightly different way. We could write it like this-- 2H2's that each have an oxidation number of plus 1. The oxidation numbers for the hydrogens has not changed. The hydrogens has neither been oxidized nor reduced. Let's think about the oxygens. On the left-hand side, you have 2O2's neutral oxidation number. And on the right-hand side, what do you have? You have 4 total oxygens. I'll combine these together. I'll just write this as 4 total oxygens. And what's each of their oxidation numbers? Well, we see it's a negative 2. So what happened to each of these 4 oxygens? I could have written 4O here instead of 2O2. Either way, I'm just really trying to account for the oxygens. Here I have 4 oxygens with a neutral oxidation number-- with an oxidation number of 0. And here I have 4 oxygens with a negative oxidation number. How do you go from 0 to negative? Each of them must have gained 2 electrons. If you have 4 oxygens, each of them gained 2 electrons, we could actually write the reaction like this. Actually, let me write it like this. Let me move this part. So cut and paste. Let me move it over to the right a little bit because what I want to show is the gaining of the electrons. So plus 8 electrons. So what happened to oxygen? Well, oxygen gained electrons. What is gaining electrons? Reduction is gaining-- RIG. GER-- gaining electrons is reduction. Oxygen has been reduced. Now, what oxidized carbon? Well, carbon lost electrons to the oxygen. So carbon oxidized by the oxygen, which is part of the motivation for calling it "oxidation." And what reduced the oxygen? Well, oxygen took those electrons from the carbon. So oxygen reduced by the carbon. And this type of reaction-- where you have both oxidation and reduction taking place, and really they're two sides of the same coin. One thing is going to be oxidized if another thing is being reduced, and vice versa. We call these oxidation reduction reactions. Or sometimes "redox" for short. Take the "red" from "reduction" and the "ox" from "oxidation," and you've got "redox." This is a redox reaction. Something is being oxidized. Something else is being reduced. Not everything is being oxidized or reduced, and we can see that very clearly when we depict it in these half reactions. And one way to check that your half reactions actually makes sense is you can actually sum up the two sides. If you take this left side right over here and you sum them up, you should get all the constituents right over here minus the electrons. And the same thing over here. You should get all the constituents that you have on the right-hand side minus the electrons. So one way to think about it, if you added all this stuff on the left-hand side, you would get this plus this 8 electrons on the left-hand side. If you added all of this up, you would have the plus 8 electrons on that side as well. And so the 8 electrons you could kind of say, oh, let's take them away from both sides. You would get your original reaction. And so I'm going to leave you there. This is a very, very powerful tool in chemistry because it really helps us think about what's actually going on inside of some type of a reaction.