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Video transcript

Let's see how to identify the oxidizing and reducing agents in a redox reaction. So here, we're forming sodium chloride from sodium metal and chlorine gas. And so before you assign oxidizing and reducing agents, you need to assign oxidation states. And so let's start with sodium. And so the sodium atoms are atoms in their elemental form and therefore have an oxidation state equal to 0. For chlorine, each chlorine atom is also an atom in its elemental form, and therefore, each chlorine atom has an oxidation state equal to 0. We go over here to the right, and the sodium cation. A plus 1 charge on sodium, and for monatomic ions, the oxidation state is equal to the charge on the ion. And since the charge on the ion is plus 1, that's also the oxidation state. So plus 1. We're going to circle the oxidation state to distinguish it from everything else we have on the board here. And for chloride anion, a negative 1 charge. Therefore, the oxidation state is equal to negative 1. And so let's think about what happened in this redox reaction. Sodium went from an oxidation state of 0 to an oxidation state of plus 1. That's an increase in the oxidation state. 0 to plus 1 is an increase in oxidation state, so therefore, sodium, by definition, is being oxidized. So sodium is being oxidized in this reaction. We look at chlorine. Chlorine is going from an oxidation state of 0 to an oxidation state of negative 1. That's a decrease in the oxidation state, and therefore, chlorine is being reduced. So each chlorine atom is being reduced here. Now, before we assign oxidizing and reducing agents, let's just go ahead and talk about this one more time, except showing all of the valence electrons. So let's also assign some oxidation states using this way because there are two ways to assign oxidation states. So let's assign an oxidation state to sodium over here. So if you have your electrons represented as dots, you can assign an oxidation state by thinking about how many valence electrons the atom normally has and subtracting from that how many electrons you have in your picture here. So for sodium, being in group one, one valence electron normally, and that's exactly what we have in our picture. Each sodium has a valence electron right here. So 1 minus 1 gives us an oxidation state equal to 0, which is what we saw up here, as well. So sodium has an oxidation state equal to 0. Notice that I have two sodium atoms drawn here, and that's just what the two reflects in the balanced equation up here. Let's assign an oxidation state to each chlorine atom in the chlorine molecule. And so we have a bond between the two chlorine atoms, and we know that bond consists of two electrons. Now, when you're assigning oxidation states and dot structures, you want to give those electrons to the more electronegative elements. In this case, it's the same element, so there's no difference. And so we give one electron to one atom and the other electron to the other atom, like that. And so assigning an oxidation state, you would say chlorine normally has seven valence electrons, and in our picture here, this chlorine atom has seven electrons around it. So 7 minus 7 gives us an oxidation state equal to 0. And of course, that's what we saw up here as well, when we were just using the memorized rules. And so it's the same for this chlorine atom over here, an oxidation state equal to 0. So sometimes it just helps to see the electrons. We'll go over here for our products. We had two sodium chlorides, so here are two sodium chlorides. And let's see what happened with our electrons. So the electron in magenta, this electron over here in magenta on this sodium, added onto one of these chlorines here. And then this electron on this sodium added onto the other chlorine, like that, and so sodium lost its valence electron. Each sodium atom lost its valence electron, forming a cation. And when we calculate the oxidation state, we do the same thing. Sodium normally has one valence electron, but it lost that valence electron. So 1 minus 0 is equal to plus 1 for the oxidation state, which is also what we saw up here. And then when we do it for chlorine, chlorine normally has seven valence electrons, but it gained the one in magenta. So now it has eight around it. So 7 minus 8 gives us an oxidation state equal to negative 1. And so maybe now it makes more sense as to why these oxidation states are equal to the charge on the polyatomic monatomic ion here. And so now that we've figured out what exactly is happening to the electrons in magenta, let's write some half reactions and then finally talk about what's the oxidizing agent and what's the reducing agent. So let's break down the reaction a little bit more in a different way. So you can see we have two sodium atoms over here. So we're going to write two sodiums. And when we think about what's happening, those two sodium atoms are turning into two sodium ions over here on the right. And so we have two sodium ions on the right. Now, those sodium atoms turned into the ions by losing electrons, so each sodium atom lost one electron. So we have a total of two electrons that are lost. I'm going to put it in magenta here. So those 2 electrons are lost, and this is the oxidation half reaction. You know it's the oxidation half reaction, because you're losing electrons here. So remember, LEO the lion. So Loss of Electrons is Oxidation. So this is the oxidation half reaction. We're going to write the reduction half reaction. The chlorine molecule gained those two electrons in magenta. So those two electrons in magenta we're going to put over here this time. The chlorine molecule gained them, and that turned the chlorine atoms into chloride anions. And so we have two chloride anions over here. And so those are, of course, over here on the right, our two chloride anions. And so here we have those two electrons being added to the reactant side. That's a gain of electrons, so this is our reduction half reaction, because LEO the lion goes GER. Gain of Electrons is Reduction. And so if we add those two half reactions together, we should get back the original redox reaction, because those two electrons are going to cancel out. It's actually the same electrons. These two electrons in magenta that are lost by sodium are the same electrons that are gained by chlorine, and so when we add all of our reactants that are left, we get 2 sodiums and Cl2, so we get 2 sodiums plus chlorine gas. And then for our products, we would make 2 NaCl, so we get 2 NaCl for our products, which is, of course, our original balanced redox reaction. So finally, we're able to identify our oxidizing and reducing agents. I think it was necessary to go through all of that, because thinking about those electrons and the definitions are really the key to not being confused by these terms here. And so sodium is undergoing oxidation, and by sodium undergoing oxidation, it's supplying the two electrons for the reduction of chlorine. Therefore, you could say that sodium is the agent for the reduction of chlorine, or the reducing agent. So let's go ahead and write that here. So sodium, even though it is being oxidized, is the reducing agent. It is allowing chlorine to be reduced by supplying these two electrons. And chlorine, by undergoing reduction, is taking the electrons from the 2 sodium atoms. That allows sodium to be oxidized, so chlorine is the agent for the oxidation of sodium, or the oxidizing agent. Let me go ahead and write that in red here. Chlorine is the oxidizing agent. And so this is what students find confusing sometimes, because sodium is itself being oxidized, but it is actually the reducing agent. And chlorine itself is being reduced, but it is actually the oxidizing agent. But when you think about it by thinking about what happened with those electrons, those are the exact same electrons. The electrons that are lost by sodium are the same electrons gained by chlorine, and that allows sodium to be the reducing agent for chlorine, and that is allowing chlorine at the same time to oxidize sodium. And so assign your oxidation states, and then think about these definitions, and then you can assign oxidizing and reducing agents.