Current time:0:00Total duration:5:03
0 energy points
Assigning oxidation states for the decomposition of hydrogen peroxide, a disproportionation reaction. . Created by Jay.
Video transcript
In most redox reactions something is oxidized and something else is being reduced. But in some redox reactions, one substance can be both oxidized and reduced in the same reaction. And those are called disproportionation reactions. And here we have one the most famous examples, the decomposition of hydrogen peroxide. And so over here on the left is hydrogen peroxide, H202. Which when you add a catalyst like potassium iodide will turn into water and oxygen. So let's go ahead and assign some oxidation states or oxidation numbers in this reaction. And we'll start with the oxygen and the water molecule. And there are a couple ways to assign oxidation states. The simplest is just, of course, to memorize the rules that you'll see in any general chemistry textbook. And when you have oxygen in water, the oxidation state is negative two for oxygen. When you have O2, so over here on the right, the oxidation state of oxygen is zero. And over here on the left, hydrogen peroxide is one of those weird examples where oxygen has an oxidation state of negative one. So you could, of course, memorize these. Or there's, of course, a different way of figuring out the oxidation state, and that involves drawing out your dot structures and thinking about electronegativity. And so let's start with, once again, the water molecule here. And we know that these bonds consists of two electrons, right? So I'm going to go ahead and draw in these electrons here. We know that oxygen is more electronegative than hydrogen. So when you're thinking about it that way, you could think about those electrons going towards the oxygen. So you treat it like an ionic bond even though it isn't an ionic bond. And now you can see the oxygen's going to get all of those electrons. Oxygen normally has six valence electrons. So oxygen normally has 6 valence electrons. Here it's being surrounded by eight. So 6 minus 8 gives us negative 2. And so that is, of course, the same number that we figured out we memorized it. But thinking about the electronegativity helps you understand oxidation states a little bit more. Let's next do oxygen in the O2 molecule here. And so once again, we think about these bonds. Each bond consists of two electrons, and so we have a double bond in this case. All right. So in this case, we have oxygen bonded directly to another oxygen atom, and of course they both have this exact same electronegativity. And so we have to share all those electrons. And so since we have four total electrons, each oxygen is going to get two. And so we can go and divide up those electrons that way. So once again, oxygen normally has six electrons around it. It is now surrounded by six. So 6 minus 6 gives you 0. So an oxidation state of 0, which is, of course, exactly what we got up here. All right, let's go ahead and do the hydrogen peroxide example over here. So once again, we draw in our electrons in these bonds. So we have it look like that. And we consider the oxygen. Let's go ahead and do the oxygen on the left here. So oxygen versus hydrogen, oxygen is more electronegative. Oxygen gets those electrons. And then over here, the two electrons between those oxygen atoms, once again, an equal electronegativity, and so therefore each oxygen gets one. And so you can see that is our situation. So this time we have 6 minus-- and if you count up those electrons in there that I've circled, you'll get seven. So 6 minus 7 gives you an oxidation state of negative 1 for oxygen in hydrogen peroxide. And so either way, the memorization way is, of course, faster, but sometimes the dot structure way is very useful. So now we have our oxidation states and we can analyze this a little bit better. And you can see that we have a case where oxygen is on the left, has an oxidation state of negative 1. And let's say this oxygen atom turned into one of the O2 oxygen atoms. So it's going from an oxidation state of negative 1 to an oxidation state of 0. So you have oxygen going from negative 1 to 0. And that is an increase in the oxidation state. That's an increase in the oxidation state, therefore by definition, we know that oxygen being oxidized here. So this is an example of an oxidation. Once again, just look at the numbers. Negative 1 is increasing to 0, so that's oxidation. All right, let's think about what else is happening, right? So we could start out with an oxygen state of negative one. And let's say that oxygen this time is going to an oxidation state of negative two, like the oxygen and water. And so that, of course, is a reduction or a decrease in the oxidation state. You're going from negative 1 to negative 2. So a decrease in the oxidation state is, of course, reduction. So in this case, oxygen is being reduced. And so you have a substance that is being both oxidized and reduced in the same reaction. And so once again, that's called a disproportionation reaction.