- Oxidation–reduction (redox) reactions
- Worked example: Using oxidation numbers to identify oxidation and reduction
- Balancing redox equations
- Dissolution and precipitation
- Precipitation reactions
- Double replacement reactions
- Single replacement reactions
- Molecular, complete ionic, and net ionic equations
- 2015 AP Chemistry free response 3a
By assigning oxidation numbers to the atoms of each element in a redox equation, we can determine which element is oxidized and which element is reduced during the reaction. In this video, we'll use this method to identify the oxidized and reduced elements in the reaction that occurs between I⁻ and MnO₄⁻ in basic solution. Created by Sal Khan.
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- After seeing Saraive7's question here, I am confused as to why MnO2 is not called manganite. And similarly, why is it carbon dioxide instead of carbonite? Since at least carbonite seems to exist, I can see that it has a -2 charge compared to the 0 charge for carbon dioxide; but I'm not able to understand the consistency with the naming...could someone explain?(3 votes)
- The "-ite" and "-ate" suffixes are reserved for oxyanions, or anions where another element is bonded to oxygen. MnO2 wouldn't be considered an oxyanion because it isn't an anion, it's a neutral compound. So it's just an oxide of manganese, or a neutral a compound where an element is bonded to oxygen. Now we can't just call it manganese oxide since there are several manganese oxides so we have to specify that it is manganese dioxide (1 manganese and 2 oxygens). Or manganese (IV) oxide where we specify the oxidation number of manganese which allows us to work out the number of oxygens.
The same reason goes for CO2 being carbon dioxide instead of carbonite, CO2 isn't an oxygen anion because it's neutral. So CO2 would be carbon dioxide, a neutral carbon oxide compound, and CO2^(2-) would be carbonite since it's an oxyanion.
Hope that helps.(9 votes)
- So I suppose ionic charge has the priority in determining oxidation number? Iodine has an ionic charge of 1- but isn't a free element supposed to have an oxidation number of 0?(4 votes)
- Oxidation number on a molecule must total to charge on the molecule. So, I-, which has a charge of -1, must have an oxidation number of -1, since there are no other atoms in the "molecule".(1 vote)
- How do you find the oxidation of the element?(2 votes)
- I think you mean the oxidation number of an atom.
KA has an article talking about this: https://www.khanacademy.org/science/ap-chemistry-beta/x2eef969c74e0d802:chemical-reactions/x2eef969c74e0d802:oxidation-reduction-redox-reactions/a/oxidation-number(2 votes)
- 3:16I thought the oxidation number for peroxides were -1 for oxygen. why did Sal write -2?(0 votes)
- What do I do if my redox reaction is 4O + 3Na5 + 3Mg?(1 vote)
- Where's the rest of your equation? Are those products, are they reactants? We don't know. We need to at least what our reactants and products are if we want to balanced redox reactions.(2 votes)
- when doing oxidation numbers does the number of atoms/molecules affect it ?(1 vote)
- Yes you have to conscious about the number of atoms of a particular element in a molecule when working out the oxidation numbers. If we consider water (H2O), the oxygen is going to have an oxidation number of -2. And for the oxidation numbers to add to 0 (since water is neutral) that must mean the hydrogen has an oxidation number of +2. But there's two hydrogens which should both get the same oxidation number, meaning that each individual hydrogen has to have an oxidation number of +1.
Hope that helps.(2 votes)
- [Instructor] What we have here is a reaction that involves iodine, manganese, oxygen, and hydrogen. And what we wanna 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. And 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. And 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 action 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, each iodine has a negative one charge. And 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. Now 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 and when we think about hypothetical charge with oxidation numbers, oxygen is going to have eight negative two oxidation number because it likes to hog those two extra electrons. And so if each of these four oxygens has a hypothetical charge of negative two, that would be negative eight total and 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 wanna review that one again because this is a little bit involved. We said oxygen, we're gonna go with the negative two 'cause it likes to hog two electrons. We have four of them. So if you add all that together, you're at negative eight and the whole ion has a negative one. So what plus a 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. And then let's look at the water. Well, water, both the hydrogen and oxygen, 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. 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 and 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 have an oxidation number of negative two. And so what would be manganese's oxidation number? Well, the compound is neutral. Two oxygens at negative two is gonna 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 and we can confirm that that makes sense. Negative two plus one is going to be negative one for each of these ions. So now, let's just think about who's been oxidized and who's been reduced. And 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 one to zero. So to go from an oxidation number of negative one to zero, you need to lose electrons. So it has been oxidized. 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.