- [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.