Hello, this is a lecture topic video discussing oxidation reduction reactions. All right, so what is an oxidation reduction reaction? This is a reaction where what we're going to have is something, all right, one of our reactants losing an electron and another one gaining those electrons. Okay, so this is a reaction where we are transferring, all right, transferring electrons. Electrons are being transferred.
And so oxidation reduction reactions, these are also known as redox reactions. So this is usually how you'll hear me refer to it. So redox reactions, because oxidation reduction can be a mouthful at times, right? We're talking about the transferring of electrons. And these reactions go hand in hand with each other, okay?
So if you have an oxidation event, you also have to have a reduction event, all right? So that's, again, why we put this all in a single name. again called oxidation reduction reactions. And so what we want to do when we're talking about these types of reactions is essentially come up with some way to track where our electrons are going. Okay.
And in order to do that, all right, there's this concept, all right, this is merely a concept because it's not like we can have these molecular compounds again, like HF, right? That's a molecular compound. We're sharing electrons here.
Okay. So it's not like that fluorine, even though it's much more electronegative than hydrogen, it's not like it has a whole extra electron around it, all right? It's a partial charge, okay?
But what we can do is we can still assign an oxidation state, all right? And again, this is just a concept. It provides a way for us to keep track of electrons in a redox reaction according to some rules that we're going to talk about in just a second, okay? Again, these values, these... oxidation states or these oxidation numbers, all right, they do not necessarily represent a complete transfer of an electron, okay?
It's merely just for record keeping, okay? And so what we can do is we can start to think about assigning oxidation states to the different atoms within our compounds, both within our reactants and within our products, and essentially gauge to see how these oxidation states that we've assigned, all right, change from the reactant side to the product side, okay? And again, there's a set of rules for us to help us assign these oxidation states.
So the first rule is that an atom in an element is always zero, okay? So if you have an element in its purest form, all right, it doesn't have to be a single atom of an element, all right? It can be a bunch of atoms of that particular element, like, in a solid form, so like sodium, for example. in its solid form, right?
It's pure, all right? There's only sodium atoms around, okay? And there are many of them. But what we would do is we would assign an oxidation state, okay, or an oxidation number of zero to that, okay? And so it doesn't have to be just a solid, all right, with a simple single element.
It can also be compounds as well, all right, that contain only a single element. And again, this can make some sense, all right, because again, for example, O2, right, oxygen has a particular electronegativity, all right, but when you bind it to itself, it's not like one of those electrons or one of those atoms of oxygen wants the electrons more than the other, okay? So again, we can have something like a diatomic molecule like O2 or Cl2 or F2 or H2, it doesn't matter what it is. some compound that has a single element or atoms of a single element that make it up, all right, those also get oxidation numbers of zero.
It doesn't have to be a diatomic molecule. It could also be something like ozone, ozone gas, all right? Again, ozone is made up of three different oxygen atoms, okay? And again, each one of those oxygen atoms would be assigned an oxidation state of zero. Or we could even have...
single atoms of a particular element that are in the liquid form, all right? So like mercury liquid, all right? Again, that would also be assigned an oxidation state of zero, okay?
So it doesn't matter what physical state we're in, solid gas or liquid, okay? We can assign things, all right, that are of a pure element as having an oxidation state of zero. Okay. And so this is a really, really important rule because I think some of these rules should carry some sort of weight.
All right. Or some sort of priority. This one is essentially up at the top.
Okay. So there's no other because there are some exceptions. Now, we don't deal a lot with these exceptions in this class.
All right. But there are exceptions. And but generally, all right, there are some hard rules that we're going to follow.
And that's one of them. Okay. Now, monatomic. ions. All right.
So like any one of our group one or group two metals, right? Group ones love to form plus one cations, right? Group twos love to form plus two cations, right? They always, monatomic ions, their oxidation state is always the same as its charge.
Okay. So again, when we're dealing with monatomic cations of our group one metals, their oxidation state all right, is always going to be a plus one. And those of our group two metals is always going to be a plus two. This is again, another really hard rule. Okay.
And so again, we can carry this over to even some of our halogens. All right. So like chloride, for example, all right, minus one.
All right. Remember our chlorides, our chlorine, our halogens in general love to form these minus ones, right? They love to accept an electron. Because if they do accept a whole electron, they're going to have that electronic configuration of the noble gas that follows them.
So again, that's just another example of a monatomic anion. So like chloride or fluoride. All right. So the monatomic ions always adopt an oxidation state that is the charge on that monatomic ion. Now, fluorine is special.
Fluorine is always a minus one. in terms of its ability when it's in a compound, okay? So its oxidation state is always a minus one. Now there's one exception to that, all right?
And the only exception is if like we're dealing with F2, okay? Because again, here's a compound and it's made up of only a single type of element, all right? So of course this is going to get, all right?
an oxidation state assignment of zero, okay? But when fluorine is in any other compound, like hydrofluoric acid or phosphorus trifluoride, for example, all right, it's always going to be a minus one. And that, again, is another one of our hard rules. How about oxygen, okay? Oxygen is almost always going to be a minus two, all right?
We're gonna only see one possible exception in this class, and that's when we're... dealing with peroxide. So peroxide is O2 2-.
And in this case, each oxygen is going to have a minus 1. Each oxygen is going to have a minus 1. Other than that, again for this class, oxygen is always going to be assigned a minus 2, a minus 2 oxidation state, unless again it is in a peroxide, which is O2 2-. And in that case, Both of those oxygens have a minus one oxidation state. Now, hydrogen, okay? Again, hydrogen is always going to form a plus one when we're talking about our covalent compounds. Now, hydrogen can also, there's exceptions to this, all right?
Especially when we get something called a hydride, a metal hydride, okay? Hydrogen could adopt a minus one, but we're not going to see that in this class, all right? So for all intents and purposes, hydrogen for this class, all right, is going to have a plus one oxidation state. Now, why do we care about all this?
Well, with these kind of rules in place, what you'll see is that there are many elements that are not on this table. They're not represented on this table. But what we can do, hopefully, is use these rules, these hard rules, and we can use those to help us solve for oxidation states of elements that are not represented on this table. Okay.
Now, there's just a couple of other rules that we need to think about. All right. The sum of oxidation states always equals zero when we're talking about an electrically neutral compound. OK, so in other words, when we tally up all these oxidation states for a neutral compound, they should equal zero.
OK, and the other rule is the sum of the oxidation states of an ion. All right. Equals the overall charge of the ion.
Okay, so with those rules, let's try some practice questions. So what we want to do here. is we want to assign oxidation numbers of each element in the following compounds. All right, so we have XeO4, XeOF4, excuse me. So how are we going to go about doing this?
All right, well, oxygen was represented on our table. Remember, this is not a peroxide. This is not O2 2-. All right, it's just a single oxygen. So we said the oxygen is going to get a minus 2 oxidation state assigned to it.
We said fluorine, unless it's F2, all right, is fluorine is always going to be a minus one. And with that bit of information, we can essentially assign or we can determine what the oxidation state on that xenon is. All right. And so how do I like to do this?
Well, I find this to be really useful. I write out, OK, my compound formula, space it out a little bit so you have some room. Okay, and what I do is I make circles under each of the atoms, okay, and then I put squares underneath each.
And what we're going to do is we're going to put the oxidation states in the circles. And then what we're going to do, because we have to think about the overall charge on the molecule, we think about its total contribution of charge to the whole molecule. And again, why is that important? Well, if we go back one slide, what you can see is...
The sum of oxidation states has to be equal zero for an electrically neutral compound. And so this is what we have. We don't have any charge on that molecule.
So essentially, when we add up the sum of all these oxidation states, they should tally zero. So let's assign our oxidation states. Well, we said fluorine always gets a minus one.
So it has a minus one oxidation state. And what is the total contribution of charge from that particular set of atoms? So we got four fluorines.
They're each minus one. And so essentially, if we're thinking about this in terms of charge, again, we're massaging things here. We would say that we get a contribution of minus four. And you can see I put in the minus.
All right, just like if we have a plus. plus one or plus two oxidation state, we have to put in the plus. Pluses and minuses, all right, are absolutely critical when you're reporting oxidation numbers, of course, unless you get one that is assigned to zero, all right? So make sure you always put your plus or your minus when you're reporting oxidation numbers. Okay, so now how about the oxygen?
Well, oxygen, we said is minus two, okay? And we only have one of those. And so it gets a minus two in terms of its total contribution of the charge to the whole molecule.
All right. So now what can you do? Well, remember, this is representing our total charge. And so it should equal zero at the end of the day. So what should the charge on that xenon be?
All right. Or what contribution of charge should the xenon provide? All right.
to essentially cancel out what amounts to minus six. All right, well, in that case, it has to be a plus six. And there's only one xenon, which means that it also gets a plus six in terms of its oxidation state. So we can assign these oxidation states for each of these elements within this compound, all right?
Where xenon has a plus six oxidation state, oxygen has a minus two, and... each of those four fluorines has a minus one, okay? And again, the contribution to the total charge in this case is an electrically neutral molecule. All those oxidation states, when you take into account the number of each of those atoms, all right, the total charge, and I put charge in quotation marks here, should essentially tally to give you zero. Now, if you like to think about this in terms of math, that's fine too, all right?
Because what we can do, is we can assign our unknown variable in this case, this xenon. So we can say X, all right? And we can say we have one oxygen.
one oxygen, and we know it's got a minus two oxidation state. And then we can say we have fluorine has a minus one or four of those guys times minus one oxidation state. And again, those have to tally to be zero. Okay, again, we want all these oxidation states.
When we take into account the number of the different types of atoms that we have, that charge should essentially equal zero. And what we can do is we can do a little bit of math. So we can shorthand this or simplify this. Minus two plus minus four equals zero.
We can say x plus minus six. equals zero, and we can say x equals plus six. Again, it works out the exact same way, okay? I prefer this little diagram because it allows me to visualize it a little bit better. But you, if you want to do a little bit of math, all right, and not do the circles and the boxes, all right, that's fine too, okay?
But this is how I'm going to do it from here on out. All right, so now what's the next one? UO2 2 plus, okay? So we got a polyatomic cation.
So let's leave ourselves a little bit of space. UO22+. Now again, it's a polyatomic cation. So hopefully we see that we have some net charge at the end of the day after we take into account the number of the different atoms and their oxidation states. So I draw some circles here.
These again are my oxidation states, right? Uranium was not represented on those rules that we talked about on the previous slide. All right, so that's essentially going to be our unknown. All right, so again, if you want to think about math, this would be our X, all right? And then we put our boxes in for our total contribution to charge, and we start assigning our oxidation states.
We know oxygen, again, is minus two, unless it's a peroxide, or if it's just like diatomic oxygen or triatomic oxygen, whatever. All right, that's not this, okay? It is part of a compound, so it gets a minus 2, and we have two of them.
So we get a total contribution of charge of minus 4. Well, in this case, we have a polyatomic cation. So when we add these charges together, they should equal a plus 2. They should equal plus 2. And so this helps us essentially figure out. All right.
Figure out what the oxidation state is on the uranium. Well, in this case, we have a minus four from oxygen. So we can carry that over to the side here. And you can see that uranium has to contribute a plus six.
All right. Which also in this case, since there's only one of them, it also has a plus six oxidation state. OK. And again, so what.
what can we do? We can tally up the oxidation states, multiply those by the number of the different atoms, right? So uranium, again, a plus six oxidation state. There's only one of them, so we get a plus six, all right, contribution to total charge. Our oxygens, we have an oxidation state of minus two for each of those, and there are two of them, so we get a minus four total charge that's being contributed from those oxidation states of the oxygen.
We tally those together, and we get a plus two, and that plus two, again, matches the overall charge on that polyatomic cation. All right. So now those were fairly simple examples. All right. But we can also apply this to some compounds that are much, much larger.
All right. So let's do this. So we got sodium.
There's four sodiums. All right. And so what do we do?
Let's start assigning our oxidation states. Again, iron, we have no idea, right? Because iron is a transition metal. It can adopt lots of different oxidation states.
It likes to deal in multiple different numbers of electrons. Okay. So we can use all these other guys to help us figure this out.
And so we said our group ones, all right, like sodium is always going to form. plus one cations All right, so therefore, they're always going to get a plus one oxidation state. And what's the total contribution to charge? There's four of them, so we get a plus four.
Our oxygens, we said, are always going to be minus two. And our hydrogens, we always said, are going to be plus one. And then what you can see here is that we have six, okay, six oxygens, okay? And so we get a minus 12 contribution.
And we have also six hydrogens. So we get a plus six. And now the question is... well, what is the oxidation state of that iron?
Okay. Well, again, if you want to think about it as X, all right, we can put X here if you'd like. All right. And then when we essentially tally up all the total charge, okay, we should equal zero. And so what you can see is we get a minus six.
Right there, we can say plus x equals 0. And we can see that we have a x plus a minus 2 equals 0. And our x equals 2. Or you can just visualize and see what it takes to essentially make 0. Okay, up above, all right, we got minus 12 plus 6, that's minus 6, all right, and then we have minus 6 plus 4, all right, that's minus 2. For this to be 0, it has to be a plus 2, okay? So let's not forget our plus there, and the oxidation state of the iron would be plus 2. All right, so that's that question, all right? Now we have something...
That's a little bit more complex. All right. And this, again, is where it could be useful to remember your polyatomic cations and your polyatomic anions. So if you remember, ammonium is NH4 plus. OK.
So we can kind of break this guy up. And our sulfate. is two minus, all right? Because again, solving for the cerium, all right?
This is one of our first lanthanides, okay? Can adopt multiple different oxidation states or charges, okay, depending on how you like to think about it. And so this would be our X, but sulfur is also not in our table, okay? So breaking these guys up and recognizing these polyatomic anions can help you essentially solve for... these unknowns.
All right. So let's do our sulfate first. And if you look here, we said our oxygens are always minus two. So let's put a minus two here.
And we have four of them. So if we're doing our little boxes over here. just our kind of isolating that sulfate, we get a minus two.
We got four of them, so we get minus eight. And our total charge on the sulfate has to be a two minus, right? And so what do we need to do that? That means our sulfur should be a plus six total contribution to our charge, which also means it's a plus six in terms of its oxidation state.
So we can fill that in down here below. And again, if you want to just tally these up, you can see that we get a minus two overall charge, which matches that on our polyatomic anion. So we can do plus six.
And we can do our total contribution of charge here. So there are three groups of this. So we got minus two.
For oxygen, there's four of them, so that's minus eight. And then there are three groups of the sulfate, so we take eight times three, so it's minus 24. Minus 24. And our sulfurs were each a plus six, and there are three of them. Okay, three of them. So that's a plus 18, plus 18. And now we can do our ammoniums over here. We said that hydrogens are always plus one, and there are four of them.
So we get a plus four. And nitrogen was not on that table, but this can help us figure it out. All right.
So if we get this plus four coming from our hydrogens and our ammonium cation has a plus one charge. Right. That means that the nitrogen has to have a minus three.
There's only one of them. Okay. So we can put in our plus one for our hydrogen and we can put in our minus three for the nitrogen.
And then what you can see is that there are two nitrogens, all right, because we got this subscript out here. And so we get a minus six. I'm going to have to shift over here. It's not letting me write. And we can do minus 3, plus 1, plus 6, minus 2. And then we can say plus 18, total charge contribution, minus 24. And we get minus 6 over here.
And we get a plus 1 times 4, so a plus 8. And so... Because again, the question is, what's the cerium? And so what you can see is we have a plus two from that.
And we have a minus six over here. And plus two minus six is minus four. And we want something that's neutral.
There's no charge on this compound. All right, so the oxidation state of the cerium should be a plus four, and it would contribute a plus four to the overall charge. Okay, so again, what you can do is get familiarized, all right, or familiarize yourself with those rules for assigning the oxidation states, which essentially cover only a fraction of the elements on the periodic table.
But with those rules, All right. And a little bit of math. OK, you can figure out the oxidation states on any other element that falls within a given compound. So here's a participation question or three for you guys. First one is what is the oxidation number of manganese in KMN04?
So that's potassium permanganate. All right. So, again, go back to those rules.
All right. Manganese in this case would be like our unknown because we have rules that cover potassium and the oxygen. OK.
And therefore, you can use those to figure out what the oxidation state or the oxidation number is of the manganese. Second one, what is the oxidation number of carbon and the oxalate ion? OK, so C2O4, 2 minus. Again, we have a rule.
For the oxygen, this is not peroxide, okay? So you can use that to help you figure out what the oxidation state or the oxidation number is on the carbon. And then the question is, what is the oxidation number of the nickel in K2, Ni, C, and 4? Now, I didn't say the name here because that would give away the oxidation number of the nickel, all right?
But again, you can go back, you can follow those rules, okay? And again, you can then figure out. what the oxidation state is on that nickel. Okay.
So let's talk about the characteristics of a redox reaction. All right. And so here's a good example of one.
All right. Two sodium solid plus two or plus chlorine gas, a Cl2 gas gives us two NaCl. All right. And As you see here, and I'll show you guys in just a second, right, this is an oxidation reduction reaction.
There is something being oxidized, all right, and subsequently there is also something being reduced, all right? So let's figure out what that means, all right? What does oxidation mean?
What does reduced mean? So the first step is let's go ahead and sign our oxidation states. So remember, anything in its pure form, all right, so like a bunch of sodiums within a metal or diatomic chlorine. Okay.
These both have oxidation states of zero. Okay. So that's that first rule.
So go back to that table and take a look. And then what's going on over here on the other side, on our product side? Well, what you can see now is that we have a compound that's been made. This is an ionic solid.
All right. And we have to now assign oxidation states. Well, the chlorine, again, it's a halogen.
Minus one. Okay. And the sodium is a plus one.
All right. So group one metal group, one metals love to form these plus one cations. All right. Just like the halogens love to form these minus one anions. All right.
And remember those charges of those monatomic ions equal the oxidation state. Okay. So what's going on here?
Like in terms of like our redox chemistry. All right. So what you can think about is electrons being transferred.
Right. That's what these charges or these oxidation states are a result of. OK, so we always say when you see an anion, a minus one anion. All right. Why is it a minus one?
Because it's essentially gained an extra electron. OK, what do we say when we have a plus charge on a polyatomic cation? All right.
Or on a monovalent cation? Doesn't matter what it is. We would say that it. donated an electron, all right? It's lost an electron.
And so what we can do is we can start to assign a word, all right, to what these definitions are, okay? Or what these processes are. Like losing an electron, okay?
What does it mean to lose an electron? Well, there's a word for that, okay? That's oxidation. Oxidation is the loss of an electron. And again, when you lose an electron, what have you done?
You've lost a negative. point charge, right? And so essentially your oxidation state, all right, or your charge essentially becomes positive, okay?
And so what you see is an increase in oxidation state when something gets oxidizes, okay? And that should make some sense, all right? Things become more positive if you're losing something that's carrying negative character, like an electron, right? So oxidation is loss of an electron. And so if we're looking up here at this reaction, all right, what's essentially being oxidized?
What has lost an electron? Okay. And what you can see here is that sodium has gone from an oxidation state of zero to now on the product side having a plus one. All right. And so you can see that it has lost an electron.
Okay. And if you go all the way back to the front of the lecture topic video, I told you that these things come in pairs. For every oxidation that happens, there should be a reduction that happens, all right? For every oxidation that happens, something has lost an electron or donated an electron.
Therefore, there should be something that essentially gains or accepts an electron, and that is called reduction. Reduction is gaining of electrons, okay? And if we're gaining electrons, you should see a... decrease in the oxidation state. We should be approaching more negative values.
Okay. Because what are we doing? We're gaining something that carries along with it negative character. Okay.
And so in this case, you can see that the diatomic chlorine gas, each of those chlorines carries an oxidation state of zero. Okay. It's a compound made up of a single type of element.
All right. They all get an oxidation state of zero. And on the product side, they are now each a minus. one.
Okay. So they have gained that electron that the sodium has lost or that the sodium has essentially donated. Okay. So again, how are we going to remember this?
All right. There's a couple of mnemonics that you can use, and I don't care which one you guys use one. Uh, you probably already know relative to the other. So, uh, the one I like to think about is oil rig.
All right. Oil rig. What does oil rig?
mean? Why am I bringing this up? Well, OilRig stands for Oxidation is losing electrons, okay?
Whereas RIG means reduction is gaining electrons, okay? So that's a mnemonic to help you remember what oxidation and reduction are. Oxidation is losing electrons.
Reduction is... gaining electrons. Okay. How about this other one? All right.
Again, it's getting at the exact same thing, but this time it's Leo the lion. So Leo the lion goes grr. All right.
So what does Leo mean? L-E-O. So L, this is losing electrons. is oxidation.
Okay, so that's what the LEO stands for. And GER is gaining electrons, is reduction. Okay, gaining electrons. is reduction. That's the GER.
And I'll just underline these guys over here too, just so you can see these. So it doesn't matter which pneumonic you use. All right. They both get at the exact same thing. Oxidation is losing.
Reduction is gaining. That's again, oil rig. Or losing electrons is oxidation.
Gaining electrons is reduction. And again, remember our friend Leo the lion, because Leo the lion goes GER. Okay. Now, What we can further do is classify, all right, what we would call an oxidizing agent and a reducing agent, okay?
Now, what are these guys, okay? Because I find that this is probably the trickiest thing, all right, because it's easy to get these confused, all right? But an oxidizing agent, all right, this is a reactant that accepts electrons from another reactant, okay? That is an oxidizing agent. So if we're thinking about this, this chlorine gas, those chlorines, all right, what are they doing?
They are forcing the sodiums, all right, to oxidize, all right? They're forcing those sodiums to give up their electrons, okay? So they're forcing those sodiums to oxidize, to give up those electrons, all right? And in the process, okay, those chlorines… get reduced, okay?
So that's the oxidizing agent. The oxidizing agent, all right, forces the oxidation of something else, and in the process of that oxidation, gets reduced. Remember, oxidation reduction reactions come hand in hand. You can't have one without the other, all right? So our oxidizing agent in this case, okay, It's getting reduced.
It's forcing the hand, essentially, of the sodium, having them give up their electrons. So it's forcing the oxidation of those sodiums. And in the process, the chlorines are accepting those electrons. So this would be our oxidizing reagent.
Now, on the other hand, we also have a reducing agent. All right. This is a reactant that.
donates electrons to another substance to reduce the oxidation state of one of its atoms, okay? So you can, again, you can, how can you think about this? You can think about this sodium essentially strong-arming this chlorine, those chlorines, to reduce or gain in electrons, all right? And in the process of doing that, all right, what has happened to it, okay? It has given up its electrons.
It's been... oxidized. Okay. So the reducing agent, all right, is forcing the reduction of something else.
Okay. It's forcing the reduction of something else. And in the process, it gets oxidized. Okay. And again, how does that manifest itself?
The reducing agent essentially gets an increase in its oxidation state. All right. And that should make some sense again, because when you oxidize, all right, you are getting rid of something.
or donating something that is negative in character, that's our electrons, therefore you should go up to higher or more positive oxidation states. So in this case, our sodium would be our reducing agent. And again, so really take some time to kind of wrap your brain around the oxidizing agent and the reducing agent. Because again, I do find that these are probably the most confusing parts for students or confusing topics for students. So it does lend itself to getting a little bit more attention.
All right. So here's a participation question. All right. These are unbalanced reactions and that's fine.
We're not going to have you guys balance these redox reactions. Okay. That's something you'll do next semester in Chem 120. All right.
But for the unbalanced reaction below, which element is actually being oxidized? Okay. So what you can do is use those rules, go through for both your reactants and your products. and determine the oxidation states.
Again, some of these things you may not be able to figure out, like the manganese and the permanganate anion. But you can use the oxidation state of the oxygens to determine the oxidation state on that manganese. So go through both reactants and products and assign your oxidation states.
And then figure out which element is being oxidized. Okay. And so again, go back, review, what does it mean to be oxidized?
Okay. And then for the second question, all right, for the unbalanced reaction below, what is the oxidizing agent? Okay.
What is the oxidizing agent? Okay. So again, what you need to do is go to your reactants and your products, make sure you can assign all the oxidation states. All right.
Figure out what's being reduced, what's being oxidized. Remember, they go hand in hand. If you don't see one or the other, there's issues.
You need to go back to the drawing board and reassign your oxidation states. All right. And then remember what it means to be an oxidizing agent. Okay. All right.
Again, that's a reactant that accepts electrons from another reactant. All right. So it's something, okay, that's getting reduced. in the process of oxidizing something, all right?
So it promotes the oxidation of something else and in the process gets reduced, okay? Now, one last thing I wanna say about oxidizing agents and reducing agents, they will never be products, okay? Remember, the chemistry happens before the products are formed, all right? So oxidizing agents and reducing agents. Your choices for assigning these guys is always your reactants.
It's always the reactants. They will never be your products. Okay. All right.
Awesome. So take a look at this. If you guys have questions, don't hesitate to reach out. I hope you guys have a great rest of your afternoon and I'll be seeing you soon.