Lecture 16. Electrochemistry Pt. 1.
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Arasasingham, Ramesh D.
 UCI Chem 1C is the third and final quarter of General Chemistry series and covers the following topics: equilibria, aqueous acid-base equilibria, solubility equilibria, oxidation reduction reactions, electrochemistry; kinetics; special topics. 
 Prior to looking at electrochemistry, we want to understand what it means when we say a reaction is an oxidation reduction reaction. So you guys remember we said virtually all reactions can be categorized into three groups. The first group are the acid-base reactions and we've looked at that in the past. We know acid-base reactions involve proton transfer. So if you look at a balanced equation and you look at two reactants and you see that a proton is being transferred from one to the other, then you know that there is a proton donor, that's the acid, and the substance that accepts that proton is the proton acceptor. That will be the base and what you end up with is an acid-base reaction. The second class of reactions that we looked at were the precipitation reactions and we said we used solubility rules to help us and if you start with two substance that are completely soluble and you combine them and you end up with an insoluble substance, that class of reactions are called the precipitation reactions. And then we said the third class of reactions are the oxidation reduction reactions or redox reactions. They meet the same thing. The abbreviated form of the oxidation reduction reaction is called redox. And these involve electron transfer. So in these reactions, one substance gives off electrons and the other substance picks up that electron and we end up by having electrons being transferred. One or more electrons are transferred and that is an oxidation reduction reaction. And as all of you can recognize, unlike the previous two where you can look at the reactants, like in proton transfer, you can see that the proton is one substance and then it moves to another, by looking at the balanced equation, it's easy to figure out whether it's an acid-base reaction or a precipitation reaction. In the case of redox reactions, just by looking at a balanced equation, it's hard for us to figure out whether electrons are being transferred because the balanced equation does not have electrons. Is that clear to everybody? So it's very difficult to look at the equation itself and figure out whether one substance is losing electrons and whether the other substance is picking up electrons. So we have to have a system of keeping track of electrons. We call that electron counting or electron bookkeeping. And so if we're going to keep track of electrons, then we have to have an easy system or set of rules that we apply to assist us in figuring out the number of electrons that are being transferred. And the system that we've come up with is called oxidation numbers. So one way to express the number of electrons lost or gained by a substance in a redox reaction is by assigning an oxidation number to each atom. So this is much like a game. You know, when you play games, you set up rules. And as long as you stick to the rules, everything kind of works out well. It's the same thing here. We're going to come up with a system of rules. And as long as we stick to those rules, then we can figure out whether electrons are being lost or gained or there's any change in the number of electrons. We're going to look at every atom in a substance individually to figure out whether there's a change in the number of electrons. So that's what an oxidation number is. An oxidation number is a number. And this number can be zero, positive, or negative, assigned to each atom in any substance in an oxidation reduction reaction that reflects whether the atom has fewer. And if it has fewer electrons than it started, it has a positive sign. It has more. The oxidation number will have a negative sign. Or the same, that will be zero as the number of electrons as an uncombined neutral atom. So this is the oxidation number is just a way for us to keep track of electrons. And so an oxidation number can be positive, negative, or zero. And we use Arabic numbers. The Arabic numbers are the one, two, three that we usually use. And the online homework, it'll say, give your answer in Arabic numbers. And that means that's what we're using. Later on, you'll see that we can also assign oxidation states, and for that we use Roman numerals to distinguish between the two. So right now we're looking at oxidation numbers, and we're going to use Arabic numbers. So if you look at the class website, I've put the rules up on the class website. And these are the rules for assigning oxidation numbers. Before I start this, usually like solubility rules, in KA values, stuff like that I will provide on the exam. But this I will not give to you, because this is something that you need to understand, internalize, and be able to use it intuitively. And so these rules, you need to understand them and be able to apply them. And these rules will not be provided on an exam. So let's start with the first one. An oxidation number of an element in its natural state is zero. So if you take sodium, in its natural state, it's metallic sodium, sodium metal. So it will have zero. If you take metallic silver, that's how the element exists in its natural state. It'll be zero. If you take mercury, the liquid state of the mercury, as all of you know, is the natural state, that will have an oxidation number of zero. Oxidation in its elemental state exists as O2, hydrogen exists as H2, chlorine exists as Cl2. So as long as it exists in its elemental state, it has an oxidation state of zero. The second is the oxidation number of a monoatomic ion is the same as its ionic charge. So if it's a cation or anion, it's monoatomic cation or anion, then its oxidation number corresponds to the charge on it. So Na plus would be plus one. Cl minus would be negative one. Fe2 plus would be plus two. F minus will be negative one. So the oxidation number corresponds to the charge. The third is when we deal with oxygen. In most compounds containing oxygen, the oxidation number of oxygen is negative two. So if it's O2, oxygen in its elemental state will be zero. But if oxygen is found in a substance, then the common state is that it has a minus two oxidation number. So if you take magnesium oxide, it has negative two. SO2, it has negative two. SO3, negative two. CO, CO2, any formula where the common form is where it has an oxidation number of negative two. The only exception is the peroxides. The peroxides are pretty rare. And a peroxide is O2, two minus. So there are two oxygens, and overall it has a two minus charge. So if you take, if there are two oxygens that have an overall two minus charge, what is the charge on each oxygen? Minus one. And that's why if you take the exception to the rule of the peroxides in which the oxygen addition number has a minus one. How do you tell whether it's a peroxide? Look at the formula, if it's like H2O2, or if you have a metal like Na2O2, all right? Look at the charge. If you take Na2O2, you can see there are two Na pluses. So that means overall O2 must be two minus. Therefore each oxygen has to be minus one. But peroxides are not very common. The most common ones you can encounter are the ones that have minus two. The fourth rule is that in most compounds containing hydrogen, the oxidation number of hydrogen is plus one. So if you take HCl, NH3, H2O, CH4, these all have plus one. The exception to this rule are the metal hydrides in which the hydrogen has an oxidation number of negative one. So now if you have like lithium, sodium, potassium, a metal with a hydrogen, then it's a metal hydride. All right, so if you have a metal attached to hydrogen, it would be a metal hydride, and its oxidation number would be negative one. Now last is that the sum of the oxidation numbers must be zero for electrically neutral compounds. For polyatomic ions, the sum of the oxidation numbers must be equal to the charge of the ion. Remember we're assigning oxidation numbers for every atom in a substance. So if you add up the oxidation numbers, if it's electrically neutral compound like water has no charge on it, that means when you add up all the oxidation numbers, if you add up the oxidation numbers for the two hydrogens and the oxidation number for that one oxygen and add them together, it needs to add up to zero. If it is charged, it needs to correspond to the charge. So if you take chromium, the oxidation number of chromium plus the oxidation number corresponding to four oxygens, if you add them up, they have to equal negative two. And if you take NH4, if you take nitrogen and add the oxidation number of this to the oxidation number of four of these, you should end up with a plus one oxidation state, or number. Is that clear to everybody? So now let's apply this because we're going to use this often to figure out whether a compound is losing electrons or gaining electrons. So we'll start by taking an example where you're asked to calculate the oxidation number. So we'll start with a simple one. Calculate the oxidation number for carbon in CO2. So your starting point would be to take the compound and we know that overall there's an electroneutral compound, so the sum of the oxidation numbers corresponding to each of the atoms should add up to zero. So since we need to find the oxidation number for carbon, we're going to call that N. And we have two oxygens, so two times. What does a rule tell us? What is the oxidation number for oxygen? Negative two. All right? So this will be negative two. So we know the sum of these two should correspond to the overall charge of the molecule, so we know N plus two minus two should equal zero. Therefore N equals plus four. So the oxidation number that corresponds to that carbon will be four. We always assign oxidation numbers per one atom. All right? So we know that even though there are two oxygens there, each oxygen individually has an oxidation number of negative two and because you have two of those, you're going to multiply it by two. Okay? So let's take another example like H3PO4. So let's say calculate the oxidation number for phosphorus in H3PO4. So once again, we have H3PO4. The one that we need to figure out, we'll call it N. We have hydrogen and we have oxygen. All right? We have three hydrogens and what does the rule tell us that the oxidation number for hydrogen is plus one. All right? This will be there four oxygens and each oxygen we know from the rules will have negative two. So that means the sum of all of this should equal the overall charge. So we know that three times plus one plus N plus four times negative two should equal zero. Therefore N should equal eight minus three which is plus five. All right? So the oxidation number for phosphorus in that compound is plus five. Now after all I need you to be able to do this kind of mentally in your head and quickly calculate these but for now we'll go through the details. Okay? Now let's take the third example. Calculate the oxidation number for chlorine in ClO3 minus. Okay? So once again, if you take this formula, this is what we need to find. So we're going to call that N. We can see that the other component is three so we know that each oxygen is negative two. There are three of those so that will be negative two. But now you can see there's a polyatomic ion that has a charge of negative one. So that means the sum of these two should equal negative one. So that means N plus three times negative two should equal negative one. Or N equals six minus one which is plus five. All right? And so once again you can see that when you assign oxidation numbers you have to put signs. It either has to have a negative sign in front if it's a number, a positive sign or it has to be zero. Okay? Now lastly let's take something like thiosulfate which is S2O3. So calculate the oxidation number for sulfur in S2O3 two minus. All right? So now remember oxidation numbers are always assigned per one atom. So there are two sulfurs. So we're going to say two N. All right? And there are three oxygens and we know from the rule that oxygen has a negative two oxidation number. So the sum of these two should equal two minus. All right? So we know two N plus three times negative two should equal negative two which is the overall charge. Therefore two N equals six minus two which is plus four. And then we always assign oxidation numbers per individual atom. So each sulfur atom will have plus two. Divide that by two. Each atom has plus two. Got it? So for any formula you should be able to figure out what the oxidation number for all the elements are. Usually most compounds we deal with have hydrogen or a-to-oxygen on it so we use those elements to figure out what the oxidation number for the other one is. If you have cations or anions you know that they correspond to the charge. So monotomic and monotomic cations or monotomic anions will correspond to the charge as well. Now let's apply this in an example where we're looking at a redox reaction. So if I take a redox reaction and we said a redox reaction involves electron transfer. So let's just take an example where we have two reactants and let's say in this reaction a transfers one electron to b. Alright so if a transfers one electron to b then what you end up with is a has lost electrons so it would be a plus plus b minus. Alright? So you have, so can everybody see that you can't directly see the electrons being transferred so I'm telling you that one electron is being transferred. Alright? Now whenever you look at redox reactions there's electron transfer and a convenient way to kind of look at what's going on is to look at what we call half reactions. Alright? So if I take half reactions you can see that a goes to a plus plus an electron. So in this case when you write half reactions you can see that when a goes to a plus it's losing one electron. Okay? Whenever you write balanced equations remember you have to maintain charge balance. So to begin with on the a side everything is electrically neutral. You end up with a plus so to balance that charge you can figure out how many electrons you need to balance that charge. So you can see a plus just needs one electron to balance that charge and you know one electron has been given up. Alright? Now on the other, so we call this the oxidation half reaction. Remember why is this the oxidation half reaction because oxidation is electron loss. And last time we looked at the definition of oxidation, oxidation involves the loss of electrons. When a substance loses electrons we call that oxidation. When a substance picks up electrons we call that a reduction. So now oxidation reduction reactions we said last time have to go hand in hand. You can't have one without the other. If something gives off electrons something has to pick it up. So you need both of those components. Alright? You can't have one substance giving out electrons and those electrons are sitting out there in thin air. Alright? You can never have electrons just sitting in thin air. If one is giving it off another substance needs to take it and they need to take place at the same time. And so if you look at the reduction half reaction, now this electron that's given off is picked up by B and it goes to B minus. Once again charge balance. You have neutral minus one. So on this side I have a minus one charge. If on this side I have to have a minus one charge as well. This is called the reduction half reaction. Why? This is where electron gain occurs. Alright? Now I want to remind you that even though we write half reactions because they're convenient for us to kind of rationalize what's going on, in reality reactions don't take place in half reactions. Do you understand that? Because the instant the electron is given off the other is taken up. And so you don't have this process where one electron comes out and the other electron is then picked up. It's immediately transferred from one to the other. So in oxidation reduction reactions they take place instantaneously and you don't have half reactions occurring. But half reactions is purely a mental process for us to kind of separate it out to understand what is going on. So if I want to make a little table I can say that if I take A, so let's combine, let's look at what happens at A and what happens at B. Remember we have to look at the reactants. Alright? So you can see A loses electrons. In other words it's losing one electron, one or more electrons. So A loses electrons, B gains electrons. Alright? A undergoes oxidation or it is oxidized. And we say B undergoes reduction or it is reduced. Alright? Now last time we looked at, when you look at redox reactions you need to be able to identify what is undergoing oxidation or what is being oxidized, what is undergoing reduction, what is being reduced. Alright? So the next is that you need to know which is the agent. So since this is undergoing oxidation this is the, A is the reducing agent. This is the agent that's responsible for reducing this substance which is undergoing a reduction. And B is a oxidizing agent. Alright? Whenever a substance undergoes oxidation the oxidation number increases. So the oxidation number of A increases. Alright? So the oxidation number of A increases. And if it's undergoing a reduction the oxidation number of B decreases. So these are important points to remember. If you're looking at an oxidation reduction reaction I need you to be able to look at each one of these and be able to identify for each reactant you need to know which one is losing electrons. You need to figure out which is gaining electrons. You need to figure out which one is being oxidized, which is being reduced. You need to figure out which is the oxidizing age and which is a reducing agent. Whenever you have an oxidation the oxidation number increases, alright? When a substance loses electrons its oxidation number increases. When a substance gains electrons its oxidation number decreases. So in an oxidation reduction reaction if the oxidation number changes you know that its oxidation reduction reaction. One substance oxidation number will go up. That means it's oxidized. And another substance oxidation number will decrease. That's the one that's being reduced. Okay, so let's take an example. And we'll start with some simple examples and then we'll go to more complicated ones. So let's take some examples. And the first example I'm going to take is where you have potassium solid. And so this is metallic potassium reacting with chlorine gas to give you an ionic compound which is potassium chloride. Okay? So quickly check where this equation is balanced. You have two potassiums, two potassiums, you have two chlorines, two chlorines. Okay? Now for you to figure out whether it's a redox reaction or not, we have to look at oxidation numbers. Okay? So to begin with we always look at it per one atom. Alright? So in this there are two potassiums in this balanced equation. And so if I take potassium this is an element in its natural state. Therefore its oxidation number has to be zero. And there are two coefficients. I'm going to put the coefficients outside. Okay? Over here chlorine is, there are two chlorine atoms. But this is an element in its natural state. Therefore the oxidation state of the two chlorine atoms in Cl2 each one will have an oxidation number of zero. Okay? So there will be two times zero. Now on this side we end up with an ionic compound. Alright? This is a metal, non-metal combination. If it's an ionic compound we have to have a cation and an anion. So if I take the potassium you can see that that's K plus. Therefore each potassium will have a plus one and each chlorine will have a minus one. So what's outside is the coefficients in the balanced equation or the number of atoms that's there in the balanced equation. Okay? So now we have to look at this equation and we want to figure out does the oxidation numbers change? So if you look at potassium each potassium atom goes from zero to plus one. So is there a change? Yes. Now you ask yourself, so that's got to be a redox reaction. Is the oxidation number increasing or decreasing? So if you take potassium it goes from zero to plus one. So it increases. Remember if the oxidation number increases what is that? Oxidation. Alright? So we start looking at the two reactants and we're going to say that when this goes to that oxidation number increases and we know that's an oxidation. And by how much does it change? It goes, each one goes from zero to negative one. So that means that if there's a change by one, alright, it goes from zero to plus one rather. If there's a change of one that means one electron is being transferred. But there are two atoms so overall you can see two electrons are being transferred. So overall two electrons are transferred. As each potassium atom goes from an oxidation number of zero to plus one there's a net change of one. The net change of one means that one electron is being lost. But overall we have two so you multiply it by two and so in this reaction two electrons are being lost. Alright? Now if you go to what's being reduced over here, now your chlorine, if you look at the oxidation number goes from zero to negative one. So you know that oxidation number decreases. And if it decreases it means it's a reduction. And if you take each chlorine atom it changes by one. Okay? So since electrons are being lost I'm going to put a negative sign here to show me that electrons are being lost. Alright? A negative sign means electrons are lost. Positive signs means electrons are being gained. So if you look at this each chlorine atom goes from zero to negative one. So it's a net change of one so it decreases by one. So I'm going to say one electron but since there are two atoms involved all together two electrons are being lost. Okay? So we figured out from this by looking at oxidation numbers that two electrons are lost and two electrons are gained. Alright? Now the cardinal rule is that the number of electrons given off have to equal the number of electrons taken up. If that doesn't work out then you know there's something wrong with your rationale. Do you understand that? Because you can't have ten electrons given off and only five taken up. Alright? The number of electrons given off always have to equal the number of electrons taken up. Alright? And you can see two electrons are given off, two electrons are picked up. Now, so what is undergoing oxidation? Potassium, this is oxidized. You can see chlorine is reduced. Alright? So the substance being oxidized is potassium because it's undergoing oxidation. The substance being reduced is chlorine because its oxidation number goes down. So that's the substance being reduced. If this is being oxidized this is the reducing agent and this is the oxidizing agent and we know overall two electrons are transferred. Alright? So it's very, very important that every example that you look at you have to be able to go through this process. You've got to figure out what's being oxidized, what's being reduced, how many electrons are being transferred and you make use of oxidation numbers to do that. Okay? So let's take another example. So we're going to look at the combustion of methane. So what happens if you burn methane in oxygen? Now a combustion reaction is also an oxidation reduction reaction. So if I take methane plus 202 gas giving me carbon dioxide plus 2, okay? So we're looking at the combustion of methane. Make sure the equation is balanced. You can see there's one carbon, one carbon, four hydrogens, four hydrogens, four oxygens, two plus two, four oxygens. Okay? So I'm going to tell you ahead of time that this is a redox reaction. Okay? So if you take a redox reaction and you've got to figure out what's being oxidized, what's being reduced, how many electrons are being transferred. Okay? So for carbon now, by now you should be able to do this really fast. The hydrogens each have plus one. There are four hydrogens with plus one. So what's the oxidation number of carbon? Negative four. Okay? If you take hydrogen, there are four times plus one. So that will be the oxidation number assigned to each of the hydrogens. Each oxygen, can you see that there are two oxygens here times two? There are four oxygens. It's an element in its natural state. Each oxygen atom will be zero. Okay? Now let's turn to this carbon. To figure out what the oxidation state of this carbon is, we turn to oxygen. Remember, each oxygen atom has, what is the oxidation number? Negative two. So there's negative two times two gives me negative four. So if oxygen is negative four, the whole compound is electroneutral. What is the oxidation number of carbon over here? Plus four. Okay? And we know there are two oxygens. That's going to be negative two. Now over here we have hydrogen. Each hydrogen is plus one. How many hydrogens do I have? Two times two, four. And then if I take the oxygen, there are two oxygens and I know that's negative two. Can everybody see I have assigned oxidation numbers for all of the atoms? Now I look at one atom at a time and figure out which ones are undergoing changes. So we start with carbon. Carbon starts from negative four to plus four. Alright? So that's one. Now does hydrogen change? The hydrogen atoms here are plus one. The hydrogen atoms here are plus one. So hydrogen does not change. Let's look at oxygen. Oxygen goes from zero to negative two over here and negative two over there. Do you see that? So that means oxygen is changing and it's changing the both sets of oxygens undergoing a change. Okay? So let's start with carbon. Now look at the oxidation number for carbon. Does that increase or decrease? It increases. It goes from negative four to plus four. So since the oxidation number increases, what do I know? This is the oxidation and methane is being oxidized. Alright? Now look at by how many does it change? Remember there's one carbon atom at the beginning, there's one carbon at the end. So if I go from negative four to plus four, what is the change? Eight. So that means there are eight electrons involved. Alright? And so this is oxidation, electrons are being lost so we know that there are eight electrons involved. So overall, eight electrons are being transferred. Okay? Now let's go to the oxygen side. It goes from zero to negative two. Does the oxidation number increase or decrease? It decreases. Remember if the oxidation number decreases, we call that process reduction. Okay? Now let's look at how many oxygen atoms there are. There is two, two times two, four oxygen atoms. There's two plus two, four oxygen atoms. So there are four oxygen atoms and each oxygen atom goes from zero to negative two. So what is the change? Two electrons. If it's a reduction, our electrons picked up or electrons lost, picked up. So you know that there are times plus two, which gives me plus eight electrons. Now how many electrons are being transferred here? Eight electrons. So in this oxidation reduction reactions you can see that the substance that's undergoing oxidation is methane. So that's being oxidized. The substance being reduced over here is oxygen. Alright? So if this is being oxidized, this would be the reducing agent. If this is being reduced, that would be the oxidizing agent. And how many electrons are transferred? Eight electrons are transferred. Okay? I need you guys to be able to do this in your sleep, okay? Because if you can't figure out how many electrons are being transferred, then the whole exercise is a waste of time. Do you get it? So, and later on when we're going to look at electrochemistry, I need you to be able to just look at a reaction, figure it out, and know how many electrons are being transferred. Okay? Now let's take another example where, let's say I have SICL4 plus, let's say that's liquid, plus 2H2O liquid, gives me 4HCl, Equus, plus SiO2 solid. Okay? So once again, let's take a look at this. So now we can go pretty fast. There are four chlorines, each chlorine has minus one, so what is the oxidation state of silicon? Plus four. There are four chlorides, and each one is negative one, because remember chlorine, Cl minus has a negative one oxidation number. Here, there are four hydrogens, each hydrogen is plus one, there are two oxygens, each oxygen will be negative two. Alright? Now on the product side, we have four hydrogens, so it will be four times plus one, there are four chlorides, four times negative one. If I take the silicon over here, you can see it's plus four, this is two times negative two. Now I'm starting to move pretty fast, because that's sort of the pace at which you need to be able to analyze these reactions. Okay? Now look at this and tell me, is anything undergoing any change? Let's tell it with oxygen. So look at this plus four, the oxidation number is plus four, at the end of the reaction it's plus four. Does that change? No. Let's look at chlorine. Each chlorine atom to begin with is negative one, on the product side each chlorine is negative one. Does that change? No. Now let's turn to water. Each hydrogen is plus one, on the product side the hydrogen is plus one. If I turn to oxygen, each oxygen is negative two, on the product side each oxygen is negative two. So do oxidation numbers change here? No, there are no changes. So is this a redox reaction? No. Okay, there's no electron transfer. Alright? So what kind of reaction is this? Remember it has to be their acid base, precipitation or oxidation reduction reaction. Look at the hydrogen atom. Do you see hydrogen atoms moving from one to the other? Where does hydrogen start with water? Is the hydrogen moving away from water? Is it staying together with water? So can you see there's a proton transfer, there are proton transfers taking place. If it's a proton transfer what kind of reaction is that? Acid base. Okay? So this is not a redox reaction. So not every reaction that you look at has to be redox and the way to figure it out is to look for oxidation numbers and whether it's a change. Okay? So now we're going to look at something that's very, very important and that is balancing oxidation reduction reactions. Alright? Now up to this point, so I'm going to give you one example. So let's take an example and let's say you want to balance the equation copper plus HNO3 giving you copper nitrate plus nitrogen plus H2O. Now it turns out that up to this point if you want to balance a chemical equation, alright, you've only looked at trying to balance atoms. Alright? And you're pretty well versed at balancing equations. But it turns out that there is a class of oxidation reduction reactions that are virtually impossible to balance. Alright? And this is one such example. So if you wanted to use sort of the general form that you've used in the past, remember usually we say when you balance equations the number of atoms at the beginning should equal the number of atoms at the end and you just play around with this and there's no hard and fast rule on how to balance it. It turns out if you try to balance this equation you can work on it for a month and unless you follow a general strategy you will not be able to balance this equation. Alright? And so these are challenging equations to balance and the reason is that you have these are redox reactions and in redox reactions when you balance redox reactions the underlying principle that you want to remember is the number of electrons given off have to equal the number of electrons taken up. Okay? So that's going to be the underlying principle. So if you want to balance this equation I'm going to write them out so that it's in one line because so we have HNO3 giving you copper nitrate plus NO2 plus H2O. The first step is we're going to assign oxidation numbers. Okay? In order to balance this equation you have to assign oxidation numbers. So I'm going to do this fast. This will be zero. Each hydrogen is plus one. Each oxygen will be three times negative two. Alright? So can you do this? This will be six. This will be plus one. So what should nitrogen come out to be? Plus five. Each copper, this is copper two plus because there are two nitrates so you know that the oxidation number of copper is plus two and therefore you can see the charges plus two oxidation numbers too. Each nitrogen, there are two nitrogens, each nitrogen will be plus five, each oxygen, there are six oxygens will be negative two. Now over here I have one nitrogen. What is the oxidation number for that nitrogen? Plus four because there are two oxygens over here. Am I going too fast? I need you to work at this pace. Okay? So people who complain that they don't have time on the exam have not mastered this skill or haven't practiced enough to be able to do it kind of intuitively. Okay? So water is there are two hydrogens, each hydrogen is plus one and then there's a regular rhythm. So what I've done is I've just assigned oxidation numbers for all the elements. Okay? Now my next step is to figure out which ones are undergoing changes. Alright? So what is the first one that undergoes a change? Copper. Alright? What else changes? Nitrogen. But can you see that the nitrogen is kind of tricky because some of the nitrogens don't change and some of them do change. Do you see that? This is why this equation is so hard to balance because can you see there are NO3, some of the NO3s, remember NO3, the nitrogen here has a plus five oxidation state. Some of them remain plus five and some of them go to plus four. Alright? And therefore we need to figure out how many of these actually change and how many of those nitrogens don't change at all. Okay? So remember what is the underlying principle? The number of electrons transferred has to be the same. So the number of electrons given off has to be the same as the number of electrons taken up. So now let's turn to this. If you take each copper, copper goes from zero to plus two. Is the oxidation number increasing or decreasing? Increasing. If it increases, what do we know? It's an oxidation and each copper atom changes from zero to plus two. Therefore two electrons are lost. Alright? So if two electrons are lost, how many electrons need to be taken up? So we know here it goes from plus five to plus four. Do you see that? So that means this is the reduction and we know that each nitrogen goes from plus five to plus four. So that decreases. That's why it's a reduction. By how much does it change? By one. So I know that each nitrogen is picking up one electron, but if I want to make these equal to each other, I have to multiply it by two so that two electrons are picked up. Alright? So that means that I have to have two nitrogens here. Do you see that? Because two of these are the ones that are being reduced. Okay? So now let's go back to this equation. I have copper, one copper, one copper. So I can't change that now because if I double this by two, then four electrons will be given up. So I'm going to stick to two electrons being given up. I can't change that. Okay? Now I have two NO2s that do change, but remember there are two NO3s. There are two nitrogens here that don't change. So what is the number I need to put here? Can you see I need to put four there? Because two, I put two here to balance the two electrons. So I have two nitrogen sources here, plus there are two here. So this two plus this two gives me four nitrogens on this site. So I have two nitrogen atoms coming from here, two nitrogen atoms coming from here. Overall, I should have four nitrogen atoms on this site. Did you get that? Once you get that, your equation is balanced. Otherwise, if you're just trying to do atom balance, you'll be sitting there for days and you will not be able to figure it out. If you do, it's purely by chance. It's just pure luck that you got it. So once you've figured that out, now I can go back and say, so I have four hydrogens. If I need two waters on this side, okay? Now let's make sure our equation is balanced. And I'm going to now confirm. I have one copper, one copper. I have four hydrogens. I have two times two, four hydrogens. Now let's go to oxygen. I have four nitrogens here. So I have four nitrogens here. I have two nitrogens and two here, four nitrogens. Now let's me do the math for oxygen. Four times three gives me 12 oxygens. So here I have two times three gives me six plus four ten plus two twelve, so I know all my oxygens are balanced and I know now I've balanced the equation. Did you guys get that? And so this way you can balance the equation by looking at oxidation numbers. Okay? For discussion next week, I will give you lots of examples to practice. Now within the last three minutes, I'm going to do another example and so we still have three minutes. So if I have, if this is my balanced equation, NaCl plus H2SO4 plus MnO2 giving me Mn2SO4 plus MnCl2 plus H2O plus Cl2. Okay? Now this question comes directly from an exam that I gave two years ago. Alright? So I'm just pulling an old exam question and you're asked to balance this equation. Alright? If you were to try to balance this equation by sort of the methods that, that you've used before which is just looking at aden balance, you would not be able to balance this equation. Alright? So it's a redox reaction. I have to figure, balance this equation by looking at the number of electrons given off and the number of electrons taken off. So quickly I'm going to assign oxidation numbers. Plus one, minus one. Two times plus one, this will be negative two times four. So remember this is eight minus two plus, so this has to be plus six. Alright? Now look at manganese. It's a transition metal. So we've got to rely on oxygen. Oxygen is two times negative two, therefore each manganese will be plus four. On this side I have two manganese, so it'll be two times. Now the sulfate is SO4. Oh wait, my formula is wrong. What did I write? Na2, oh, this should be Na2, sorry. Alright? There's no charge of balance there, so, and I copied this wrong. Okay? So there are two sodiums, each one is plus one. Each oxygen is negative two times two. Each sulfur will be plus six. Over here, each chloride is two times negative one, so this will be plus two. This will be plus one times two, negative two, zero. Over here? Over here. Yes, thank you. Alright? Now I need to figure out what is undergoing changes. Alright? Sodium is plus one, sodium is plus one, so sodium didn't change. Okay? Chloride is negative one, it goes to zero. So we know chlorine has changed, but there are two types of chlorine. There are some chlorines that don't change and some that do. Alright? And then the other thing that changes is manganese goes from plus four to plus two. Alright? So let's start with manganese. It goes from plus four to plus two. Is that an oxidation or is that reduction? Increases or decreases? So if it decreases, it's a reduction. Okay? How many electrons are being, if it's a reduction, it gains electrons, it goes from plus four to plus two. Two electrons are picked up. Now let's look at chlorine. Each chlorine goes from negative one to zero, so it's increasing. So that means that's an oxidation. And it increases by one, so that means that one electron is being picked up or lost. I'm sorry, one electron is being lost. Now to balance this, I need two. So I'm going to multiply that by two. So that means I need two chlorines on this side. So if I have two chlorines on this side, two that undergo change, but there's two that don't change. So how many do I put here? Four. Alright? Because there are two chlorines that go from negative one to zero, and there are two chlorines that go from negative one to negative one. Alright? So I need four in ACL. Once I have got that, now I need four sodiums, so I'm going to put two sodiums there. Now if you look at the sulfate, I have two entities of the sulfates. So I need two sulfates over here. Now I have four hydrogens, so I need four hydrogens over here, and now my equation is balanced. I will repeat this example. I want you to take a look at this example at home and think about it. I will repeat this example at the beginning of class next time and go through this once again, because it's very, very important that you understand how to balance redox reactions. Okay?
 Prior to looking at electrochemistry, we want to understand what it means when we say a  reaction is an oxidation reduction reaction.  So you guys remember we said virtually all reactions can be categorized into three groups.  The first group are the acid-base reactions and we've looked at that in the past.  We know acid-base reactions involve proton transfer.  So if you look at a balanced equation and you look at two reactants and you see that  a proton is being transferred from one to the other, then you know that there is a proton  donor, that's the acid, and the substance that accepts that proton is the proton acceptor.  That will be the base and what you end up with is an acid-base reaction.  The second class of reactions that we looked at were the precipitation reactions and we  said we used solubility rules to help us and if you start with two substance that are completely  soluble and you combine them and you end up with an insoluble substance, that class of  reactions are called the precipitation reactions.  And then we said the third class of reactions are the oxidation reduction reactions or redox  reactions.  They meet the same thing.  The abbreviated form of the oxidation reduction reaction is called redox.  And these involve electron transfer.  So in these reactions, one substance gives off electrons and the other substance picks  up that electron and we end up by having electrons being transferred.  One or more electrons are transferred and that is an oxidation reduction reaction.  And as all of you can recognize, unlike the previous two where you can look at the reactants,  like in proton transfer, you can see that the proton is one substance and then it moves  to another, by looking at the balanced equation, it's easy to figure out whether it's an acid-base  reaction or a precipitation reaction.  In the case of redox reactions, just by looking at a balanced equation, it's hard for us to  figure out whether electrons are being transferred because the balanced equation does not have  electrons.  Is that clear to everybody?  So it's very difficult to look at the equation itself and figure out whether one substance  is losing electrons and whether the other substance is picking up electrons.  So we have to have a system of keeping track of electrons.  We call that electron counting or electron bookkeeping.  And so if we're going to keep track of electrons, then we have to have an easy system or set  of rules that we apply to assist us in figuring out the number of electrons that are being  transferred.  And the system that we've come up with is called oxidation numbers.  So one way to express the number of electrons lost or gained by a substance in a redox reaction  is by assigning an oxidation number to each atom.  So this is much like a game.  You know, when you play games, you set up rules.  And as long as you stick to the rules, everything kind of works out well.  It's the same thing here.  We're going to come up with a system of rules.  And as long as we stick to those rules, then we can figure out whether electrons are being  lost or gained or there's any change in the number of electrons.  We're going to look at every atom in a substance individually to figure out whether there's  a change in the number of electrons.  So that's what an oxidation number is.  An oxidation number is a number.  And this number can be zero, positive, or negative, assigned to each atom in any substance  in an oxidation reduction reaction that reflects whether the atom has fewer.  And if it has fewer electrons than it started, it has a positive sign.  It has more.  The oxidation number will have a negative sign.  Or the same, that will be zero as the number of electrons as an uncombined neutral atom.  So this is the oxidation number is just a way for us to keep track of electrons.  And so an oxidation number can be positive, negative, or zero.  And we use Arabic numbers.  The Arabic numbers are the one, two, three that we usually use.  And the online homework, it'll say, give your answer in Arabic numbers.  And that means that's what we're using.  Later on, you'll see that we can also assign oxidation states, and for that we use Roman  numerals to distinguish between the two.  So right now we're looking at oxidation numbers, and we're going to use Arabic numbers.  So if you look at the class website, I've put the rules up on the class website.  And these are the rules for assigning oxidation numbers.  Before I start this, usually like solubility rules, in KA values, stuff like that I will  provide on the exam.  But this I will not give to you, because this is something that you need to understand,  internalize, and be able to use it intuitively.  And so these rules, you need to understand them and be able to apply them.  And these rules will not be provided on an exam.  So let's start with the first one.  An oxidation number of an element in its natural state is zero.  So if you take sodium, in its natural state, it's metallic sodium, sodium metal.  So it will have zero.  If you take metallic silver, that's how the element exists in its natural state.  It'll be zero.  If you take mercury, the liquid state of the mercury, as all of you know, is the natural  state, that will have an oxidation number of zero.  Oxidation in its elemental state exists as O2, hydrogen exists as H2, chlorine exists  as Cl2.  So as long as it exists in its elemental state, it has an oxidation state of zero.  The second is the oxidation number of a monoatomic ion is the same as its ionic charge.  So if it's a cation or anion, it's monoatomic cation or anion, then its oxidation number  corresponds to the charge on it.  So Na plus would be plus one.  Cl minus would be negative one.  Fe2 plus would be plus two.  F minus will be negative one.  So the oxidation number corresponds to the charge.  The third is when we deal with oxygen.  In most compounds containing oxygen, the oxidation number of oxygen is negative two.  So if it's O2, oxygen in its elemental state will be zero.  But if oxygen is found in a substance, then the common state is that it has a minus two  oxidation number.  So if you take magnesium oxide, it has negative two.  SO2, it has negative two.  SO3, negative two.  CO, CO2, any formula where the common form is where it has an oxidation number of negative  two.  The only exception is the peroxides.  The peroxides are pretty rare.  And a peroxide is O2, two minus.  So there are two oxygens, and overall it has a two minus charge.  So if you take, if there are two oxygens that have an overall two minus charge, what is  the charge on each oxygen?  Minus one.  And that's why if you take the exception to the rule of the peroxides in which the oxygen  addition number has a minus one.  How do you tell whether it's a peroxide?  Look at the formula, if it's like H2O2, or if you have a metal like Na2O2, all right?  Look at the charge.  If you take Na2O2, you can see there are two Na pluses.  So that means overall O2 must be two minus.  Therefore each oxygen has to be minus one.  But peroxides are not very common.  The most common ones you can encounter are the ones that have minus two.  The fourth rule is that in most compounds containing hydrogen, the oxidation number of hydrogen  is plus one.  So if you take HCl, NH3, H2O, CH4, these all have plus one.  The exception to this rule are the metal hydrides in which the hydrogen has an oxidation number  of negative one.  So now if you have like lithium, sodium, potassium, a metal with a hydrogen, then it's a metal  hydride.  All right, so if you have a metal attached to hydrogen, it would be a metal hydride,  and its oxidation number would be negative one.  Now last is that the sum of the oxidation numbers must be zero for electrically neutral  compounds.  For polyatomic ions, the sum of the oxidation numbers must be equal to the charge of the  ion.  Remember we're assigning oxidation numbers for every atom in a substance.  So if you add up the oxidation numbers, if it's electrically neutral compound like water  has no charge on it, that means when you add up all the oxidation numbers, if you add up  the oxidation numbers for the two hydrogens and the oxidation number for that one oxygen  and add them together, it needs to add up to zero.  If it is charged, it needs to correspond to the charge.  So if you take chromium, the oxidation number of chromium plus the oxidation number corresponding  to four oxygens, if you add them up, they have to equal negative two.  And if you take NH4, if you take nitrogen and add the oxidation number of this to the  oxidation number of four of these, you should end up with a plus one oxidation state, or  number.  Is that clear to everybody?  So now let's apply this because we're going to use this often to figure out whether a  compound is losing electrons or gaining electrons.  So we'll start by taking an example where you're asked to calculate the oxidation number.  So we'll start with a simple one.  Calculate the oxidation number for carbon in CO2.  So your starting point would be to take the compound and we know that overall there's  an electroneutral compound, so the sum of the oxidation numbers corresponding to each  of the atoms should add up to zero.  So since we need to find the oxidation number for carbon, we're going to call that N. And  we have two oxygens, so two times.  What does a rule tell us?  What is the oxidation number for oxygen?  Negative two.  All right?  So this will be negative two.  So we know the sum of these two should correspond to the overall charge of the molecule, so  we know N plus two minus two should equal zero.  Therefore N equals plus four.  So the oxidation number that corresponds to that carbon will be four.  We always assign oxidation numbers per one atom.  All right?  So we know that even though there are two oxygens there, each oxygen individually has  an oxidation number of negative two and because you have two of those, you're going to multiply  it by two.  Okay?  So let's take another example like H3PO4.  So let's say calculate the oxidation number for phosphorus in H3PO4.  So once again, we have H3PO4.  The one that we need to figure out, we'll call it N.  We have hydrogen and we have oxygen.  All right?  We have three hydrogens and what does the rule tell us that the oxidation number for hydrogen  is plus one.  All right?  This will be there four oxygens and each oxygen we know from the rules will have negative  two.  So that means the sum of all of this should equal the overall charge.  So we know that three times plus one plus N plus four times negative two should equal zero.  Therefore N should equal eight minus three which is plus five.  All right?  So the oxidation number for phosphorus in that compound is plus five.  Now after all I need you to be able to do this kind of mentally in your head and quickly  calculate these but for now we'll go through the details.  Okay?  Now let's take the third example.  Calculate the oxidation number for chlorine in ClO3 minus.  Okay?  So once again, if you take this formula, this is what we need to find.  So we're going to call that N.  We can see that the other component is three so we know that each oxygen is negative two.  There are three of those so that will be negative two.  But now you can see there's a polyatomic ion that has a charge of negative one.  So that means the sum of these two should equal negative one.  So that means N plus three times negative two should equal negative one.  Or N equals six minus one which is plus five.  All right?  And so once again you can see that when you assign oxidation numbers you have to put signs.  It either has to have a negative sign in front if it's a number, a positive sign or it has  to be zero.  Okay?  Now lastly let's take something like thiosulfate which is S2O3.  So calculate the oxidation number for sulfur in S2O3 two minus.  All right?  So now remember oxidation numbers are always assigned per one atom.  So there are two sulfurs.  So we're going to say two N.  All right?  And there are three oxygens and we know from the rule that oxygen has a negative two oxidation  number.  So the sum of these two should equal two minus.  All right?  So we know two N plus three times negative two should equal negative two which is the overall  charge.  Therefore two N equals six minus two which is plus four.  And then we always assign oxidation numbers per individual atom.  So each sulfur atom will have plus two.  Divide that by two.  Each atom has plus two.  Got it?  So for any formula you should be able to figure out what the oxidation number for all the  elements are.  Usually most compounds we deal with have hydrogen or a-to-oxygen on it so we use those elements  to figure out what the oxidation number for the other one is.  If you have cations or anions you know that they correspond to the charge.  So monotomic and monotomic cations or monotomic anions will correspond to the charge as well.  Now let's apply this in an example where we're looking at a redox reaction.  So if I take a redox reaction and we said a redox reaction involves electron transfer.  So let's just take an example where we have two reactants and let's say in this reaction  a transfers one electron to b.  Alright so if a transfers one electron to b then what you end up with is a has lost  electrons so it would be a plus plus b minus.  Alright?  So you have, so can everybody see that you can't directly see the electrons being transferred  so I'm telling you that one electron is being transferred.  Alright?  Now whenever you look at redox reactions there's electron transfer and a convenient way to  kind of look at what's going on is to look at what we call half reactions.  Alright?  So if I take half reactions you can see that a goes to a plus plus an electron.  So in this case when you write half reactions you can see that when a goes to a plus it's  losing one electron.  Okay?  Whenever you write balanced equations remember you have to maintain charge balance.  So to begin with on the a side everything is electrically neutral.  You end up with a plus so to balance that charge you can figure out how many electrons  you need to balance that charge.  So you can see a plus just needs one electron to balance that charge and you know one electron  has been given up.  Alright?  Now on the other, so we call this the oxidation half reaction.  Remember why is this the oxidation half reaction because oxidation is electron loss.  And last time we looked at the definition of oxidation, oxidation involves the loss  of electrons.  When a substance loses electrons we call that oxidation.  When a substance picks up electrons we call that a reduction.  So now oxidation reduction reactions we said last time have to go hand in hand.  You can't have one without the other.  If something gives off electrons something has to pick it up.  So you need both of those components.  Alright?  You can't have one substance giving out electrons and those electrons are sitting out there  in thin air.  Alright?  You can never have electrons just sitting in thin air.  If one is giving it off another substance needs to take it and they need to take place  at the same time.  And so if you look at the reduction half reaction, now this electron that's given off is picked  up by B and it goes to B minus.  Once again charge balance.  You have neutral minus one.  So on this side I have a minus one charge.  If on this side I have to have a minus one charge as well.  This is called the reduction half reaction.  Why?  This is where electron gain occurs.  Alright?  Now I want to remind you that even though we write half reactions because they're convenient  for us to kind of rationalize what's going on, in reality reactions don't take place in  half reactions.  Do you understand that?  Because the instant the electron is given off the other is taken up.  And so you don't have this process where one electron comes out and the other electron  is then picked up.  It's immediately transferred from one to the other.  So in oxidation reduction reactions they take place instantaneously and you don't have  half reactions occurring.  But half reactions is purely a mental process for us to kind of separate it out to understand  what is going on.  So if I want to make a little table I can say that if I take A, so let's combine, let's  look at what happens at A and what happens at B. Remember we have to look at the reactants.  Alright?  So you can see A loses electrons.  In other words it's losing one electron, one or more electrons.  So A loses electrons, B gains electrons.  Alright?  A undergoes oxidation or it is oxidized.  And we say B undergoes reduction or it is reduced.  Alright?  Now last time we looked at, when you look at redox reactions you need to be able to identify  what is undergoing oxidation or what is being oxidized, what is undergoing reduction, what  is being reduced.  Alright?  So the next is that you need to know which is the agent.  So since this is undergoing oxidation this is the, A is the reducing agent.  This is the agent that's responsible for reducing this substance which is undergoing a reduction.  And B is a oxidizing agent.  Alright?  Whenever a substance undergoes oxidation the oxidation number increases.  So the oxidation number of A increases.  Alright?  So the oxidation number of A increases.  And if it's undergoing a reduction the oxidation number of B decreases.  So these are important points to remember.  If you're looking at an oxidation reduction reaction I need you to be able to look at  each one of these and be able to identify for each reactant you need to know which one  is losing electrons.  You need to figure out which is gaining electrons.  You need to figure out which one is being oxidized, which is being reduced.  You need to figure out which is the oxidizing age and which is a reducing agent.  Whenever you have an oxidation the oxidation number increases, alright?  When a substance loses electrons its oxidation number increases.  When a substance gains electrons its oxidation number decreases.  So in an oxidation reduction reaction if the oxidation number changes you know that its  oxidation reduction reaction. One substance oxidation number will go up. That means it's  oxidized. And another substance oxidation number will decrease. That's the one that's  being reduced. Okay, so let's take an example. And we'll start with some simple examples  and then we'll go to more complicated ones. So let's take some examples. And the first  example I'm going to take is where you have potassium solid. And so this is metallic  potassium reacting with chlorine gas to give you an ionic compound which is potassium chloride.  Okay? So quickly check where this equation is balanced. You have two potassiums, two  potassiums, you have two chlorines, two chlorines. Okay? Now for you to figure out whether it's  a redox reaction or not, we have to look at oxidation numbers. Okay? So to begin with  we always look at it per one atom. Alright? So in this there are two potassiums in this  balanced equation. And so if I take potassium this is an element in its natural state. Therefore  its oxidation number has to be zero. And there are two coefficients. I'm going to put the  coefficients outside. Okay? Over here chlorine is, there are two chlorine atoms. But this  is an element in its natural state. Therefore the oxidation state of the two chlorine atoms  in Cl2 each one will have an oxidation number of zero. Okay? So there will be two times  zero. Now on this side we end up with an ionic compound. Alright? This is a metal, non-metal  combination. If it's an ionic compound we have to have a cation and an anion. So if  I take the potassium you can see that that's K plus. Therefore each potassium will have  a plus one and each chlorine will have a minus one. So what's outside is the coefficients  in the balanced equation or the number of atoms that's there in the balanced equation.  Okay? So now we have to look at this equation and we want to figure out does the oxidation  numbers change? So if you look at potassium each potassium atom goes from zero to plus  one. So is there a change? Yes. Now you ask yourself, so that's got to be a redox reaction.  Is the oxidation number increasing or decreasing? So if you take potassium it goes from zero  to plus one. So it increases. Remember if the oxidation number increases what is that?  Oxidation. Alright? So we start looking at the two reactants and we're going to say that  when this goes to that oxidation number increases and we know that's an oxidation. And by how  much does it change? It goes, each one goes from zero to negative one. So that means that  if there's a change by one, alright, it goes from zero to plus one rather. If there's a  change of one that means one electron is being transferred. But there are two atoms so overall  you can see two electrons are being transferred. So overall two electrons are transferred.  As each potassium atom goes from an oxidation number of zero to plus one there's a net change  of one. The net change of one means that one electron is being lost. But overall we have  two so you multiply it by two and so in this reaction two electrons are being lost. Alright?  Now if you go to what's being reduced over here, now your chlorine, if you look at the  oxidation number goes from zero to negative one. So you know that oxidation number decreases.  And if it decreases it means it's a reduction. And if you take each chlorine atom it changes  by one. Okay? So since electrons are being lost I'm going to put a negative sign here  to show me that electrons are being lost. Alright? A negative sign means electrons are  lost. Positive signs means electrons are being gained. So if you look at this each chlorine  atom goes from zero to negative one. So it's a net change of one so it decreases by one.  So I'm going to say one electron but since there are two atoms involved all together  two electrons are being lost. Okay? So we figured out from this by looking at oxidation numbers  that two electrons are lost and two electrons are gained. Alright? Now the cardinal rule  is that the number of electrons given off have to equal the number of electrons taken  up. If that doesn't work out then you know there's something wrong with your rationale.  Do you understand that? Because you can't have ten electrons given off and only five  taken up. Alright? The number of electrons given off always have to equal the number  of electrons taken up. Alright? And you can see two electrons are given off, two electrons  are picked up. Now, so what is undergoing oxidation? Potassium, this is oxidized. You  can see chlorine is reduced. Alright? So the substance being oxidized is potassium because  it's undergoing oxidation. The substance being reduced is chlorine because its oxidation  number goes down. So that's the substance being reduced. If this is being oxidized this  is the reducing agent and this is the oxidizing agent and we know overall two electrons are  transferred. Alright? So it's very, very important that every example that you look  at you have to be able to go through this process. You've got to figure out what's  being oxidized, what's being reduced, how many electrons are being transferred and you  make use of oxidation numbers to do that. Okay? So let's take another example. So we're  going to look at the combustion of methane. So what happens if you burn methane in oxygen?  Now a combustion reaction is also an oxidation reduction reaction. So if I take methane plus  202 gas giving me carbon dioxide plus 2, okay? So we're looking at the combustion of methane.  Make sure the equation is balanced. You can see there's one carbon, one carbon, four  hydrogens, four hydrogens, four oxygens, two plus two, four oxygens. Okay? So I'm going  to tell you ahead of time that this is a redox reaction. Okay? So if you take a redox reaction  and you've got to figure out what's being oxidized, what's being reduced, how many electrons  are being transferred. Okay? So for carbon now, by now you should be able to do this  really fast. The hydrogens each have plus one. There are four hydrogens with plus one.  So what's the oxidation number of carbon? Negative four. Okay? If you take hydrogen, there are  four times plus one. So that will be the oxidation number assigned to each of the hydrogens.  Each oxygen, can you see that there are two oxygens here times two? There are four oxygens.  It's an element in its natural state. Each oxygen atom will be zero. Okay? Now let's  turn to this carbon. To figure out what the oxidation state of this carbon is, we turn  to oxygen. Remember, each oxygen atom has, what is the oxidation number? Negative two.  So there's negative two times two gives me negative four. So if oxygen is negative four,  the whole compound is electroneutral. What is the oxidation number of carbon over here?  Plus four. Okay? And we know there are two oxygens. That's going to be negative two. Now  over here we have hydrogen. Each hydrogen is plus one. How many hydrogens do I have?  Two times two, four. And then if I take the oxygen, there are two oxygens and I know that's  negative two. Can everybody see I have assigned oxidation numbers for all of the atoms? Now  I look at one atom at a time and figure out which ones are undergoing changes. So we start  with carbon. Carbon starts from negative four to plus four. Alright? So that's one. Now  does hydrogen change? The hydrogen atoms here are plus one. The hydrogen atoms here are  plus one. So hydrogen does not change. Let's look at oxygen. Oxygen goes from zero to  negative two over here and negative two over there. Do you see that? So that means oxygen  is changing and it's changing the both sets of oxygens undergoing a change. Okay? So let's  start with carbon. Now look at the oxidation number for carbon. Does that increase or decrease?  It increases. It goes from negative four to plus four. So since the oxidation number increases,  what do I know? This is the oxidation and methane is being oxidized. Alright? Now look  at by how many does it change? Remember there's one carbon atom at the beginning, there's  one carbon at the end. So if I go from negative four to plus four, what is the change? Eight.  So that means there are eight electrons involved. Alright? And so this is oxidation, electrons  are being lost so we know that there are eight electrons involved. So overall, eight electrons  are being transferred. Okay? Now let's go to the oxygen side. It goes from zero to negative  two. Does the oxidation number increase or decrease? It decreases. Remember if the oxidation  number decreases, we call that process reduction. Okay? Now let's look at how many oxygen atoms  there are. There is two, two times two, four oxygen atoms. There's two plus two, four oxygen  atoms. So there are four oxygen atoms and each oxygen atom goes from zero to negative two.  So what is the change? Two electrons. If it's a reduction, our electrons picked up or electrons  lost, picked up. So you know that there are times plus two, which gives me plus eight electrons.  Now how many electrons are being transferred here? Eight electrons. So in this oxidation  reduction reactions you can see that the substance that's undergoing oxidation is methane. So  that's being oxidized. The substance being reduced over here is oxygen. Alright? So if  this is being oxidized, this would be the reducing agent. If this is being reduced, that  would be the oxidizing agent. And how many electrons are transferred? Eight electrons  are transferred. Okay? I need you guys to be able to do this in your sleep, okay? Because  if you can't figure out how many electrons are being transferred, then the whole exercise  is a waste of time. Do you get it? So, and later on when we're going to look at electrochemistry,  I need you to be able to just look at a reaction, figure it out, and know how many electrons  are being transferred. Okay? Now let's take another example where, let's say I have SICL4  plus, let's say that's liquid, plus 2H2O liquid, gives me 4HCl, Equus, plus SiO2 solid.  Okay? So once again, let's take a look at this. So now we can go pretty fast. There  are four chlorines, each chlorine has minus one, so what is the oxidation state of silicon?  Plus four. There are four chlorides, and each one is negative one, because remember chlorine,  Cl minus has a negative one oxidation number. Here, there are four hydrogens, each hydrogen  is plus one, there are two oxygens, each oxygen will be negative two. Alright? Now on the  product side, we have four hydrogens, so it will be four times plus one, there are four  chlorides, four times negative one. If I take the silicon over here, you can see it's plus  four, this is two times negative two. Now I'm starting to move pretty fast, because  that's sort of the pace at which you need to be able to analyze these reactions. Okay?  Now look at this and tell me, is anything undergoing any change? Let's tell it with  oxygen. So look at this plus four, the oxidation number is plus four, at the end of the reaction  it's plus four. Does that change? No. Let's look at chlorine. Each chlorine atom to begin  with is negative one, on the product side each chlorine is negative one. Does that change?  No. Now let's turn to water. Each hydrogen is plus one, on the product side the hydrogen  is plus one. If I turn to oxygen, each oxygen is negative two, on the product side each oxygen  is negative two. So do oxidation numbers change here? No, there are no changes. So is this  a redox reaction? No. Okay, there's no electron transfer. Alright? So what kind of reaction  is this? Remember it has to be their acid base, precipitation or oxidation reduction  reaction. Look at the hydrogen atom. Do you see hydrogen atoms moving from one to the  other? Where does hydrogen start with water? Is the hydrogen moving away from water? Is  it staying together with water? So can you see there's a proton transfer, there are proton  transfers taking place. If it's a proton transfer what kind of reaction is that? Acid  base. Okay? So this is not a redox reaction. So not every reaction that you look at has  to be redox and the way to figure it out is to look for oxidation numbers and whether  it's a change. Okay? So now we're going to look at something that's very, very important  and that is balancing oxidation reduction reactions. Alright? Now up to this point, so I'm going  to give you one example. So let's take an example and let's say you want to balance  the equation copper plus HNO3 giving you copper nitrate plus nitrogen plus H2O. Now it turns  out that up to this point if you want to balance a chemical equation, alright, you've only  looked at trying to balance atoms. Alright? And you're pretty well versed at balancing  equations. But it turns out that there is a class of oxidation reduction reactions that  are virtually impossible to balance. Alright? And this is one such example. So if you wanted  to use sort of the general form that you've used in the past, remember usually we say  when you balance equations the number of atoms at the beginning should equal the number of  atoms at the end and you just play around with this and there's no hard and fast rule  on how to balance it. It turns out if you try to balance this equation you can work  on it for a month and unless you follow a general strategy you will not be able to balance  this equation. Alright? And so these are challenging equations to balance and the reason is that  you have these are redox reactions and in redox reactions when you balance redox reactions  the underlying principle that you want to remember is the number of electrons given  off have to equal the number of electrons taken up. Okay? So that's going to be the  underlying principle. So if you want to balance this equation I'm going to write them out  so that it's in one line because so we have HNO3 giving you copper nitrate plus NO2 plus  H2O. The first step is we're going to assign oxidation numbers. Okay? In order to balance  this equation you have to assign oxidation numbers. So I'm going to do this fast. This  will be zero. Each hydrogen is plus one. Each oxygen will be three times negative two. Alright?  So can you do this? This will be six. This will be plus one. So what should nitrogen  come out to be? Plus five. Each copper, this is copper two plus because there are two nitrates  so you know that the oxidation number of copper is plus two and therefore you can see the  charges plus two oxidation numbers too. Each nitrogen, there are two nitrogens, each nitrogen  will be plus five, each oxygen, there are six oxygens will be negative two. Now over  here I have one nitrogen. What is the oxidation number for that nitrogen? Plus four because  there are two oxygens over here. Am I going too fast? I need you to work at this pace.  Okay? So people who complain that they don't have time on the exam have not mastered this  skill or haven't practiced enough to be able to do it kind of intuitively. Okay? So water  is there are two hydrogens, each hydrogen is plus one and then there's a regular rhythm.  So what I've done is I've just assigned oxidation numbers for all the elements. Okay? Now my  next step is to figure out which ones are undergoing changes. Alright? So what is the  first one that undergoes a change? Copper. Alright? What else changes? Nitrogen. But can  you see that the nitrogen is kind of tricky because some of the nitrogens don't change  and some of them do change. Do you see that? This is why this equation is so hard to balance  because can you see there are NO3, some of the NO3s, remember NO3, the nitrogen here has  a plus five oxidation state. Some of them remain plus five and some of them go to plus  four. Alright? And therefore we need to figure out how many of these actually change and  how many of those nitrogens don't change at all. Okay? So remember what is the underlying  principle? The number of electrons transferred has to be the same. So the number of electrons  given off has to be the same as the number of electrons taken up. So now let's turn  to this. If you take each copper, copper goes from zero to plus two. Is the oxidation number  increasing or decreasing? Increasing. If it increases, what do we know? It's an oxidation  and each copper atom changes from zero to plus two. Therefore two electrons are lost.  Alright? So if two electrons are lost, how many electrons need to be taken up? So we  know here it goes from plus five to plus four. Do you see that? So that means this is the  reduction and we know that each nitrogen goes from plus five to plus four. So that decreases.  That's why it's a reduction. By how much does it change? By one. So I know that each  nitrogen is picking up one electron, but if I want to make these equal to each other,  I have to multiply it by two so that two electrons are picked up. Alright? So that means that  I have to have two nitrogens here. Do you see that? Because two of these are the ones  that are being reduced. Okay? So now let's go back to this equation. I have copper, one  copper, one copper. So I can't change that now because if I double this by two, then  four electrons will be given up. So I'm going to stick to two electrons being given up.  I can't change that. Okay? Now I have two NO2s that do change, but remember there are  two NO3s. There are two nitrogens here that don't change. So what is the number I need  to put here? Can you see I need to put four there? Because two, I put two here to balance  the two electrons. So I have two nitrogen sources here, plus there are two here. So this  two plus this two gives me four nitrogens on this site. So I have two nitrogen atoms  coming from here, two nitrogen atoms coming from here. Overall, I should have four nitrogen  atoms on this site. Did you get that? Once you get that, your equation is balanced. Otherwise,  if you're just trying to do atom balance, you'll be sitting there for days and you will  not be able to figure it out. If you do, it's purely by chance. It's just pure luck that  you got it. So once you've figured that out, now I can go back and say, so I have four  hydrogens. If I need two waters on this side, okay? Now let's make sure our equation is  balanced. And I'm going to now confirm. I have one copper, one copper. I have four hydrogens.  I have two times two, four hydrogens. Now let's go to oxygen. I have four nitrogens  here. So I have four nitrogens here. I have two nitrogens and two here, four nitrogens.  Now let's me do the math for oxygen. Four times three gives me 12 oxygens. So here I  have two times three gives me six plus four ten plus two twelve, so I know all my oxygens  are balanced and I know now I've balanced the equation. Did you guys get that? And so  this way you can balance the equation by looking at oxidation numbers. Okay? For discussion  next week, I will give you lots of examples to practice. Now within the last three minutes,  I'm going to do another example and so we still have three minutes. So if I have, if  this is my balanced equation, NaCl plus H2SO4 plus MnO2 giving me Mn2SO4 plus MnCl2 plus  H2O plus Cl2. Okay? Now this question comes directly from an exam that I gave two years  ago. Alright? So I'm just pulling an old exam question and you're asked to balance this  equation. Alright? If you were to try to balance this equation by sort of the methods that,  that you've used before which is just looking at aden balance, you would not be able to  balance this equation. Alright? So it's a redox reaction. I have to figure, balance  this equation by looking at the number of electrons given off and the number of electrons  taken off. So quickly I'm going to assign oxidation numbers. Plus one, minus one. Two  times plus one, this will be negative two times four. So remember this is eight minus  two plus, so this has to be plus six. Alright? Now look at manganese. It's a transition  metal. So we've got to rely on oxygen. Oxygen is two times negative two, therefore each  manganese will be plus four. On this side I have two manganese, so it'll be two times.  Now the sulfate is SO4. Oh wait, my formula is wrong. What did I write? Na2, oh, this  should be Na2, sorry. Alright? There's no charge of balance there, so, and I copied this  wrong. Okay? So there are two sodiums, each one is plus one. Each oxygen is negative two  times two. Each sulfur will be plus six. Over here, each chloride is two times negative  one, so this will be plus two. This will be plus one times two, negative two, zero. Over  here? Over here. Yes, thank you. Alright? Now I need to figure out what is undergoing  changes. Alright? Sodium is plus one, sodium is plus one, so sodium didn't change. Okay?  Chloride is negative one, it goes to zero. So we know chlorine has changed, but there  are two types of chlorine. There are some chlorines that don't change and some that do.  Alright? And then the other thing that changes is manganese goes from plus four to plus two.  Alright? So let's start with manganese. It goes from plus four to plus two. Is that an  oxidation or is that reduction? Increases or decreases? So if it decreases, it's a reduction.  Okay? How many electrons are being, if it's a reduction, it gains electrons, it goes from  plus four to plus two. Two electrons are picked up. Now let's look at chlorine. Each chlorine  goes from negative one to zero, so it's increasing. So that means that's an oxidation. And it  increases by one, so that means that one electron is being picked up or lost. I'm sorry, one  electron is being lost. Now to balance this, I need two. So I'm going to multiply that  by two. So that means I need two chlorines on this side. So if I have two chlorines on  this side, two that undergo change, but there's two that don't change. So how many do I put  here? Four. Alright? Because there are two chlorines that go from negative one to zero,  and there are two chlorines that go from negative one to negative one. Alright? So I need four  in ACL. Once I have got that, now I need four sodiums, so I'm going to put two sodiums  there. Now if you look at the sulfate, I have two entities of the sulfates. So I need two  sulfates over here. Now I have four hydrogens, so I need four hydrogens over here, and now  my equation is balanced. I will repeat this example. I want you to take a look at this  example at home and think about it. I will repeat this example at the beginning of class  next time and go through this once again, because it's very, very important that you  understand how to balance redox reactions. Okay?
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