If you watch my weight. >> All right. Good afternoon. Couple of comments about the exam to things that I enjoy saying whenever I get a chance to. Here's one of them. Nobody failed this exam. That certainly wasn't the case on the first exam, but it was the case on this. So I don't know is everybody getting motivated or what, but congratulations. I like to be able to say that whenever I can. Here's something else that's fun to say. You'll notice that the median score here is 71 out of 80. The significance of that is, of course, the median is the score that half of the class did better than and half of the glass did worse than they may have also noticed that I tend to put little smiley faces at the top of any exam where somebody scores more than 70 out of 80 points. So here's the smiley face cut-off line. And anytime we have more people in the smiley face section then not in a smiley face section. >> I think that's a good thing. >> So congratulations to the group as a whole and especially to those of you who are in the smiley face zone now were going to do is just keep it up for the rest of the semester. For those of you who are concerned about your freshmen midterm grade, you may ignore it now because the freshman midterm grade was calculated just on the basis of whatever your first exam score was. Now that we have to exam scores for you, we can update we well, because freshman midterm grades are meaningless, but you can calculate your own updated grain by putting your two scores together and compare it with what's on the syllabus and see how everything stance. And I also didn't have time to factor in your lab grade synchronous somebody by collaborating to factor that in as well. But I'm pleased with how the group as a whole was doing 75% for the semester so far, is right around the b minus c plus borderline. And we'll see if we can keep pushing it north of that line. So a good first half of the semester for the group as a whole. But I want to talk about today is the subject matter in Chapter 19 The title of which is electrochemistry. Now I just came from a meeting with your TAs and one of the things that apparently somebody heard from one of you folks, maybe not somebody who's in this room. I don't know. But they had the complaint was why don't we do anything cool in lab? >> Okay, go home and make clocks out of potatoes. In fact, read this segment right here. >> Your nutrition majors, making batteries out of food. That we might be talking too much about how to make batteries at a fiddle, but we will be talking about batteries before our discussion of electrochemistry in Chapter 19. >> Is that what I want to talk about today? >> Which kind of is the fundamental stuff that occurs earlier. Your textbook before we can have a serious conversation about electrochemistry is the kind of chemical reactions involved. And actually it's all way back in chapter four, specifically for 0.4, section 4.4, the title of which is oxidation-reduction reactions. And I want to start there just to introduce what some of that terminology means. Maybe I want to talk about today is how one goes about balanced equations for oxidation-reduction reactions, which is not as trivial as perhaps you might think. So let me introduce these two reactions right here, which we've seen before. The first reaction is the rusting of iron. The second reaction is the oxidation or combustion carbon, for example, in the form of charcoal. What a charcoal grill or something like that. We saw these earlier in the context of talking about kinetics. These two reactions do not occur at the same speed. But today we're going to look at these from a different perspective. The point is, this is a reaction between neutral iron and neutral oxygen, both in their elemental form to form ferric oxide, which is an ionic compound. So by the time we have this compound, which is the positive ion and which is the negative ion. Now this you should know from Chem 101, yeah, oxygen's negative, irons, positive, sure, in general. And, and write the formulas for ionic compounds. The positive ion first, slightly more challenging. What's the charge on a negative oxygen ion? >> Yes, negative two. >> So therefore, what must be the charge on each iron ion in this molecule? Yes. And I shouldn't say molecule because it's an ionic compound. But you're right, it's positive three. In other words, as this reaction takes place, we go from each of these elements having a charge of 0 to each of them having some sort of a charge that's not 0. And of course, iron, which is a metal, acquires a positive charge, and oxygen, which is a nonmetal, acquires a negative charge. Now, the only way that can happen for electrons to be transferred as iron goes from 0 to a positive three ion, what it has to do is lose three electrons. And the point is each individual atom of iron has to lose three electrons to do that. Likewise, when a molecule of oxygen forms to oxide ions, each with a charge of negative two. It goes from a charge of 0 to a total charge of make it at four. Therefore, each oxygen atom must gain two electrons. The O2 molecule as a whole gains for electrons in the process of becoming the o double minus i. I will go back to that slide just a moment. But this is the point at which I want to introduce some terminology. Because a moment ago we pointed out that the title of chapter or section 4.4 is oxidation-reduction reactions. And what oxidation-reduction reactions are all about is a transfer of electrons from one atom to another atom. And more specifically, oxidation means the loss of electrons. Reduction means the gain of electrons. So in the example we were just looking at, the iron atoms are losing electrons to form positive ions. Therefore, iron undergoes oxidation. The oxygen atoms are gaining electrons to form negative ions. Therefore, oxygen is being reduced here. Now hopefully the terminology makes sense. But one thing that as a challenge sometimes in the context of all of this, and you'll see why shortly is keeping straight which one is oxidation and which one is reduction? And there were a couple of mnemonic devices that have been developed to try to help students remember these things. Although I'm not sure if the cultural references are still there. >> But let's find out one such mnemonic device. >> Who can tell me what happens at the beginning of every MGM movie? >> I'm not sure when the last time you saw at MGM movie, but hoping that if I tell you the answer, it might sound familiar. The lion roars, right? >> Okay. >> Some of you have seen MDM movies, okay, probably another year too. And that cultural references guy. But for the moment, at the beginning of every MGM moving, LEO the lion says GER loss of >> The electrons is oxidation. >> The gain of electrons is commonly written as e negative is reduction. If it helps, it helps. Some people prefer another mnemonic device, OIL RIG. >> Oxidation is loss of electrons, reduction is gain of electrons, whatever works for you. >> But the main point is we will be talking about oxidation reduction a lot in Chapter 19. So hopefully it's clear what the words mean. Now back to this reaction for a moment. Okay, so iron is undergoing oxidation and becoming positive. Oxygen is undergoing reduction and becoming negative. But the only reason either of those things is happening is that this reaction is taking place. So what causes something to be oxidized or reduced? The answer is it's undergoing a chemical reaction. Oxidizing agent is the other reactant that causes something else to be oxidized. The reducing agent is the reactant that causes something else to be reduced. And part of what makes this concept complicated for some people anyway, the oxidizing agent is the thing that gets reduced and the reducing agent is the thing that gets oxidized. The only way the oxidizing agent can be an oxidizing agent and cause something else to lose electrons is if it itself is willing to accept electrons. But in the act of accepting electrons, it becomes reduced. And the only way the reducing agent can be a reducing agent and cause something else to accept electrons is to lose electrons itself. So the reducing agent is the thing that gets oxidized. So in the context of this reaction, I'll put this back up in a moment. Iron is oxidized and oxygen is reduced. But that also means that iron is the reducing agent because it causes oxygen to be reduced. And oxygen is the oxidizing agent because it causes iron to be oxidized. And let people finish writing that down, but I hope it's clear what the terminology means. By the way, for those of you following along in the lecture notes, we kind of picked up where we left off or on page 53 as we talk about these things. Anybody need more time with this slide? All right, now let's look at this reaction up here, which actually has a lot in common with this reaction over here. In both cases, oxygen is reacting with some other element to combat the difference is over here, oxygen is reacting with the metal to form an ionic compound. Here, oxygen is reacted with a non-metal carbon and forming a molecular compound, carbon dioxide. >> The point is, even though this is not an ionic compound, let's pretend for the moment that it is. >> If oxygen, as usual, has a negative two charge on each oxygen atom here, what would be the charge on the carbon? Bear in mind that the whole molecule must be neutral. Yeah, positive four. Now here's the thing that's not really happening, which is to say that carbon is not a plus four i in that molecule. But the similarity between these two reactions is sufficiently great that one way that it's sometimes convenient to think about it is to pretend that really it is happening. When we pretend that things like that are really happening, we refer to carbon F a plus four, not as a charge because it's really not, but as an oxidation state, or some people use the term oxidation number. So what I want to talk about for a few moments, It's how we assign oxidation states or oxidation numbers to particular atoms in particular compounds, it turns out to be maybe not quite as difficult as you might think. For starters, for any monatomic ion, that is for any i, and that is composed of one atom with either a positive or negative charge. >> The oxidation state is the same thing as the charge on the ion. >> So in the case of a positive sodium ion, the oxidation state is simply plus one. So we can refer to a sodium can't high in, as also sodium in its plus one oxidation state. Likewise for a chloride ion, negative one charge, sometimes referred to as chlorine, and it's minus one oxidation state. It's the different terminology that means pretty much the same thing for elements in their elemental form that have not reacted with anything else to form compounds. The oxidation state is 0, so metallic iron or gaseous oxygen, or any other element, the way it normally occurs as an element >> A charge of 0, oxidation state of 0 for molecules and polyatomic ions. >> And we lump these two together because molecules and polyatomic ions are held together by covalent bonds, even though that's true, we still kind of assume that they are ionic. And therefore, if you have to make a decision about which electrons in the bond belong to which atom in the bond. You just assume that the electrons belong to whatever the more electronegative atom happens to be. Who remembers what the most electronegative atom is? Can one-on-one. Anybody remember which is the most electronegative elements? Fluorine is the correct answer. Therefore, in any compound that features fluorine, fluorine must be in its minus one oxidation state. What's the second most electronegative element? Oxygen. And we talked before about the fact that oxygen usually forms a minus two. I, and I'm going to say usually here because before the semester is over, we will see some other examples where oxygen is in a different oxidation state, but most of the time oxygen is minus 21. Other thing we talked about in Chem 101, but it bears repeating. Hydrogen can be either positive one or negative one depending on what it finds itself having reacted with. But those are the only two possible oxidation states for hydrogen. So the point is, if you follow these rules, and there's one other rule that I didn't state on this slide, but I think you have a sense of what that rule is from the examples we've looked at so far. And that is the sum of all the oxidation states must be equal to the charge on the species, which means if it's a neutral molecule, the syllable, the oxidation states must be 0. But if it's a polyatomic ion, the sum of the charges must be the same as whatever the charge on the ion is. Now the same is true for oxidation states set in above. It will look at a couple of specific examples of assigning oxidation states. I want you to see if you can use these rules to tell me what the oxidation state is of particular atoms, in particular molecules or polyatomic ions. Everybody got what they need from this line. Okay, lets look at those examples. Let's look at the nitrate ion, which has a charge of minus one. Now, bearing in mind that the most common oxidation state for oxygen is minus two. What would be the oxidation state of nitrogen in the nitrate? >> I heard somebody say something, No, would not be negative one. >> This is not a monatomic ion, this is a polyatomic ion. So you can't just go by the charge. The charge is negative one. But the question was, what's the oxidation state of nitrogen? >> Yes. >> Right? And you get positive five? >> Yeah, absolutely, absolutely correct. >> If each oxygen has negative 23 of them would be negative six. And negative one charge tells you that the sum of all the oxidation states should be negative one. >> To make that happen, nitrogen must be at its plus five oxidation state with the logic make sense? >> All right, k, m and 04 is the formula of a compound called Potassium permanganate. See if you can figure out what the oxidation state of manganese is in this compound. Now it may be helpful to consult the periodic table, but I'm kinda hoping that maybe you already know what the charge is typically on potassium in any compound in which it finds itself. But if not, that's why we have a periodic table just over my right shoulder. Unfortunately, you folks watching this video at home don't. But you can get out a periodic table that's in your textbook. By the way, if you have your iPhone, you can get a periodic table.com and get one that way. >> So any guesses for the oxidation state of Maine and guineas and K M n O four, who does what potassium's charges Say again, plus one. >> And if oxygen is its usual minus two and the whole compound is neutral, manganese must be louder. >> Positive seven is correct. >> Four oxygens minus two each, that's minus a. Therefore, the sum of all the positive ions must be positive. Eight potassiums, positive one >> Manganese must be positive seven. >> A clue to this would be to look at the periodic table. And notice the nitrogen happens to be located in group five a, and manganese happens to be located in group seven b. However, as we'll see later on, that trick doesn't always work for all compounds because nitrogen has different oxidation states and so does manganese. But that also might help figuring out the oxidation states of everything in this compound. What's the usual oxidation state of lithium? Find a lithium on the periodic table and see how it behaves compared to the other elements in the same column. But now yep, positive one because it's in group 1A and the periodic table, just like potassium is about aluminum, that's typical show are done. >> An aluminum ion spring break is 25 minutes away. >> That's a it can be ten minutes away the quicker you get limited, typically plus three. Therefore, what must be the oxidation state of hydrogen in this combat? Well, negative four totals of each individual hydrogen is negative. >> What? >> Yep, hydrogen can only have two oxidation state is plus one or minus one. And in this case it's everything else is positive. >> Hydrogen must be in its minus one oxidation state concept makes sense. >> Okay? The point is, if those things make sense, then we can get to the topic that is discussed in Section 19.1, the title of which is balancing redox equations. >> Redox is short for reduction and oxidation, so you'll see that term redox a lot. And it just refers to the kind of reactions that we were just talking about, the balancing redox equations. It's a little bit more challenging than the straightforward balancing of equations that you've seen so far up until now. The only thing you've really had to worry about when you write balanced equations for anything is making sure the coefficients come out right. That's one factor that does come into play. But there are others. For one thing, since electrons are being transferred back and forth, we have to make sure that when we're done and the total number of electrons lost by the reducing agent is the same as the total number of electrons gained. By the oxidizing agent were not allowed when we're done to have free electrons floating around, every electron lost by one thing must be gained by something else. Redox equations are commonly written as net ionic equations, which means you may not see the spectator ions in there. Which means you may have charges to deal with, which means the total charge on the reactant side of the equation must be the same as the total charge on the product side of the equation. Another factor that comes into play has to do with whether you're doing these reactions under acidic or basic conditions. So all of those things turn out to be important. It's now in the lecture notes for those of you following along, I show two different ways of doing this. The method on page 55 in the lecture notes is based on the method that I learned when I was an undergraduate more than four decades ago. And it has to do with using oxidation states that you're comfortable with working with oxidation states, that's fine. >> You can use this method. >> Many people are not. So alternative method is shown on page 56 in the lecture notes. And the method on page 56 is actually very similar to the method that's used in the textbook. So for today, that's the method I'm going to show you. And here are the basic steps in the process. The idea is to first separate the reaction, whose equation you're trying to balance into 2.5 reactions, one of which will be the oxidation half reaction and the other will be the reduction half reaction. Then you're going to balance each half reaction by itself. First, by balancing the atoms if they're not already balanced, except we're not going to worry about hydrogen and oxygen when we do that part. Then balance the oxygens by adding water molecules to one side or the other. Then bandwidth hydrogens by adding protons to one side or the other. And then finally, balancing charge by adding electrons to one side or the other. Once we have balanced each individual half reaction, that way we can recombine the half reactions. Just bearing in mind that when we do, the total number of electrons gained must equal the total number of electrons lost. The example we're going to do today involves an acidic reaction taking place under acidic conditions. So at that point we stop. But there is a simple correction that can be applied if you're doing this under basic conditions. And maybe once everybody comes back from the break is a bit more focused, we'll take a look at an example of how to do this under basic conditions. So for the moment punishing finished jotting down the basic ideas. And then we'll show you an example of how there Everybody got what they need. Alright, let's take a look at an example of how to write a balanced redox equation for this reaction is the reaction between ammonia NH3, and the dichromate ion CR 207 with a minus two charge to form elemental nitrogen N2 and chromium three oxide CR2, O3, happening under acidic conditions. Alright, so we're going to follow that step-by-step pathway that we just outlined. Step one, separate the half reactions. Now one of the nice things about this method is that you don't have to worry for the moment about what's being oxidized, what's being reduced. You just need to realize that there are two main elements involved here, other than hydrogen and oxygen, and nitrogen and chromium. So we're going to separate this. We're going to actually draw a line down the middle of the page and separate this into the nitrogen half reaction NH3 going to N2, and the chromium half reaction CR2 seven double minus going to CR2 O3. That's all you have to do for step one. Any questions about step one? Step two, balance each individual half equation first by balancing Adams, then oxygen, then hydrogen, then finally electrons. Okay, in this equation, are the nitrogen atoms balanced? Now, what do we have to do to this equation to make the nitrogen atoms balanced two in front of the NH3. >> Ok. >> Now, nitrogens balanced in this equation, are the chromiums Balanced? Yes, we have two chromiums on each side of the arrow, so we're good there. >> Okay, next step. Balanced oxygen atoms. >> No problem here. There aren't no oxygen atoms. Is oxygen balanced over here? >> No. Ok. >> The ballots, oxygen. What we're going to do is add water molecules to one side or the other to balance the oxygen atoms in this half equation, how many water molecules do we need to add? And to which side, left side or right side? >> Right side, four. >> So we're going to add four water molecules over there. And now oxygen's balanced, still with me. >> Okay, now we're going to balance the hydrogen by adding protons in nitrogen half reaction to balance hydrogen atoms. How many protons do we add? And to which side? Left to right? It's going to be the right side. How many protons? Somebody say three more than 36, because we now have two times 36 hydrogen atoms on the left side. So we need six hydrogen atoms on the right side. But it's important that your pencil them in with the positive charge. Again, just write six h. It has to be six protons. You'll see why later in the Chromium half reaction, how many protons do we need? And on which side? Eight on the left side, four times 28 hydrogens on the right side already. So we need eight hydrogens on the left side, but we have to include the positive charge. Here's why we have to include the positive charges. Well, a, this is happening under acidic conditions. This has gotta be protons floating around. Anyway, let me now it's time to add electrons and balanced the charges. Let's start with the nitrogen half reaction. What's the total charge on the left side? >> Right now, not oxidation state charge 0. >> Do you see any charges on the left side? Ido. So total charge on the left side at the moment is 0. What's the total charge on the right side? >> Yeah, positive six. Okay. >> When I say balanced charge, by adding electrons, I mean we have to make those two charges come out equal to each other. Bearing in mind that the electrons are negative, how many electrons do we need? And on which side to make that happen? I was afraid somebody would say, if I put six electrons on the left side. It is a very easy mistake to make. If I put six electrons on the left side, that makes this side negative six and this side positive six. And that's not balanced, so you're right about the six, but where we want them is on the right side, hopefully. Now it's obvious that the sum of all the charges on this side is 0, and that matches the 0 on this side. Okay? >> Chromium half equation, total charge on the right side, 0. Total charge on the left side saying in plus six is correct. >> We have eight plus one's right here, so that's plus eight. >> But we also have a minus two on the dichromate, so plus six on the left side. >> Therefore, how many electrons do we add and to which side to make those two charges come out equal to each other? Six to the left side. And as the time for six to the left side. Yep, because now the sum of all the charges over here, 0, which matches the 0 over there. >> Still following all this, okay? >> By the way, for what it's worth, electrons on the right side, oxidation half reaction, electrons on the left side, reduction half reaction. That's not crucial. What is crucial is that we, we combine these together. Again, the electrons must all disappear, which in this case can conveniently do because six electrons on the right side, six electrons left side are going to cancel out. But suppose this had come out to be three electrons. What will we have to do before we combine these two equations? Yep, multiply the entire equation by two to get the number of electrons come out equal to each other. In this case, we don't have to do that. We can just bring everything down because the electrons will cancel out. So let's look at what all else we get to NH3. And then on the left side over here, plus eight H plus plus Cr 2072 minus becomes everything on the right side into six H plus. >> And then right side over here, CR2032 for H2O. Okay. >> That's what we get if we just break everything down and make one big equation out of it, what else can we cancel out here? We do have some things that are the same on both sides that we can get rid of. We should always be able to cancel the electrons. >> But in this case, we can cancel something else to what is it? >> The longer you go without answering, the farther away spring break gets? How many? >> Yep. >> We have eight H pluses on the left side, six H pluses on the right side. So we can cancel six H pluses from both sides, which means this term goes away and this coefficient simply becomes two, which is eight minus six. >> That in fact the correct balanced equation for this reaction taking place under acidic conditions. >> Because we have reached the point on our big scheme where we says Step four year, if it's happening under acidic conditions, we stop. Okay, it's happening under acidic conditions. >> So we stopped. >> There is the complete balanced redox equation for this, but obviously there's a bit more to it than just fiddling with coefficients. Basic steps make sense. >> Okay? >> When I see you again in ten days from now, we'll try this example by a similar method. In the meantime, I have a wonderful break. >> Rest up, relax, combat motivated, and we'll see you in ten days yeah. Hey, I got 98.8 Pham. >> Okay, so that's an average of 45 points per exam, which is passing me obviously not quite where you want to be, but it's better than where you were going after one exam? Yeah. Right. >> Okay. >> And as lab doing okay to find good 90% higher, lower like maybe like an 18 ARE okay, it's close enough to 90%. Ok, realize that lab counts as much as 2.5 exams. Labs going to have a significant impact. >> Like I mean, I know I've I mean, I've got 59 on this exam. >> I mean, it's important. >> The important thing to realize is that's a big step in the right direction. >> Rather dixon, Yeah. >> I mean, I did a lot better. Like I like I'm proud of myself. >> Like I know guy, not hold it, hold that thought because here's the thing. >> I think you and I talked before about how friendly you are with math, which is to say not Barry. >> Okay. >> If you take a look at the upcoming practice exams that I looked at, the if you looked at number three, that's five. Take a look at number four, number 52. You're going to see that the amount of math goes down insignificantly. And this is one place where people who struggle with math tend to start improving as the semester goes up. I'm remembering the case of one of my students in this course long, long time ago who got a six on the first exam, humbling, but she stayed with the core shear. And once you got past her math problems in the first couple of exams, every exam score was higher than the one before it. She just kept on getting better. >> The keypads. Oh, yeah. >> So bottom line, you can do that and you've taken a big step in the right direction, come, come back reasonably prepared for the third exam and we'll see how things go there. And then after we're beyond that, hopefully the math would be a big deal. You can get past that particular heard. I just want to see, well, if lambda is going as well as you're saying it is, that I would say you probably have about a C average now. And so the main thing to do is keep doing well in lab, but hopefully keep improving on exams or can have a good spring break. >> You do the same board. No.

chem102-010-20170324-122000.mp4

From Dana Chatellier October 11, 2018

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