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Chemical Science - Acid Base Titrations Oxidation/Reduction

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Principles of Chemical Science/n * Email this page/nVideo Lectures - Lecture 24/nTopics covered: /nAcid Base Titrations and Oxidation/Reduction/nInstructor: /nProf. Catherine Drennan/nTranscript - Lecture 24/nWe are going to finish up acid-base titration stuff today./nAnd I think actually most of you have probably ventured into the Problem Set, which is great./nI think the best way to learn acid-base titrations is usually you hear a little bit about acid-base titrations, then sit down and try to do problems, go back and revisit how to do the problems and then try it again./nThis is one of the units where you really cannot learn how to do it by just sort of listening to a lecture. You have to sit down and work numerous problems. That may be true in other units./nBut I think it is particular true in this one./nThat it is all about practicing doing the problems so it is real important to work through a number of them. And you will have other opportunities. Of course, this will be material on Exam 3. But also acid-base titration always makes an appearance on the final./nIt will be well worth your time to work these problems. They will be worth a number of points for the rest of the semester./nAnd, as I said, you will see all this information again in organic chemistry, in biochemistry and a lot of other courses that you take./nYou will talk about pKas and buffers in biology. You will see all this again. All right. We were talking about the titration of a weak acid with a strong base, and we went through this figure on Monday./nAnd the particular problem that we were working, we are going to go through and look at different time points along the titration and calculate what the pH is./nAnd the problem we were working was 25 mls of a 0.1 molar solution of a weak acid titrated with a strong base./nIn this case, NAOH. And that was 0.15 molar. And you were given the Ka for the weak acid in this problem. We had done one time point, and that was the initial time point when no strong base had been added yet./nLet me just put that down./nWe are plotting the volume of the strong base that is added versus pH. And we had one time point, and that was a pH of 2.38. And that was when there was zero mls of the strong base added. So just the pH of the weak acid in water was 2.38 for pH./nAnd then we had started calculating what happens at 5 mls, when 5 mls of the strong base is added, and that is this time point here./nHere we are in between zero and the volume at the equivalence point, and we have the 5 mls. Let me just review. We talked about this right at the end of class./nThe first step in doing this type of problem is to figure out how many moles of both the weak acid that you have in the problem and the number of moles of strong base that have been added so far./nFor the acid, there was 25 mls originally of 0.1 molar solution, so 2.5 times 10 to the minus 3 moles of the weak acid. And for the strong base now we have added 5 mls of a 0.15 molar concentration./nSo far we have added 0.75 times 10 to the minus 3 moles./nThis strong base will react with the acid. It will pull off a hydrogen ion or a proton from the weak acid and convert it to its conjugate base./nNow we need to figure out how much we have left after this reaction. We had 2.5 times 10 to the minus 3 moles of the weak acid. We have now converted over 0.75 times 10 to the minus 3 moles./nWe have 1.75 times 10 to the minus 3 moles of the acid left./nAnd now we formed 0.75 times 10 to the minus 3 moles of the conjugate, so 0.75 times 10 to the minus 3 moles have been converted from acid to the conjugate base. Now we need to know the molarities of those amounts, so we take the weak acid, the number of moles divided by the total volume now, which is 25 plus the 5 we just added, or 30 milliliters./nAnd we have a concentration./nAnd then we do the same with a weak base, number of moles that we have divided by the new volume gives you the molarity. Right at the end of class you said, OK, the next thing you can do, there are two different options./nOne option, which we talked about, was set up this little equilibrium table. And the important point to remember, and when you set this up, is that you have the weak acid and its conjugate over here./nAnd the other problem is when you just have a weak acid and water problem you just have the weak acid, or if you have a weak base in water problem you just start with a weak base./nHere, this is really like a buffer problem. You have both the acid and its conjugate to some amount, so you need to put things on both sides. We had come up with the pH of 3.38. And you have to set up using Ka, check your assumptions and calculate pH./nThere is another way that you can do this particular problem, and that is option two./nThe first part is the same. You have to do all the calculations with a number of moles and calculate molarity. The second part is different. Instead of setting up that equilibrium table, you can use the Henderson-Hasselbalch equation./nIn this case, you need to know pKa. And you know Ka. You take the minus log of Ka to get pKa. And then you can fill in your concentration of the weak acid and the concentration of the conjugate base that is formed after the addition of 5 mls./nAnd then you can calculate./nAnd you should definitely get the same pH out of this 3.38. Now, this equation, the Henderson-Hasselbalch equation only works with the assumption that X is small, so that Five Percent Rule that we talked about./nAnd that is because in the Henderson-Hasselbalch equation, if you were to remember back, we are making an assumption./nWe are assuming that the initial concentrations that we have are the same as the concentrations we have at equilibrium, which is basically saying X is small./nYou can only use it if X is, in fact, small. The way that you would check, in this case, is you have to back calculate. You know your pH answer, and then you have to calculate then what the hydronium ion concentration was, which is easy because pH just equals minus log of hydronium ion concentration./nAnd so you can get this answer out./nThen you take this and divide by either of these two. You want to use really the smaller one or both of them and see is X small? And, in this case, both of those are less than 5%. 4.2 times 10 to the minus 4 is less than 5% of this number and of this number, so that assumption is OK./nIf it was not OK, you would have to have used option one and used the quadratic equation./nIf X is not small, the only way you can do it is option one with quadratic. But most of the time X is small because we are talking about weak acids and bases./nAnd a lot of people find the Henderson-Hasselbalch a faster way to get the answer. And as you are doing these problem sets, you realize these problems take a while to do. And on a 50 minute exam, part of the process in thinking about how to do these problems should be how can I do the problem the fastest to get it done?/nAnd a lot of people find this to be a little bit of a faster way to do it./nHenderson-Hasselbalch only works for buffers, so it only works under conditions where you are starting with an acid and its conjugate base or base and its conjugate acid. Only in the buffer region will Henderson-Hasselbalch apply, but when it does apply you should think about using it./nAll right./nLet's put our other pH up here. After we have added 5 mls then we have a pH of 3.38./nOK. Now we have reached a special point called the half equivalence point. This is right in the middle of this buffering region./nAgain, the buffering region is the part of the curve that is flat where you are adding your strong base but it is not changing the pH very much because the solution is being buffered./nThe pH is staying the same, so it is the flat part of the curve. And right in the middle of the flat part you find the half equivalence point. That is when you have added, in this case, half the amount of the strong base that you need to reach the equivalence point./nAt the equivalence point, the number of moles of strong base are equal to the number of moles of weak acid you had to begin with./nAt the half equivalence point, you have converted half over to the conjugate base. You have of the acid you started with and half of it is being converted. This is in the buffering region, and so you can use the Henderson-Hasselbalch equation./nAnd you see that if you plugged in values into here and you would have equal amounts of these two, you have converted half the amounts, you half this and half that and you solved the log, you would find this term drops out./nAnd so pH just equals pKa./nIf you know the pKa, and usually the Ka is given so you can easily calculate the pKa then that is the pH. The pKa is 3.75, so that is then also the pH at the half equivalence point./nHere would be 3.75 at the half equivalence point./nNow, when I was mentioning that on an exam you are going to look for fast ways to solve the problem, if you are asked about the pH at a half equivalence point, you can write pH equals pKa equals 3.75./nThat is it. It might be a number of points question, but that is all you are looking for and you give the answer. This is a fast problem./nMake sure that you remember this trick. You can work it out the long way and figure out the same answer, but this is a perfectly acceptable answer to say that pH equals pKa equals 3.75./nThat is a shortcut for doing these problems. Then we get to the equivalence point. I mean, sometimes you might have here, this has point D also in the curve, but we will skip that./nYou would do that problem the same way you did the problem on the other side of the half equivalence point, so we can skip over that and move right to the equivalence point./nAgain, the equivalence point, also called the stoichiometric point is when you have added the amount of strong base, the moles of strong base that are equal to the moles of the weak acid you had to begin with./nYou have converted all of your weak acid to its conjugate./nIn terms of the pH, the pH is going to depend on the salt that is formed as a result of this addition. In the case when you are titrating a weak acid with a strong base, you are going to get a pH greater than 7 because all of your weak acid will be converted to the conjugate base./nAll you are going to have around is conjugate base, and base is basic so you are going to be greater than 7./nIn this case, you will get the conjugate base in the form of a salt and water. The NA+ doesn't do anything for you./nIt has no effect on pH. All Group 1 and Group 2 have no effect on pH, but we have a conjugate base so we are going to have a pH greater than 7 at the equivalence point. And this is a good way to quickly check your math./nIf you know what kind of problem it is and you don't get a pH that is greater than 7 for this type of problem, you know you have made some kind of math mistake somewhere along./nAnd you can go back and try to find that./nAll right. How do you do the problems? You want to find the pH then at the equivalence point. The exact pH. You know it is greater than 7, but what is it exactly? The first thing you need to do is find the total volume, and you need to know the volume of NAOH that you added to the solution to get to the equivalence point./nYou know how many moles of acid you had to begin with./n2.5 times 10 to the minus 3. That is how many moles then of conjugate base you formed. And that must be how many moles of OH you needed to add to get the equivalence point./nYou need this number of moles. This was your concentration. And so that is how many liters needed to be added to get the equivalence point. And in terms of your total volume then you had 25 milliliters to begin with./nYou are adding 16.7 of the base./nNow you are at 41.7 milliliters of volume. All right. Just to show this again in cartoon form. The point to remember, at this equivalence point, is that this is all you have./nYou just have that conjugate base. If you had four molecules of weak acid to begin with and you added four molecules of NAOH then each molecule of NAOH or of OH will pluck one of these hydrogen ions off and will leave you then just with a conjugate base./nYou are going to have equal moles here to what you had here, so you have these equal moles./nIn terms then of the molarity of this, you have to have the same number of moles now, or the weak base is what you had of the weak acid to start with. You have converted all of it to the weak base./nYou have that many moles and you calculated the total volume so you know your new molarity./nThis is just a weak base in water problem. And we talked already about how to do this. We can set up this table here. Now all we have is the weak base. We don't have any of this left. It has all been converted so all we have is this./nWe have 0.0600 molar minus X, X and X./nNow, again, you cannot use the Henderson-Hasselbalch here. It is not a buffer problem. We have to solve it using Kb because it is a base in water. We cannot use Ka./nWe have to use Kb for a base in water. Kb is the base ionization constant. If I was given a Ka, how would I find the Kb for the conjugate? What would I use? Kw, right. That is an easy conversion, using Kw to calculate Kb./nAnd then that would be equal to X squared over 0.06 minus X./nWe can simplify that if X is small, double check that that is OK, and if we do that we find that the OH concentration is 1.83 times 10 to the minus 6 molar. Then we can calculate a pOH, and from that we can calculate the pH./nIt would be 14 minus this and the pH is 8.226./nThat is greater than 7 so that makes sense. We can draw that up here. We have 8.26 at the equivalence point. And that is greater than 7, which is what we expect./nIt should be basic./nNow, a lot of people forget because they are going quickly, to convert from pOH to pH. And that should be a way to double-check. If you find out that you are the equivalence point and the pH is less than 7, you realize that something is wrong./nAnd hopefully you will remember to convert then to pH. Greater than 7. All right. Beyond the equivalence point, so the last kind of point that you may have in a curve up here./nBeyond the equivalence point, you still have your conjugate base hanging out and you still have all that you had here, but that was not making the solution really very basic./nThere wasn't much. It was like 10 to the minus 6 molar. There was not much of it. And so the pH is really going to be determined only by the amount of excess OH that is added to the solution. Now you can forget all about your weak acids and weak bases at this pint and just look at this problem as a strong base problem./nYou are swamping the system with your excess strong base./nThis kind of problem at the end, past the equivalence point, is going to be a strong acid, strong base type problem. In this case, a strong base problem. We know how to do this already. We talked about this before./nIf you are 5 mls beyond, and you say 5 mls times the concentration, so you have 7.5 times 10 to the minus 4 moles of excess base./nThe only trick to this problem is remembering the entire volume that you have in there./nYou know how many moles you have and it is 5 mls past, you had 25 mls to begin with, you added 16.7 to get to the equivalence point, add those together and you can get the concentration. Now, remember the concentration we had at the equivalence point of OH was like 10 to the minus 6./nThat is still around, but if you add something of 10 to the minus 6 to this number it is not significantly going to change it at all./nYou can ignore that and only consider the concentration of the OH that is due to that excess strong base added and calculate the pOH, and then from that the pH. Our final point on the curve then somewhere up here 12.21./nAnd this is 5 mls beyond the equivalence point./nThere we have a titration curve where we have calculated various points along the way. The point to remember in doing these problems is that you just need to break down each part of the curve and define it as a different type of problem./nHere we started out and we did a weak acid problem./nThen we were in the buffering region and we did a buffer problem. We had this special category of buffer when we were at the half equivalence point./nAnd then we moved up here. All the weak acid was converted to weak base at this point, so it is just a weak base problem. And then beyond the equivalence point it is a strong base problem. The titration curve, this is why it takes a long time to do these, because it is really many problems all wrapped up into one problem./nAnd the other thing to remember, especially if you are given different subsections, you are not always going to be working through it systematically./nYou may be asked to start here or start down here. And so you always have to ask yourself what kind of problem is this? And I tell people who are grading the exam that if you write on the top of the problem this is really a buffer problem or this is a weak base problem at the equivalence point, if you write something down that tells me that you know what type of problem it is that is worth some points right there, even if you get lost in the math./nIf you think about these in a way that will help you do them, it also might give you a little partial credit on the test./nAll right. Again, there are just these types of problems. And it is a matter of picking and choosing and putting them all together. There are not a huge number of types of problems, but it is hard sometimes for people to figure out where they are which one of these things to apply./nAll right. That moves into today's handout./nI am going to start today's handout finishing acid base. And I did this for the first time last year, this kind of little cheat sheet on how to solve these problems./nAnd I will write this stuff on the board. And as maybe some of you remember, I mentioned that last year we had the best performance on acid base that I think we have seen in 5.111 for a really long time./nPeople did really, really well on this./nAnd so I would like to continue that trend of having fantastic acid-base titration answers where there are little smiley faces drawn on the exams where everything is really perfect./nLet's see if we can keep this going. All right. Let's work on our cheat sheet and how you go about solving all the different kinds of problems. First we will just quickly review this type, the weak acid being titrated with a strong base./nLet's consider now, we will work backwards through this, when we are beyond the equivalence point./nAnd I guess I should say this is weak acid, strong base. OK. Beyond the equivalence point then is a strong base problem, as we just did./nIt is really due to the excess strong base that we had. We would solve this problem saying, OK, the extra strong base, and that is the number of moles, over the total volume./nAnd that is going to give you the concentration of hydroxide ion. And then you can find pOH and then pH./nThat is how you would do that category of problem here. Let's move then to volume equals the volume at the equivalence point./nAnd this is a salt problem that gets broken down into a weak base problem, because now we have converted all our weak acid into its conjugate base./nThis is an A- plus H2O going to HA plus OH- type problem./nAnd we would use Kb, because we are talking about a base in water, to find the concentration of hydroxide ions. Then you would calculate pOH and then pH./nAnd you should know how to write an equation out for all the types of problems you are doing./nIt will help you figure out what you are doing in the problem. When your volume equals the volume at the half equivalence point, pH equals pKa. And that is really it./nYou might have to find pKa from Ka, but that is pretty straightforward./nNow, when you are in the buffering region, again, the buffering region is where the pH titration curve is in the flat region./nThis is a buffer problem./nAnd here mostly you want to try to use the Henderson-Hasselbalch equation, pH equals pKa minus the log of HA divided by A- of your weak acid and the conjugate base. And you can use this if the Five Percent Rule applies, but that usually does so that is probably going to be your first route to do these calculations./nAgain, you have to find the concentrations of these./nYou have to remember that the amount of strong base you are adding is converting some of the weak acid to the conjugate. You need to find out how much gets converted, find these concentrations and then plug it into this equation./nThis equation only works for a buffer problem./nAnd I also just want to say that every once in a while people decide instead of pKa, let's rewrite this equation for a pKb instead. Don't do that. That is my little tip./nIt has been tried, and it has never actually succeeded in yielding the correct answer. If someone wants to say on this test I will come up with this new equation using pKb and I will prove that we can get it, you can do that if you want, but I think it is best to stick with this equation just from previous experience./nOK./nThen lastly volume equals zero, so this is a weak acid problem. And a weak acid problem, you have your acid in water going to hydronium ion concentrations and your conjugate base. You are going to use Ka, the acid equilibrium constant to find hydronium ion concentration./nAnd then from that you can calculate pH./nThose are the different ways that that breaks down. Now let's do the same thing for a weak base strong acid./nAnd I will move that onto the next. To start with, start at the top of the chart. Here, in the beginning, when volume equals zero, all you have is your weak base./nAnd then, as you add acid, the pH is going to drop. In the beginning, when volume equals zero, this is a weak base problem./nAnd so you can draw that as your base in water going to the conjugate acid and hydroxide ion concentration./nNow you are going to use Kb, the base ionization equilibrium constant to find the concentration of OH-. Then you can calculate pOH and from that calculate pH. Next we will consider the buffering region when the volume is in between zero and the volume at the equivalence point./nAnd so this is a buffer problem. Again, I recommend using Henderson-Hasselbalch as written with pKa./nAnd so here you need to know the concentration of the conjugate acid over the concentration of the weak base that you had./nAnd so you can use this if the Five Percent Rule applies./nAnd you can check that by calculating./nThis will also work. Now, remember you are starting then with your weak base. You have converted some of it to the conjugate acid depending on how much strong acid you have added./nBefore you can plug in numbers here, you have to subtract how many moles of the strong acid were added, so that means how many moles of the weak base do you have left, how many moles of the conjugate acid have been formed./nAgain, we are in this region./nIf you had, say, four molecule to begin with and you added one molecule of the strong acid then you would be left with three and you would have one converted. These problems all involve subtractions before you can plug in your numbers to the Henderson-Hasselbalch equation./nThen we have the special category of buffer where your volume equals the volume at the half equivalence point. Again, you can just solve this. pH equals pKa./nIf you are given a Kb, you will have to convert to a Ka to find your pKa, but this will still work./nMaybe I will write on this next board./nThen your volume at the equivalence point. And this is, again, a salt problem./nBut the salt problem breaks down to a weak acid problem because when you have a strong acid its conjugate is neutral./nAnd if you have a weak base its conjugate is also a weak acid. The salt that is formed in this problem is going to be acidic. We can write that down./nWe have our conjugate acid that has formed as a result of the titration in water forming the conjugate base and hydronium ion concentrations./nNow we can use Ka to find hydronium ion concentration, and then we can calculate pH. Again, at the equivalence point you converted all of the weak base you had down to too weak acid./nYou have added enough acid that it has protonated all of the weak base you have and now all you have left is acid, so it is really just an acid in water problem./nAnd then, finally, if you are beyond the equivalence point, now you are just pouring excess acid in. And so this is a strong acid problem./nYou still will have a little acid solution left from the titration, but that is going to be overwhelmed by the amount of strong acid./nYou can solve that by just saying how many extra moles of strong acid over the total volume. And that is going to give you hydronium ion concentration and then you can calculate pH. At this point it is just a very simple calculation./nThe only tricky part is to figure out the total volume to get that number right./nThat is the cheat sheet which should help lead you through the rest of the problem set and also problems for the exam. Let's do a little bit, we have just a little bit that I want to do today on oxidation reduction reactions./nAnd so this is the next unit that we are moving into./nAnd today we are just going to talk in the last 15 minutes about some simple rules for doing oxidation reduction. We are also moving away from the acid base, but you will find there are a lot of parallels between these units./nWe are still going to be talking about equilibrium constants. We are still going to be talking about delta G as we go along. This is another application of thermodynamics and chemical equilibrium./nThere are some rules to start with./nMost of them are actually pretty simple. And you need to know what these rules are for assigning oxidation numbers. Here are the rules. If you have a free element, the oxidation number is going to be zero./nIf you have ions then your oxidation number, here is lithium plus one./nYou have an oxidation number of plus one, so the oxidation number is equal to the charge on the atom. And we find that Group 1 are always going to have plus one, Group 2 metals on the Periodic Table are always going to be plus two and aluminum is always going to be plus three./nThose are the ones that will help guide you through doing these problems./nOxygen is a little bit more tricky, but it almost always minus two. The only exception is in peroxides where you have a minus one state./nThis will also help you in solving a lot of these problems, but you are actually going to have a couple of examples where oxygen is a bit different. Hydrogen also is usually plus one, but there are exceptions./nAnd you will actually see a number of exceptions in this unit./nYou should know that in these particular cases it will be minus one. Here we have Group 1 and Group 2 metals which will always be plus. And so in this case, hydrogen is minus one. Fluorine is always minus one./nOther halogens are usually minus one, but there are some exceptions./nWhen you combine it with oxygen you have an exception to that. And so then you could have a different oxidation state when it is combined in these oxoacids./nAnd so then you would have a positive oxidation state or oxidation number. All right. Things need to add up. In neutral molecules the sum of all the oxidation numbers needs to be equal to zero./nIf you have a polyatomic anion, the sum of the oxidation numbers needs to add up to the net charge on that particular ion./nYou need to be able to sum to get the overall charge of the molecule. All right. Let's look at an example, NH4+. What would the oxidation number of hydrogen be in this case?/nIt will be plus one here./nWhat about nitrogen?/nAnd so the sum needs to add up. And so if you have four plus one minus three gives you plus one. There is one example. All right./nAnother rule is that they don't always have to be integers./nMost of the time they are, but the oxidation number of oxygen in O2 minus one would be what?/nMinus a half for each individual oxygen atom in there. That is kind of one of the only exceptions to this./nYou won't see this very often. All right./nLet's look at a couple more examples to get you used to doing these problems. And I will just erase this part right here./nOK. Let's look at Li2O. What is oxygen going to be in this compound? It will be minus two and lithium will be plus one./nLithium is always going to be plus one. And so if we have plus two minus two that gives us zero, which is the overall charge. There was no charge. That was a neutral molecule so that worked out./nLet's look at PCl5. What is chloride going to be in this case? Minus one. We have five times minus one. What does that give us for phosphorus?/nPlus five. You will actually find in this unit on oxidation reduction, we are going to go into balancing oxidation reduction equations, too, and oxidation numbers, it is counting./nThis unit, lots of counting. It may be a little break. I think people find the acid based titrations a little bit of thinking more here, you count, so that might be a little nice relief coming up./nOK./nHNO3, what would be oxygen? Minus two, and we have three of those. What is hydrogen going to be here? Plus one. What does that leave us for nitrogen? Plus five, right, to get our overall sum equal to zero./nAnd one last one./nN2O. What is oxygen here? Minus two. What does that mean nitrogen has to be? Plus one. Two plus one to give us zero. See, nitrogen can have lots of different numbers. And a number of atoms can./nThat is why knowing some things in the rules about oxygen, chloride and other things will help you figure out then what that other molecule has to be./nAll right. Here are the whole set of rules. OK./nOops, going in the wrong direction. All right. Let's look at a few definitions. Oxidation, most of these should be familiar to people. Oxidation means that you are losing an electron. Reduction means that you are gaining an electron./nAnd then we have an oxidizing agent, and an oxidizing agent is something that accepts electrons and gets reduced itself./nAnd a reducing agent is something that is going to donate the electrons, give electrons away, so it itself becomes oxidized./nAnd these two terms often confuse people because an oxidizing agent goes and oxidizes other things, but it itself is reduced so it is an agent of oxidation. It is running around trying to oxidize other things./nThat is the part to remember./nSometimes people get these parts backwards. These are really a lot of the definitions you need to know. There are not so many definitions in this particular unit. One other term that you need to be familiar with in here is these disproportionation reactions./nAnd so, like we saw with acid base where we could have something that could serve as an acid in a base is amphoteric./nWe also find that you can have an element that can be both oxidized and reduced in a particular reaction./nWe saw that with acid-base we had water plus water going to hydronium ion and hydroxide ion where one molecule of water was acting as an acid and the other was acting as a base. A similar kind of idea here, but in this case we are having one thing be oxidized and reduced./nLet's take a quick look at that. We can break this down to two different equations./nAnd sodium is a spectator ion. Poor sodium really doesn't get to do much of anything in both this unit and the last unit./nAnd we are only focusing on these species here. You are going to have one ClO- going to ClO3- and another one going to Cl-. What is happening here?/nLet's take a look at what is happening. Let's look at the oxidation numbers./nOxygen here will be minus two. What does that leave for this chloride? If we write that out, it all has to be equal to one. That would be plus one. And let's consider what is going on over here./nAgain, this minus two, three of them here. And it all needs to be equal to minus one over here./nWhat does that leave for this? We have plus five. We are going from plus one here to plus five. And so this particular guy is being oxidized./nIt is losing electrons here. Now let's look down here. We already did this one. It is the same. We are at plus one. And here we are going to minus one. We have this particular guy going from plus one to minus one, so it is being reduced./nYou have the same chloride in both cases, but in one reaction it is being oxidized and in the other it is being reduced. That is an example of a disproportionation reaction and also another example of how you look at what the oxidation numbers are./nAll right. Let's see if we can actually get through some of these balancing reactions./nHow many people have seen balancing oxidation reduction reactions before? A lot of people. Again, it is counting./nI think we can go through this very fast. You have to consider whether an acid solution or basic solution at least some of the time. Let's do acid solution first. You can take this and write your unbalanced equations out for this./nBreak it down into two halves and figure out what is happening./nAgain, here we have minus two times seven. And it all has to be minus two. This oxidation would be what? Six times two. We are plus six to plus three over here./nPlus six to plus three, and so we have a reduction. We are going down in number here./nAnd so this one is the reduction, and that is good because over here we have plus two to plus three so we have our oxidation in the other part./nYou are always going to have oxidation reduction. You will always have something being oxidized and something else being reduced. Now we have to balance this. And so the first thing you are going to do, you are going to look and insert coefficients to balance the elements, except for oxygen and hydrogen, on both sides./nDo we need to do something up here or what? Multiply by two./nWhat about down here? Nothing. Now we are going to add water to balance oxygen. Again, we only need to worry about one side. There are seven oxygens here, so we have to add seven waters on the other side./nWe have our seven waters here./nNow we want to balance hydrogens. And you can add H+. The book uses hydronium ion, which makes it much harder, so I suggest you don't do that and just add H+. If you just add H+, how much do you need to add? You need to add 14 to the other side./nAgain, nothing down here. And so now you need to balance charge by adding electrons./nIn this case it is going to be plus six because we have 14 minus two on this side and we have this six on the other side./nAnd then over here what do we have to add? We just have to add one. Now we take this, and we need to multiply the reactions so the number of electrons are equal. We would multiply by six here./nAnd now our electrons are equal./nThe electrons have to cancel out. If they don't you have done something wrong. Now we add and make the appropriate cancellations. Here is the whole equation. And we should be able to cancel out electrons./nIf we don't we did something wrong. And can we cancel anything else out? Not really. You want to double-check that it is balanced. 14H is here./nTwo chromiums. The seven oxygens and the six ions, that is the same on the other side, the same charge on both sides./nAnd then finally if it is basic solution all we are going to do, the same steps, is adjust the pH by adding OH. If we had 14 +, we would add 14OHs and then come along and you would add those up. And then you can cancel again./nAnd so now, instead of having H+, we have OH./nAnd so in acidic solution you would have H+ and in basic solution you have OH. That is balancing. It is pretty simple. You just have to make sure that everything adds up. (less)

Channels: Lectures Chemistry (General)
Tags: Oxidation Reduction

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