Understanding Redox and Electrode Potentials

Hi there, my name is Chris Harris and I'm from Allery Chemistry and welcome to this video on OCRA, Redox and Electrode Potentials. So in this video we're going to look at Redox and Electrode Potentials specifically for OCRA. So if you're studying OCR chemistry at A-level then this video is specifically designed for you. There are a full range of videos for all the major UK exam boards on Allery Chemistry YouTube channel, they're all free to access so please subscribe and show your support because there's no charge for them other than just... hit the subscribe button that'd be brilliant and all the videos here are designed to be revision based so you might have gone through this content already at school or college or you might be self-teaching yourself in which case it contains the main points that are tailored to OCRA so it is important though to practice exam technique as well there are videos on Allergy Chemistry that looks at exam technique so have a look at them as well because that is just as important as knowing the content It's a bit like if you're learning to drive, it's a bit like picking up the highway cord. And you may know the highway cord back to front, but it doesn't mean you need to hop in the car and start to drive. So it's exactly the same approach you should take to your exams as well. All these slides here, if you want your own copies of these slides, they're available to purchase in my Tez shop. So if you click on the link in the description box, you'll be able to get a hold of them there as well. They're great for revision, great value, and you can use them as and when you require. And it's a good way to supplement your revision that you get from school or college. Of course, we get a lot of people who are doing it and being self-taught effectively. Okay, so like I say, this is specifically designed for OCR-A and it matches the specification. So these are the points taken from the specification and everything you see in here will adhere to them rules. Okay, so let's look at reduction and oxidation. So you might have seen bits of this in year one. one chemistry um so this is just um confirming what you may may already know and extending it into this topic because it is all about redox you're going to see quite a lot obviously because that's the name of the title of the of the video so electrons are transferred when reduction oxidation occurs and we call we use an acronym oil vague so oxidation is loss and reduction is gain of electrons um so you might have seen that gcse you might have heard of it already but it's a really good acronym and it's It's very fitting for chemistry as well, calling it oil rig. So this helps us to understand what's going on. So for example, this reaction is when calcium completely is burnt in oxygen. It involves reduction and oxidation, and we call this a redox reaction. So we've got reduction and oxidation at the same time, not to be confused with the shower gel. In this reaction, calcium is being oxidized and it is losing... electrons so we can see here that calcium is forming calcium 2 plus and two electrons are also being produced as well In this reaction, this is the other half of the reaction for oxygen. So in this reaction, oxygen is being reduced and is gaining electrons. So you can see here that the oxygen is gaining two electrons, a half or two, is gaining two electrons to form O2-. So this is a very simple way of showing redox reactions. We're showing the oxidised side first, then the reduction side. So it's just breaking that down. And if we combine them and cancel the electrons out, it's just... you'll see later, then we form the full equation. And it's very important as well, you'll come across reducing agents and oxidizing agents. So these are, don't get them confused with oxidized and reduced. Although they go hand in hand, they mean different things. So for example, reduction is the gain of electrons, but reducing agents lose electrons and are themselves oxidized. You've got to remember that. So it's just like the opposite. So reduction is gain, reducing agents lose electrons. Okay. And oxidizing agents gain electrons and are reduced themselves. So you've got to know the difference between a reducing agent and an oxidizing agent. Because you'll use them or you'll see them quite a lot in A-level chemistry. Okay, so let's look at how we balance half equations. So half equations show reduction and oxidation stages in two equations. All half equations must have electrons in this. So the rules. The first thing is you write down the species before and after the reaction. Number two is we balance any atoms apart from oxygen and hydrogen. These will be dealt with later. Third rule is we balance any oxygens with water. The fourth rule is we balance any hydrogens with H plus ions. And the fifth rule is we balance charges with electrons, which is E minus. Now, you might have a different method of balancing equations. You might be taught a different method. I find that this method... works really well with even some of the obscure balance and half equations. So this is just the method that I think is the simplest method. There are other ways in which you can balance it, and you may have been taught something else, but this is just another method. It all comes to the same result anyway. Okay, so you'll see that I've put asterisks near 3 and 4. This only should be done if we need to balance oxygen and hydrogens. It might not be the case that we need to balance oxygen and hydrogens. Not all half equations have them in. But these rules are fundamental because you're going to see this all the way through the topic. So make sure you're familiar with this and all the methods that I'll do will be using these rules. So we're going to write half equations showing the conversion of iron 2 plus to iron 3 plus. So there's the first thing is to write down the species before and after. So it's Fe2 plus, Fe3 plus. We then need to balance any atoms apart from those with oxygen, hydrogen as these will be balanced later. Well, they're already balanced. We've already got one Fe and one Fe on the left and one Fe on the right. Balance any oxygens. Well there isn't any oxygens so we don't need to do anything with that. And the same again with hydrogen. There's no hydrogens in here that we need to be concerned about. So we go straight on to step 5 which is balance any charges with electrons. So you can see we have a 2 plus on the left, a 3 plus on the right. So an electron is always negatively charged. It's got a minus 1 charge. So we put the electron on the right hand side. So that means we've got a 2 plus on the left and 2 plus on the right. This is shown in oxidation state. So oxidation is loss of electrons, Fe2+, losing electron to form Fe3+. Okay, so we need an oxidizing agent to allow this to happen. Okay, so remember this has been oxidized, but we need something to actually allow this to happen. So let's have a look. So we're going to write a half equation showing the conversion of MnO4-, that's manganate, to Mn2+. So the first thing... we need to write the species before and after the reaction so it's MnO4-going to Mn2+. We need to balance any atoms apart from oxygen and hydrogen and deal with these later. So it's already balanced remember we ignore the oxygen we know there's four oxygen left but we ignore them we're just looking at manganese so there's manganese one Mn on the left and one Mn on the right. We now need to balance oxygens and we use water to do this now we know there's four oxygens on the left So it means we must need four oxygens on the right. And we know that water has one oxygen in there, so we need four waters. There we are. So we need four waters on the right. Then we need to balance any hydrogens with H plus ions. So you can see we have eight hydrogens now because we've chucked that water in there. We've got eight hydrogens on the right, so we need eight hydrogens on the left as well. So there it is. Okay. So you can see it's fairly straightforward. Balance charges with electrons. So now we need to look at the charges on the left. Total charge on the left. And total charge on the right. So on the left we have a total charge of plus 7. Because we've got a minus and 8 positives. And on the right we've got plus 2. We've got 2 plus. So the electrons go on the left hand side. And we need 5 of them. And that will bring the charges equal. Left and right. Remember the charges don't have to be 0 either side. They just have to be balanced. And we use electrons to do that. So there's our half equation for that. This shows reduction, so we've got a reduction step here, because reduction is the gain of electrons, so we've got gaining of electrons on the left-hand side. So reduction will always show electrons on the left-hand side. Oxidation will always show electrons on the right-hand side. So what we can do with these two half equations, so the iron one that we've seen before, and the manganese one and the manganate one that we've seen here on here, we can... combine these two equations and form a full ionic equation because these equations are half equations because they contain electrons. Okay so let's look at how we combine these half equations. So these two half equations can be combined to make that full ionic equation. We're just going to make sure our electrons balance. So remember this one shows oxidation so we've got Fe2 plus going to Fe3 plus plus an electron it's an oxidation reaction and remember the manganate one Shows reduction because it's gaining electrons. So electrons on the left-hand side is reduction. Electrons on the right-hand side is oxidation. So you can see the electrons don't balance. So we've got to make sure, and if you've done, you might have done this in maths, which is simultaneous equations, this is exactly the same principle. So we're finding in this case the commonality is always the electrons in this case. There might be other elements, but it must be the electrons to start with. So electrons don't balance, so we need to multiply the top row by 5. So here we go. So we've got 5 Fe2+, we'll form 5 Fe3+, and 5 electrons. Okay, so we cancel the electrons out and we combine the two equations together. So there we are. So we've got MnO4-, 8H+, 5Fe2+, we'll form Mn2+, 5Fe3+, and 4H2O. So we can see we've cancelled the electrons and your ionic equation, your full ionic equation, should never have electrons inside there. Also, your ionic equation also shows reduction. and oxidation because we've combined the two separate steps and we formed a redox reaction. So that's the the full ionic equation is your redox reaction. Okay so let's look at redox titrations. So you would have seen titrations before already. So the titrations are actually showing you the titrations you would have seen are acid based titrations. These ones are going to be slightly different. These ones are going to look at redox reactions instead. So we're not using acid in the base. We're using redox substances, so reduction in oxidation. So transition elements can be used to oxidise and reduce other substances. And we'll look at transition metals later because they're quite a special group of chemicals. So transition elements have variable oxidation states, which means they can accept or donate electrons easily. Hence the word transition, they can transition between various different states, unlike other elements which generally have fixed oxidation states. So transition elements are coloured depending on their oxidation state at the time, and this makes them particularly useful in titrations because we can spot an end point. So remember in acid-base titrations where you have to add an indicator to spot an end point, these ones generally you do this without the use of an indicator, so you just use the... the colour of the transition element to do that. So let's look at a specific example. You might have seen this example as well in lab work if you've done this. But we use acidified potassium permanganate solution, which is KMnO4, is an oxidising agent. Okay, so remember what an oxidising agent is. So oxidising agent is something that gains electrons. It dissociates to produce MnO4-ions, where... Mn has an oxidation state of plus seven in there Okay, and they are reduced to Mn2 plus during the titration now You've got to remember your oxidation numbers and how you work out oxidation numbers And if you're not sure are you a little bit rusty on that area? Then there are videos that I've done in the year one playlist and that show oxidation numbers specifically for OCR as well So they they are there so have a look, but I'm going to assume that you're fairly confident with it. I'm going to go through some of the basics as well, but just to skip past that bit. Okay, so they're reduced to Mn2 plus ions during the titration. Okay, so let's have a look at the first half equation. So we've got MnO4 minus plus 8H plus plus 5 electrons will form Mn2 plus. plus 4H2O. And so the MnO4-is reduced. Obviously, that's supposed to be a little n rather than a capital N. So that's been reduced. And obviously, we know how to make these half equations because we've just seen them before. What's been oxidised in this titration is iron. So we've got iron 2 plus, plus iron 3 plus. We'll produce iron 3 plus, plus 5 electrons. This has already been balanced for us. So the Fe2 plus is being oxidised. So if we combine them we get this overall equation here just like we've seen before. Now what we see with this reaction when we do the titration is we start from permanganate which is a lovely purple colour and it goes colourless as we progress through the titration if we're doing it this way. So this colour change from purple to colourless will show us an end point. So another one acidified potassium dichromate is an oxidising agent and it dissociates to produce dichromate ions, so Cr2O72-ions, where chromium has an oxidation state of plus 6, and these are reduced to Cr3 plus ions during a titration. very similar to manganate but a different substance so here we've got cr207 to minus has been reduced so you can see here that it's going from cr207 to cr3 plus so it's been reduced looking at the oxidation numbers and then using zinc for this reaction zinc has been oxidized so to zinc 2 plus so combining them we get this and the color change we see it goes from orange to green So this is a classic sign again it's self-indicating we haven't had any indicator here but we can detect an end point as we form a product and from orange to green. Okay so let's look at calculations because it wouldn't be a titration without a calculation of course. So titrations can be used to work out the concentration of a reducing or oxidizing agent. So in this example we're going to look at finding the concentration of manganate. Now we looked at that in the previous slide. by titrating against a reducing agent like Fe2+. So this example looks at finding a concentration of a reducing agent. All we do is we just reverse it for oxidizing agents. There's no difference. So what we've got here is our classic titration setup. You've got your burette. You've got your burette here, which is at the top. There's your burette. And then we've got a conical flask at the bottom. What we'll have in this conical flask... a reducing agent so we have Fe2 plus solution with an unknown concentration but we know the volume okay so we don't know how strong it is but we know we probably have say 25 centimeters cubed in that conical flask so what we've got to do is got to add dilute sulfuric acid into there as well so this is all dissolved in an acidic solution and so this is to ensure that you have sufficient H plus ions in that solution to allow the reduction of the oxidizing agent so it's just making sure there's a backup supply there and to allow this to happen so we it's basically acidified it's an acidified solution okay so in our burette we have an oxidizing agent in uh in the burette with a known concentration so we know the strength of it um which is in this case is going to be manganate ions um so mn o4 minus and then We're going to add the manganate ions into the burette, in the burette, into the conical flask until we see a faint colour of permanganate, so the manganate ion appear. And this is known as the end point. So there's no indicator added here. We're going to use the manganate to provide the end point for us. And we add it drop by drop near the end. Again, you would have done titrations before. It's vitally important to make sure you get an accurate reading. And we do that by adding it really slowly near the end if we go too much one drop can make a difference And if we go too quickly you'll get a result that is inaccurate Okay, so and there is a sharp color change okay like I say it's very dramatic And so the manganate ions form aqueous potassium permanganate solution are purple okay and they'll immediately react with the reducing agent until all the reducing agent is used up. So it'll react with the Fe2+, and it will keep on doing that and keep on doing that until there's no more Fe2+, ions left, because they've all reacted. So one drop at the end, like I say, can turn the solution purple, which is the colour of the oxidising agent. And you could use a colourless oxidising agent and a coloured reducing agent, and you could look out for the colour disappearing. But in this example, what we're looking for is the colour to appear. in the conical flask but it can be done the other way it just depends on on the setup of your titration so what we're going to do is read how much oxidizing agent was added and we're going to read the bottom of the beniscus and we always read at eye level so you get right down to eye level right where you can see um the the burette um the the level and you read from the bottom which is the the semicircle the bottom of the semicircle there and you record your results to two decimal places at all times okay and you repeat until you get two results that are concordant so they've got to be within 0.1 centimeter cubed or 0.10 centimeters cubed of each other and that just ensures that um your you can your results are valid you can depend on your results if your results are all over the place you it's very difficult to say what the actual value is if you don't have any that are um that are concordant okay so um so Like I say, they can be used to work out the concentration of a reagent. So let's look at a specific example. So we've looked at the practical side of it, about how you can conduct a titration. But here we're going to look at how you actually calculate it. So we're going to take the same, exactly the same reagents. We're going to have 18.3 centiHg cubed of 0.025 moles per diem cubed. of potassium permanganate, okay, which is KMnO4, and that's reacted with 25 centimeters cubed of acidified iron 2 sulfate solution. Remember it's got to be acidified to produce them H plus ions. And then we calculate the concentration of Fe2 plus ions. So that's what we want to work out. So the first thing here is we need to write out the equation and balance it. So we've learned how to do that already. So we know how to create a half equation, and we know how to combine it to form our ionic equation. So this is the ionic equation here. We're going to put the potassium permanganate into the burette and we're going to put the acidified iron sulfate solution into the conical flask. And that iron sulfate solution is going to be the source for iron 2 plus ions. Because you can't just get a pot of Fe2 plus, it has to be as a substance like that, so iron sulfate. So we write it out and bounce it. So we've seen that before. Then what we need to do is calculate the number of moles of manganate. If in doubt, always use the saying, if in doubt, work out the moles. So moles equals concentration times volume. Concentration is 0.025 because we've been given that. Times by 18.3 times 10 to the minus 3. So I like to use, you've seen my videos before, I like to just, instead of putting divide by 1,000, I like to just put times 10 to the minus 3 on the end. That just means the same as dividing by 1,000. And that gives us the total number of moles of manganate. So the total moles of manganate are 4.58 times by 10 to the minus 4. So the third thing is we use the equation to find out molar ratio in order to work out the moles of Fe2+. So we can see from the equation there's a 1 to 5 ratio between MnO4-and Fe2+. So the moles of Fe2+, is basically going to be the number of moles that we've calculated before, multiplied by 5. So that's going to give us 2.29 times by 10 to the minus 3 moles. So now we know the number of moles of Fe2+, in that conical flask. So then what we need to do is you then calculate the concentration using concentration equals mass over volumes. So the concentration of moles per dm cubed is 2.29 times by 10 to the minus 3 divided by 25 times by 10 to the minus 3 because we've got 25 centimeters cubed of that. So remember convert to decimates cubed and that gets us 0.092 moles per dm cubed of our concentration of our iron 2 plus. Okay. So let's look at another example here, which is iodine sodium thiosulfate titration. So these are a little bit more complicated. Now, I have done another video on this, and I'll put a link on there all the way through this. I'll put a link, so if you have a look on the, I think it'll probably be on the top of the video, and then you'll be able to have a look at the whiteboard version of this, where I actually go into a little bit more detail. The titration is useful for finding out the concentration of an oxidizing agent. The titration work is conducted in three broad steps. The first step is we use the oxidizing agent, which is KIO3, to oxidize iodide ions to iodine. That's step one. Step two is we then carry out the titration to work out the moles of iodine produced in step one. Then the third step is we use the moles of iodine in step two. to work out the concentration of IO3 minus ions. Okay, so there's three broad steps. So we're going to look at step one first. Okay, so step one, remember, is use the oxidizing agent KIO3 to oxidize iodide ions to iodine. So what we need to do is measure out a volume of KIO3. That's potassium iodate 5, because it's got an oxidation state of 5. And that's your oxidizing agent. And this will produce the IO3 minus ion needed. So usually we'd use 25 centimeters cubed, but it could be any volume. 25 centimeters cubed is normally the volume you'll probably see. So we're going to add an excess acidified potassium iodide solution, Ki, to the KiO3 solution. And so here's the equation. So we've got Io3 minus, which is your aqueous solution. And we're going to react this with our iodide ions. So this is our oxidizing agents. H plus so it's an acidic solution okay so we've got H plus ions here and that's going to form water and there's our iodine that's that's produced here now the I minus ions are oxidized effectively to I2 and so the more concentrated the oxidizing agent is the more I minus ions are oxidized so that's critical okay so the more concentrated this is then the more of these ions are actually converted into iodine That's going to be pivotal when we look at the next steps for the titration. Okay, so let's look at step two, which is the method. So remember step two, we've kept the main steps up there at the top, just as a reminder. So we've used the oxidizing agents now, and now we're going to carry out the titration to work out how many moles of iodine were produced from that previous step. So here's the method. So we're going to add the solution from step one into the conical flask. That's going to go into the bottom there. Then what we're going to do is we're going to add sodium thiosulfate Into the conical flask and look out for a pale yellow color, okay? So we're going to add that into there so you sodium thiosulfate goes in here And we're going to drip feed that into the speaker here And we're looking out for that pale yellow color so as the color change is really difficult to see if you don't know about you If you try and look at something colorless to pale yellow it's really difficult to see so what we're going to do is going to add 2 centimeters cubed of starch into there And starch turns a deep blue colour if iodine is still present in the flask. And we're going to keep adding until that blue colour disappears. In other words, we want to make sure the iodine has completely reacted. And then at this point, all of the iodine has reacted, like I say. And then we can use the volume of sodium thiosulfate that we've added in here to work out the number of moles of iodine that were present from the first step. Okay, are you following? There's a lot of steps here. So that's what we do with that one there. And then the third step, so we're going to do the calculation and we'll look at the third step. Of course we're going to do the calculation. So for example, let's look at the calculation example here. So all the iodine reacted with 10.5 cm3 of 0.14 molar thiosulfate solution. So that's what we've added in there. That's the amount that was required. So here's the reaction. So thiosulfate is here. That was in the burette. And the iodine... That was the iodine that was produced from this reaction here. And the amount of iodine was dependent on the concentration of your oxidizing agent. So this is the reaction, the overall reaction here. So the equation shows what's happening in the titration. So the moles of thiosulfate is concentration times volume. And we need to divide that by 1,000 to get decimetres cubed. So the moles of thiosulfate is 0.14 because we've worked it out there. times by 10.5, because that's the volume we used to add into there, divide it by 1,000, and that gets us the total number of moles of thiosulfate. So the moles of thiosulfate react in a 2 to 1 ratio. So we basically divide that by 2 to get the number of moles of iodine. So 2 to 1 ratio, and then we get this answer here. 7.35 times by 10 to the minus 4 moles of iodine. were in that conical flask. So this is where we look at step three. So we use the moles of iodine that we've just worked out in step two to work out the concentration of I or three minuses that we used in step one. So just as a reminder, the moles of iodine from step two was that, and we used 25 centimetres cubed of oxidising agent in step one. So the equation from step one was that, so we've seen that already. So in step three, we have to look back at our original equation from step one, because we'll use this to work out the number of moles of IO3, and hence the concentration. So we've got three moles of I2, and these are produced, so this is it here, three moles of I2, and these are produced from one mole of IO3-, because we have a three to one ratio. Remember, we know that. bit because we've just worked that out from step two so the moles of io3 minus how many is in here is divided by three so it's seven point three five times ten to the minus four number of moles of iodine divide that by three and that tells us the number of moles of io3 minus so you see we're kind of working backwards here and so the concentration of io3 minus is moles divided by volume okay so that's no difference remember that from year one and remember we've got to convert the volume to centimeters cubed to decimeters cubed by divided by a thousand so we do 2.45 times 10 to the minus 4 which is the number of moles divide that by 25 divided by a thousand that converts it into decimeters cubed and then finally we get the concentration of the original oxidizing agent that we used which is 9.8 times by 10 to the minus 3 moles per diem cubed so There's a lot of information there and if you weren't too sure I have done another video. It's one of my whiteboard videos with a white thumbnail and like I said you've seen the information come at the top of the video and you just click on that and it'll take you to that video. It is a long process, but be methodical, break it down. So you've got to work out your oxidizing agent. So you need to oxidize that first to produce the iodine. Then we use the titration, work out the number of moles of iodine, and then revert back to step one again to then work out how much oxidizing agent you've used. So we're now going to look at half cells. So half cells is to do with electrochemistry. It's... very dependent though on redox so it's it's all linked together that's why we go through redox first um to understand what half cells are so a half cell is just one half of an electrochemical cell and they can be constructed of a metal dipped into its ions or platinum electrode with two aqueous ions so what we've got here here's the first one so this one is an example of the of a half cell but this time we're going to be using a metal and its iron dipped in a solution of its own iron. So if we had an iron electrode which is here, this is dipped in a solution of Fe2+, then what will actually happen is we will have a reaction. I know it doesn't sound very exciting because if you just put a chunk of metal into its ions of solution, you might think well I don't see a lot and you probably won't. But there is a reaction and the reaction is this. Basically we have an equilibrium system. So we have Fe2+, which is the solution in here, picking up two electrons. That will form iron to get a deposit of iron on here So this is and the reaction is in equilibrium so it can flip from one side to the other But this is just if you just had a beacon you chucked a bit of metal iron and metal In there, so let's look at another example where we don't have a metal electrode dipped into its own iron So this is where we can use platinum electrode and we can dip that into a solution of ions but we must have two ions in there, so in this example we have a platinum electrode dipped in a solution of Fe2 and Fe3 plus ions we've got two ions in there and the half equation is Fe3 plus plus an electron will form Fe2 plus that's the half equation for this again it's reversible all these are reversible the reason why we use platinum here is platinum is inert it doesn't react with anything in here so we're not going to get interference from platinum atoms interfering with whatever's happening in this beaker and From an electricity point of view, because we're looking at electrochemicals, it's a really good conductive electricity, but it's really expensive. So you might have some of this at school or college, like a very thin wire. It doesn't look very impressive, to be honest, but it is really expensive. Platinum, obviously, it's used in jewellery. It's a precious metal. So, yes, this is what you can set up a half cell like that. So an electrochemical cell... is created by joining two different half cells together. And we can have loads of combinations of these half cells, loads of different types. Imagine it's like letters in the alphabet, loads of different letters, and you can join one letter with another letter to form a word. And it's the same with an electrochemical cell. We can pick a library of different half cells, put them together, and see if we can make some electricity from it, basically. Not all of them work, though. Some of them don't work. So there are certain rules, and we're going to look into what... what is an electrochemical cell and how that works. Okay, so an electrochemical cell is made, like I say, of two half-cells that are joined together, and we need a wire that joins the two electrodes together, we need a voltmeter in between the wire to measure the potential difference, and a salt bridge. Now, if you do physics, this will probably be quite straightforward. Luckily, there isn't a lot of physics in this, although I don't mind physics personally, I quite enjoy it, but really the emphasis here is on the chemical side, so the chemistry. So here's a setup of an electrochemical cell. So we connect two half-stills together. We get one side undergoing a reduction process and we get another side undergoing an oxidation process. So essentially we have a redox reaction but we're not actually mixing the chemicals together like we've seen before. We're effectively keeping them separate. So we're isolating these two chemicals but we're kind of connecting them with a wire. So you can see here The setup now we have a voltmeter and that is to measure the voltage or potential difference is the proper word for physicists start screaming It is and between two half cells now. This is called the EMF or E cell okay? So this is what we call the EMF value you're going to see E cell values later on so that's gonna be quite important Electrons will always flow from a more reactive metal to a less reactive one. That's very important and The zinc half cell, as you can see, we've got a zinc and a copper one, but the zinc half cell shows the loss of electrons as zinc loses electrons easier than copper and oxidation has occurred. And we'll look at more about this later on in the next slide. But the reaction here is zinc forming zinc 2 plus, plus 2 electrons. And so what we will see is because zinc is effectively. disintegrating, it's moving, well it's not disintegrating but it's reacting to form Zn2+, we have less zinc in our electrode. So this electrode will look thinner because we're producing more electrons. Okay, so there we are. Okay, so the electrons whizz round, there we are. Okay, so zinc forms, we get more Zn2+, the electrons from that reaction go through the wire and they end up moving around to the copper. And so the copper half cell accepts the electrons produced by zinc. Reduction has occurred here. So we've got copper 2 plus picking up them two electrons. These have come from zinc and we're forming copper solid. And what we see, the observation here, is actually we see a buildup of copper around that electrode and it gets thicker. We also have a salt bridge in between. The salt bridge is normally made up of potassium nitrate. Potassium nitrate is a type of salt but we saturate it. It's basically a salt bridge. And again, you might have done this in school or college, but a salt bridge is pretty much just a bit of filter paper folded up and you dip it into a highly saturated solution of potassium nitrate. It just completes the cell. There we are. So it just allows the ions to flow through which balance the charges out. You don't need to be too concerned about the salt bridge, but it is imperative. If you didn't have a salt bridge in there, it just wouldn't work. Okay, so each half cell has an electrode potential. We call that an E0, okay? And that E0 is a value, and a specific value, under standard conditions and temperature, okay? We measure that in volts, and it tells us how easily the half cell gives up electrons. Very, very important. This gets a little bit complicated, so... Just bear with me two moments. So you've noticed from that previous slide where we had zinc and copper that were two half cells connected to form a full cell. So we have two half cells and they each have a reversible reaction. So in electrochemical cells, we always write the equation that is reduced in the reduced form. Okay, so that means electrons are always on the left hand side and you will see you'll get an electrochemical series. and you'll be given the information for this. You don't need to remember the numbers for these, but you'll always see them written in the reduced form. That's the standard way in which they're shown. Irrespective of whether they're being reduced or not is irrelevant. They're always written like that, okay? So we always have electrons on there. So what this means is we always show electrons with reduction in the forward direction. In other words, zinc is gaining electrons. Zinc 2 plus, sorry, is gaining electrons to form zinc, okay? And that's a reduction process. But remember, when we connect these two half cells together, we always have one cell undergoing reduction and one cell that's undergoing oxidation. What we need to work out is which one is being reduced and which one is being oxidised. We do that by looking at their electrode potential values, their E0 values. A data book does show you this, but you will be either given the information in the exam or it will be on your...... data sheet okay it's more likely going to be written written in your exam so don't worry about the values here so let's have a look we've got two values here so we know zinc and this is a standard value zinc zinc two plus half cell has an e naught value of minus 0.76 and a copper two plus has an e naught value of plus 0.34 now what we've got to what we've got to do is use that information to work out if we connect these two which one's going to be reduced and which one's going to be oxidized. So what we've got to remember is a rule. And again, this is just an acronym that I've made up. But we remember a rule which is no problem. So the most negative half cell will undergo oxidation. The most positive half cell will undergo reduction. So no problem. Remember that because these can be really tricky when you've got oxidation and reduction and enod numbers and everywhere. So if we can try and remember that, that will make this whole process so much easier. So in this case, zinc 2 plus zinc is the most negative. So we'll be the half cell where oxidation takes place. Okay, so oxidation is a loss of electrons. So we flip the equation to show this. Okay, so in this cell, we have zinc giving up the electrons. So here it is here. So zinc giving up the electrons. and copper two plus ions are accepted them. Okay, so the overall equation will be zinc, okay, plus copper two plus, because remember we flipped that the other way around, copper two plus. will form zinc 2 plus which is over here and copper so that's the overall equation because remember we've got to have one that's been oxidized and one that's been reduced so we've got to remember that rule no problem negative half cell will undergo oxidation the positive half cell will undergo reduction so we flip we flip that one round to zinc and zinc 2 plus so there it is there okay so just apply that principle and then you'll know what reaction you know what's being reduced and what's being oxidized Okay, to help us measure the standard electrode potential for each half-cell, we've got to measure it against a reference. And luckily, we've got a reference called a SHE, which is a standard hydrogen electrode. And this is, like I say, it's used as a reference to measure against these half-cells. Because you might think, well, how do you know that's minus 0.72? We measure it against this. Now, the... They can't be measured on their own because you've got to connect it with something to work out the potential difference. So the standard hydrogen electrode is calibrated to equal zero. So it's got zero volts and so that means any reading that it shows on the voltmeter will be because of the other cell, not because of the standard hydrogen electrode. It is a standard setup as you might expect. So this is the setup here and what it has is it has hydrogen going in and you need to know this information. But hydrogen goes in at 298 Kelvin and 100 kilopascal, so that's standard conditions. That goes into this pretty glass tube here with a hole in the side of it. Okay, so we have one moles per dm cubed of H plus ion, so this is acid, in the beaker. And for copper, we also have one moles per dm cubed of copper ions in there, so we keep everything the same. We're just looking for a fair... Voltmeter result here, so we've got to keep the concentrations the same so like I say the word standard is important We've got to have these at temperature of 298 Kelvin a pressure of 100 kilopascals and the concentrations must be 1 moles per dm3 You must remember these conditions for a standard cell Anything deviates away from this will give you a different voltmeter reading so it is important So the diagram on the left like I say it shows the standard hydrogen electrode connected to a copper cell So assuming that all the conditions above are met, then what we can work out is the standard electrode potential for copper, which is the value that we'd seen in the previous slide. So it's just a benchmark. It allows us to work out what the values are. So to get 1 moles per dm cubed of H plus ions, you've got to be really careful about this, because sometimes the example can be a little bit sneaky. So to get 1 moles per dm cubed of H plus ions, we need 1 moles per dm cubed of HCl. or half a mole of H2SO4. Now, can you think why? Well, the reason is H2SO4 is diprotic, and you would have seen that from a previous... If you've seen the previous video on acids, bases, buffers, you would have seen that H2SO4 is diprotic, and it gives up two protons for every one molecule of H2SO4. So we need half the amount of H2SO4, and that will still give us one mole of H plus ions. Okay, so let's look at the electrochemical series. So this is nothing like a Netflix series or anything exciting like that. It's just basically a table with loads of electrode potentials in. But the electrochemical series is a list of half-cell reactions and their standard electrode potentials. So that's E0. So this is just an example of the type of thing that you might say. They're all different. It depends on what you're going to work out. But there is just simply nowhere near enough space. to fit the full series on here so this is just an extract from an example so the table shows the e naught value in descending order okay in this example it could be the other way around so don't take this for granted and it could be flipped the other way around it depends on how the exam board wants to wants to portray that so in this case as we go up we get the stronger oxidizing agent so the one with the most positive e naught value and the agents on the left hand side of the equation more easily reduced so these ones here so in other words these ones are the most easily reduced compared to on the right hand side so they have an increase in tendency to gain electrons and so they are the most most powerful oxidizing agent what the exam board can do is it can give you a table and ask you what which is the most powerful oxidizing agent in this case it's the one that's the most positive and it's the one on the left hand side here so this one's got the highest tendency to gain electrons so therefore is the most powerful oxidizing agent So in this case, the most powerful oxidizing agent is Cl2, and the weakest oxidizing agent is Mg2+. So it's this one on the left here. And likewise, we can go the other way. So the stronger reducing agent is the one towards the bottom, the one with the most negative value in this case. So agents on the right-hand side of the equation are more easily oxidized. So they have an increase in tendency to lose electrons. So they're more powerful reducing agents. So here the most powerful reducing agent is magnesium. And the weakest reducing agent is Cl-which is at the top. here so you've got to be able to you've got to be able to identify from um an electro electrochemical series what's the most powerful oxidizing agent and the most powerful reducing agent just make sure you memorize these different areas because it's um it's it may only be worth one mark but it could um you know every mark counts Okay, so let's look at some calculations involving standard cell potential. So standard electrode potentials, E0, as we've seen before, they can be used to work out a standard cell potential, which is basically the value of the cell. We use E0 cell equals E0 reduced minus oxidized. Okay, now you can remember this as the acronym REDOX. Yeah, so it's just the same. REDOX, reduced, oxidized, and you put a minus in between. So. The half cell, what we've got to remember is the half cell equation with the most negative E naught value is being oxidized. Okay, so that's the key thing. So the most negative goes on the right hand side. And obviously the most... positive goes on the left you might see you might be taught a different equation and it might be slightly different depends on what the view you know what what you've been taught but this is a way in which I remember it using the analogy that I used before which is no problem Okay, so it just ties it all in really. But if you have two positives or two negatives, then it's the most negative that's oxidized. Okay, so that's the most important thing. Okay, so let's look at an example of how we work this out. So we're going to use the data in the electrochemical series to calculate the E-naught of the cell when Cl2 and Cl-minus half-cell and a Zn2 plus and Zn half-cell are connected. Okay, so first of all, we need to identify which is being oxidized. Okay. Okay, so this is where you use your no problem. Okay, so zinc 2 plus and zinc half cell is the most negative, so is oxidized. Okay, so most negative, oxidized, no. And then problem is positive reduced. So that's the most oxidized. Yeah, that's the one that's been oxidized. So number two, we calculate the E naught of a cell. So E naught cell is 1.36. Okay, because that's our chlorine one there. Okay, and the most oxidized is minus. minus 0.76 which is our zinc one which is down here so if we put them in there we get plus 2.12 volts let's look at another example so we're going to use the data in the electrochemical series to work out still using a chlorine half cell but we're going to use copper instead so again identify which one's been oxidized so it's the copper 2 plus copper half cell that's the most negative so is oxidized okay so what we're going to do is we're going to put all these in here so we've got one point three six which is obviously our chlorine and the copper one the reason why I've picked this one is copper is positive as well so you've got two positives but because that is the most positive that goes on the left and the least positive goes on the right that's been oxidized so the totally not the cell here is one point zero two volts Okay, so you've seen the diagram of the cells, how we draw cells with the beakers and salt bridges, etc. That's, you know, this is chemistry. We don't faff around and we don't draw diagrams and silly things like that with beakers and salt bridges. We need a simplified version. it's so much quicker so thankfully there's something called a cell notation and this allows us to simplify the structure of a cell and but there is a standard way of representing them in chemistry okay so a standard notation is represented like this and you can see on here it's just so load of lines with words in between so the most negative half cell potential goes to the left of the double line okay so you've got to remember that most negative on the left okay most positive on the right so we always have um this structure here so most negative half cell goes to the left of that double line so we have certain features here so a single line single line shows a physical state change so that could be metal to metal iron solution like we've just seen before. A solid double line shows a salt bridge obviously that bridges the two half cells. And so if we use the zinc and copper cell that we've seen before, here's an example here. So we've got zinc, which is the most negative half cell on the left. Copper is the most positive on the right. But you see we have a copper solid and the aqueous ion. So there's a line there to show, to signify the difference in state. Okay, so zinc 2 plus has an oxidation state of plus 2. Whereas zinc has an oxidation state of zero. So because that is the most oxidised form out of the two, that's the one that sits closest to the salt bridge. So the most oxidised form goes closest to the salt bridge. So what if you have two aqueous ions? So what we do, we follow the same rules as above, but instead of having the solid line, we use a comma. So we separate with a comma because they're in the same physical state. So that might be, for example... Fe2 plus Fe3 plus and platinum so between Fe2 plus and Fe3 plus we have a comma so there's the example there so you can see here Fe3 plus Fe2 plus the comma separates these two they're not they're in the same state so we use a comma but that platinum is a different state to either of these two so we use a solid line there and obviously that's shown in the state symbol so remember with these type of cells where we have two ions in solution we've got to use a platinum electrode and because there's no solid material there we need something to clip to clip onto the the voltmeter so we use platinum as the electrode here okay so what we're going to do is going to predict reaction feasibility so if you've seen the video on Gibbs free energy so the one with enthalpy and entropy before then it's similar to this except what we're going to do is we're going to use standard electrode potentials to predict So it's a bit like predicting, well, if only this could predict Lockley numbers, that would be great, wouldn't it? But it's a bit like predicting from Gibbs 3. So standard electrode potentials, so E0, they can be used to predict if a stated reaction is likely to proceed under standard conditions. OK, so let's have a look at an example. I'm going to bring up our electrode potential series here. So example one, use the data in the electrochemical series to predict whether solid magnesium... will react with copper two plus ions in solution under standard conditions okay so what we're doing is we're saying right if i get magnesium and i've got some copper two plus ions like copper sulfate for example a blue solution and i took that magnesium in will it will i see anything will it react so first of all what we've got to do and we've got i'll remember i'll remember a bit there as well so about half cell equation where the most negative e naught value is being oxidized okay so just remember that but the first thing you is we need to identify which has been oxidized. So the Mg2 plus Mg half cell equation has the most negative value, which is this bit here. So that is being oxidized. Okay, so remember that it's no problem. Okay, so the most negative is oxidized. So we take the oxidized equation and reverse it. Remember, we flip that round. Yeah, we write the other two equations next to each other like this. So we write it magnesium and copper. So we then combine the two equations to obtain the feasible reaction. So whatever that is, that's the feasible one. Mg plus CO2 plus will form Mg2 plus plus Cu. And then what we do is we compare that equation with the one in the question. So we can see that magnesium will react with copper 2 plus ions as they match. So this is telling us the feasible reaction is magnesium reacting with copper to form Mg2 plus and copper. So yes, it will react. And we can always confirm this is always worth doing this by calculating the E0 of the cell and all feasible reactions will always have a positive E0 cell value. So we put that in our equation, we can see that that does actually give us a positive voltage. Okay, so let's look at a second example. So use the data in the electrochemical series to justify why iron nails become rusty when in contact with air and moisture. Now this is a slightly different type of question. because they're asking us to justify it they tell us yeah it does work we know it works but you tell us why you know prove it so this is just as important in science to prove something um so here we go so we've got our electrochemical series there's the two um half equations that we need so we need the um iron one because it's made from iron but we know that rust is caused by aerial oxidation in other words we've got oxygen in the air that reacts with it you So that's the most appropriate question, the most appropriate half equation, which is the one towards the top there. It's this one. So we need to identify which has been oxidized and so the Fe2 plus Fe has the most negative value so that one's been oxidized. No problem. Remember that. So the second one is we take the oxidized equation, we flip it around the other way and we write the two equations side by side. Exactly the same as what we've done before. And so here we multiply the top equation by two because we've got to balance it. OK, to make sure the number of electrons in both equations are the same. So here it is here. So we combine the two equations and this obtains the feasible reaction, which we can see here. Then we compare the equation to the reaction state in the question and we can say that obviously iron will react with O2 and H2O as they match. But we know that already because they've told us they just want us to justify it, prove it. And so we can confirm this again further by using an E0 cell calculation, and we can see that it is positive. It's only just positive, so it's a slow reaction. Obviously, we're not going to see something rust instantly, you know, not in these conditions anyway, but nonetheless, it is feasible. It does work. We see it. We observe it. We see iron going rusty. Okay, so let's look at a third example. Now, what the key thing here, and this is what I'm saying here, is just because we calculate E0, E naught to say if something's feasible doesn't mean it will necessarily go a little bit like the Gibbs free example we use. So for example you might have non-standard conditions so if we change the concentration the temperature etc this can cause the electrode potential to change. So for example we might have a really cold environment and that may not cause it to to work it might not actually be feasible. So let's take that rusty nail example from the last slide. So here's the E naught value and we have the half half equations and pull them across and then we have the e naught cell value that we've calculated if we count if we increase the concentration of oxygen okay equilibrium will shift right which means it's easier for the oxygen to gain electrons okay and so the electrode potential for this half cell will become more positive and the and the cell potential will be higher okay so if we increase the concentration of oxygen. In other words, if we've got more oxygen here, then the equilibrium will shift to the right. So it'll move to this side. Okay? So if we increase the concentration of iron, equilibrium will shift left. Less electrons will be used up. And so the electrode potential of the Fe2 plus and Fe becomes less negative. And the full cell potential will be lower. So it's important to look at these in terms of equilibria as well. And that's rightly important. And another example might be if the kinetics is not favourable. So for example, the rate of reaction is slow. So it appears though there's no reaction. especially with rusting as I mentioned before and if the reaction is a high activation energy it may stop the reaction from happening at all so and catalyst can help to use bring the activation energy down but there's a lot of factors at play here so it's not necessarily that it means that just because it says it is feasible that it it's definitely going to happen it depends on the conditions Okay, so let's look at the final part of this topic, which is cells and batteries. Okay, so energy storage cells. So what we need to do, obviously we looked at half cells, and we looked at connecting them together to form an electrochemical. cell a full cell and there's got to be some use to this and the use is is clear it's batteries and we have it in everyday life okay so this is very very relevant chemistry so to establish the overall reaction in in a lithium-ion battery what we need to know know uh the half equations at each electrode so we've got um a lithium ion battery is something you would see in a mobile phone okay so they're very light uh lightweight batteries and used to power uh quite complicated devices like say phones and laptops and tablets etc but the half equation for this type of reaction is this so it's lithium um your lithium ion at the top and then you've got your cobalt oxide uh reaction here so um you'll be given these it's about using the information you so Li plus Li this one has the most negative e naught value so oxidation occurs here so this means we flip the equation and this shows the electrons being produced ie it's a negative electrode so it's exactly the same as what we've done before so the negative electrode is Li will produce Li plus an elect and plus the electron so we flip that equations that's it flipped and obviously the positive electrode is just the same reaction here because that's positive so anything it's most negative we flip it around so there's our two new half equations And so the overall equation on discharge, so this is when we're using the cell. So we're using the battery. So it's lithium reacting with the cobalt oxide that's in there to produce your lithium ion or your lithium salt product. So to work out the E naught of the cell, we do exactly the same as what we've done before. So it's reduced minus oxidized. So it's 0.56, which is this value here, minus minus 3.04, which is the most negative because that's your oxidized. And that's 3.6 volts. Now, you might be asked to say, right, what would the reaction be if we plug the thing in? Because obviously the battery will run out eventually. So this is a rechargeable battery. It's just the opposite way. So all we have to do is reverse the equation. So this is the equation on discharge. If it's on recharge and it back up again, then we just flip the equation the other way. It's exactly the same. All we're doing when we're charging a battery up is we're introducing more electrons. into, or electrons into the system, and that moves it back the other way, and it recharges the battery up again. Okay, fuel cells are an interesting one. So electricity can be generated from fuel cells, but it needs a continuous external supply. So unlike a battery in a phone where it doesn't need a continuous supply, it will last for a few hours, and then you'll have to recharge it, fuel cells need a continuous supply of fuel. They need to be constantly connected to something, okay? So they're not a ready store like batteries. So an alkaline hydrogen oxygen fuel cell is an example of a fuel cell. And the diagram, you might be familiar with it, but the diagram shows a hydrogen oxygen fuel cell. And what we're going to do is, you see there's numbers on there. We're going to go through each number on there and explain what's happening in these. So the first one is a hydrogen feed. So hydrogen is fed in. So you need a constant supply of hydrogen. And what this does is it reacts with the OH minor signs in solution in 9. So in section 9, which is this bit here. So that's section 9. But we'll look at this later on because obviously we're going to do this in numerical order. But the reaction that does occur here is hydrogen comes in, reacts with the OH minor signs in here because it's an alkaline cell, and that forms water and four electrons. But we'll come back on that. So the second one is the flow of electrons. So these electrons here that are produced. These travel through a platinum electrode through here. It's a good conductor. It's inert remember and it goes through Through the wire and then three is the components that depends on what you're powering And so this could be a car you can have a fuel cell car. It could be a house You might have a fuel cell outside the house and generate in power. So that depends on what you're powering Okay, so the fourth one is the oxygen feed. So we've got that feeding through the other side of the cell. So this reacts with water and the four electrons made from step one to make OH minor signs and the reaction that occurs is basically oxygen comes in reacts with water and the four electrons that's just come through here on that side and that produces the four H minor signs which then drifts around in in section nine so the fifth one is the negative electrode so this is it here this is the cathode here so electrons float to the negative electrodes you can see them flowing in there and that's made from platinum as well and six is the electrolyte so this is the um the uh electrolyte that's held in in this compartment here and that's made from potassium hydroxide because it is an alkaline cell and what this does is this carries the oh minor signs that were produced from this reaction here from um from the cathode to the anode which is along on the side there So the seven is the positive electrode. This is the anode. This is this bit here and electrons will flow from this section across into the into the cathode which is the other side so this is made from platinum and There is a byproduct that's emitted and this case it is water and the water that was produced here Remember is now being kicked out at this end here. So the water that was produced at this side. That's then moved out And then it's released into the surrounding so that's the only byproduct so that's better than burning a fossil fuel which would produce carbon dioxide Sulfates for example nitrates and another pollutants, so it's quite a green way of generating electricity And number nine is the movement of the OH minus sign so that's what I mentioned in the first step So OH minus signs that are produced from the reaction in step four here I've actually carried through the electrolytes to the other side of the cell Um And what we have as well between the electrodes, between the two sections, are ion exchange membranes. And what these do is these align the platinum electrodes on each side. And they allow the OH-ions to pass through, but not hydrogen or oxygen gas. So it's very selective what can go through. Because what you don't want is the hydrogen and oxygen mixing in the middle here. You just want the OH-ions to drift through. So you can exploit the electrons that are being transferred and use that as electricity. Okay, so to summarise the reactions that's happening in this cell, so we have a hydrogen feed, so remember that's on number one, so it reacts with the OH minus ions in solution in nine, and this reaction occurs, which is H2 plus H2. four OH minus will produce four H2O and four electrons now the electrons obviously go round the circuit and they're received at the other side which is this bit here and so that forms oxygen there's your four electrons plus your water that's produced as well so some of the water's moved moved across and then that produces your 4-OH-ions which then drift through the membrane and so the overall equation here if we combine both of them is we produce hydrogen plus oxygen and that forms water so two molecules of water are pushed out in this reaction so the only emission is water Okay, so let's look at the advantages and disadvantages of fuel cells. You might think that sounds like a pretty good idea. Why don't we generate all electricity like that? Well, the advantages are they're much more efficient than internal combustion engines, so they convert a lot more energy. And in fact, we do have cars now that are hydrogen powered. There are some hydrogen powered cars. It's mainly in, I think, California have a lot of hydrogen powered cars. But they do exist. More energy is converted into kinetic energy. whereas in combustion engines it's thermal energy and many electric vehicles are battery powered however so fuel cells don't need to be recharged so they're much much better in terms of a continuous supply effectively you put hydrogen into the car stored in the tank and as long as that tank is full you can't keep running you don't need to keep plugging it in but you do need that supply of oxygen hydrogen oxygen comes from the air but the hydrogen comes from your fuel so the only way this product is water so no carbon dioxide emitted so that's much better for the environment than a combustion engine so the disadvantages um hydrogen is very very flammable um it has to be stored correctly in a pressurized tank in the back of your car um so that can put some people off and it's really expensive it's literally you're transporting thin air um so it's expensive to transport and store hydrogen um so it has to be stored correctly and safely of course um and energy is required to make that hydrogen and the oxygen in the first place. So fossil fuels are normally used to pass water through an electrolysis process to make that. So you take brine, electrolysis of brine, and you make your hydrogen that way, and that can actually contribute to carbon dioxide, so it's an indirect emission. And that's it. So that's the video on redox and electrode potentials. It's quite a difficult topic, but there's a few acronyms in there to help you navigate it through that. Please remember to subscribe to the channel. There's loads more videos on OCR dedicated to OCR. there is for other exam boards as well they're all tailored to each exam board and so please subscribe they're all free and that would be great and remember you can have a copy of this as well if you click on the link below you can purchase the the slides as well and you can use it for revision it's great for that and but that's it bye bye

hit the subscribe button that'd be brilliant and all the videos here are designed to be revision based so you might have gone through this content already at school or college or you might be self-teaching yourself in which case it contains the main points that are tailored to OCRA so it is important though to practice exam technique as well there are videos on Allergy Chemistry that looks at exam technique so have a look at them as well because that is just as important as knowing the content It's a bit like if you're learning to drive, it's a bit like picking up the highway cord. And you may know the highway cord back to front, but it doesn't mean you need to hop in the car and start to drive. So it's exactly the same approach you should take to your exams as well. All these slides here, if you want your own copies of these slides, they're available to purchase in my Tez shop.

So if you click on the link in the description box, you'll be able to get a hold of them there as well. They're great for revision, great value, and you can use them as and when you require. And it's a good way to supplement your revision that you get from school or college. Of course, we get a lot of people who are doing it and being self-taught effectively.

Okay, so like I say, this is specifically designed for OCR-A and it matches the specification. So these are the points taken from the specification and everything you see in here will adhere to them rules. Okay, so let's look at reduction and oxidation.

So you might have seen bits of this in year one. one chemistry um so this is just um confirming what you may may already know and extending it into this topic because it is all about redox you're going to see quite a lot obviously because that's the name of the title of the of the video so electrons are transferred when reduction oxidation occurs and we call we use an acronym oil vague so oxidation is loss and reduction is gain of electrons um so you might have seen that gcse you might have heard of it already but it's a really good acronym and it's It's very fitting for chemistry as well, calling it oil rig. So this helps us to understand what's going on. So for example, this reaction is when calcium completely is burnt in oxygen.

It involves reduction and oxidation, and we call this a redox reaction. So we've got reduction and oxidation at the same time, not to be confused with the shower gel. In this reaction, calcium is being oxidized and it is losing...

electrons so we can see here that calcium is forming calcium 2 plus and two electrons are also being produced as well In this reaction, this is the other half of the reaction for oxygen. So in this reaction, oxygen is being reduced and is gaining electrons. So you can see here that the oxygen is gaining two electrons, a half or two, is gaining two electrons to form O2-.

So this is a very simple way of showing redox reactions. We're showing the oxidised side first, then the reduction side. So it's just breaking that down. And if we combine them and cancel the electrons out, it's just... you'll see later, then we form the full equation.

And it's very important as well, you'll come across reducing agents and oxidizing agents. So these are, don't get them confused with oxidized and reduced. Although they go hand in hand, they mean different things. So for example, reduction is the gain of electrons, but reducing agents lose electrons and are themselves oxidized. You've got to remember that.

So it's just like the opposite. So reduction is gain, reducing agents lose electrons. Okay. And oxidizing agents gain electrons and are reduced themselves.

So you've got to know the difference between a reducing agent and an oxidizing agent. Because you'll use them or you'll see them quite a lot in A-level chemistry. Okay, so let's look at how we balance half equations. So half equations show reduction and oxidation stages in two equations.

All half equations must have electrons in this. So the rules. The first thing is you write down the species before and after the reaction.

Number two is we balance any atoms apart from oxygen and hydrogen. These will be dealt with later. Third rule is we balance any oxygens with water. The fourth rule is we balance any hydrogens with H plus ions.

And the fifth rule is we balance charges with electrons, which is E minus. Now, you might have a different method of balancing equations. You might be taught a different method. I find that this method... works really well with even some of the obscure balance and half equations.

So this is just the method that I think is the simplest method. There are other ways in which you can balance it, and you may have been taught something else, but this is just another method. It all comes to the same result anyway.

Okay, so you'll see that I've put asterisks near 3 and 4. This only should be done if we need to balance oxygen and hydrogens. It might not be the case that we need to balance oxygen and hydrogens. Not all half equations have them in. But these rules are fundamental because you're going to see this all the way through the topic. So make sure you're familiar with this and all the methods that I'll do will be using these rules.

So we're going to write half equations showing the conversion of iron 2 plus to iron 3 plus. So there's the first thing is to write down the species before and after. So it's Fe2 plus, Fe3 plus.

We then need to balance any atoms apart from those with oxygen, hydrogen as these will be balanced later. Well, they're already balanced. We've already got one Fe and one Fe on the left and one Fe on the right.

Balance any oxygens. Well there isn't any oxygens so we don't need to do anything with that. And the same again with hydrogen. There's no hydrogens in here that we need to be concerned about. So we go straight on to step 5 which is balance any charges with electrons.

So you can see we have a 2 plus on the left, a 3 plus on the right. So an electron is always negatively charged. It's got a minus 1 charge. So we put the electron on the right hand side. So that means we've got a 2 plus on the left and 2 plus on the right.

This is shown in oxidation state. So oxidation is loss of electrons, Fe2+, losing electron to form Fe3+. Okay, so we need an oxidizing agent to allow this to happen. Okay, so remember this has been oxidized, but we need something to actually allow this to happen.

So let's have a look. So we're going to write a half equation showing the conversion of MnO4-, that's manganate, to Mn2+. So the first thing... we need to write the species before and after the reaction so it's MnO4-going to Mn2+. We need to balance any atoms apart from oxygen and hydrogen and deal with these later.

So it's already balanced remember we ignore the oxygen we know there's four oxygen left but we ignore them we're just looking at manganese so there's manganese one Mn on the left and one Mn on the right. We now need to balance oxygens and we use water to do this now we know there's four oxygens on the left So it means we must need four oxygens on the right. And we know that water has one oxygen in there, so we need four waters.

There we are. So we need four waters on the right. Then we need to balance any hydrogens with H plus ions. So you can see we have eight hydrogens now because we've chucked that water in there. We've got eight hydrogens on the right, so we need eight hydrogens on the left as well.

So there it is. Okay. So you can see it's fairly straightforward.

Balance charges with electrons. So now we need to look at the charges on the left. Total charge on the left. And total charge on the right. So on the left we have a total charge of plus 7. Because we've got a minus and 8 positives.

And on the right we've got plus 2. We've got 2 plus. So the electrons go on the left hand side. And we need 5 of them. And that will bring the charges equal.

Left and right. Remember the charges don't have to be 0 either side. They just have to be balanced.

And we use electrons to do that. So there's our half equation for that. This shows reduction, so we've got a reduction step here, because reduction is the gain of electrons, so we've got gaining of electrons on the left-hand side.

So reduction will always show electrons on the left-hand side. Oxidation will always show electrons on the right-hand side. So what we can do with these two half equations, so the iron one that we've seen before, and the manganese one and the manganate one that we've seen here on here, we can...

combine these two equations and form a full ionic equation because these equations are half equations because they contain electrons. Okay so let's look at how we combine these half equations. So these two half equations can be combined to make that full ionic equation.

We're just going to make sure our electrons balance. So remember this one shows oxidation so we've got Fe2 plus going to Fe3 plus plus an electron it's an oxidation reaction and remember the manganate one Shows reduction because it's gaining electrons. So electrons on the left-hand side is reduction.

Electrons on the right-hand side is oxidation. So you can see the electrons don't balance. So we've got to make sure, and if you've done, you might have done this in maths, which is simultaneous equations, this is exactly the same principle. So we're finding in this case the commonality is always the electrons in this case.

There might be other elements, but it must be the electrons to start with. So electrons don't balance, so we need to multiply the top row by 5. So here we go. So we've got 5 Fe2+, we'll form 5 Fe3+, and 5 electrons. Okay, so we cancel the electrons out and we combine the two equations together.

So there we are. So we've got MnO4-, 8H+, 5Fe2+, we'll form Mn2+, 5Fe3+, and 4H2O. So we can see we've cancelled the electrons and your ionic equation, your full ionic equation, should never have electrons inside there.

Also, your ionic equation also shows reduction. and oxidation because we've combined the two separate steps and we formed a redox reaction. So that's the the full ionic equation is your redox reaction.

Okay so let's look at redox titrations. So you would have seen titrations before already. So the titrations are actually showing you the titrations you would have seen are acid based titrations. These ones are going to be slightly different.

These ones are going to look at redox reactions instead. So we're not using acid in the base. We're using redox substances, so reduction in oxidation. So transition elements can be used to oxidise and reduce other substances.

And we'll look at transition metals later because they're quite a special group of chemicals. So transition elements have variable oxidation states, which means they can accept or donate electrons easily. Hence the word transition, they can transition between various different states, unlike other elements which generally have fixed oxidation states.

So transition elements are coloured depending on their oxidation state at the time, and this makes them particularly useful in titrations because we can spot an end point. So remember in acid-base titrations where you have to add an indicator to spot an end point, these ones generally you do this without the use of an indicator, so you just use the... the colour of the transition element to do that.

So let's look at a specific example. You might have seen this example as well in lab work if you've done this. But we use acidified potassium permanganate solution, which is KMnO4, is an oxidising agent.

Okay, so remember what an oxidising agent is. So oxidising agent is something that gains electrons. It dissociates to produce MnO4-ions, where...

Mn has an oxidation state of plus seven in there Okay, and they are reduced to Mn2 plus during the titration now You've got to remember your oxidation numbers and how you work out oxidation numbers And if you're not sure are you a little bit rusty on that area? Then there are videos that I've done in the year one playlist and that show oxidation numbers specifically for OCR as well So they they are there so have a look, but I'm going to assume that you're fairly confident with it. I'm going to go through some of the basics as well, but just to skip past that bit. Okay, so they're reduced to Mn2 plus ions during the titration.

Okay, so let's have a look at the first half equation. So we've got MnO4 minus plus 8H plus plus 5 electrons will form Mn2 plus. plus 4H2O. And so the MnO4-is reduced.

Obviously, that's supposed to be a little n rather than a capital N. So that's been reduced. And obviously, we know how to make these half equations because we've just seen them before. What's been oxidised in this titration is iron. So we've got iron 2 plus, plus iron 3 plus.

We'll produce iron 3 plus, plus 5 electrons. This has already been balanced for us. So the Fe2 plus is being oxidised. So if we combine them we get this overall equation here just like we've seen before.

Now what we see with this reaction when we do the titration is we start from permanganate which is a lovely purple colour and it goes colourless as we progress through the titration if we're doing it this way. So this colour change from purple to colourless will show us an end point. So another one acidified potassium dichromate is an oxidising agent and it dissociates to produce dichromate ions, so Cr2O72-ions, where chromium has an oxidation state of plus 6, and these are reduced to Cr3 plus ions during a titration.

very similar to manganate but a different substance so here we've got cr207 to minus has been reduced so you can see here that it's going from cr207 to cr3 plus so it's been reduced looking at the oxidation numbers and then using zinc for this reaction zinc has been oxidized so to zinc 2 plus so combining them we get this and the color change we see it goes from orange to green So this is a classic sign again it's self-indicating we haven't had any indicator here but we can detect an end point as we form a product and from orange to green. Okay so let's look at calculations because it wouldn't be a titration without a calculation of course. So titrations can be used to work out the concentration of a reducing or oxidizing agent.

So in this example we're going to look at finding the concentration of manganate. Now we looked at that in the previous slide. by titrating against a reducing agent like Fe2+.

So this example looks at finding a concentration of a reducing agent. All we do is we just reverse it for oxidizing agents. There's no difference.

So what we've got here is our classic titration setup. You've got your burette. You've got your burette here, which is at the top. There's your burette. And then we've got a conical flask at the bottom.

What we'll have in this conical flask... a reducing agent so we have Fe2 plus solution with an unknown concentration but we know the volume okay so we don't know how strong it is but we know we probably have say 25 centimeters cubed in that conical flask so what we've got to do is got to add dilute sulfuric acid into there as well so this is all dissolved in an acidic solution and so this is to ensure that you have sufficient H plus ions in that solution to allow the reduction of the oxidizing agent so it's just making sure there's a backup supply there and to allow this to happen so we it's basically acidified it's an acidified solution okay so in our burette we have an oxidizing agent in uh in the burette with a known concentration so we know the strength of it um which is in this case is going to be manganate ions um so mn o4 minus and then We're going to add the manganate ions into the burette, in the burette, into the conical flask until we see a faint colour of permanganate, so the manganate ion appear. And this is known as the end point.

So there's no indicator added here. We're going to use the manganate to provide the end point for us. And we add it drop by drop near the end.

Again, you would have done titrations before. It's vitally important to make sure you get an accurate reading. And we do that by adding it really slowly near the end if we go too much one drop can make a difference And if we go too quickly you'll get a result that is inaccurate Okay, so and there is a sharp color change okay like I say it's very dramatic And so the manganate ions form aqueous potassium permanganate solution are purple okay and they'll immediately react with the reducing agent until all the reducing agent is used up. So it'll react with the Fe2+, and it will keep on doing that and keep on doing that until there's no more Fe2+, ions left, because they've all reacted.

So one drop at the end, like I say, can turn the solution purple, which is the colour of the oxidising agent. And you could use a colourless oxidising agent and a coloured reducing agent, and you could look out for the colour disappearing. But in this example, what we're looking for is the colour to appear. in the conical flask but it can be done the other way it just depends on on the setup of your titration so what we're going to do is read how much oxidizing agent was added and we're going to read the bottom of the beniscus and we always read at eye level so you get right down to eye level right where you can see um the the burette um the the level and you read from the bottom which is the the semicircle the bottom of the semicircle there and you record your results to two decimal places at all times okay and you repeat until you get two results that are concordant so they've got to be within 0.1 centimeter cubed or 0.10 centimeters cubed of each other and that just ensures that um your you can your results are valid you can depend on your results if your results are all over the place you it's very difficult to say what the actual value is if you don't have any that are um that are concordant okay so um so Like I say, they can be used to work out the concentration of a reagent. So let's look at a specific example.

So we've looked at the practical side of it, about how you can conduct a titration. But here we're going to look at how you actually calculate it. So we're going to take the same, exactly the same reagents. We're going to have 18.3 centiHg cubed of 0.025 moles per diem cubed.

of potassium permanganate, okay, which is KMnO4, and that's reacted with 25 centimeters cubed of acidified iron 2 sulfate solution. Remember it's got to be acidified to produce them H plus ions. And then we calculate the concentration of Fe2 plus ions.

So that's what we want to work out. So the first thing here is we need to write out the equation and balance it. So we've learned how to do that already.

So we know how to create a half equation, and we know how to combine it to form our ionic equation. So this is the ionic equation here. We're going to put the potassium permanganate into the burette and we're going to put the acidified iron sulfate solution into the conical flask. And that iron sulfate solution is going to be the source for iron 2 plus ions.

Because you can't just get a pot of Fe2 plus, it has to be as a substance like that, so iron sulfate. So we write it out and bounce it. So we've seen that before. Then what we need to do is calculate the number of moles of manganate. If in doubt, always use the saying, if in doubt, work out the moles.

So moles equals concentration times volume. Concentration is 0.025 because we've been given that. Times by 18.3 times 10 to the minus 3. So I like to use, you've seen my videos before, I like to just, instead of putting divide by 1,000, I like to just put times 10 to the minus 3 on the end. That just means the same as dividing by 1,000.

And that gives us the total number of moles of manganate. So the total moles of manganate are 4.58 times by 10 to the minus 4. So the third thing is we use the equation to find out molar ratio in order to work out the moles of Fe2+. So we can see from the equation there's a 1 to 5 ratio between MnO4-and Fe2+.

So the moles of Fe2+, is basically going to be the number of moles that we've calculated before, multiplied by 5. So that's going to give us 2.29 times by 10 to the minus 3 moles. So now we know the number of moles of Fe2+, in that conical flask. So then what we need to do is you then calculate the concentration using concentration equals mass over volumes. So the concentration of moles per dm cubed is 2.29 times by 10 to the minus 3 divided by 25 times by 10 to the minus 3 because we've got 25 centimeters cubed of that.

So remember convert to decimates cubed and that gets us 0.092 moles per dm cubed of our concentration of our iron 2 plus. Okay. So let's look at another example here, which is iodine sodium thiosulfate titration.

So these are a little bit more complicated. Now, I have done another video on this, and I'll put a link on there all the way through this. I'll put a link, so if you have a look on the, I think it'll probably be on the top of the video, and then you'll be able to have a look at the whiteboard version of this, where I actually go into a little bit more detail. The titration is useful for finding out the concentration of an oxidizing agent.

The titration work is conducted in three broad steps. The first step is we use the oxidizing agent, which is KIO3, to oxidize iodide ions to iodine. That's step one. Step two is we then carry out the titration to work out the moles of iodine produced in step one. Then the third step is we use the moles of iodine in step two.

to work out the concentration of IO3 minus ions. Okay, so there's three broad steps. So we're going to look at step one first. Okay, so step one, remember, is use the oxidizing agent KIO3 to oxidize iodide ions to iodine. So what we need to do is measure out a volume of KIO3.

That's potassium iodate 5, because it's got an oxidation state of 5. And that's your oxidizing agent. And this will produce the IO3 minus ion needed. So usually we'd use 25 centimeters cubed, but it could be any volume. 25 centimeters cubed is normally the volume you'll probably see. So we're going to add an excess acidified potassium iodide solution, Ki, to the KiO3 solution.

And so here's the equation. So we've got Io3 minus, which is your aqueous solution. And we're going to react this with our iodide ions. So this is our oxidizing agents.

H plus so it's an acidic solution okay so we've got H plus ions here and that's going to form water and there's our iodine that's that's produced here now the I minus ions are oxidized effectively to I2 and so the more concentrated the oxidizing agent is the more I minus ions are oxidized so that's critical okay so the more concentrated this is then the more of these ions are actually converted into iodine That's going to be pivotal when we look at the next steps for the titration. Okay, so let's look at step two, which is the method. So remember step two, we've kept the main steps up there at the top, just as a reminder.

So we've used the oxidizing agents now, and now we're going to carry out the titration to work out how many moles of iodine were produced from that previous step. So here's the method. So we're going to add the solution from step one into the conical flask.

That's going to go into the bottom there. Then what we're going to do is we're going to add sodium thiosulfate Into the conical flask and look out for a pale yellow color, okay? So we're going to add that into there so you sodium thiosulfate goes in here And we're going to drip feed that into the speaker here And we're looking out for that pale yellow color so as the color change is really difficult to see if you don't know about you If you try and look at something colorless to pale yellow it's really difficult to see so what we're going to do is going to add 2 centimeters cubed of starch into there And starch turns a deep blue colour if iodine is still present in the flask. And we're going to keep adding until that blue colour disappears. In other words, we want to make sure the iodine has completely reacted.

And then at this point, all of the iodine has reacted, like I say. And then we can use the volume of sodium thiosulfate that we've added in here to work out the number of moles of iodine that were present from the first step. Okay, are you following?

There's a lot of steps here. So that's what we do with that one there. And then the third step, so we're going to do the calculation and we'll look at the third step.

Of course we're going to do the calculation. So for example, let's look at the calculation example here. So all the iodine reacted with 10.5 cm3 of 0.14 molar thiosulfate solution.

So that's what we've added in there. That's the amount that was required. So here's the reaction. So thiosulfate is here. That was in the burette.

And the iodine... That was the iodine that was produced from this reaction here. And the amount of iodine was dependent on the concentration of your oxidizing agent. So this is the reaction, the overall reaction here.

So the equation shows what's happening in the titration. So the moles of thiosulfate is concentration times volume. And we need to divide that by 1,000 to get decimetres cubed. So the moles of thiosulfate is 0.14 because we've worked it out there. times by 10.5, because that's the volume we used to add into there, divide it by 1,000, and that gets us the total number of moles of thiosulfate.

So the moles of thiosulfate react in a 2 to 1 ratio. So we basically divide that by 2 to get the number of moles of iodine. So 2 to 1 ratio, and then we get this answer here.

7.35 times by 10 to the minus 4 moles of iodine. were in that conical flask. So this is where we look at step three. So we use the moles of iodine that we've just worked out in step two to work out the concentration of I or three minuses that we used in step one. So just as a reminder, the moles of iodine from step two was that, and we used 25 centimetres cubed of oxidising agent in step one.

So the equation from step one was that, so we've seen that already. So in step three, we have to look back at our original equation from step one, because we'll use this to work out the number of moles of IO3, and hence the concentration. So we've got three moles of I2, and these are produced, so this is it here, three moles of I2, and these are produced from one mole of IO3-, because we have a three to one ratio. Remember, we know that. bit because we've just worked that out from step two so the moles of io3 minus how many is in here is divided by three so it's seven point three five times ten to the minus four number of moles of iodine divide that by three and that tells us the number of moles of io3 minus so you see we're kind of working backwards here and so the concentration of io3 minus is moles divided by volume okay so that's no difference remember that from year one and remember we've got to convert the volume to centimeters cubed to decimeters cubed by divided by a thousand so we do 2.45 times 10 to the minus 4 which is the number of moles divide that by 25 divided by a thousand that converts it into decimeters cubed and then finally we get the concentration of the original oxidizing agent that we used which is 9.8 times by 10 to the minus 3 moles per diem cubed so There's a lot of information there and if you weren't too sure I have done another video.

It's one of my whiteboard videos with a white thumbnail and like I said you've seen the information come at the top of the video and you just click on that and it'll take you to that video. It is a long process, but be methodical, break it down. So you've got to work out your oxidizing agent.

So you need to oxidize that first to produce the iodine. Then we use the titration, work out the number of moles of iodine, and then revert back to step one again to then work out how much oxidizing agent you've used. So we're now going to look at half cells.

So half cells is to do with electrochemistry. It's... very dependent though on redox so it's it's all linked together that's why we go through redox first um to understand what half cells are so a half cell is just one half of an electrochemical cell and they can be constructed of a metal dipped into its ions or platinum electrode with two aqueous ions so what we've got here here's the first one so this one is an example of the of a half cell but this time we're going to be using a metal and its iron dipped in a solution of its own iron. So if we had an iron electrode which is here, this is dipped in a solution of Fe2+, then what will actually happen is we will have a reaction.

I know it doesn't sound very exciting because if you just put a chunk of metal into its ions of solution, you might think well I don't see a lot and you probably won't. But there is a reaction and the reaction is this. Basically we have an equilibrium system.

So we have Fe2+, which is the solution in here, picking up two electrons. That will form iron to get a deposit of iron on here So this is and the reaction is in equilibrium so it can flip from one side to the other But this is just if you just had a beacon you chucked a bit of metal iron and metal In there, so let's look at another example where we don't have a metal electrode dipped into its own iron So this is where we can use platinum electrode and we can dip that into a solution of ions but we must have two ions in there, so in this example we have a platinum electrode dipped in a solution of Fe2 and Fe3 plus ions we've got two ions in there and the half equation is Fe3 plus plus an electron will form Fe2 plus that's the half equation for this again it's reversible all these are reversible the reason why we use platinum here is platinum is inert it doesn't react with anything in here so we're not going to get interference from platinum atoms interfering with whatever's happening in this beaker and From an electricity point of view, because we're looking at electrochemicals, it's a really good conductive electricity, but it's really expensive. So you might have some of this at school or college, like a very thin wire. It doesn't look very impressive, to be honest, but it is really expensive. Platinum, obviously, it's used in jewellery.

It's a precious metal. So, yes, this is what you can set up a half cell like that. So an electrochemical cell... is created by joining two different half cells together.

And we can have loads of combinations of these half cells, loads of different types. Imagine it's like letters in the alphabet, loads of different letters, and you can join one letter with another letter to form a word. And it's the same with an electrochemical cell. We can pick a library of different half cells, put them together, and see if we can make some electricity from it, basically. Not all of them work, though.

Some of them don't work. So there are certain rules, and we're going to look into what... what is an electrochemical cell and how that works. Okay, so an electrochemical cell is made, like I say, of two half-cells that are joined together, and we need a wire that joins the two electrodes together, we need a voltmeter in between the wire to measure the potential difference, and a salt bridge. Now, if you do physics, this will probably be quite straightforward.

Luckily, there isn't a lot of physics in this, although I don't mind physics personally, I quite enjoy it, but really the emphasis here is on the chemical side, so the chemistry. So here's a setup of an electrochemical cell. So we connect two half-stills together.

We get one side undergoing a reduction process and we get another side undergoing an oxidation process. So essentially we have a redox reaction but we're not actually mixing the chemicals together like we've seen before. We're effectively keeping them separate.

So we're isolating these two chemicals but we're kind of connecting them with a wire. So you can see here The setup now we have a voltmeter and that is to measure the voltage or potential difference is the proper word for physicists start screaming It is and between two half cells now. This is called the EMF or E cell okay?

So this is what we call the EMF value you're going to see E cell values later on so that's gonna be quite important Electrons will always flow from a more reactive metal to a less reactive one. That's very important and The zinc half cell, as you can see, we've got a zinc and a copper one, but the zinc half cell shows the loss of electrons as zinc loses electrons easier than copper and oxidation has occurred. And we'll look at more about this later on in the next slide.

But the reaction here is zinc forming zinc 2 plus, plus 2 electrons. And so what we will see is because zinc is effectively. disintegrating, it's moving, well it's not disintegrating but it's reacting to form Zn2+, we have less zinc in our electrode. So this electrode will look thinner because we're producing more electrons.

Okay, so there we are. Okay, so the electrons whizz round, there we are. Okay, so zinc forms, we get more Zn2+, the electrons from that reaction go through the wire and they end up moving around to the copper.

And so the copper half cell accepts the electrons produced by zinc. Reduction has occurred here. So we've got copper 2 plus picking up them two electrons. These have come from zinc and we're forming copper solid.

And what we see, the observation here, is actually we see a buildup of copper around that electrode and it gets thicker. We also have a salt bridge in between. The salt bridge is normally made up of potassium nitrate. Potassium nitrate is a type of salt but we saturate it. It's basically a salt bridge.

And again, you might have done this in school or college, but a salt bridge is pretty much just a bit of filter paper folded up and you dip it into a highly saturated solution of potassium nitrate. It just completes the cell. There we are.

So it just allows the ions to flow through which balance the charges out. You don't need to be too concerned about the salt bridge, but it is imperative. If you didn't have a salt bridge in there, it just wouldn't work.

Okay, so each half cell has an electrode potential. We call that an E0, okay? And that E0 is a value, and a specific value, under standard conditions and temperature, okay? We measure that in volts, and it tells us how easily the half cell gives up electrons. Very, very important.

This gets a little bit complicated, so... Just bear with me two moments. So you've noticed from that previous slide where we had zinc and copper that were two half cells connected to form a full cell. So we have two half cells and they each have a reversible reaction. So in electrochemical cells, we always write the equation that is reduced in the reduced form.

Okay, so that means electrons are always on the left hand side and you will see you'll get an electrochemical series. and you'll be given the information for this. You don't need to remember the numbers for these, but you'll always see them written in the reduced form. That's the standard way in which they're shown.

Irrespective of whether they're being reduced or not is irrelevant. They're always written like that, okay? So we always have electrons on there.

So what this means is we always show electrons with reduction in the forward direction. In other words, zinc is gaining electrons. Zinc 2 plus, sorry, is gaining electrons to form zinc, okay?

And that's a reduction process. But remember, when we connect these two half cells together, we always have one cell undergoing reduction and one cell that's undergoing oxidation. What we need to work out is which one is being reduced and which one is being oxidised. We do that by looking at their electrode potential values, their E0 values. A data book does show you this, but you will be either given the information in the exam or it will be on your......

data sheet okay it's more likely going to be written written in your exam so don't worry about the values here so let's have a look we've got two values here so we know zinc and this is a standard value zinc zinc two plus half cell has an e naught value of minus 0.76 and a copper two plus has an e naught value of plus 0.34 now what we've got to what we've got to do is use that information to work out if we connect these two which one's going to be reduced and which one's going to be oxidized. So what we've got to remember is a rule. And again, this is just an acronym that I've made up.

But we remember a rule which is no problem. So the most negative half cell will undergo oxidation. The most positive half cell will undergo reduction. So no problem.

Remember that because these can be really tricky when you've got oxidation and reduction and enod numbers and everywhere. So if we can try and remember that, that will make this whole process so much easier. So in this case, zinc 2 plus zinc is the most negative.

So we'll be the half cell where oxidation takes place. Okay, so oxidation is a loss of electrons. So we flip the equation to show this. Okay, so in this cell, we have zinc giving up the electrons.

So here it is here. So zinc giving up the electrons. and copper two plus ions are accepted them.

Okay, so the overall equation will be zinc, okay, plus copper two plus, because remember we flipped that the other way around, copper two plus. will form zinc 2 plus which is over here and copper so that's the overall equation because remember we've got to have one that's been oxidized and one that's been reduced so we've got to remember that rule no problem negative half cell will undergo oxidation the positive half cell will undergo reduction so we flip we flip that one round to zinc and zinc 2 plus so there it is there okay so just apply that principle and then you'll know what reaction you know what's being reduced and what's being oxidized Okay, to help us measure the standard electrode potential for each half-cell, we've got to measure it against a reference. And luckily, we've got a reference called a SHE, which is a standard hydrogen electrode.

And this is, like I say, it's used as a reference to measure against these half-cells. Because you might think, well, how do you know that's minus 0.72? We measure it against this.

Now, the... They can't be measured on their own because you've got to connect it with something to work out the potential difference. So the standard hydrogen electrode is calibrated to equal zero.

So it's got zero volts and so that means any reading that it shows on the voltmeter will be because of the other cell, not because of the standard hydrogen electrode. It is a standard setup as you might expect. So this is the setup here and what it has is it has hydrogen going in and you need to know this information. But hydrogen goes in at 298 Kelvin and 100 kilopascal, so that's standard conditions.

That goes into this pretty glass tube here with a hole in the side of it. Okay, so we have one moles per dm cubed of H plus ion, so this is acid, in the beaker. And for copper, we also have one moles per dm cubed of copper ions in there, so we keep everything the same.

We're just looking for a fair... Voltmeter result here, so we've got to keep the concentrations the same so like I say the word standard is important We've got to have these at temperature of 298 Kelvin a pressure of 100 kilopascals and the concentrations must be 1 moles per dm3 You must remember these conditions for a standard cell Anything deviates away from this will give you a different voltmeter reading so it is important So the diagram on the left like I say it shows the standard hydrogen electrode connected to a copper cell So assuming that all the conditions above are met, then what we can work out is the standard electrode potential for copper, which is the value that we'd seen in the previous slide. So it's just a benchmark.

It allows us to work out what the values are. So to get 1 moles per dm cubed of H plus ions, you've got to be really careful about this, because sometimes the example can be a little bit sneaky. So to get 1 moles per dm cubed of H plus ions, we need 1 moles per dm cubed of HCl.

or half a mole of H2SO4. Now, can you think why? Well, the reason is H2SO4 is diprotic, and you would have seen that from a previous... If you've seen the previous video on acids, bases, buffers, you would have seen that H2SO4 is diprotic, and it gives up two protons for every one molecule of H2SO4.

So we need half the amount of H2SO4, and that will still give us one mole of H plus ions. Okay, so let's look at the electrochemical series. So this is nothing like a Netflix series or anything exciting like that. It's just basically a table with loads of electrode potentials in. But the electrochemical series is a list of half-cell reactions and their standard electrode potentials.

So that's E0. So this is just an example of the type of thing that you might say. They're all different. It depends on what you're going to work out.

But there is just simply nowhere near enough space. to fit the full series on here so this is just an extract from an example so the table shows the e naught value in descending order okay in this example it could be the other way around so don't take this for granted and it could be flipped the other way around it depends on how the exam board wants to wants to portray that so in this case as we go up we get the stronger oxidizing agent so the one with the most positive e naught value and the agents on the left hand side of the equation more easily reduced so these ones here so in other words these ones are the most easily reduced compared to on the right hand side so they have an increase in tendency to gain electrons and so they are the most most powerful oxidizing agent what the exam board can do is it can give you a table and ask you what which is the most powerful oxidizing agent in this case it's the one that's the most positive and it's the one on the left hand side here so this one's got the highest tendency to gain electrons so therefore is the most powerful oxidizing agent So in this case, the most powerful oxidizing agent is Cl2, and the weakest oxidizing agent is Mg2+. So it's this one on the left here. And likewise, we can go the other way. So the stronger reducing agent is the one towards the bottom, the one with the most negative value in this case.

So agents on the right-hand side of the equation are more easily oxidized. So they have an increase in tendency to lose electrons. So they're more powerful reducing agents.

So here the most powerful reducing agent is magnesium. And the weakest reducing agent is Cl-which is at the top. here so you've got to be able to you've got to be able to identify from um an electro electrochemical series what's the most powerful oxidizing agent and the most powerful reducing agent just make sure you memorize these different areas because it's um it's it may only be worth one mark but it could um you know every mark counts Okay, so let's look at some calculations involving standard cell potential.

So standard electrode potentials, E0, as we've seen before, they can be used to work out a standard cell potential, which is basically the value of the cell. We use E0 cell equals E0 reduced minus oxidized. Okay, now you can remember this as the acronym REDOX.

Yeah, so it's just the same. REDOX, reduced, oxidized, and you put a minus in between. So. The half cell, what we've got to remember is the half cell equation with the most negative E naught value is being oxidized. Okay, so that's the key thing.

So the most negative goes on the right hand side. And obviously the most... positive goes on the left you might see you might be taught a different equation and it might be slightly different depends on what the view you know what what you've been taught but this is a way in which I remember it using the analogy that I used before which is no problem Okay, so it just ties it all in really.

But if you have two positives or two negatives, then it's the most negative that's oxidized. Okay, so that's the most important thing. Okay, so let's look at an example of how we work this out. So we're going to use the data in the electrochemical series to calculate the E-naught of the cell when Cl2 and Cl-minus half-cell and a Zn2 plus and Zn half-cell are connected.

Okay, so first of all, we need to identify which is being oxidized. Okay. Okay, so this is where you use your no problem. Okay, so zinc 2 plus and zinc half cell is the most negative, so is oxidized. Okay, so most negative, oxidized, no.

And then problem is positive reduced. So that's the most oxidized. Yeah, that's the one that's been oxidized. So number two, we calculate the E naught of a cell.

So E naught cell is 1.36. Okay, because that's our chlorine one there. Okay, and the most oxidized is minus. minus 0.76 which is our zinc one which is down here so if we put them in there we get plus 2.12 volts let's look at another example so we're going to use the data in the electrochemical series to work out still using a chlorine half cell but we're going to use copper instead so again identify which one's been oxidized so it's the copper 2 plus copper half cell that's the most negative so is oxidized okay so what we're going to do is we're going to put all these in here so we've got one point three six which is obviously our chlorine and the copper one the reason why I've picked this one is copper is positive as well so you've got two positives but because that is the most positive that goes on the left and the least positive goes on the right that's been oxidized so the totally not the cell here is one point zero two volts Okay, so you've seen the diagram of the cells, how we draw cells with the beakers and salt bridges, etc. That's, you know, this is chemistry.

We don't faff around and we don't draw diagrams and silly things like that with beakers and salt bridges. We need a simplified version. it's so much quicker so thankfully there's something called a cell notation and this allows us to simplify the structure of a cell and but there is a standard way of representing them in chemistry okay so a standard notation is represented like this and you can see on here it's just so load of lines with words in between so the most negative half cell potential goes to the left of the double line okay so you've got to remember that most negative on the left okay most positive on the right so we always have um this structure here so most negative half cell goes to the left of that double line so we have certain features here so a single line single line shows a physical state change so that could be metal to metal iron solution like we've just seen before.

A solid double line shows a salt bridge obviously that bridges the two half cells. And so if we use the zinc and copper cell that we've seen before, here's an example here. So we've got zinc, which is the most negative half cell on the left.

Copper is the most positive on the right. But you see we have a copper solid and the aqueous ion. So there's a line there to show, to signify the difference in state. Okay, so zinc 2 plus has an oxidation state of plus 2. Whereas zinc has an oxidation state of zero. So because that is the most oxidised form out of the two, that's the one that sits closest to the salt bridge.

So the most oxidised form goes closest to the salt bridge. So what if you have two aqueous ions? So what we do, we follow the same rules as above, but instead of having the solid line, we use a comma. So we separate with a comma because they're in the same physical state. So that might be, for example...

Fe2 plus Fe3 plus and platinum so between Fe2 plus and Fe3 plus we have a comma so there's the example there so you can see here Fe3 plus Fe2 plus the comma separates these two they're not they're in the same state so we use a comma but that platinum is a different state to either of these two so we use a solid line there and obviously that's shown in the state symbol so remember with these type of cells where we have two ions in solution we've got to use a platinum electrode and because there's no solid material there we need something to clip to clip onto the the voltmeter so we use platinum as the electrode here okay so what we're going to do is going to predict reaction feasibility so if you've seen the video on Gibbs free energy so the one with enthalpy and entropy before then it's similar to this except what we're going to do is we're going to use standard electrode potentials to predict So it's a bit like predicting, well, if only this could predict Lockley numbers, that would be great, wouldn't it? But it's a bit like predicting from Gibbs 3. So standard electrode potentials, so E0, they can be used to predict if a stated reaction is likely to proceed under standard conditions. OK, so let's have a look at an example.

I'm going to bring up our electrode potential series here. So example one, use the data in the electrochemical series to predict whether solid magnesium... will react with copper two plus ions in solution under standard conditions okay so what we're doing is we're saying right if i get magnesium and i've got some copper two plus ions like copper sulfate for example a blue solution and i took that magnesium in will it will i see anything will it react so first of all what we've got to do and we've got i'll remember i'll remember a bit there as well so about half cell equation where the most negative e naught value is being oxidized okay so just remember that but the first thing you is we need to identify which has been oxidized.

So the Mg2 plus Mg half cell equation has the most negative value, which is this bit here. So that is being oxidized. Okay, so remember that it's no problem.

Okay, so the most negative is oxidized. So we take the oxidized equation and reverse it. Remember, we flip that round.

Yeah, we write the other two equations next to each other like this. So we write it magnesium and copper. So we then combine the two equations to obtain the feasible reaction.

So whatever that is, that's the feasible one. Mg plus CO2 plus will form Mg2 plus plus Cu. And then what we do is we compare that equation with the one in the question.

So we can see that magnesium will react with copper 2 plus ions as they match. So this is telling us the feasible reaction is magnesium reacting with copper to form Mg2 plus and copper. So yes, it will react.

And we can always confirm this is always worth doing this by calculating the E0 of the cell and all feasible reactions will always have a positive E0 cell value. So we put that in our equation, we can see that that does actually give us a positive voltage. Okay, so let's look at a second example.

So use the data in the electrochemical series to justify why iron nails become rusty when in contact with air and moisture. Now this is a slightly different type of question. because they're asking us to justify it they tell us yeah it does work we know it works but you tell us why you know prove it so this is just as important in science to prove something um so here we go so we've got our electrochemical series there's the two um half equations that we need so we need the um iron one because it's made from iron but we know that rust is caused by aerial oxidation in other words we've got oxygen in the air that reacts with it you So that's the most appropriate question, the most appropriate half equation, which is the one towards the top there. It's this one. So we need to identify which has been oxidized and so the Fe2 plus Fe has the most negative value so that one's been oxidized.

No problem. Remember that. So the second one is we take the oxidized equation, we flip it around the other way and we write the two equations side by side.

Exactly the same as what we've done before. And so here we multiply the top equation by two because we've got to balance it. OK, to make sure the number of electrons in both equations are the same. So here it is here. So we combine the two equations and this obtains the feasible reaction, which we can see here.

Then we compare the equation to the reaction state in the question and we can say that obviously iron will react with O2 and H2O as they match. But we know that already because they've told us they just want us to justify it, prove it. And so we can confirm this again further by using an E0 cell calculation, and we can see that it is positive.

It's only just positive, so it's a slow reaction. Obviously, we're not going to see something rust instantly, you know, not in these conditions anyway, but nonetheless, it is feasible. It does work.

We see it. We observe it. We see iron going rusty. Okay, so let's look at a third example. Now, what the key thing here, and this is what I'm saying here, is just because we calculate E0, E naught to say if something's feasible doesn't mean it will necessarily go a little bit like the Gibbs free example we use.

So for example you might have non-standard conditions so if we change the concentration the temperature etc this can cause the electrode potential to change. So for example we might have a really cold environment and that may not cause it to to work it might not actually be feasible. So let's take that rusty nail example from the last slide.

So here's the E naught value and we have the half half equations and pull them across and then we have the e naught cell value that we've calculated if we count if we increase the concentration of oxygen okay equilibrium will shift right which means it's easier for the oxygen to gain electrons okay and so the electrode potential for this half cell will become more positive and the and the cell potential will be higher okay so if we increase the concentration of oxygen. In other words, if we've got more oxygen here, then the equilibrium will shift to the right. So it'll move to this side. Okay?

So if we increase the concentration of iron, equilibrium will shift left. Less electrons will be used up. And so the electrode potential of the Fe2 plus and Fe becomes less negative.

And the full cell potential will be lower. So it's important to look at these in terms of equilibria as well. And that's rightly important.

And another example might be if the kinetics is not favourable. So for example, the rate of reaction is slow. So it appears though there's no reaction. especially with rusting as I mentioned before and if the reaction is a high activation energy it may stop the reaction from happening at all so and catalyst can help to use bring the activation energy down but there's a lot of factors at play here so it's not necessarily that it means that just because it says it is feasible that it it's definitely going to happen it depends on the conditions Okay, so let's look at the final part of this topic, which is cells and batteries.

Okay, so energy storage cells. So what we need to do, obviously we looked at half cells, and we looked at connecting them together to form an electrochemical. cell a full cell and there's got to be some use to this and the use is is clear it's batteries and we have it in everyday life okay so this is very very relevant chemistry so to establish the overall reaction in in a lithium-ion battery what we need to know know uh the half equations at each electrode so we've got um a lithium ion battery is something you would see in a mobile phone okay so they're very light uh lightweight batteries and used to power uh quite complicated devices like say phones and laptops and tablets etc but the half equation for this type of reaction is this so it's lithium um your lithium ion at the top and then you've got your cobalt oxide uh reaction here so um you'll be given these it's about using the information you so Li plus Li this one has the most negative e naught value so oxidation occurs here so this means we flip the equation and this shows the electrons being produced ie it's a negative electrode so it's exactly the same as what we've done before so the negative electrode is Li will produce Li plus an elect and plus the electron so we flip that equations that's it flipped and obviously the positive electrode is just the same reaction here because that's positive so anything it's most negative we flip it around so there's our two new half equations And so the overall equation on discharge, so this is when we're using the cell. So we're using the battery. So it's lithium reacting with the cobalt oxide that's in there to produce your lithium ion or your lithium salt product.

So to work out the E naught of the cell, we do exactly the same as what we've done before. So it's reduced minus oxidized. So it's 0.56, which is this value here, minus minus 3.04, which is the most negative because that's your oxidized.

And that's 3.6 volts. Now, you might be asked to say, right, what would the reaction be if we plug the thing in? Because obviously the battery will run out eventually.

So this is a rechargeable battery. It's just the opposite way. So all we have to do is reverse the equation. So this is the equation on discharge.

If it's on recharge and it back up again, then we just flip the equation the other way. It's exactly the same. All we're doing when we're charging a battery up is we're introducing more electrons. into, or electrons into the system, and that moves it back the other way, and it recharges the battery up again.

Okay, fuel cells are an interesting one. So electricity can be generated from fuel cells, but it needs a continuous external supply. So unlike a battery in a phone where it doesn't need a continuous supply, it will last for a few hours, and then you'll have to recharge it, fuel cells need a continuous supply of fuel.

They need to be constantly connected to something, okay? So they're not a ready store like batteries. So an alkaline hydrogen oxygen fuel cell is an example of a fuel cell. And the diagram, you might be familiar with it, but the diagram shows a hydrogen oxygen fuel cell. And what we're going to do is, you see there's numbers on there.

We're going to go through each number on there and explain what's happening in these. So the first one is a hydrogen feed. So hydrogen is fed in.

So you need a constant supply of hydrogen. And what this does is it reacts with the OH minor signs in solution in 9. So in section 9, which is this bit here. So that's section 9. But we'll look at this later on because obviously we're going to do this in numerical order. But the reaction that does occur here is hydrogen comes in, reacts with the OH minor signs in here because it's an alkaline cell, and that forms water and four electrons.

But we'll come back on that. So the second one is the flow of electrons. So these electrons here that are produced. These travel through a platinum electrode through here. It's a good conductor.

It's inert remember and it goes through Through the wire and then three is the components that depends on what you're powering And so this could be a car you can have a fuel cell car. It could be a house You might have a fuel cell outside the house and generate in power. So that depends on what you're powering Okay, so the fourth one is the oxygen feed.

So we've got that feeding through the other side of the cell. So this reacts with water and the four electrons made from step one to make OH minor signs and the reaction that occurs is basically oxygen comes in reacts with water and the four electrons that's just come through here on that side and that produces the four H minor signs which then drifts around in in section nine so the fifth one is the negative electrode so this is it here this is the cathode here so electrons float to the negative electrodes you can see them flowing in there and that's made from platinum as well and six is the electrolyte so this is the um the uh electrolyte that's held in in this compartment here and that's made from potassium hydroxide because it is an alkaline cell and what this does is this carries the oh minor signs that were produced from this reaction here from um from the cathode to the anode which is along on the side there So the seven is the positive electrode. This is the anode.

This is this bit here and electrons will flow from this section across into the into the cathode which is the other side so this is made from platinum and There is a byproduct that's emitted and this case it is water and the water that was produced here Remember is now being kicked out at this end here. So the water that was produced at this side. That's then moved out And then it's released into the surrounding so that's the only byproduct so that's better than burning a fossil fuel which would produce carbon dioxide Sulfates for example nitrates and another pollutants, so it's quite a green way of generating electricity And number nine is the movement of the OH minus sign so that's what I mentioned in the first step So OH minus signs that are produced from the reaction in step four here I've actually carried through the electrolytes to the other side of the cell Um And what we have as well between the electrodes, between the two sections, are ion exchange membranes.

And what these do is these align the platinum electrodes on each side. And they allow the OH-ions to pass through, but not hydrogen or oxygen gas. So it's very selective what can go through.

Because what you don't want is the hydrogen and oxygen mixing in the middle here. You just want the OH-ions to drift through. So you can exploit the electrons that are being transferred and use that as electricity. Okay, so to summarise the reactions that's happening in this cell, so we have a hydrogen feed, so remember that's on number one, so it reacts with the OH minus ions in solution in nine, and this reaction occurs, which is H2 plus H2. four OH minus will produce four H2O and four electrons now the electrons obviously go round the circuit and they're received at the other side which is this bit here and so that forms oxygen there's your four electrons plus your water that's produced as well so some of the water's moved moved across and then that produces your 4-OH-ions which then drift through the membrane and so the overall equation here if we combine both of them is we produce hydrogen plus oxygen and that forms water so two molecules of water are pushed out in this reaction so the only emission is water Okay, so let's look at the advantages and disadvantages of fuel cells.

You might think that sounds like a pretty good idea. Why don't we generate all electricity like that? Well, the advantages are they're much more efficient than internal combustion engines, so they convert a lot more energy. And in fact, we do have cars now that are hydrogen powered. There are some hydrogen powered cars.

It's mainly in, I think, California have a lot of hydrogen powered cars. But they do exist. More energy is converted into kinetic energy. whereas in combustion engines it's thermal energy and many electric vehicles are battery powered however so fuel cells don't need to be recharged so they're much much better in terms of a continuous supply effectively you put hydrogen into the car stored in the tank and as long as that tank is full you can't keep running you don't need to keep plugging it in but you do need that supply of oxygen hydrogen oxygen comes from the air but the hydrogen comes from your fuel so the only way this product is water so no carbon dioxide emitted so that's much better for the environment than a combustion engine so the disadvantages um hydrogen is very very flammable um it has to be stored correctly in a pressurized tank in the back of your car um so that can put some people off and it's really expensive it's literally you're transporting thin air um so it's expensive to transport and store hydrogen um so it has to be stored correctly and safely of course um and energy is required to make that hydrogen and the oxygen in the first place.

So fossil fuels are normally used to pass water through an electrolysis process to make that. So you take brine, electrolysis of brine, and you make your hydrogen that way, and that can actually contribute to carbon dioxide, so it's an indirect emission. And that's it.

So that's the video on redox and electrode potentials. It's quite a difficult topic, but there's a few acronyms in there to help you navigate it through that. Please remember to subscribe to the channel.

There's loads more videos on OCR dedicated to OCR. there is for other exam boards as well they're all tailored to each exam board and so please subscribe they're all free and that would be great and remember you can have a copy of this as well if you click on the link below you can purchase the the slides as well and you can use it for revision it's great for that and but that's it bye bye

Transcript for:Understanding Redox and Electrode Potentials

Transcript for:
Understanding Redox and Electrode Potentials