Transcript for:
Redox and Electrode Potentials

[Music] [Music] hi there my name is Chris Harris and I'm from Allawi chemistry and welcome to this video on OCR a redox and electrode potentials so in this video we're going to look at redox and electrode potentials specifically for OCR so if you're studying OCR chemistry in a level then this video is specifically designed for you and there are full range of videos for all the major UK exam boards on a Larry 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 and or you might yourself teaching yourself in which case it contains the the main points that are tailored to OCR ape so it is important though to practice exam technique as well there are videos on a Larry chemistry that looks at exam technique so have a look at them as well because that is just as important as knowing the content and it's a bit like if you're learning to drive it's about like picking up the the hybrid cord and you may know the highway code back-to-front that doesn't mean you hop in the car and start to drive so it's exactly the same approach you should take to your exams as well and all these slides here if you want your own copies of these slides that are available to purchase in my test shop so if you click on the link in the description box you'll be able to get ahold 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 and your revision that you that you get from from school or college or if you'll of course we get a lot of people who who are doing it and being self-taught effectively ok so I can say this is specifically designed for OSHA a and it matches the specification so these are the points taken from the specification and everything you see in here will adhere to their rules ok so let's look at reduction and oxidation so you might have seen bits of this in year one chemistry and so this is just confirming what you may may already know and extending it into this topic because it is all about redox you're going to see it quite a lot obviously since 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 Vake so oxidation is loss and reduction is gain of electrons so you might have seen that GCSE you might have heard of it already but it's a really good acronym and it's very fitting for chemistry as well calling it oil lake and so this helps us to understand what's going on so for example the reduction this reduction when calcium is complete or this reaction sorry is when calcium completely is burnt oxygen 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 and calcium has been oxidized and it is losing electrons so we can see here that calcium is forming calcium 2 plus and 2 electrons are also being produced as well and in this reaction this is the other half of the reaction for oxygen so this reaction oxygen is being reduced and is gaining electrons so you can see here that the oxygen is gaining two electrons 1/2 or two is gaining two electrons to form o to minus so this is a this is a very simple way of showing redox reactions we are showing the reduction site at the oxidized oxidized side first then the reduction science is just breaking that down and if we combine them and cancel the electrons out as you'll see later and then we form the full 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 and ox reduction is the gain of electrons but a reducing agents lose electrons and all 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 rich 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 in a liberal 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 hydrogen's with H+ ions and the fifth rule is we balance charges with electrons which is e - now you might have a different method of balance and equations and you might be taught a different method and I find that this method works really well with even some of the obscure balance and 1/2 equations so this is just the method that I think it's simplest method there are other ways in which you can balance it and you may have been taught something else and but this is just another method it all comes to the same result anyway ok so you'll see that I put Asterix near and three and four and this only should be done if we need to balance oxygen hydrogen as it may not be the case that we need to balance oxygen hydrogen it's not all half equations have the main 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 I'll all the methods that I'll do will be using these rules so we're going to write half equations show in the conversion of iron two plus two is reap Loess so there's the first thing is to write down the species before and after so it's fe 2 plus Fe 3 plus we then need to balance in the atoms apart from those with oxygen hydrogen as these we balance later well they're already balanced we've already got one Fe and one Fe on the left and one I feed on the right balance any oxygens well there isn't any oxygen so we don't need to do anything with that and the same against with hydrogen there's no hydrogens in here that we need to be concerned about so we go straight into step 5 which is balancing the charges with electrons so you can see we have a 2 + on the left Plus on the right so an electron is always negatively charged it's got a minus one charge so you put the electron on the right hand side so that means you've got a two plus on the left and two plus on the right okay this is shown an oxidation state so oxidation is loss of electrons Fe two persons using electron to form Fe three-plus 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 and 1/2 equation so in the conversion of mno4 - that's Mangan it to em n 2 plus so the first thing we need to write the species before and after the reaction so it's mno4 - go into m n2 plus 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 this fortune left we ignore them we're just looking at manganese so there's manganese 1 MN on the left and 1 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 fork students 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 hydrogen's with H+ ions so you can see we have eight hydrogen's now because we've took that water in there but 8 hydrogens on the right so we need 8 hydrogens on the left as well so there it is okay so you can see it's fairly straightforward balanced 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 +7 because you've got a minus and eight positives and on the right we've got +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 a half equation for that this shows reduction so we've got a reduction step here but 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 okay so what we can do with these two half equations so the iron one that we've seen before and the manganese the manganese one 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 are 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 Fe 2 plus go into Fe 3 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 and you might have done this in maths which is Tim your taneous equations this is exactly the same principle okay so we're finding in this case the common the commonality is always the electrons in this case there might be other elements but must be the electrons to start with so electrons don't balance so we need to multiply the top row by five so here we go so we've got five Fe 2 plus will form five Fe 3 plus and five electrons okay so we cancel the electrons out and we combine the two equations together so there we are so we've got mno4 minus eight h plus five Fe 2 plus well form mn 2 plus five Fe 3 plus and 4 h2o so we can see we've cancelled electrons and your ionic equation your full-on 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 titration so you would have seen titrations before already so the titrations are actually showing you the titrations you would have seen our acid based titrations these ones are going to be slightly different so these ones are going to look at redox reactions instead so we're not using us in the base weezing redox substances of reduction and oxidation so transition elements can be used to oxidize and reduce of the substances and can you look at we'll look at transition metals later cause they're quite a special group of chemicals so transitional elements have variable oxidation states which means they can accept or donate electrons easily hence the word transition they can they can transition between various different states unlike other elements which generally have fixed oxidation states so transition elements are colored depending on their oxidation state at the time and this makes them particularly useful in titrations because we can spot an endpoint so remember an acid-base titrations where you have to add an indicator to spot an endpoint these ones generally you do this without the use of an indicator so you just use the color of the transition element to do that so let's look at a specific example you might have seen this example as well in in lab work if you if you've done this but we use acidified potassium permanganate solution which is kmno4 is an oxidizing agent okay so remember what an oxidizing agent is so oxidizing agent is something that gains electrons it dissociates to produce mno4 minus ions where MN has an oxidation state of plus 7 in there okay and they are reduced to MN 2 + 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 CR as well so they they are there so have a look but I'm going to assume that you know that you'll be pretty feel fairly confident with it I'm going to go through some of the some of the basics as well but just to skip past that bit okay so they reduced to M and two plus ions do Jo in the titration okay so let's have a look at the first half equation so we're mno4 minus plus h 8 h plus plus 5 electrons or form MN and well from mn 2 plus plus 4 h2o and so the mno4 minus is reduced obviously that's supposed to be a a little n rather than the capital n ok so that's been reduced and obviously we know how to make these 1/2 equations because we've just seen them before what's been oxidized in this titration and is iron so we've got iron 2 plus plus iron three-plus will produce iron 3 plus plus 5 electrons this has already been balanced for us so the fe 2 plus is being oxidized 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 color and and it goes colorless as we as we progress through the titration if we're going to doing it this way so this color change from purple to colorless will show as an endpoint so another one a certified potassium dichromate is an oxidizing agent and it dissociates to produce dichromate ion so cr2 o7 to minus ions where chromium has an oxidation state of plus six and and these are reduced to c r3 plus ions during a titration ok so very similar to manganate but a different substance so here we've got cr2 o7 to- has been produced so you can see here that it's going from cr2 o7 to CR 3 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 endpoint as we form a product on from orange to green okay so let's look at em 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 agents 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 Fe 2 plus okay so this example looks at finds in a concentration of a reducing agent okay and all we do is you just reverse it for oxidizing agencies no there's no difference okay so so what we've got here is our classic titration setup you got your burette you get 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 is a reducing agent so we have Fe 2 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'll probably have say a 25 centimeters cubed in that conical flask so what we've got to do is go to add dilute soft sulfuric acid into there as well now so this is all dissolved in an acidic solution and so this is to ensure that you have sufficient H+ 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 in the burette with a known concentration so we know the strength of it which is in this case is going to be manganese ions and so mno4 minus and then we're going to add the manganese ions into the burette in the burette into the conical flask until we see a faint color of permanganate of so the manganese iron appear and this is known as the endpoint so there's no indicator added here we're going to use the manganese to provide the endpoint 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 and if we go too much one drop can make a difference and if we go too quickly you'll get a result that isn't accurate okay so and there was a sharp color change okay like I said it's very dramatic and so the manganese ions form aqueous potassium permanganate solution or purple okay and they'll immediately react with the reducing agent until all the reducing agent is used up so what the reacts with the fa2 plus and it will keep on doing that and keep on doing that until there's no more Fe to prosigns left because they've all they've all reacted so one drop at the end like it's a can turn the solution purple which is the color of the oxidizing agent when you could use a colorless oxidizing agent and a colored reducing agent and you could look out for the color disappearing but in this example what we're looking for is the color to appear in the conical flask but it can be done the other way it just depends on on the set up of your titration so what we can do is read how much oxidizing agent was added and we're going to read to the bottom of the meniscus and we always read at eye level so you get right down to eye level right where you can see the the burette and the the level and you read from the bottom which is that 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 to results that concordance so they've got to be within point one centimeter cubed or point one zero or centimeters cubes of each other and that just ensures that your you can your results are valid so 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 concordant okay so and 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 and have eighteen point three centimeters cubes of point naught to five moles per diem cubed of potassium permanganate okay which is kmno4 and that's reacted with 25 centimeters cubed of acidified iron ii sulfate solution remember it's got certified to produce them h+ ions and then we calculate the concentration of fe 2 plus ion 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 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 two-plus ions okay so you can't just get a pot of fe 2 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 manganese okay if in doubt always use this in if in doubt work out the moles so moles equals concentration times volume concentration is naught point naught 2 5 because we've been given that times by eighteen point three times ten to the minus three so I like to use and you've seen my videos before I like to just sort of put and divide by a thousand I like to just put times ten to the minus three on the end that just means the same as dividing by a thousand and that gives us a total number of moles of manganese may be out so total moles of manganese of four point five eight times by ten to the minus four so the third thing is we use the equation to find out molar ratio in order to work out the moles of Fe 2 plus so we can see from the equation as I want to five ratio between mno4 minus and Fe 2 plus so the moles of Fe 2 plus is basically going to be the number of moles that we've calculated before multiplied by five so that's going to give us two point two nine times by ten to the minus three moles so now we know the number of moles of iron 2 plus in that conical flask so then what we need to do is we then calculate the concentration using concentration equals mass over volume so the concentration of moles per diem cubed is 2.2 nine times by 10 to the minus 3 divided by 25 times by 10 to the minus 3 because about 25 centimeters cubed of that to remember convert to decimeter cubed and that gets us naught point naught 9 to moles per diem cubed of our concentration of our iron two-plus okay so let's look at another example here and 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 and all the way through this I'll put a link review 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 but the titration is useful for finding out the concentration of an oxidizing agent okay so the titration work is conducted in three broad steps so the first step if we use the oxidizing agent which is ki O 3 to oxidize iodide ions to iodine okay so that's step one step two is we then carry out the titration to work out the moles of iodine produced in step one and then the third step as we use the moles of iodine in step two to work out the concentration of I or three 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 carryover 3 to oxidize iodine ions to IAD so what we need to do is measure a volume of ki O three that's potassium iodide five because it's got an oxidation state at five and that's your oxidizing agent and this will produce the I or 3 minus I 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 that excess acidified potassium iodide solution K I to the ki O three solution and so here's the equation so we've got io 3 - which is your aqueous solution so and we're going to react this with our iodine so this is our oxidizing agents yeah and this is going to be in hich plus so it's it's in acidic solution okay so we've got H+ 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 they're more concentrated the oxidizing agent is the more I - ions are oxidized so that's critical okay so the more concentrate that 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 okay so let's look at step two which is the method so remember step two we've kept the main steps up there the top just as a reminder so we've made we've made 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 so that's going to go into the bottom there and 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 gonna add that into there see sodium thiosulfate goes in here and we're going to drip feed that into the speaker here and we're looking at that pale yellow color so as the color change is really difficult to see you have to 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 two centimeters cubed of starch into there and starch and turns a deep blue color if iodine is still present in the flask okay and we're going to keep adding until that blue color disappears in other words we want to make sure the iodine has completely reacted and then at this point all of the iodine is reacted back to 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 okay and then the third step oh so we're going to do the calculation and we'll look at the third step of course we can do the calculation so for example let's look at the calculation example here so all the iodine reacted with ten and a half centimeters cubed of 0.14 molar that 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 when 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 Mo's the moles of thiosulfate is concentration times volume and we need to divide that by a thousand to get decimetres cubed so the moles of thiosulfate is nought point one four because we worked it out there times by ten point five because that's the volume we used to add into there divided by a thousand and that gets us the total number of moles of thiosulfate so the malls of thiosulfate reacts in a two to one ratio so we basically divide that by 2 to get the number of moles of iodine so two to one ratio and then we get this answer here seven point three five times by ten to the minus four moles of iodine were in that conical flask okay so this is where we look at step three okay 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 okay so just as a reminder the moles of iodine from step two was that and we use 25 centimeters cubed of oxidizing agent in step one so the equation from step one was that so we've seen that already okay 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 Io or three 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 I or three minus because we have a 3 to 1 ratio remember we know that bit could you just work that out from step two so the moles of Aisle 3 minus heroin is in here is divided by three so it's seven point three five times 10 to the minus 4 number of moles of iodine divide that by 3 and that tells us the number of moles of I or 3 minus kind of working backwards here and so the concentration of io3 - 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 and two centimeters cubed to decimeters cube I divide by a thousand so we do two point four five times ten to the minus four which is number of moles divide that by 25 divided by a thousand that converts it into decimates cubes and then finally we get the concentration of the original oxidizing agent that we used which is nine point eight 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 videos from one of my whiteboard videos with a white thumbnail and like I said you've seen the the the information come at the sumers 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 them work out how much and how much oxidizing agent you've used okay so we're now going to look at and 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 what I will go through redox first and 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 it it's ions or platinum electrode with two equus ions so what we've got here is the first one so this one is an example of the other of a half cell but this time we're going to be using a metal and it's 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 Fe 2 plus then what will actually happen is we will have a reaction I know it doesn't sound very exciting because we just put a chunk of metal into its ions of solution you know anything well it'll see a loss and you probably won't but there is a reaction and the reaction is this basically have an equilibrium system so we have Fe 2 plus which is the Lucian in here picking up two electrons and that will form iron so you'll 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 beak and you talked 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 it so an 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 Fe 2 and Fe 3 plus ions we've got two ions in there and the half equation is fe 3 plus plus an electron will form fe 2 plus that's the half equation for this again it's reversible all these 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 and from an electricity point of view because you're looking at electro chemicals it's a really good conductor of electricity but it's really expensive so you might have some of this at school or college and make a very thin wire doesn't look very impressive to be honest and but it is really expensive platinum of seats using jewelry it's that's a precious metal so yes so 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 and loads of different letters and you can join one letter with another letter to form a word and it's the same with 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 okay some of them some of them don't work so there are certain rules and we're going to look into you know what what is an electrochemical cell and how that works okay so an electrochemical cell is made like a say of two half cells that are joined together and we need a wire that join the two electrons together we need a voltmeter in between the wire to measure the potential difference and a salt bridge now if you do physics this this will probably quite straightforward and luckily there isn't a lot of physics in this although I don't mind physics personally I quite enjoy it but you you really the emphasis here is on the chemical size of the chemistry so here's a setup of an electrical electrochemical cell so we connect two half-cells together we get one side undergo in a reduction process and we get another side and to go in 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 what ISIL ating these two chemicals but we're kind of connecting them with a wire okay so you can see here and the set up now we have a voltmeter and that is to measure the voltage or potential differences the proper word before physicists start screamin s's and between two half-cells now this is called the EMF or es al 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're going to zinc in a copper one but the zinc half cell shows the loss of electrons as zinc loses electrons easier than copper and oxidation is curtain and we'll look at more about this later on in the next slide and but the reaction here is zinc forming zinc two-plus plus two electrons and so what we will see is because zinc is effectively disintegrating it's it's move wasn't disintegrating but it's reacting to form zinc two plus we have less zinc on our electrode so this electrode will look thinner and because we're producing more electrons okay so there we are okay so electrons whiz round oh yeah okay so zinc forms we get a more Z two plus ions 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 two-plus picking up them two electrons these have come from zinc and we're forming cop 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 salt bridge is normally made up of potassium manganate potassium nitrate sorry potassium nitrate is it's a type of salt but we saturate it's basically a salt bridges just and again you might have done this in in in school or college but salt bridge is pretty much just a bit of filter paper fold it up and you dip it into a highly saturated solution of potassium nitrate it just completes the cell there we are okay so it just allows the the ions to flow through which balance the charges out so you don't need to be too concerned about the salt bridge but it is imperative you didn't have a salt bridge now just wouldn't work okay so each half cell has an electrode potential we call that an enormous if Iqbal you understand the conditions and temperature okay we measure that in volts and it tells us how easily the half cell gives up electrons very very important it's getting 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 electrical 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 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 Sean irrespective 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 a zinc two-plus sorry is gaining electrons to form zinc okay that's a reduction process so but remember when we connect these two half-cells together we always have one cell undergoing reduction and one cell that's undergoing oxidation okay what we need to work out is which one has been reduced and which one's been oxidized and we do that by looking at their electrode potential values they're in law at values and you will have the data book does show you this but you will be either given the information in the in the exam what will be on your on your data sheet okay it's more likely going to be written written in you 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 naught point seven six and a copper two-plus has an enormous naught point three four now what we've got to what we've got to do 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 real and we this again this is just an acronym that I've made up and but we remember a rule which is no problem okay so the most negative half cell will undergo oxidation the most positive half cell will undergo reduction so no problem okay remember that because these can be really tricky when you were oxidation and reduction and odd 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 two-plus zinc is the most negative so will be the half cell where oxidation takes place okay so oxidation is a loss of electrons so we flipped the equation to show this okay so in this cell we have zinc giving up the electrons so here it is here so zinc zinc giving up the electrons and copper two-plus ions are accepting 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 two plus which is over here and copper so that's the overall equation because remember when I have one that's been oxidized and one that's been reduced so I'm gonna remember that rule no problem negative half cell will undergo oxidation that positive half cell will undergo reduction so we flip we flip that one round to zinc and zinc two-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 the electrode potential for each half so we've got to measure it against a reference and luckily we've got a reference called the Shi which is a standard hydrogen electrode okay and this is like you say it's used as a reference to to measure against and these half cells because you might think well how do you know that's - naught point seven - and we measured against this now the they can't be measured on the moment 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 okay so it's got zero volts and so that means we 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 set up as you might expect so this is the set up here and what it has is it has hydrogen going in and you need to know these you need to know this information but hydrogen goes in at 298 Kelvin and a hundred kilo Pascal so that standard conditions that goes into this pretty glass tube here with a with a hole in the side of it okay so we have one moles per diem cubed of H+ ion so this is acid in - in the beaker and for copper we also have one moles per diem cubes 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 stand is important we've got to have these at temperature of 298 Kelvin a pressure of 100 kilo Pascal's and the concentrations must be 1 moles per diem cubes you must remember these conditions for a standard cell anything deviates away from this will give you a different volt meter reading so it is important so the diagram on the left like to say it shows the standard hydrogen electrode connected a copper cell so assuming that all the conditions above are met then this 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 okay so it's just a it's a benchmark it allows us to work out and what the values are okay so to get one moles per diem cube to page plus sign you've got to be really careful about this because the example can be a little bit sneaky and so to get one moles one mole bmq dove hates plus signs we need one more dmq dove HCl or half a mole of h2 so4 now can you think why well the reason is h2so4 is diprotic and you would have seen that from previous we've seen the previous video on acids bases buffers you would have seen that pitch to us as well as diprotic and it gives up two protons for every one molecule of h2 so4 so we need half the amount of h2so4 and that will still give us one mole of H+ 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 extra potentials in and but the electrochemical series is a list of half-cell reactions on this standard electrode potential so that's inert 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 and but there isn't do 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 enormous ascending order okay in this example it could be the other way around so don't take this for granted and it could be flip the other way round it depends on how the exam board wants to it wants to portray that so in this case as we go up we get the stronger oxidizing agents so the one of the most positive inor value and the agents on the left-hand side of the equation are more easily reduced to 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 examples 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 electron so therefore it's the most powerful oxidizing agent so in this case if the most powerful oxidizing agent is CL 2 and the weakest oxidizing agent is mg 2 plus 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 reduce reducing agent is magnesium and the most powerful oxidant the weakest reducing agent is CL - which is at the top here so you've got to be able to it you've got to be able to identify from and 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 it's it may only be worth one mark but it could you know every mark counts ok so let's look at some calculations involving standard cell potential so standard electrode potential Z naught as we've seen before they can be used to work out a standard cell potential which is basically the the value of the of the cell so we use e naught cell equals e naught reduced - 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 and the half cell what we've got to remember is the half cell equation with the most negative inor 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 it might be slightly different depends on what if you 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 and but and if you have two positives or two negatives then it's the most negative that's oxidized okay so that's 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 nought of the cell when CL 2 and Cl minus 1/2 cell and a zinc two-plus and zinc half cell are connected okay so first of all we need to identify which is being oxidized okay so this is where you use your no problem okay so zinc to person 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 yet that's the one that's been oxidized so number two we calculate the e nought of a cell so Y naught cell is one point three six okay because that's our chlorine one there okay and the most oxidized is minus not point seven six which is our zinc one which is down here so if we put them in there we get plus two point one two volt 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 ones been oxidized so it's the copper two 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 coppers positive as well so you 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 they're 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 fluff around that we don't draw diagrams and and silly things like that with beakers and salt bridges we need a simplified because 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 is just a 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 Moute negative on the left okay most positive on the right so we always have 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 so here that bridges the two half-cells and 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 so on the Left copper is the most positive on the right but you say we have a copper solid and the aqueous ion so this is a lying there to show to signify the difference in state okay so zinc two-plus has a noxious taste it has an oxidation state of +2 where a zinc has an oxidation state of zero so because that is the most oxidized form out with the two that's the one that sits closest to the salt bridge so the most oxidized form goes closest to the salt bridge so what if you have two aqueous ions so what we do and we follow the same rules above but instead of having the solid line we use a comma okay so we separate with a comma because they're in the same physical state so that might be for example Fe 2 plus Fe 3 plus and platinum so between Fe 2 plus and Fe 3 plus we have a comma so there's the example there so you can see here Fe 3 plus Fe 2 plus the comma separates these two they're not they're in the same state so we'll use a comma but that Platinum is a different States to either of these two so we use a solid line there and obviously that's showing 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 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 the lottery numbers that would be great wouldn't it put ten but it's a bit like predicting from Gibbs free so standard electrode potential sorry naught they can be used to predict a for stated reaction is likely to proceed under standard conditions okay 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 saying right if I get magnesium and I've got some copper two-plus ions they copper sulfate for example a blue solution and I took that magnesium in will it well I see anything will it react so first of all what we've got to do oh my god I remember I remember bit there as well so a half cell equation with the most negative V naught value is being oxidized okay so just remember that but the first thing is we need to identify which has been oxidized so the mg 2 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 reactions or whatever that is that's the feasible one mg plus CO 2 plus or for mg 2 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 two-plus ions as they match so this is telling us the feasible reaction is magnesium react with copper for mg 2 plus and copper so yes it will react and we can always confirm this is always worth doing this by calculating the e nought of the cell and all feasible reactions will always have a positive inor 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 in moisture now this is a slightly different type of question because they're asking is to just to find it the tellers 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 and so here we go so we've got our electrochemical series there's the 2 and 1/2 equations that we need to renew the iron one because it's made from iron but we know that rust is caused by arielle oxidation in other words we've got oxygen in the air that reacts with it so that's the most appropriate question the most appropriate 1/2 equation which is the one towards the top that's this one so we need to identify which has been oxidized and so the fe 2 plus fe has the most negative value so that that one's been oxidized ok no problem remember so the second one is we take the oxidized equation we flip it round 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 2 because we've got a balance it's 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 states in the question and we can see 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 this to justify it prove it and so we can confirm this again further by using a north-south calculation and we can see that it is positive it's only just positive so it's a it's a slow reaction obviously we're not going to see something lost instantly and you know not in these conditions anyway but nonetheless it is feasible it does work we see it we observe it we see iron go on 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 e nought and to save something's feasible it 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 electoral potential to change so for example and 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 and from the last slide so here's the e naught value and we have the half equations and from pulled them across and then we have the e naught cell value that we've calculated if we cope 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 electric potential for this half cell will become more positive and less and the cell potential will be higher okay so if we increase the concentration option overs if we've got more oxygen here then the equilibrium will shift to the right so I'll move to this side okay so if we increase the concentration of iron equilibrium will shift left less electrons will be used up okay and so the electrode potential of the fe 2 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 equilibrium as well and that's rightly important and another example might be if the kinetics is not favorable so for example the rate of reaction is slow so it appears though is no reaction especially with rusting as I mentioned before and if the reaction is a high activation energy it may stop the reaction from how at all so and catless 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's definitely going to happen it depends on the conditions okay so let's look at the final parts of this of this topic which is cells and batteries in the case of energy storage cells so what we need to do obviously we looked at half cells and we look at it connect them together to form a set like a electrochemical cell a full cell and there's got to be some use to this and the uses is clear with batteries and we have it in everyday life okay so this is very very relevant chemistry so to establish the overall reaction in a lithium-ion battery what we need to know the half equations at each electrode so we've got a list amaya batteries some that you would see in a mobile phone okay so they're very light lightweight batteries and used to power quite complicated devices like to say phones and laptops and tablets etc but the half equation for this type of reaction is this so it's lithium your lithium ion at the top and then you've got your cobalt oxide reaction here so you'll be given these it's about using the information 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 I it's and negative electrode so it's exactly the same as what we've done before so the negative electrode is Li will produce Li plus and elect and plus the electron so we flip that equation so that's it flips and obviously the positive electrode it's just the same reaction here because that's positive so anything it's the most negative we flip it round 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 a cobalt oxide that's in there to produce your lithium ion or your lithium salt product so to work out the e nought of the cell we do exactly the same as what we've done before so it's reduced - oxidized so it's nought point five six which is this value here - - 3.04 which is the most negative because that's you've oxidized and that's three point six volts and now you might be asked to say right what would the reaction be if we plug the thing in obviously the battery will run out eventually so this is a rechargeable battery and 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 or quiz we're just we we're introducing more electrons into 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 and but it needs a continuous external supply so in like a battery and 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 already 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 and the diagram shows a hydrogen oxygen fuel some what we're gonna 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 an oxygen is a hydrogen feed sorry okay so hydrogen is fed in so you need a constant supply of hydrogen and what this does is it reacts with the O H minus ions in in solution in nine so in section nine which is this bit here so that's section nine and but we'll look at this later on because that's obviously that's we're going to do this in the America order and but the reaction that does occur here is hydrogen comes in reacts with the O H - signs it in here cuz it's an alkaline cell and that forms water and for electrons but we'll come back on that so the second one is the flow of electrons so these electrons here that are produced and these travel through a platinum electrode through here and it's a good conductor it's an ER to remember and it goes through and 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 how you might have a fuel cell outside the house and generating power so that depends on what your parent 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 and water and the four electrons made from step one to make o h minus ions 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 minus 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 flow to the negative electrode so you can see them flowing in there and and that's made from platinum as well and six is the electrolyte so this is the the 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 that this carries the O h minus signs that were produced from this reaction here from from the cathode to the anode which is along on the site 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 as submitted in 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 surround and so that's the only by-product so that's better than as a of burning a fossil fuel which would produce carbon dioxide sulfates for example nitrates and another pollutant so it's quite a green way of generating electricity and number nine is the movement of the O h minus sign so that's what I mentioned in the first step so it's minus ions that are produced from the reaction in step four here are actually carried through the electrolyte to the other side of the cell and and what we have is well between the electrodes between the two sections are ion exchange membranes and what these do is these are aligned their platinum electrodes on on each side and they allow the which minus signs to pass through but not hydrogen or oxygen gas so and it's very selective what can go through what you don't want is the hydrogen oxygen mixing in the middle here you just want the o h minus science to drift through so you can exploit the electrons that are being transferred and use that as electricity okay so to summarize the reactions that's happening in this cell so we have a hydrogen feed so remember that song number one so it reacts with the o h minus ions in solution in nine and this reaction occurs which is H 2 plus 4 h minus will produce 4 h2o and 4 electrons now the electrons obviously go round to the circuit and they received at the other side which is this bit here and so that forms oxygen there's your four electrons plus you water that's produced as well so some of the waters move moved across and then that produces your voyage oh it's minus 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 and 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 they convert a lot more energy and 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 so but they do exist and more energy is converted to 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 and keep one you don't need to keep plug in we do need that supply of oxygen hydrants the oxygen comes from the air but the hydrogen comes from your fuel so the only waste product is water so no carbon dioxide is emitted so that's much better for the environment than a combustion engine so the disadvantages and hydrogen is very very flammable and it has to be stored correctly in a pressurized tank in the back of your car so it that could put some people off and it's really expensive if it's literally you transport in thin air so it's expensive to transport and store hydrogen so it has to be stored correctly and safely of course and energies required to make that hydrogen and the oxygen in the first place so it'll fossil flue fossil fuels are normally used you know to pass water through an electrolysis process to make that see you take Brian 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 um 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 and please remember to subscribe to the channel there's loads more videos on OC are dedicated to OCR there was four other exam boards as well they're all tailored to each exam board and so please subscribe them 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