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our goal is to calculate the equilibrium constant K for this reaction so for this reaction right here and we're going to use the standard reduction potentials to do so so in the previous video we talked about the relationship between the equilibrium constant K and the standard cell potential so e0 so if we can find easy ro for this reaction we can calculate the equilibrium constant K and we've seen how to find easier of the standard cell potential in earlier videos so let's start with this first half reaction here where we see solid iodine gaining electrons so it's being reduced to turn into iodide anions the standard reduction potential for this half-reaction is positive 0.5 4 volts and we can see that's what's happening in our redox reaction up here right we can see that we're going from solid iodine and left side and we have iodide anions on the right side so we're going to keep this we're going to keep this reduction half-reaction if we look at what's happening with aluminum we're going from solid aluminum to aluminum 3 plus so solid aluminum must be losing electrons to turn into aluminum 3 plus so aluminum is being oxidized here but this half reaction is written as a reduction half-reaction so we need to reverse it right we need to start with solid aluminum so we reverse our half reaction to start with solid aluminum so aluminum turns into aluminum 3 plus and to do that it loses 3 electrons right so loss of electrons is oxidation here is our oxidation half reaction so we reversed we reversed our half reaction right and remember what we do to the standard reduction potential we just change the sign alright so for the for the reduction half reaction the standard reduction potential is negative one point six six we've reversed the reaction so we need to change the sign so the standard oxidation potential is positive one point six six volts for our reduction half-reaction we left it how it was written alright so we're just going to write our standard reduction potential is positive 0.5 4 volts next we need to look at our balanced equation right we need to we need to make the number of electrons equal for our half reactions so for this first half reaction I'm just going to draw a line through I'm just going to draw a line through this half reaction so we don't we don't get ourselves confused for our first half reaction here we have two electrons and then over here right for our oxidation half reaction we have three electrons we need to have the same number of electrons so we need to have six electrons for both half reactions because remember the electrons that are lost are the same electrons that are gained so we need to multiply our first half reaction by 3 if we multiply our first half reaction by 3 we'll end up with 6 electrons and our second half reaction we would need to multiply the oxidation half-reaction by 2 in order to end up with 6 electrons so let's rewrite our half reactions so first we'll do the reduction half-reaction so we have let me change colors again here and let's let's do this color so we have 3 times so we have 3 I 2 now so we have 3 I 2 and 3 times 2 gives us 6 electrons so 3 I 2 plus 6 electrons and then 3 times 2 right gives us 3 times 2 gives us 6 I - all right so we multiplied our half reaction by 3 but remember we don't multiply the voltage by 3 because voltage is an intensive property so the standard reduction potential is still positive 0.5 4 volts so we have positive 0.5 4 volts for this half reaction next we need to multiply our oxidation half reaction by 2 so we have 2 al so this is our oxidation half-reaction so 2 al so 2 aluminum and then we'd have two al three plus so 2 al three plus and then 2 times 3 gives us electrons so now we have our six electrons and once again we do not multiply our standard oxidation potential by two so that we leave that so the standard oxidation potential is still positive one point six six volts next we add our two half-reactions together I mean if we did everything right we should get back our overall equation so our overall equation here we have six electrons on the reactant side six electrons on the product side so the electrons cancel out and so we have for our reactants three i2 so we have three i2 plus two al and for our products right here we have six I minus so six I minus plus two al three plus plus two al three plus so this should be our overall reaction right this should be the overall reaction that we were given in our problems let's double check that role fast so three i2 plus two al all right so right up here so three i2 plus two al should give us six I minus plus two al three plus so six I minus six I minus plus two al three plus so we got back our original reaction remember our goal was our goal was to find the standard cell potential e zero because from a zero we can calculate the equilibrium constant K so we know how to do that again from an earlier video to find the standard cell potential right so define the standard cell potential all we have to do is add our standard reduction potential and our standard oxidation potential so if we add our standard reduction potential and our standard oxidation potential we'll get the standard cell potential so that would be positive 0.5 four volts so positive 0.5 four plus one point six six plus positive one point six six volts so the standard potential for the cell so easy row cell was equal to 0.5 four plus one point six six which is equal to two point two zero volts all right now that we've found the standard cell potential we can calculate the equilibrium constant so we can use one of the equations we talked about in the last video that relates to standard cell potential to the equilibrium constant so I'm going to choose I'm going to choose one of those forms so so easy Rho is equal to zero point zero five nine two volts over n times log of the equilibrium constant so again this is from the previous video so the standard cell potential is two point two zero volts so we're going to plug that in we're going to plug that in over here alright so now we have now we have two point two zero volts is equal to zero point zero five nine two volts over n remember what n is n is the number of moles transferred in our redox reactions so we go back up here and we look at our half reactions and how many moles of electrons were transferred well six electrons were lost right and then six electrons were gained so n is equal to six so we plug in n is equal to 6 into our equation so n is equal to 6 and now we have the log of the equilibrium constant K log of the equilibrium constant K so we just need to solve for K now so let's get out the calculator and solve for K so we would have two point two two point two zero times six all right divided by point zero five nine two and that gives us let's say 223 so this gives us 223 so 223 is equal to so this is equal to log of our equilibrium constant K so we need to get rid of our log and we can do that by taking 10 to both sides all right so if we take 10 to the 223 and tend to the log of K then that gives us K the equilibrium constant so K the equilibrium constant is equal to 10 to the 2 one hundred and twenty third power which is obviously a huge number so a huge number we get a huge value for the equilibrium constant which is a little bit surprising because we only we only had two point two zero volts right what doesn't sound like that much so from only two point two zero volts we get a huge number for the equilibrium constant so the reaction goes to completion all right with a huge with a huge value for the equilibrium constant like that you pretty much don't have any for anything for your reverse reaction which is why which is why we were not drawing any kind of any kind of arrow going backwards here we only have only have this arrow here going forwards so because of our huge number for the equilibrium constant

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