here's an example of a voltaic cell simulator and i will try to embed this on our canvas page but it requires us to be able to play flash player and sometimes it doesn't work in a lot of the newer computer so we'll see whether it works or not it works then you can play around with it but if it doesn't then this is the way the video comes in so in this case we have a two half reaction that we combined together to get into voltaic cell on one side we have zinc metal and then zinc nitrate solution which the concentration is set at one molar and on the other side we have copper metal and the copper nitrate solution which is set to concentration of one molar as well so right now the chemical reaction occur at a standard condition so you can there's no temperature control here so the temperature is 298 kelvin so what should we expect the voltage will be when we turn it on this voltmeter well we can combine the two half reactions of the zinc and then the copper remember we have to flip one of them because one of them have to donate one of them have to absorb the electron right so in this case we're going to flip the zinc one so it becomes positive volt and the copper one remains at the 0.34 volt and if we combine them together we should expect that the voltage that we measure to be 1.1 volt so in this case that's exactly what we see electrons is being donated from the zinc into the copper and then the voltage that we measure is going to be 1.1 volt so that's kind of an example of the cell potential at standard condition that we practiced before by combining the two half reaction but of course when we're looking at electrochemical reaction we're not always going to work at standard concentrations and pressures so you can have copper and zinc in this case the concentration is going to be slightly different so let's say i'm going to change the concentration of the zinc to 0.01 molar and then i'm going to change the concentration of a copper to 2 molar so what would happen to the cell potential here well i don't know what it is because i haven't calculated it yet but i know the concept the voltage that we're going to measure is going to change based on the nurn's equation and that's the one that's e is equal to e naught which is our standard minus rt divided by nf times ln of q so in this case if we turn it on you can see that the voltage now measure at 1.20 volt instead of 1.10 volt so changing the concentration will change the voltage that you measure and of course what happens if we were to change the metals that we play around in this particular case well i can change for example the metals let me see if i i want to make sure that i remain positive potential here so let's see if i can change on this side is copper and the concentration of the copper is going to be 0.1 molar let's say on the other side we're going to use silver metal and we're going to use a silver nitrate solution and i'm going to keep the concentration of one molar as well so in this case i play around with concentration turn on the voltage and you can see that in this case we have a different voltage right so it's a different battery construction when we combine the two half reaction we get a different voltage and in this case it measures 0.46 volt and that's because it's a combination of the copper reaction here and then the silver but in this case the silver is going to excuse me the copper is going to get flipped so the copper is going to is going to be the electron donor and the silver ion here is going to be the electron acceptor so in this case this becomes if you flip the first reaction here becomes negative 0.34 volt and 0.80 volt so what you get when you combine them together is a positive 0.46 volts so in essence it's just a simulator and this is kind of what we look like on what we measure when we measure the voltage of our cell potential of an electrochemical cell what we're interested in in the end is connecting the experimental setup that we have with the experimental measurement that we observe in our system now i have a question for you so far we've been using two different metals right silver and copper silver and zinc what happens now if i were to use the same metal so in this case let's use a silver on one side and let's do another let's do silver and silver right so let's see if we have silver and silver solution one molar silver silver solution of one molar so if i were to have this construction both uh both reaction on the left and the right are silver but the here in this case so there the cell potential is both 0.8 the concentration of both is going to be one molar so what would happen if i were to turn on this cell potential what happened to the voltage so hopefully you can see that well the concentration is the same the metal is exactly the same so when i turn it on i should get zero right um the standard cell potential is that means i have to combine two of these one of them i have to flip so one of one of the half reaction is going to be positive 0.8 and the other one is going to be negative 0.8 so when i combine them together i should get zero now how about the concentration effect well right now the concentration is exactly the same one molar so if i were to plug it into negative rtlnq the lnq on one side is one molar divided by one molar so it should be zero so the voltage here that we measure should be zero in this case the volt meter actually doesn't turn on because there's no voltage um there's no power to power up the voltmeter so that's why it doesn't turn on in any case so what would happen now if i were to have the same chemical reaction but now the concentration is going to be different so what if now instead of having the same concentration of one molar i'm going to keep the one on the right as two molar and then the one on the left as 0.001 molar what would happen now to the voltage that we measure so in this case you can see there's actually a voltage here of 0.2 volts you see what happens is that the reactant the system is not at equilibrium from one another right so at equilibrium the concentration of the silver and the left and the right should be equal to b to one another that is if they're if one of them is one molar the other one must be one molar as well right so that's equilibrium and the voltage at equilibrium which is what uh the previous example was before supposed to be zero but in this case the concentration of the silver on both sides is not at equilibrium one of them is two molar and the other one is what is this two thousand times smaller and you can see that the concentration gradient actually gets converted to voltage we can actually convert that into voltage and that concentration guardian means that there's a potential energy and if there's a potential energy difference between the two solution here what means that the difference in concentration actually translate into energy now why do i kind of want to kind of show this particular example well one of them is just it's a very common mcat question to have two of the sides is exactly the same metal but two different concentrations which we're going to try and do on a different problem here but it's also kind of the way that our neuron works that the way that action the the action potential work and how the signal propagates is that there's an energy gradient between the inside and outside of the cell of the sodium and potassium concentration so that concentration gradient translate into energy and that's essentially there needs to be some energy in order to be able to do cause something else to happen and a constant gradient concentration gradient is the energy fuel that we're using