 Okay, so let's try this one. It says calculate the cell potential. I had 25 degrees Celsius for the reaction pictured. Two aluminum solid plus three iron two plus aqueous goes to two aluminum three plus aqueous plus three iron solid. Given that, the concentration of iron two plus is 0.020 molar. The concentration of aluminum three plus is 0.10 molar and the standard reduction potential for the aluminum reaction is negative 1.66 volts and negative 0.45 volts for the iron reaction. So again, like we've said for these ones, you can start in different places. The place I would like to start or usually like to start is to split this balance. So this is a balance redox equation. I'll split that into its half reactions. Okay? There's a number of reasons why I would do that, but initially it's going to be defined, which is the reduction reaction, which is the oxidation reaction. So two AL, let's split it, two AL solid goes to two AL three plus aqueous and the other half reaction is going to be three F e two plus aqueous goes three F e. So, like that. So these are balanced. So we know that the charges are the coefficients are balanced so that we need to balance the charges. So we see two times three, that's six plus overall here. We have zero overall here. So we're going to have to add six electrons to that side there. Okay? So like we said, this equation is balanced so we're going to have to add six electrons here. Okay, so now we can identify which is the reduction reaction and which is the oxidation reaction. Okay? So remember, reduction is gaining electrons. Okay? So if we look here, the six electrons are being added to the iron, so that's gaining the electron, so that's being reduced. So this is the reduction reaction and that happens at the cathode. So we've got the, of course, if that's the reduction then that must be the oxidation. Okay? Which happens at the anode. So let's give these half reaction potentials to these two reactions now. So what do we've got? We've got the standard potential of the anode. Well, that's going to be what the AL 1, so negative 1.66 volts there. And the standard reduction, or the standard potential of the cathode is going to be negative 0.45 volts. Okay? So why is that important? Because now we can figure out what the standard cell is. Okay? The standard potential for the cell is. So how do we do that? Well, that's going to be the standard cathode minus the standard potential of the cathode minus the standard potential of the anode. Are we cool with that, Seth? Yes. I don't know. Did you need that? Sorry about that. So let's go ahead and just plug in chugs so we can get our E cell. So the cathode 0.45 volts minus the minus 1.66 volts. So I think I can do this one in my head. So that's going to be 1.21 volts. So positive 1.21 volts. So is everybody okay with all the way up to here? Yes. So let's just write that E cell down here, the standard E cell. And hopefully you guys are thinking we're probably going to be using the Nernst equation in a second. Okay? So since we're thinking that, let's go ahead and write the Nernst equation down really quick. Okay? So the Nernst equation is asking us, well what's the potential of the cell at right now? And that's what this question was asking us. What is the E cell? Well, that's wonderful. Okay? So Nernst equation gives us the equation to figure out the E cell at standard conditions, which is 25 degrees Celsius. We're going to have 0.0592 volts divided by N times the log, okay, E cubed. So we've gotten to that point. Everybody is okay with using the Nernst equation at this point? Okay. So where does that leave us? Well, that means if we're going to use this equation to solve for this, we've got this. Well, what is N? Where does that come from? Well, that comes from the number of electrons that have been transferred. So if we see 6 electrons, 6 electrons, that means that N equals 6. Okay? The other thing we hopefully see is that we don't have keq yet. Okay? So we need to figure what is keq? So what is the equation for keq? For this particular reaction, remember it's going to be the products divided by the reactants raised to their coefficients, so concentration of Al3 plus squared, like that, divided by concentration of Fe2 plus q. So now let's just plug in those numbers. So here we get 0.10 squared divided by 0.020 q. Okay? And I'm not going to put those units in because keq doesn't have any units associated with it. Go ahead and do this in our calculator. So, okay, so when I do that, and I'm going to take this out to a bunch of 6 units, or at least 36 units, 1250, like that. So that's the keq for this reaction. So let's go back and now, well, what are we solving for E cell? So let's just, I'm going to write the keq over here and then just erase this part and then solve for E cell. Is everybody okay with that? Yes. Okay, so now really it's just a plug and chug problem because we've got all of these. So 1.21 volts minus 0.0592 volts divided by 6 times the log of 1250. So now I'm just going to do it all in my calculator, okay? So hopefully, if you're having trouble with this, you can come to me afterwards and we can see if you're plugging in a big one. So that, so I'm going to multiply that by 592 divided by that. So what I get with all of this, okay, and I'm going to just take it out to 2 decimals, or 2 after that decimal there, we're going to get 1.21 volts minus 0.03 volts, like that. Okay, so hopefully you got 0.03. And then, I think I can do this one in my head that's 1.18 volts. So that's going to be the standard potential at these concentrations, okay, and 25 degrees, or the potential, cell potential. Okay, any questions on that one? Okay, one.