 OK. So let's try to calculate the potential of this concentration cell that we have. So it says a concentration cell consists of two silver, silver one half cells, and half cell A. Electro A dips into a 0.010 molar silver nitrate solution. So remember, silver nitrate nitrates the spectator on it. We just give rhythm. So then it says in half cell B, electro B, which is a silver electrode, right, dips into a 4.0 times 10 to the negative 4th molar silver nitrate solution. What is the cell potential at 298.15 kelvin? So that's 25 degrees Celsius. So we can reduce our Nernst equation. So we're going to have to use the Nernst equation again. So remember, these things flow from concentrated to dilute. So let's write our cell equation. So it's going to look a little strange from the ones that you've been writing before, because you're going to have on the reactant side and on the product side the same thing, the silver plus. So it's going to be AG plus A plus. Remember how we do this for these concentrations, put our semicolon there, and then write our concentration like that. And that goes to, well, AG plus 4.0 times 10 to the negative 4th molar. So how many electrons does silver need to be reduced? Just one. So even though you can't really see it here, remember that potential is silver, plus the one electron goes to the silver electrode. The reason that we don't have those two silver electrodes in here is because it would be on each side of this equation. And what happens if we have something on both sides of the equation? So we cancel it out. So we've got to watch out about that. So it's not going to be shown in the equation necessarily. So we've got to just think, what's the half reaction? And it should be fairly straightforward. We see that that's 1 plus. So we could imagine that only one electron. So let's write that and down. Let me just move that. Is everybody OK with what I just said? So let's write the Nernst equation, because that tells us what the cell potential is. So what about the standard cell potential? So let's think about that. So again, like I mentioned, we have the same thing, oxidizing and reducing each other. Those potentials are going to cancel out. And so it's going to be 0 overall. But let's just put 0.00 volts to keep our significant increase, so we won't think that we have to cut this down. So we've got that. Well, log of Q. Well, what would Q be? Help me out. Thank you. AG plus over AG plus, right? But the way I like to write it is with a little C and with a little D to say concentrated and dilute. That really kind of helps me to keep things in order. So what's on the top there? Dilute. Dilute again. So AG plus D, C. And could we solve for that equation? Have any units? So let's not throw our similarities in there, even though in this case they would cancel. You should be able to solve what the cell potential is now. I want to do the equation so we can always take this to 2M for this, because these are oftentimes very small numbers. So something like if we only had 1, we would get a 0.0 and that would be 0.1. So any other questions on this one? So it's the same thing. It's almost even easy. It just looks a little more confusing. So don't let it scare you.