 So let's do another one of these electroplating problems. This one says a quantity of 0.300 grams of copper was deposited from a copper 2 sulfate solution by passing a current of 3.00 amps through the solution for 304 seconds. Calculate the value of the Faraday constant. So we're going to need to do a couple of things in this problem. So the way I would approach this problem is to balance that half reaction that we were talking about first. So the half reaction that they mentioned was copper 2 sulfate going to copper salt. So let's just write down that information. So Cu2 plus. So I'm just taking away the sulfate since it's a spectator ion. AQ plus, well, kind of giving it away, going to copper solid. So if you look at the oxidation state of copper, it goes from a 2 plus state to a 0 state. So that must mean it's being reduced. So this is the reduction half reaction. So that must mean we have to add some amount of electrons. And from plus 2 to 0, it's 2. So that's important. The other thing that's important, well, we're looking for the Faraday constant. So if our final answer is not close to 96500, then we know that we're off. The other thing we want to remember is the formula that relates current to charge. So if you don't recall, that's I equals Q divided by T. So Q is the charge there. So we're going to have to figure out, well, what is that charge? So let's rearrange this equation to solve for Q. The other thing you might want to remember is that one Coulomb is 1 amp second like that. So if you don't remember that, get that in your head, because that's a conversion that you won't be getting. So what do we got here? So Q equals I T like that, I 3.00 amps times T 304 seconds. And remember, we want charging units of Coulombs, because that's charge units is Coulomb. So for every Coulomb, we have 1 amp second like that. So now let's cancel, cancel, cancel, cancel, cancel. And we're left with Coulombs. So I guess I can do this in my head, 9, 12 Coulombs. So that's how much charge has been transferred over that unit of time. So I'm going to put that over here, just so we won't lose that information. I'm going to erase this portion of the book. So the other thing we know, well, we know the balance reaction equation, and we know the mass of the copper that was plated. So if we look up at the periodic table, the molar mass of copper is 6355, OK? So let's write that down. So that should give us the number of moles. We could get the number of moles of copper, right? So 0.300 grams of copper times 3 moles of copper divided by 63.55 grams of copper. But from there, so that would be moles of copper. We're going to get moles of electrons from that. That's how we're going to figure out Faraday's constant, OK? So let's just go one step further and do the number of moles of electrons that we've got, OK? So if we look at our reaction equation, right? We see for every one mole of copper, we've got two moles of electrons, OK? So we can write that down, OK? So just another conversion back there. So for every one mole of copper, two moles of electrons. So that's going to give us the number of moles of electrons, OK? So let's go ahead and calculate that out. So 0.3 divided by 63.55 times 2. And I get 9.44 times 10 to the negative 3 moles of electrons, like that, OK? So remember, the units of the Faraday's constant, right, are going to be coulombs per moles of electrons, OK? So do we have coulombs? Yes. So do we have the number of moles of electrons? Yes. So if you want to think of it that way, it's going to be charge per mole, f equals q per n, like that, OK? So 4 times 10 to the negative 3 moles of electrons says on my calculator, so 96, 596 coulombs per mole, right? That's pretty close. But we want to put this to the right number of sig figs, so we've got 3, 3. So let's put this to the right number of sig figs, so 9.66 times 10 to the 1, 2, 3, 4 coulombs per mole of electrons. That would be a calculated Faraday's constant, OK? Are there any questions about that? No. I think you just need to remember the stepwise process, really, just like anything.