 So here's an example question, it says during the electrolysis of multi-potassium iodide using graphite electrodes, potassium metal is produced at the cathode and iodine gas at the anode. Write an equation, state the advantage of using graphite as your electrodes, write an overall equation and state at which electrode oxidation occurs. So let's break that up a little bit. So the first part of the question says write an equation for what's occurring at each electrode. So at the cathode, okay, we're going to get reduction of the potassium metal. Potassium ions are going to be reduced, so k plus, we need one electron and that's going to produce potassium, now that'd be liquid potassium over here we'll call the potassium ions, they're not in solution either, so we'll just call them liquid too. So at the other side at the anode, we're making iodine gas, so the iodide ions are going to, we're going to need two of those, they're giving up their electrons because they're being oxidized, so they're being oxidized to iodine and that's a gas. So again, they're molten in there, so we call that liquid and we need to have our two electrons on that side. So the state of the advantage of using graphite electrodes, graphite electrodes, they conduct electricity but they won't react necessarily, they're not going to react with the potassium or the iodine gas, so they're inert, they don't react but they can still conduct electricity. So I've got an overall equation for the reaction, so what we need to do is add our two half equations over here, so I need to double everything that's with my potassium, so I'm going to have, so it's going to be 2k plus, plus 2i minus, goes to 2k plus i2, so all I'm doing is doubling what I had for my cathode reaction, I'm just going to add that to my anode reaction then cancel out the electrons because they should cancel out from both sides. So the last question says, state at which electrode oxidation occurs, oxidation occurs at the anode. Another question says, draw a diagram of an electrolytic cell which would be suitable for plating a nickel chain with silver using silver nitrate as the electrode. So I've got to set up one over here, so let's label this. So I'm going to start on this side, we're going to, so placing, so I'm going to call this, so this is going to be a silver rod, I'm going to pop that in the solution. Down here we've got silver nitrate, agn03, solution, aqueous, so down here we've got silver nitrate, that means this is going to be my nickel chain, it doesn't look much like a chain but we get the idea. Okay, we're going to label some things, so this is going to be our anode over here, anode is going to be positive in this case, over here the nickel chain is going to be negative, it's going to be the cathode, and so what's happening is the silver ions from the solution will plate onto the nickel chain, so they'll come out of solution, they'll be reduced, silver ions are reduced to silver on the surface of the nickel chain. We might plot in the direction of the electron flow, so the electrons are moving in this direction, so they're moving from positive to negative, which is the reverse of what they would normally do, we're shoving in energy, so they're pushing in the non-spontaneous direction, what else do we need, that's probably about it, we could write down the equations for what's happening as well, so at the anode we're getting oxidation, so the solid silver is going to silver ions plus the electron, so that's oxidation, so we'll write the equation on the other side as well, so over here silver ions are being reduced, so we've got the silver ion in solution is gaining an electron, and we're reducing that to silver metal, that's an S believe me or not, we're reducing that silver metal on the nickel chain down here, so the nickel chain solution is going to be plated with silver from the solution, so today we looked at electrolytic cells and looked at some example questions, that's it for this morning's class today, see ya!