 So here's our second example question. So we've got a Voltex cell constructed from a half cell which cadmium. So I'm going to highlight some important bits as we go through. So we've got a cadmium rod dipping into cadmium nitrate. So that's our electrolyte on that side. On the other half we've got a silver rod and that's going to silver nitrate. So that's important information to know. So we've got what our metals are and our electrolytes. Two half cells connected by a salt bridge so that's good to know. And the silver ions are being reduced. That is very important. So the silver is being reduced that means that the cadmium has to be oxidized. So draw a sketch to the cell, label anode and cathode, write half reactions, indicate electron flow and sign of the electrodes. So let's do that now. So here's my diagram. I've hooked up a voltmeter. So first bit of information that we were told that's very important is the silver ions are reduced. Now I like to have the oxidation half cell on this side and the reduction half cell on this side. So I'm just going to label that quickly. So that side is going to be ox and that side is going to be red. So the silver ions are reduced. So if we look back at our information we're told that there's a silver nitrate is our solution down here. So we've got our electrolyte is agn03 and on the other side we had cadmium nitrate. Okay now silver is being reduced. So we know the site of reduction is the cathode. So this is our cathode and consequently the cadmium is being oxidized. So that's our anode over here. We're going to get a build up of electrons down here as the cadmium is oxidized. So a build up of electrons is your negative charge. So the anode is going to have a negative charge. The cathode is going to have a positive charge. What else do we need to know? Let's look back at our question. We'll do the half of the occasions in a second, direction of electron flow. So build up of negative charge on this side so the electrons are going to be pushed away. So our electrons are going to move from the anode to the cathode like that. And now we can write our half equations. So our cadmium rod, so the rod there is made out of cadmium. Over here we have our silver rod. The cadmium is being oxidized. So oxidation is loss. So cadmium, so solid cadmium, that's going to cadmium ions. CD2 plus and that's aqueous and plus two electrons. So it's losing the electrons. Over here the silver is the site of reduction. So the silver ions are being reduced from the solution. So AG plus aqueous, that's dissolved in the solution. Plus one electron because silver only has a charge of plus one goes to solid silver, so AGS. If we wanted to we could then write the overall equation by times this one down here by two. So we have the same number of electrons in both sides. And then we'd be cancelling out the electrons. So just write that out quickly. So we've got 2 AG plus aqueous, so it's two silver ions in solution plus two electrons goes to two solid silvers. Now we can write the overall equation if we like. So we can set out the electrons which are on both sides of the arrow. So we've got cadmium solid plus two silver ions in solution goes to cadmium ions in solution plus two solid silvers. All right. So that's an example of electrochemical questions. That's it for the video. Thanks for watching. I'll see you in the next video. Bye-bye.