 Howdy! Today on Flipping Science we're going to do an example electrochemistry question. Here's the start of the question. A vanadium flow battery has been developed. We have a picture of the battery and we're told to state the energy conversion that occurs when the battery is discharging. What that means is when we're giving off energy because it's a battery we need to figure out what kind of conversion is happening. So this goes back to the proper year 8 science where we're talking about energy types. So there's chemical energy in the battery, chemical potential energy in the battery, and that's being turned into electrical energy. And that's the conversion that we're worried about. So we're breaking down one electrode, that's now turning the chemical energy in the bonds into electrical energy that we're using in the external circuit. Then we're on to the next bit. It says equations for the reactions that take place are shown below using sub-shell notation right in the electron configuration of vanadium. In this case it's the vanadium ion, vanadium plus 2 2 plus. So we need to find vanadium. It's number 23 here. So we're going to use that. I'm going to do vanadium first then I'll figure out what the V2 plus line is. So it's number 23. So we go 1S22S22P6 3S23P6 4S2. So we're up to 2, 4, 10, 12, 18, 20, 3D3. So to make the plus 2 ion we're going to get rid of the 4S2 electrons there. So what I'll do is I'll just get rid of the last bit here. And go 3D3. Now if I add this up we should add up to 21 because we've lost the two electrons from here. So 2, 4, 10, 12, 18, 19, 20, 21. There we go. So the next part of the question says state the oxidation number of vanadium in VO2 plus. What I'm going to do there is I'm going to do the working out down the bottom and write my answer at the top here. So we've got X for vanadium plus 2 times the oxidation number of oxygen which is minus 2 in most compounds. That's equal to the charge over here which is plus 1. So we've got X minus 4 equals plus 1. Add 3 to both sides. Add 4 to both sides. So it gets plus 4, so plus 4, so it gets plus 5. The oxidation number is going to be equal to plus 5. Next part of the question says explain why electrode 2 is the cathode when the battery is discharging. If we look at our equation that they gave us for electrode 2 we look at which side the electrons are on. They're on the left hand side here. So if it's on the left hand side that means reduction is happening and cathode reduction. So reduction happens at the cathode. So when I'm answering the question here explain why electrode 2 is the cathode. Reduction is occurring at electrode 2 and the site of reduction is the cathode in a battery when it is discharging. That's how I would sum up the answer to that question. So in the second part of the question it says draw an arrow to show the direction of electron flow in the external circuit. So to do that we need to figure out what the charge is on the electrodes. So electrode 2 we just figured out that that is the cathode. The rule is if it's a cathode in a galvanic cell that's positive so that means the other one has to be negative. Cathode positive reduction is how I remember that. So CPR and the other side is A in O. So that's the first three letters of ammo. So for a galvanic cell cathode positive reduction and O negative oxidation. So now if we look at it we need to figure out which direction the electrons are going. The electrons go from negative to positive. They're appelled from the negative side and they go to the positive side. Right so I've drawn some arrows on there to show that. It says draw an arrow. So what I'll do is I'll just do a big arrow up here and this will be my answer. That's what I want there. Last part of the question says one advantage of the vanadium flow battery is that it's rechargeable. Identify the type of cell that operates during the recharging process. So if it's a galvanic cell when it is discharging that means it's going to be an electrolytic cell when it is recharging. So our answer down here is electrolytic. I'm sure you'll have better handwriting than me. Then it says identify the product formed at the cathode during the recharging process. So in this case if we're recharging we're forcing the direction of the electrons to reverse. We're pushing it in the opposite direction. So that means essentially what's happening is the reverse of these is occurring. So that one's going in that direction and that one's going in that direction. So in that case which one would be the cathode in an electrolytic cell this side would be the cathode now because we're pushing the electrons in the other direction. So that means reduction is still occurring over here but that means it's positive. Anyway it says identify the product form so we're pushing this in the other direction so we're going to be making v plus two ions. Right so in a recharging process we've got the electrons flowing in the other direction. We're still getting reduction at the cathode. That makes this the cathode now. So we're going to be getting the cathode being regenerated in this case. So the anode overall would be being regenerated by making this the cathode when we're recharging the battery. So that's an example question from an old exam on electrochemistry. That's it for Flipping Science today. See ya.