 This is an interesting one, so let's try to write the balanced ionic equation to represent the oxidation of the iodide ion by the permanganate ion and a basic solution to yield molecular iodine and manganese for oxide. So I've taken the liberty of writing the reaction equation up there for you so you didn't know what those particular things were, I guess, get to know what they are. So this is a redox reaction and it wants us to balance it in a basic solution. So both of those things are important things to know in order to start balancing this reaction. So remember for redox reactions we're going to break it up into two oxidation and reduction reactions of the two half reactions. So remember when you're doing that look for the species that are both similar to each other on either side of the reaction arrow. So hopefully you can tell that the permanganate ion and manganese for dioxide are the things that are similar to each other and the first of those reactions and then of course what's left is going to be in the other reaction and it goes to molecular iodide. Okay so now if you wanted to you could label one of these the reduction in oxidation reaction. Hopefully we were saying the things that gained the electrons were being reduced and the things that lost the electrons were being oxidized. So eventually we'll see that but we can take a preliminary just kind of glance at this and see that the manganese atom here has lost two oxygens, right? So that's another way of thinking that thing has been reduced. So for our preliminary kind of assessment we should probably be thinking this is the reduction reaction and this is the oxidation reaction. So let's just kind of preliminarily just say R and O label one like that, okay? So let's go ahead and balance this one. So hopefully you see well we have both manganese oxygen, manganese and oxygen in both of them but they're not the same amount of oxygens, right? So we need to do something to get our oxygens balanced. So in order to do that do you remember what we add? So we add water, water molecules, okay? So which side would we add them to? To the right hand side, right? And how many would we add? Well remember water only has one oxygen so we would need two of them. Does that make sense? So two H2. Okay so now our oxygens are balanced. Do you guys see that? But what do you notice? That H's aren't balanced now, you know? So how do we balance our H's? Well we're going to have to add H's over on this side but remember we add them as protons, okay? Or H plus ions, okay? So how many of those H plus ions are we going to need? Four, exactly. So we're going to add four H plus, like that. Okay so four H's, four H's, one Mn, one Mn, four O's, four O's. So that one's balanced, okay? By atoms. But the next thing we need to do is balance the charge, okay? So on this side of the reaction, so remember count up the charges, right? So is there the same amount of charge on this side as this side? So how many pluses do we have here? Four and a minus, right? How many minuses? So what's the overall charge here? Minus, you got plus three? Plus three, right? What's the overall charge over here? Zero. So is the charges the same on both sides of the reaction? No. So the only way we can balance our charges is to add electrons. And remember, electrons are negatively charged, okay? So which side of the reaction would we be adding those electrons to? This side or this side? The left side. Yeah, the left side, right? Okay, so how many would we have? Does that make sense? We're recording, so we don't want to be in the recording for that one. Okay, with your McDonald's. I like your hair, by the way. That feels nice. Okay, so now charges are balanced, atoms are balanced, everything's cool, right? So what do we see, right? Electrons are being added to this side. So what does that mean? Gaining electrons, so it's the reduction, right? It's the reduction factor. Notice our initial assessment was correct. Just by looking at the amount of oxygens, okay? So that's another way to look at those reductions. Okay, so that must mean this is the oxidation. Okay, so first thing we have to do is what? Balance the atoms, okay? We don't have an iodine on this side. Okay, so there's no no H's to worry about, so we don't have to do that, but our charges aren't balanced, do you see that, right? So what are we going to have to do? Add these two half-reactions up, but we have to have the same amount of electrons transferred from one side of the half-reaction to the other side, okay? Is that the case? Does three electrons equal two electrons? No, of course not, right? So what are we going to have to do? We're going to have to find a common factor for these things, okay? So what would be the common factor for three and two we're recording? Six. Six, very good. So what would we have to do to this one? Multiply by two, and this one? Five, three, okay? So remember when we do that, okay, we're not just multiplying this one, we've got to multiply everything because the coefficients are, you know, they're relative to each other, okay? So two times three, what do we get? Six. Six, erase, okay, with my finger. You guys, I know you're going to have to rewrite the whole thing, but anyway, what was it? Two times four? Eight. Eight. Two times one? Two times one? Two. Two, and two times two? Four. Four, okay? So what do we have to do to this one? Multiply by three, right? So three times two? Six. Six. Three times one? Three. Three, and three times two? Six. Six, okay? So now let's go back and see, do we have the same electrons, the amount of electrons being transferred from the reactants to the problems? Yes. Yes, we do. Okay? So let's go ahead and add these equations together, right? What's on the left side of the reaction and on the right side of the reaction will cancel out with each other, okay? Just like an outbreak equation, okay? So what are we going to do? Cancel out our electrons. Like that, and like that. Okay, do y'all see anything that's on the reactant side? That's the same that's on the product side? Anymore. Oftentimes you'll see those. In this one, do you see it? Anything? I don't see anything. Everything. So like, are there any H pluses over here? No. No. Are there any permanganate ions over here? Uh-uh. Remember, that's manganese for dioxide, right? That's a different thing than permanganate. What about the iodide anions? Do we see any of those over here? Uh-uh. So we're cool. Okay? So now we can add those two reactions up. So, plus two permanganate, plus six iodides. Okay, that's everything for the reactant. Two manganese for dioxide, plus four waters, plus three ion-2. Okay, we're recording, so if you don't mind. Okay, so if it asked us to have balanced it in an acidic solution, this would have been the balanced acidic solution equation. Okay? How do I know that? Of course, we remember from our pH chapter, right, that H pluses and OH minuses are the only things that change your pH, right? And H pluses make it acidic. OH minuses. Hydroxide ions make it basic, right? Okay? So if we look here, do we have any hydroxide, OH minus ions? I don't see them. Okay? But we do have some protons or H pluses and cations, right? So, acid to it, or not like adding an acid, it is adding an acid to it. So this, if it had asked us, balanced in acidic, balance it in a basic solution. Okay? So we should be wanting some OH anions like we were just saying, right? Okay, so in order to balance it in a basic solution, we're going to have to take it another step further. Okay? So is everybody okay with me erasing this top portion? So I don't have to. I thought about bringing candy this morning since we bought two. I figured you guys would probably be candied out. Well, jeez, you know, I thought you'd be candied up. Okay? Next time I'll bring this one. When I get back to testing. Let's finish balancing this thing. Acidic, we want basic. Okay? So how do we change from acidic to basic? We're going to have to add hydroxide ions. Okay? So we have to add the same amount of hydroxide ions. Okay? So since we have eight protons, we're going to add eight hydroxides. Okay? When you have hydroxide and H pluses, they combine to make water molecules. Okay? So keep that in mind. The other thing we want to remember is what we do to one side of the equation. We have to do to the other side of the equation. So what are we going to do over here? A-O-H-O-H. Yeah, A-O-H-O-H. It's very good. Okay? So, now don't cancel them out. Okay? So let's just combine these guys now. Okay? So combine them. What do we get when we do that? Four. Okay? So that's H2O, right? So that's A-H2O, right? Okay? So A-H2O is plus 2 and plus 6 i minus goes to 2 MnO2 plus 4 H2O plus 3 i2 plus 8 O H minus. Okay? We're almost done. Okay. So do you see anything that's the same on the left side as on the right side of the reaction? H2O, right? So how many do we have here? Eight. And over here? Four. Okay? So four. We're going to cancel out there. We're going to cancel that eight. And it's going to be a four. Okay? So let's just write our final reaction down here. So four H2O plus 2 permanganate plus 6 i minus goes to 2 MnO2 plus 3 i2 plus 8 hydroxide. Okay? So let's assess our final reaction. Do we see any hydroxides in here? Yes. There's hydroxides. Okay? So this is going to be a basic solution. Right? Do we see any protons? I don't see any. Okay? So we're cool. Is everything balanced? And I'm going to do a quick assessment because I know we did everything correctly. Okay? I'm not going to count everything. But this one is the balanced. Any questions on that? It's a real steppy process, but remember just follow step by step by step you should be fine.