 Okay, let's go ahead and attempt to balance this redox chemical equation here. So, remember the first thing you want to do with these redox equations is to split them up into their reduction and oxidation half reactions. And it's good to, of course, label those things as reduction and oxidation just to know what you know what you're talking about. So, remember, oxidation, one of the definitions that we talked about was to like get more oxygen on you. So that would, in turn, you could remember that and say, well then, that means if I'm getting reduced that I'm getting less oxygens on me. Okay, so just think about that. So if I look at the two half reactions here, let's split them up. That should be kind of obvious for you by now, right? So there's the chromium one and the iodine one. So let's break those two up. So, of course, the other one would be, let's actually write it down a little bit low. Okay, so let's label one of these as reduction and the other one as oxidation. So remember what I said, if you're getting more oxygens on you, that means you're being reduced. So that means that if I'm getting less oxygens on, or oxidized, if I'm getting less oxygens on me, I'm getting reduced, right? So if I look at this guy, chromium, he's getting less oxygens on him, so he's being reduced, okay? So this is the reduction and that would mean this is the oxidation. So the next thing we want to do is balance, well, first off, we want to balance the non-oxygen atoms, okay? So here we see that we've got two chromiums and over here we've only got one, so we're going to have to put a two here. And here we've got all these oxygens, right? So O7 and here we've got zero oxygens. So whenever that happens, we're going to have to add the oxygens in the form of water, okay? So H2O, H2O liquid like that, and we say, well, there's seven oxygens here and there's only one oxygen and water, so there must be seven waters that have been formed. Does that make sense? That makes sense, okay? So now when we look at this, right, we've got two hydrogens times seven, but we've got no hydrogens over here. So there's 14 over there, none over here. Over here we've got an atom by adding protons, right? So there's going to be 14 protons over here. So are you cool up today? Okay, good. So now we want to think about the charge differential, right? So on this side, you see we've got plus three times two. So overall, this side has a plus six, okay? So over here I see, well, I've got one times 14, so that's plus 14 minus two. So that would be what? Twelve. Twelve, right? So plus 12. And remember, whenever you're doing this charge, like balancing, you can only add electrons. So electrons are negatively charged, right? So you're going to have to add them to the one that has the more positive charge. So that, in this case, yeah, would be the reactants, yeah? So we're going to add, remember, electrons are one negative, so we want to get to plus six. We're going to add six electrons. Okay, so now if you're not sure, you want to make sure that it's balanced, but it is, okay? So I'm going to rewrite that this is the reduction half reaction. Okay, so you're cool with that one? So let's do the other one, the oxidation half reaction. So again, first thing you want to do is balance the non-oxygen atoms. So two to two, and then, well, we didn't have to add any oxygen to either side, right? So we don't have to do all that balancing stuff. But the one thing we should see is that we've got two negative on this side. So minus two on this side and a zero on this side. So remember, the only thing we can add is negative charges. So to the products, right? We're going to add two electrons. Okay, so hopefully you see that everything's balanced. If you don't, go ahead and recheck yourself. Okay, so now the thing you probably have noticed is that the number of electrons is different between the two half reactions. So what we're going to do is balance those electrons, okay? So in order to do that, we're going to have to find some sort of common factor. And of course, six and two, we can multiply this thing by three. So in order to do that, we're going to have to multiply everything by three. So it's just like an algebraic equation. So in these six I minus aqueous goes to three I two solid plus six electrons like that. And then remember, well, what are we doing? We're adding these two things up. And in order to do that, we want to make sure the electrons that are transferred are the same on both sides. So we see that this has six, that has six, so we're just going to cancel those things out. So we're going to have the added overall reaction. I'm going to write it here in the middle, okay? So that would be the overall balanced reaction equation here. So in order to do that, you do aqueous. So that's all the reactants here that aren't crossed out. And then we put the reactants from here plus six, okay? So that's all the reactants from the reduction and the oxidation half reaction. So two CR three plus aqueous. And now we're onto the product plus seven water plus three I two, okay? So the overall balanced reaction equation. And you can see that this reaction would occur in an acidic solution because there's protons in there and you see that, okay? So yeah, so that's the overall balanced equation. So you think you can do that on your own then? Okay, wonderful.