 Okay, so let's try to balance another one of these redox equations in acidic solutions. So you can already tell this is a redox equation, or I can really quickly, because I see that iron here has changed its oxidation, okay, it's actually been oxidized in this reaction. So we want to remember, we want to split this thing into half reactions, okay, the oxidation half reaction and the reduction half reaction, so oxidation, half reaction is going to be the Fe2 plus aqueous, so you can see the oxidation number has changed, then the reduction is going to be BrO3 minus aqueous goes to Br minus aqueous, okay. So if you recall from what we did last time, the next thing you want to do is make sure the number of atoms is balanced on both sides, so if you look here we've got one iron and one iron there, so we're good there, and then we look down here, we've got one bromine here, one bromine there, and we're good there, but we have oxygens over here, so we need to add oxygens to this side, okay, so recall the way that we did that in these types of problems is to add water molecules, okay, so let's go ahead and do that, we're having liquid water, and we see we have three oxygens and water has one, so we're going to need three of those guys, so recall also, these are redox reactions in acidic solution, and remember we talked about the H plus being the acid, the proton, the thing that gives acid its acidity, so what we're going to do is find out how many H pluses are going to be on this side of the edge, so two times three is going to be six, so let's just make sure, right, so we've got six, six, three, three, one, one, right, so everything's balanced there, does that make sense? Okay, so the next thing we want to do is balance our charges, okay, so here we've got a plus two, and here we've got a plus three, so in order to get that down to a plus two, we're going to add an electron up to the, on the right hand side or on the product side of that equation, so now we're going to have plus three minus one equals plus two, does that make sense? So this one is balanced by number in charge, okay, so now we're going to come down here and balance this equation, so here we've got six plus, right, plus one minus, so we're going to have overall plus five on this side, right, and here we've got minus one on this side, does that make sense? Okay, so the thing we're going to have to do is put electrons on this side, okay, because we're going to have to get that plus five to be equal to minus one. In order to do that, we're going to add six electrons like that, and so if we add minus six plus five plus minus six, that is equal to minus one, does that make sense? Remember, we have to add, we have to add electrons to do this, so now, so can I erase this portion of it, the charge part? Okay, so now we're called the last portion, we have to make sure the same number of electrons here is in, that's in this half reaction is in this half reaction, okay, and in order to do that, we have to multiply all our coefficients by the same number, okay, so we have six electrons transferred here, so we have to get six electrons to be transferred here, okay, so how do we do that? We take the whole reaction equation and multiply it by six, why? Because we need to multiply one by six to get six, so now let's just multiply through, so six FB two plus a grid goes to six FB three plus a grid plus six electron, does that make sense there? Okay, now the last thing you want to do is if you did all this correctly, you should have the same number of electrons being transferred from one of your half reactions, okay, so that's one of your half reactions to the other one, okay, so when that's the case, right, just like an algebraic equation, we can cancel them out if they're on opposite sides of each other, and now what do we do? We just go ahead and add up these two equations, so some of these two equations, and that's your overall balanced redox reaction equation, so let's, it doesn't matter at all, okay, whichever one you like the best, okay, or if it's a multiple choice test, whichever one it writes it out, you know, so, okay, so let's do it, six FB two plus a grid, so that's all of the reactants on the oxidation half reaction, so now let's do the reduction half range, six H plus a grid plus VR O3 minus aqueous, okay, so it looks like that's all of the reactants, the action area, reactants of the oxidation, or the product of the oxidation, three plus aqueous plus VR minus aqueous, and lastly three waters, liquid, and water's always in these things going to be in liquid, so that's the balanced redox equation. You guys think you can do that? Now that you've seen two examples of it, I'll give you guys one more example, at least. Any questions on this one right now?