 So, like we were just saying, today we're going to learn how to balance redox reactions. And hopefully you can look up here and see that already that this is a redox reaction. The way I can tell really quickly is because the oxidation number of chlorine here is negative one and the oxidation of chlorine here is zero. So I know the oxidation number changed, so that means I have a redox reaction. So these types of redox reactions is kind of a hairy one. So some of them you can balance normally, but some of them aren't, you're unable to balance normally. You're going to have to do what we call the half reaction way of balancing these types of redox reactions. But anyways, like we were saying, we know which species got oxidized, so it's, let's write the two half reactions. So the half reaction with the Cl, and notice it's not balanced, we're not balancing it yet. We just want to write the two reactions. And the reduction reaction is the other portion of the reaction. So the Cr2O7, 2 minus aqueous, 1 goes to 3 plus. Okay, so hopefully you guys can see that neither one of those equations is balanced as in the types of particles on either side of those equations, right? So what am I saying? Here we've got one chlorine on this side, but here we've got two. Does everybody see that? So we need to balance that, right? So just like balancing a normal equation, we put a 2 as a coefficient there. Is everybody okay with that one? If you look down here, you see we've got two chromiums, but we've only got one on the product side, right? So the similar thing as we did previous, we're going to put a 2 as a coefficient there. So that's pretty straightforward, that's just like basic balancing reactions. Now the problem starts to come in when we look at these oxygens here. So on the other side of any of these reactions, we don't even have oxygen. Okay, so in these redox reactions, when you see something like that, what it's telling you is that this is going to be a redox reaction where you need to add acid to it and you're going to make water as a product, okay? So in other words, I see that I have 7 oxygens here and I know that I have to have water as a product. Water has one oxygen in it, so I'm going to need 7 waters here. Does that make sense? Okay, so yeah, now hopefully you see that the hydrogens are not balanced, right? So we're going to have to do something about that, okay? If you recall, and if you don't, that's okay because we mentioned it just briefly, right? But hydrogens, when we're talking about acids, right, they're H pluses, or the acid component of the reaction. Okay, so that's what we're going to put on this side of the equation is H pluses, like that. But hopefully you see we need 7 times 2 of them, which is 14, and we're going to make those 8. So this is still the reduction. Okay, so the last thing we have to do with these half reactions is balance the charge. Remember I said when you have a charge on one side of the reaction, you have to have the same charge on the other side of the reaction. So hopefully you see over on this side we have a negative 2 charge, right? But on this side we don't have a charge at all. Does everybody see that? So in this case, what we're going to do to give ourselves negative 2 charges, we're going to add electrons, okay? To get a negative 2 charge from electrons, we're going to have to add 2 electrons. Does that make sense to everybody? So what we're going to do is put a plus 2 electron, like that. So now hopefully you see we've got a negative 2 charge over here and a negative 2 charge over here. So let's look at the bottom here, right? 14 pluses minus 2, so overall it's going to be a plus 12, right? Over here what do we got? 3 times 2, so that's plus 6. So in order to make this charge balance, we're going to have to remove 6 pluses from here, or in other words add 6 negatives, okay? Because we're adding electrons, that's how we do this. So we're going to want to add 6 electrons here. Does everybody understand why we're doing that? Plus 6 electrons. And now the last step, so now I guess 6 minuses, 2 minuses, 14 pluses. So now the last step is that we need both of these path reactions to have the same number of electrons being transferred, okay? So this one has 6, do you guys see that? And this one up here has 2, okay? So we're going to have to make this one 6 as well. So in order to do that, we're going to have to multiply by 3 straight across, okay? So just like in an algebraic reaction, a chemical reaction, what you do to one of the things you have to do to all of them, right? So in this case, we want to multiply this by 3, we've got to multiply everything by 3. So now I'm going to write our final oxidation half reaction. So it's going to be 6Cl minus AQ goes to 3Cl2AQ plus 6. Does that make sense to everybody? Any questions on something like that? Okay, so now the last step we're going to do is figure out what the final reaction is. And in order to do that, we have to add the two balanced half reactions, okay? So what you want to know is that when you're adding these two reactions, the amount of electrons should be the same on one side as it is on the other. And remember that you have the same thing on one side as on the other, they can be cancelled out, okay? So we have six electrons here and six electrons here, so we can cancel those out. And now all we do is just add everything up to give us our balanced redox reaction. So in other words, the balanced redox is going to be what? 6Cl minus AQ plus 14H plus AQ plus CR2O7 2 minus AQ. Do you guys see where I got all that stuff from? It goes to Cl2AQ plus 2CR3 plus 7 questions. So the first thing you want to look at is, is this a redox reaction or not, okay? So remember, that was the first thing we figured out when I said, when I looked at this chlorine, I said the negative one, this one is zero, so that's a redox, right? And then when I realized, I don't have all of the atoms on both, the same type of atoms on both sides, then I knew I had to do something weird to it, okay? So it's because these oxygens were here, I realized that I had to put these waters here. Does that make sense? And there, you know, the H's had to be put over here. Does that answer, does that answer that question? Okay, are there any more questions? Okay, so you will have to balance some of these for the next exam. I'll do another example on Friday and hopefully maybe we'll have time to do one on one. Okay, good job guys.