 So, we're going to do another one of these balancing a redox equation in an acidic solution. But with this one, you're going to see that you can cancel out more things than just electrons. Okay? So, anyways, let's just get to it. So, it says write the balanced chemical equation for benzal alcohol being oxidized to benzaldehyde by dichromate if your inorganic product is the chromium 3 plus ion. Okay? So, let's just balance this equation. So, remember, the first thing you want to do is take this to 2 half reaction. So, how was I able to decide this was the reduction and this was the oxidation? Because, I mean, a number of reasons, but probably the most obvious one was there was a reduction of hydrogens in this molecule. So, remember, if you lose hydrogens, that's going to be oxidation. So, anyways, let's go ahead and now balance these equations. Okay? So, we've got two chromiums here, only one there. So, put a 2 there. Recall that when we have oxygens, we're going to, on the other side, put waters to balance them. The water and it's liquid formed and there's 7 oxygens, so we're going to need 7 waters. So, that in turn is going to make us have an unbalanced number of hydrogens. Right? And remember, we balance hydrogens by coming over to the other side of the reaction and putting those H plus ions or protons. Okay? So, we've got 2 times 7 of them. So, that's 14 and we still did 3 minus 4. And now, we want to balance the charge. Okay? So, how do we do that? Well, on this side, we've got a plus 6 charge. Why? Because it's 2 times 3 plus there. So, over here, we've got 14 minus 2. So, that's a plus 12 overall charge. Everybody okay with doing that? So, remember, in order to balance the charges, we add electrons. Okay? So, in order to take plus 12 to plus 6, we're going to add 6 electrons. So, like I was saying, I didn't even leave myself alone. So, there's the 6 electrons that we'll add. Okay? So, the reduction half equation is balanced. Now, let's go to the oxidation half equation. So, the first thing you might notice is that there is an unbalanced number of hydrogens in the bottom equation. So, how do we take care of that? Remember, we make protons out of them or H plus ions. So, we have 8 over here and 6 over here. So, we're going to add 2 over here. Now, the next thing we need to do is balance the charge on the bottom reaction. So, in order to do that, we have to, since there's a zero charge over here and a plus 2 charge over here, we have to add 2 electrons if over here. Okay? And remember, the second to final step is to... So, now we have this one all balanced, right by itself. But the second to final step is to balance the number of electrons between these two half reactions. So, in order to do that, we've got 2 here and 6 here. We've got to multiply this whole reaction equation by 3 to make this 6. Everybody okay with that? So, let's go ahead and rewrite the reaction equation. So, now remember, final step is to combine the two half reactions. So, let's do that. Remember, the reason we multiply by 3 is so we can cancel out our electrons. Okay? So, because we have 6 over here and 6 over here, we can cancel those out. So, hopefully you're noticing at this point in time that we have 14 H pluses aqueous over here and 6 H plus aqueous over here. So, that's another thing that you can cancel out. So, again, if you've got things on both sides of the reaction equation, those things can be canceled out. Okay? So, let's cancel this out. 6 H plus aqueous. And we're going to take 6 away from that. So, it's going to be 14 minus 6. It's going to be like that. Is everybody okay with doing that in the last step? So, this is just an additional step to this type that we've talked about last time. Okay? And then the last thing is to just write the balanced reaction equation. So, I'm going to erase it so I don't have to erase this part. So, I don't have to. But remember, just combine these two reaction equations. So, we've got 8 H plus aqueous plus CR 207 2 minus dichromate and plus 3 benzoyl alcohols. C7H goes to canine plus 7 waters plus 3 benzoyl. But if you know anything about it, this is how you take primary alcohol to primary alcohol cell. Are there any questions about this one? So, again, this is just throwing in one more step than you guys. Okay?