 Hello, my name is Brad Langdell, I'm here to talk to you about balancing with oxidation numbers, but today we're going to talk about what happens if you've got acidic conditions. Little different, few new things to keep in mind. While this is the same as before, your number of electrons lost in your reaction have to equal the number gained. This is the conservation of matter, big, big idea here, but new for acidic conditions. If you've got some extra oxygen kicking around, you're going to have to balance those with water molecules. And what happens if you have some hydrogens you can't account for, you're going to balance those with hydrogen ions. That's what's going to make it acidic, since the hydrogen ion here is usually found in acidic solutions. So as a bonus, I'm also going to show you another way you can kind of go through and solve this problem if the oxidation number thing isn't your cup of tea. So let's give it a shot. Here's our reaction. We're going to start off like our regular balancing problems. We're going to go and assign oxidation numbers to everything. Oxygen is really easy. It is too negative in every circumstance, and there's seven of them. So that's a total of negative 14. And here I'm going to have to make up an oxidation state for chromium that's going to give me positive 12 in total, because I want positive 12 and negative 14 together to add up to the charge of negative 2. So I guess each of the chromium, there's two of them, they're going to have to be plus 6. Okay, what's happening with chromium over here? Plus 3 is the charge, plus 3 is the oxidation number. Okay, iodine, minus 1. That was pretty easy. It's an element over here, so it's 0. Not bad. Okay, so let's see what's happening to the chromium. Each of the chromium ions is going from 6 to 3. So it's going down in oxidation numbers. So it is undergoing reduction, and it's doing so by three electrons. It's going from 6 to 3. Now in the dichromate ion, there's two chromiums. So that means if it's reducing, it's gaining six electrons. The reason I had to double that by 2 is because in dichromate, there's two chromiums. Each chromium went from 6 to 3, so each of them gained 3. But there's two of them in the compound that can react, so it's a total of 6. Now let's look at what's losing electrons here. The one that's increasing in oxidation number is the iodine. So iodine, the ion, loses one electron. Okay, so now we have to think about balancing. Here we have a gain of 6, a loss of 1. That's not possible. You can't have something gain 6 when there's only one lost. So we're going to have to go and multiply the iodine here by 6. That way we'll have 6 lost, 6 gained, and everything will balance out. So I know I have to have a 6 in front of that iodine. Now I'm going to take a look at balancing everything else out. So I know my coefficients in front of the dichromate and the iodine. I can go and balance out the chromium, so that's got to be a 2. That'll balance the chromiums on either side, and I can balance my iodines. Okay, so those are good. But I've got a problem, I don't even have oxygen on the product side. So here's where the acidic conditions are going to come in. I'm going to add a water molecule over to the product side. Now I can do that because if it's in acidic conditions, like everything's aqueous, there's water here that could react, right? It means dissolved in water, that's what the aqueous state means. So that's where our oxygen is going to come in. And I can balance that out here, I've got seven oxygens. So I guess I'm going to have to put a 7 here in front of that water to balance those out. But now I've got a whole heap of hydrogens. Well, that's no big deal, I can add some hydrogens as well. So because this is under acidic conditions, there is going to be hydrogen around. And I'm going to need 14 to balance it out. So you can check it now and it's just balanced. If you want to double check, a really good idea is also to check the charges. So for charges, what do I have here? I've got plus 14, and I've got minus 2, and then I've got, so that's 12 in total. And then I'm going to take away 6, 6 times 1, so that's negative 6 in total here. And on this side, I've got, sorry, positive 6, and on this side I also have positive 6. So it's balanced, I'm set, I'm finished, I'm done. Now if you want to check your work or think of this in another way, you can also do this reaction by adding together the appropriate half reactions. Here we have a half reaction for dichromate, and here we have a half reaction for, where is it? Iodine, there we go. Now we can take those two and add them together, we can get the same balanced equation. So it's the same sort of idea here, we're going to start off with that dichromate. So CR2, oh, 7, 2, negative, I'll omit the states for speed, 14 hydrogens, and 6 electrons, and that forms 2 of the dichromates, and 7 of the waters. Okay, there's one half reaction. Now the other half reaction I'm going to put this with is the bottom one here, the iodine forming the iodine ion, but I've got to flip this one around. I'm going to write the reactant first, or the product first, because what's happening is that's what's actually reacting with the dichromate in our original question. So I'm going to flip this one around, and that's also going to make it so that the electrons are on opposite sides, which is good because we want them to cancel out. There's gain, there's a loss. And again, we see we don't have the same amount on either side, so we have to multiply this equation through by 3. So what do I get in total? I'm going to get one of the dichromates. I'm going to have 14 hydrogens, 6 electrons, cancels with 3 times 2, 6 electrons, so those cancel. And I also have the 2 of the iodines over here, and on the other side, we have 2 chromium ions, 7 waters, and our 3 iodines, because we have to multiply that by 3. That's the same equation as we had a second ago. 14, 1, 6, 2, 3, 7, 14, 1, 6, 2, 3, 7. And it works out. Of course, here I went 3 times 2 was 2. There's a 6 now. If you're looking for some more help with balancing with oxidation numbers, check out the website, www.ldindustries.ca.