 Hello, my name is Brad Langdell and I want to talk to you today about balancing equations with oxidation numbers You know back in the day you used to balance equations just kind of by trial and error and sometimes it worked and sometimes it didn't and mostly We gave you ones that worked really nicely So it didn't seem that hard, but in reality you can balance equations particularly redox equations using this method It works really nicely. Here's the big thing to keep in mind It's the conservation of matter the number of electrons that are lost in any sort of chemical reaction has to equal the number of electrons that are gained in that same reaction and that's the big idea We're going to exploit here in order to balance this equation. Here's our example Here we have three species on either sides The first step we're going to do is we're going to go and determine the oxidation number of each of these different Species each of these different ions and atoms so the first one I'm going to start off with here I got iodine it's in the element form it gets an oxidation number of zero because all elements have an oxidation number of zero Oxygen's are always negative two, so that's an easy one to fill in and The oxidation total of the oxidation numbers has to equal the charge on the ion So here if the oxygen is a negative two then chlorine's got to be positive one that way negative two and positive one We'll give the negative charge of the chlorine and oxygen ion here All right here. We've got negative two again for oxygen and positive one for hydrogen that again Will total up to give me the negative one charge on hydrogen. They're on the hydroxide ion negative one for chlorine chlorine's a mono atomic ion, so it's charge matches this oxidation number We've got three Oxygen's here, and so that's three times negative two is negative six in total for the oxygen And that means I need to have this is going to be a little weird positive five for the iodine Why positive five for that iodine because the total charge is negative one So negative six plus five negative one and again, you know the routine by now here. We've got negative two for oxygen and there's two of The hydrogens each of those is positive one so that balances out to be neutral So we have our oxidation states figured out Now the second thing here is I'm going to look at what is losing electrons what's gaining electrons in this reaction So let's look at the iodine first iodine is going from an oxidation state of zero up to five What's happened? There's the increase of the oxidation number is increased. So this is an oxidation Anytime you see that number go up. It's an oxidation reaction, and that's a loss of electrons and in this case we had Go we'd gone from zero to five. So That's a loss of five electrons per iodine ion or ten electrons per iodine The element and since I'm dealing with element Elemental iodine here. This is the one. I'm going to be worried about I'm losing ten electrons here in this reaction Where are they going? Where's the gain? Well, let's take a look here at the other Species that has an oxidation number changes chlorine This is a reduction because we're going down an oxidation number We went down from negative one to zero pardon me from positive one to zero to negative one That's a change of two so we gained Two electrons here Now remember the number of electrons gained has to equal the number lost Which means if I want this equation to balance what I'm going to need to do here is go and put a Five in front of that chlorine and that way five times the two electrons that are gained will give ten total electrons gained By this ion and that is going to balance out the ten lost by the iodine So now that I have the number of electrons gained and lost the same I know the two coefficients that are going to go in front of those those species and that's going to allow me to go and balance The rest of it without too much trouble. I know that those are two are correct You put a little check mark next to them if you want now Let's go balance like we always did so we have two iodines. We need two iodines. There we go Those are balanced. I need five chlorines put a five there. I know those are all correct Now I'm going to look at the oxygens and hydrogens. These are always the ones that are pain Here I've got five six oxygens here. I've got six seven. Okay, so I need to get seven on the left-hand side I'll put a two in front here Five plus two is seven oxygens now. Let's check over here. I've got six seven oxygens perfect and two hydrogens Two hydrogens. I got a one here. It's balanced. That wasn't so bad You need more help with balancing with oxidation numbers you check out my website. There's more examples there at LD industries dot CA