Balancing Redox Using Half-Reaction Method - Concept

So let's talk about the balancing of chemical equations. When dealing with chemical equations and balancing them, we have to make sure we adhere to a simple rule. That's the conservation of matter. You might know it it as a conservation of mass. That is saying that matter is not created, or destroyed. It's only changed.



We want to make sure that we don't magically make up new elements, as we're dealing with chemical equations. We want to make sure whatever we have on the reactant side, is exactly the number of particles we have on the product side.



So let's put into action what I'm talking about. Let's deal with this. We have this written chemical equation. I'm going to put that into a skeletal equation, or the actual chemicals. So we have liquid carbon disulfide. So that's CS2(l). It's a liquid so we have to denote that. Reacts with oxygen gas. Oxygen gas is one of our diatomics. So we're going to say it's O2, and it's a gas, yields or produces carbon-dioxide gas CO2(g) and sulfur-dioxide gas; CS2(l)+O2(g)’!CO2(g)+SO2(g).



We want to make sure when we're dealing with chemical equations, that the number of particles, or the number of elements you have on the reactant side, equals the number of elements you have on the product side. So let's go ahead, and check that. We have Carbon, Sulfur, and Oxygen here, which means we should have strictly Carbon, Sulfur, and Oxygen on the product side.



We want to make sure when we're dealing with these, we do not change the actual molecules themselves. We only change the number of molecules that we have. So the only place I actually want to put numbers, and make changes is the place before that. We call those coefficients.



So on the reactant side, we have 1 Carbon. We have 2Sulfur atoms. We have 2Oxygen atoms. On the product side, we have 1Carbon, 1Sulfur, and 2+2 is 4Oxygen. So right now before I even balanced it, we notice that we have uneven number of elements on either sides. So make sure they're even and balanced.



So right now the Carbons are balanced. We have 1 Carbon on the reactant side, and one on the product side, So we're good there. Look at Sulfur, we have 2 Sulfur atoms, and we have 1 Sulfur atom over here. So we have to change that.



The place we change it is before the compound, right here. So I have 2 in that side, I want to make 2 in this side. So I'm going to put a 2 here. That tells me I produced 2 Sulfur-dioxide particles. So 2 Sulfurs. But then I also changed the Oxygen. So we have 4 now, plus that other 2 which is 6.



So now our Carbons are balanced, check. Our Sulfurs are balanced; check. We want to make sure our Oxygens then are balanced. We have two in the reactant side, and 6 on the product side. I'm going to change that.



So over here, we want to say we need 3 Oxygen gas particles to react with one Carbon Disulfide particle. So we're going to change that to 6. Now it's 1, 2, 6 and 1, 2, 6. We are good to go. This is fully balanced. Great reaction.



Let's use something a little more complicated. Going over here, before we dive into this equation. This reaction we have on the board which potassium chromate plus lead 2 nitrate yields potassium nitrate plus lead 2 chromate. We want to make sure we balance this properly, but there's a lot of atoms in this. So we want to make sure it's easy on ourselves. Before we jump right into it, we should notice a pattern.



We have a Chromate polyatomic ion here. We also have a Chromate polyatomic ion here. I can just keep this together. I also have a nitrate polyatomic ion here, and nitrate here. I don't want to separate the Nitrogen, and the Oxygen. I want to keep them together, since they're on the reactants, and the product side both.



So I'm going to separate everything else out. Potassium, Chromate, Lead, and Nitrate. I'm going to do the same thing on the product side. So I have 2Potassiums, 1Chromate, 1Lead, and 2Nitrates. I have 1Potassium, 1Chromate, 1Lead, and 1Nitrate.



So let's deal with the first one first. Potassium we have 2 over here, we have 1 over here. I do not want to put a 2 within the compound itself. I do not want to change this at all. I want to put it in front. Some thing like that. We have now 2Potassium over here, but that also changes my nitrates to 2. Let's see. So we have 2, 1, 1, 2. 2, 1, 1, 2. It looks like the whole thing is done. We are good to go.



Let's look at something a little bit more complicated. Let's put this reaction here. So we have C3H6 reacts with Oxygen gas to produce CO2 plus H2O. Let's do what we've been doing; writing our elements out. C, H, and O. We have 3Carbons. We have 6Hydrogens, and 2Oxygens. We have 1Carbons. We have 2Hydrogens, and we have 3Oxygens.



So first things first we're going to balance our carbons. So you have 3 on the reactant side, one on the product side. I want to put a 3 in front, so that changes this to 3, and it changes this to 6 plus 1 is 7. Hydrogens I have 6 on the reactant side, 2 on the product side. So I want to put a 3 here making this 6. That also changes the Oxygen, so we now have 3 plus 6 which is 9.



So then we go to our Oxygens. Our Oxygens we have 2 on the reactant side, and 9 on the product side. How are we going to make this change to 9? So we'll put a 4.5 there. 4.5 times 2 is 9. But that looks terrible. We don't want to ever put a decimal or fraction as a coefficient. So how are we going to get rid of that? Well the good solution is to multiply the whole reaction by 2. So that means I'm going to change all my coefficients to multiply it by 2.



I then I'm going to put this 2, because because 1 times 2 is 2. 4.5 times 2 is 9. 3 times 2 is 6, we're going to change that to a 6. 3 times 2 is 6.



So then let's check it. We have 6carbons, 6carbons check. We have 12hydrogens, 12hydrogens awesome. We now have 18Oxygens, we now have 12 plus 6, great 18Oxygens. We are done. That is how you balance chemical reactions.