Which of the following represents an oxidation-reduction reaction? I. PCl 3 (aq) + Cl 2 (g) → PCl 5 (aq) II. 2 AgNO 3 (aq) + Cu (s) → Cu(NO 3 ) 2 (aq) + 2 Ag (s) III. CO 2 (g) + 2 LiOH (aq) → Li 2 CO 3 (aq) + H 2 O (l) IV. FeCl 2 (aq) + 2 NaOH (aq) → Fe(OH) 2 (aq) + 2 NaCl (aq)

Here in this practice question it says, which of the following represents an oxidation reduction reaction? So remember, to be a redox reaction, you need a reactant to be reduced and one to be oxidized. You can't have one and not the other. So if we take a look at the first option, we have p c l 3 plus c l 2 gives us p c l 5. Well, here when chlorine is by itself with no charge in its natural state, its oxidation number is 0. When it's with other elements that are not oxygen, it's minus 1. Phosphorus, we don't know what phosphorus is, there's no rule for it so it's our x. So for p c l 3 we would do x plus 3 chlorines, each one is minus 1, equals the charge of our compound. So x would equal plus 3. So phosphorus would start off as plus 3. Then over here as a product, we'd say x+5chlorines, each one is minus 1, and that equals the charge of the compound. X minus 5 equals 0, x equals plus 5. So if we look, we can see that c l 2 went from being 0 to minus 1 as a product. So its oxidation number decreased, so it was reduced. Then we can say that phosphorus that's in p c l 3 went from being plus 3 to plus 5. Its oxidation number went up so it was oxidized. So since we have both reduction and oxidation occurring this is a redox reaction. So that means that c can't be an answer and e can't be an answer. Let's look at the next one too. So here we have silver nitrate plus copper solid gives us silver, gives us copper 2 nitrate plus silver solid. Alright. So realize first that we have nitrate here and we have nitrate here. If the form of your polyatomic ion does not change, it stays consistent, that means that its oxidation numbers would stay the same. So all the elements that make up n o 3 have the same oxidation numbers on both sides. So ignore that. Here we're going to say that copper oxidation number here is 0, silver here is 0, because they're in their natural states. Here, we're gonna say that Nitrate is minus 1 for its charge, which would mean that silver here would have to be plus 1. Remember it's neutral overall so we we can say that it's plus 1. If we did x plus 1 Nitrate which is minus 1 equals 0, we'd still see that we got that plus 1. Here, this again is minus 1. We can make copper here x if we want. Making copper x here, we'd see that its oxidation number would be plus 2. Seeing again, we see that silver within silver nitrate, it goes from plus 1 to 0. So its oxidation number went down, so it was reduced. And then we can see that copper solid goes from being 0 to plus 2. Its oxidation number increased, so it was oxidized. So this is a redox reaction as well. So we know 1 and 2 definitely. Now we just have to figure out if 3 is an oxidation a redox reaction or not, and then we'll be able to get our answer at the end. So here if we look, we have carbon dioxide. Oxygen here is minus 2 because it's not a superoxide or a peroxide. Oxygen here is minus 2. Lithium here is plus 1. Hydrogen here is plus 1. Remember, we're going by the rules that we've learned in terms of oxidation numbers. Lithium here is plus 1, oxygen here is minus 2, plus 1 minus 2. Notice I didn't do it for carbon because we don't have a rule for carbon. We're gonna make it our x. We have to solve for this carbon. So we're gonna say here for carbon dioxide, it's x+2timesminus2 equals 0. So x minus 4 equals 0. Copper starts off as plus 4. Then in lithium carbonate, we're gonna say that we have 2 lithium ions, that's 2 times plus 1 plus x plus, we have what? 3 oxygens, each one is minus 2 equals 0. So 2+x-6equals0.x-4equals0.x equals+4. So we can see here that carbon not so we can see here that carbon stays exactly the same in terms of oxidation number. It was plus 4 in both cases. We can see that lithium, oxygen, and hydrogen also have consistent oxidation numbers. Nothing is changing. So nothing is being oxidized or reduced. That means that reaction 3 cannot be a redox reaction. So automatically a is out and b is out. There's no need to do 4 because d here would be our final answer. So just remember, we calculate oxidation numbers if we don't have it in within our rules. But if we remember oxidation rule and the element is present, provide that oxidation number to determine who's been oxidized and who's been reduced.