Balancing Oxidation-Reduction Reactions

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concept

## Balancing Redox Reactions

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Hey, guys, gonna continue with our discussion of Redox reactions by now learning how to balance them, whether they're in acidic solutions or basic solutions. So we know the ins and outs of Redox reactions. We know oxidation. We know the reduction from the past videos. But now it's Our goal is to learn how to balance them when they're in acidic solutions or basic solutions. And to do that, we're gonna follow a different set of rules. They almost go hand in hand together, except when we get to a basic solution, we have to add an additional step. So first, let's suppose to balance it out in an acidic solution. So here are going to say Step one. We have to write the equation, the full Redox equation first into two half reactions, and we'll see exactly how do we do this? But basically, we look for elements different from oxygen or hydrogen, first on either sides of the arrows and bring those down, separating them into two separate equations. Once we do that, we balance out these elements that are not oxygen or hydrogen is first. Once that's done, we can move on to step three when we have to balance out oxygen's by adding molecules of water to the opposite side so that both sides have the same number of oxygen. What we do next after that is we have to balance out hydrogen by adding H plus. Now here comes the tricky part. After we do this, we have to balance out the overall charge of that half reaction by adding electrons to the mawr positive side. And we have to make sure that when we do this, the electrons in both half reactions must match. And if they don't, that means we're gonna have to multiply either one of our half reactions or maybe even both. So we're gonna say electrons from both half reactions must match. If not, that means we have to balance either one of half reactions or both of them again. All right, after we do that, once our electrons matchup in both half reactions weaken. Then combine those two half reactions back into our full redox reactions. And this time we're gonna cross out intermediates. Remember, intermediates are species that look alike, except one will be a reactant and one will be a product. Those will cancel out number if they're on the same side, let's say they're both reactions. Both products, they don't cancel out at all. They actually add up together. So remember intermediates ones are reacting. When the product they look alike, they will cancel out once we've done this in acidic solution. Basic solution is basically almost all the same rules except for one additional rule at the end. So for in balancing and basic solutions, we follow rules. And once we've accomplished this, we're gonna we're gonna balance H plus by adding O h minus ions to both sides of the chemical reaction when we get to basic solutions, will see exactly what this really means. So remember, it's imperative that you guys first remember the basic rules that we went over because this tells us what's being reduced and what's being oxidized. Remember, if you have been reduced your the oxidizing agent, But if you've been oxidized, you're the reducing agent. Mastering that helps you to move on to this aspect of Redox reactions or asked to balance them out. And remember, you have to remember all the guidelines all the steps for either acidic or basic in order to get the full credit for that question. So hopefully you guys can remember these rules and will apply them on the next series of videos on how to balance Redox reactions.

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concept

## Balancing Acidic Redox Reactions

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Hey, guys, In this new video, we're gonna put to practice some of the rules that we learn to balance read office reactions in a six solution. So let's take a look at this first one here. We need to balance out this Redox reaction in an acidic solution. So let's go over the rules that we know. First, we're gonna break this up into two half reactions. So we look for the elements different from oxygen or hydrogen first. Here we have an end here. Here we have another end right here. So what we're gonna do is we're gonna bring them both down and say that represents our first half reaction. So we're gonna say N o to minus gives us an 03 minus. Here we have mn and here we have mn So those will also give us another half reaction. M n 04 minus gives us mn two plus. Alright. Next we have to balance on elements different from oxygen or hydrogen. So what we need to see is on the first half reaction, we only have one nitrogen, so we don't have to worry about balancing there already balanced on the other side we have one mag unease on each side, so we don't need to balance that either. Now we move on to balancing out oxygen's by adding water on this side, we have to Oxygen's on this side. We have three. So we need to put one mole of water or yeah, one mole of water on the left side. So now that both sides have three let's look at the other half reaction here. This house four. But the other side has none. So we have to add four moles of water. Yeah. Now we balanced on oxygen. Now it's time to balance out H plus. So looking back on the left half reaction, we're going to say we have to. H is right here, but we have none on the product side. So we have to add to h Plus is to this side, then looking at the other half reaction, we have four times too. We have eight ages there, so we have to add eight. H plus is here. Now, this is the tricky part. We have to balance out overall charge. So what we're gonna say here is we're gonna say water is neutral. There's no charge on this. So we're gonna say it's charges zero, but nitride ion No two minus has a negative one charge. So we're gonna say, overall, this side is negative one. Let's look at the other side. Here we have HPE plus for H, but there's two of them, so that really counts as plus two and then we're gonna say are nitrate ion and all three minus has a minus one charge, so that's minus one. So overall, the charge on this side is plus one. Remember, we have to add electrons to the more positive side so that both sides have the same exact charge. So we're gonna add electrons to the plus one side. Now we need that plus one to become a negative one, just like on the left side. So how many electrons do we need to add? We need to add two electrons because adding two electrons is equivalent to adding a negative too. So here, we're gonna say, plus one minus two equals minus one. So both sides are now negative one. Let's go to the other side. Here we have eight pluses. So that's plus eight. We have a negative one here so that's minus one. So this side is plus seven here on the other side. Water again is neutral, has no charge. So it's zero here we have plus two. So this side is plus two overall. Now we need to add electrons of the more positive side. So we're gonna add them to the left side because it's plus seven. We need both sides to be the same exact charge. So how many electrons do I need to add? Two plus seven. So that drops down two plus two. We're gonna have to add five electrons because remember, each electron is negative. So adding five electrons is equivalent to doing negative five plus seven minus five equals plus two. Both sides are now plus two. Now, finally, we have to check to make sure our electrons are equal. Here we have two electrons, but here we have five. They're not the same number. So we're gonna do here is we have to multiply both half reactions by a number so that they both have equal number of electrons. So the common multiple between two and five is 10. So we multiply this whole thing here. Times five, we multiply this whole thing, Times two. And remember, we're doing this because we need to have the same number of electrons on in both half reactions. If we don't, then it can't be a valid way of balancing it in acidic solution. All right, so everything gets multiplied by five. So we're gonna have five and o to minus plus five waters, Yes. Gives me five and all. Three minus plus 10 h plus, plus 10 electrons. And while you should remember here, just remember this for later on. We have electrons as products. If you have electrons as products, then this is an oxidation. So remember, if you have electrons products, it's an oxidation. Let's look at the other one. Everyone is getting multiplied by two. So this would be to M N 04 minus plus eight times to 16 h plus plus electrons Gives me two mn two plus plus eight h +20 And look, we have electrons in the second one as reactant. So electrons react INTs reactant means it's reduction. Okay, so remember, if you're electrons are products, it's oxidation. But if you're electrons are reacting, it's reduction. R and R. All right, so We have to make sure that the electrons cancel out and they do. One's a product. Ones are reacting there. Intermediates. Let's see who else could be intermediates? We can say that here, these 10 h plus that our products, all of them get canceled out by 10 from here, leaving us with six left. We're gonna say all five of these waters that are reacting to get canceled out by five from here, leaving us with three left. And it looks like those are our only intermediates that we have everything left. We bring it down. So we have left at the end, we're gonna have to m n 04 minus plus five and o to minus plus six h plus Give me 503 minus plus to him and to plus plus three h two up. So that would be our final answer when balancing this Redox reaction in acidic solution. I know it's a lot of moves a lot of steps to remember, but remember, go at it slowly if you're getting lost. I think the worst part of this whole thing is really looking at the overall charge. But just remember, if you don't see the compound with any type of charge. Zero. It has a coefficient in front of it that would multiply the charge whatever it ISS, making it bigger. And as long as you can get those basics down, you'll know how toe add the number of electrons you need so that both sides have the same overall charge. And again, remember your electrons in both half reactions must match up. If they don't, then it's done incorrectly. Now that we've done this one, I want you guys to attempt to do the next one on your own. So it's the same basic concepts we're going over the same rules. Break it up into half reactions first, and from there we take it on to do the rest of the steps to bounce it in acidic solution. If you get lost, it's okay. Just click on the explanation button, and a video of me will show you how best to approach this next problem. Good luck, guys.

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Problem

Balance the following reaction in an **acidic** solution.

Cl_{2}(g) + S_{2}O_{3}^{2-}(aq) ----> Cl^{-}(aq) + SO_{4}^{2-}(aq)

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4

concept

## Balancing Basic Redox Reactions

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Hey, guys, in this new video, we're finally gonna look at how do we balance the Redox reaction in a basic solution? So let's take a look at the first example. So here we're gonna have to balance out this massive reaction in a basic solution. So let's just follow the rules that we know first when following in acidic solution. Because, remember, the first six rules are the same for basic solutions. So here we have m. O. And here we have M o. So let's bring that down as its own half reaction. And here we have BR and here we have BR Let's bring those down as their own half reaction. Next, we have to make sure we balance out elements that are different from oxygen or hydrogen. So here we have three miles, but over here we only have one. So we're gonna throw a three in front of that and then on the other side, both bro means air just one, so we don't have to worry about anything there. Next we have to balance out oxygen's by adding water on this side. We have nine oxygen's on the other side. We have none, so we have to add nine waters. Here we are, four oxygen's, But on the other side, we have none. So we're gonna add four waters next, we're gonna balance out H plus. So here we have nine times to that gives us 18. So we're gonna put 18 h plus on this side, and then here we have four times to that's eight. So we have to add eight h plus to this side. Now for the tricky part Overall charge. Now the overall charge. Here we have 18 times plus one. So that's plus 18 minus three. It will be plus 15 overall on this side on the other side, M o and water of both neutral. We don't show any charges for them. So overall it zero on this side, then on the other side, we have water, is neutral again. But we have a negative with the BR, So this side is negative one overall. Then we're gonna say we have eight times plus one, so that's plus eight. And then here this is a minus two. So here this is plus six on this side. So we need to balance out the overall charge by adding electrons with a more positive side. So on this side, we're gonna add 15 electrons so that both sides are zero and then here, plus six. We're gonna add how maney electrons we need to get to negative one. We eat at seven electrons. Now we're gonna say the electrons definitely don't match. So we have to think of a number. We can multiply them both by so that that they both match. So we have to do here is we have to multiply this one by seven. We multiply this one by because the common number between them is 105. So now we're gonna have really big numbers we have to deal with. So we're gonna have seven m 03093 minus plus, we have 18 times seven. So it's 126 h plus plus 105 electrons. Gives me 45 m o plus. We're gonna have 15. I'm actually sorry we're multiplying by seven. I was looking at the 15 so it's seven times three, which is 21 and then we have seven times nine, which is 63. Okay, so remember to multiplied by seven. I was looking at the 15 for a moment, but it's time seven. Now we look at the other one, So here we're gonna have 15 multiplying everyone. So we're gonna have 15 br minus plus 60 h two all gives me 15 b r 04 to minus. Plus, we're gonna have 15 times eight, which is 120 plus 105 electrons. So there's a lot of big numbers going on here. Now we have to cross that intermediates. So who looks alike? But ones are reacting. One is a product. All 120 of these h plus cancel out with 120 from here leaving us with six. All 105 electrons, of course, have to cancel out all 60 of these waters. Cancel out what's 60 from here, leaving us with three. Bring down everything. So at the end, we're just gonna have a balanced redox reaction in an acidic solution. But we needed in a basic solution, so we have to do one additional step. So we said we have to balance out H plus by adding O H minus to both sides. So How many age plus is Do we have? We have six here, so that means we have to add 60 h minus here, plus six o h minus. Here. And remember what happens when you have HPE Plus in all which minus together, they're opposite charges. So they're going to combine to give us water. So 60 H minus six h plus combined to give a six waters, but we can have water on both sides of the equation. So what's gonna happen here? They're gonna cancel each other out. So all three of these would counsel out with three from here, leaving us with three waters. So we'd say the final answer. We'll just bring bring down everyone. Okay, so we get we get this for our final answer. So that last steps a little bit different from what we're accustomed to seeing. But that's the last step you'd have to use in order to balance it in a basic solution. So look at how maney h plus you have left. Add a which minus two. Both sides equal to that number. I remember H plus o h minus together, gives us water and then you have to cancel out the waters from both sides of the reacting and products. Whoever is left is whoever's left at the end. Nothing happens to this O H minus on this side, so we just bring it down. So that would be our final answer. Now that you guys have seen this, I want you guys to attempt the next one and I'll give you guys some help on the last one because it's different than all the others here. The common element found in both is xenon. So I actually help you out here. We're gonna say Zen on So co two gives me hte two x e and Co two gives me X e 04 Since xenon is found in both, Xena will be part of both half reactions in ANO too. So now that I've set that up for you, just follow the rules that we've learned from all the other examples in order to balance it first in an acidic solution. Second, in a basic solution. Good luck, guys.

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Problem

Balance the following reaction in a **basic** solution.

XeO_{2} (aq) ----> H_{2}Xe (aq) + XeO_{4} (aq)

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