Classification & Balancing of Chemical Reactions

# Balancing Redox Reactions: Basic Solutions

**Balancing Basic Redox Reactions** requires all the same steps as balancing in an acidic solution plus an additional step.

Balancing Basic Redox Reactions

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## Balancing Redox Reactions: Basic Solutions

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Now, before we start talking about balancing Redox reactions and basic solutions, I highly suggest, if you haven't done so yet to go back and take a look at my videos on balancing redox reactions in acidic solutions. Now that's because balancing basic Redox reactions requires all the same steps as balancing in acidic solution, plus one additional step. So if you've mastered how to balance a Redox reaction in acidic solution, all this is is adding one last step a Step seven to get your balanced Redox reaction. Now we're going to say here that for basic Redox reactions, we generally have the presence of hydroxide ion. Remember hydroxide ion is O. H. Minus. All right, so if you watch my videos in terms of balancing in acidic solutions, you'll get most of this right off the bat. It's just that Step seven, that's gonna be a little bit different. So when we get to that point, we'll see what we need to do to balance our Redox reaction. For now, click on to the next video and let's take a look at the example question

**Basic Redox Reactions **generally have the presence of an OH^{–} ^{}ion.

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## Balancing Redox Reactions: Basic Solutions Example 1

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in this example, it says balance the following Redox reaction if it is found to be in a basic solution, All right, so again, many of the steps that we're going to see we've already employed when dealing with acidic solutions if you haven't watched those videos are highly recommend, you go and take a look at them again. So for step one, we're gonna break the full Redox reaction into two half reactions. We do this by focusing on the elements that are not oxygen or hydrogen. So here we have mango knees with mango knees, nitrogen with nitrogen. So are to have reactions. Would be M N 04 minus gives us mn two plus and to h four gives me and 03 minus. Now let's look at the other steps. Step two for each half. Reaction balance elements that are not oxygen or hydrogen. So here we have come one mag unease. One. Magan ease their balance and fine. We have to nitrogen one nitrogen. So throw it to here for each half reaction balance the number of oxygen's by adding water. So come back up here. We have four oxygen's here, but none on the product side. So we have to put four waters. Then we have, what, two times? Three. That's six Oxygen's. Here is products, so I have to put six waters here. Next for Step four balance for each half reaction. You're going to balance the number of hydrogen by adding H plus. So if we come back up here, we have four times to which is eight hydrogen. So I have to put eight h plus here. Now both sides have eight hydrogen. Then we have six times to which is 12 plus four, which is 16. So I need to add 16 H plus ions here. Now both sides have 16 hydrogen. Then we go to step five. We balance the overall charge by adding electron Well, you're all charged by adding electrons to the more positively charged side off each path. Reaction. So how do we do that? Well, here we're going to say we have minus one and then we have eight times plus one. So that's plus eight. Which means the overall charge on this side is plus seven. For the product side, we have to manganese ions. So overall charge here is plus two because water is neutral. It has no charge. So this side has a charge overall of plus seven. This side here has an overall charge of plus two. I have to add enough electrons to the plus seven side so that it is also plus two. So how many electrons we need to add To go from plus seven to plus two, you would need to add five electrons. Let's go to the other side both and to H four and H two are neutral. They have no charges that we can see. So the overall charge here is zero. Then we have two times minus one, which is minus two 16 times plus one is plus 16. So this is plus 14. So we have plus 14 overall here. Now we have to have enough electrons of the plus 14 side. So it has the same overall charge as a side with zero. So how many electrons do I need to add To go from plus 14 to 0, we have to add 14 electrons. So here now the number of electrons differ. So then you multiply to get the lowest common multiple between them. So here we would say that the lowest common multiple that they have between them is 70. So how do we get that? Well, we're going to say I have to multiply this here by 14 and multiply this one here by the five. When I do that, I'll get 70 electrons for each half reaction. Now, at this point, I'm gonna bring down everything. Everything is getting multiplied by 14. So 14 m n o for minus plus. So we're gonna do eight times 14 here. So when we do eight times 14 here, that's gonna give us 112 h plus plus 70 electrons. Product side would be 14 mn two plus plus 56 h +20 go to the other half reaction. Everything is getting multiplied by five. So five end to H four plus 30 h 20 gives me 10 minus. Plus, we're gonna have here 80 h plus plus 70 electrons here. We're gonna combine the half reactions and cross that reaction intermediates. So remember reaction intermediates are things that look the same except ones of product, and one's reacted. Electrons are always reaction intermediates, and they must always completely cancel out. We have all 30 of these H 20 is canceling out with 30 from here. So that gives us 26 left. All 80 of these h plus is cancel out with 80 from here, which leaves us with 32 now there's nothing else to cancel out. Nothing else is a reaction. Intermediate. Bring down everything. So here bring down all the reactant. Since you can see the process is pretty long. So 14 mn two plus plus 10 n + minus gives me 26 h +20 At this point, all we've done is balancing if it were in an acidic solution. Now, with the basic solution, we do Step seven balance any remaining H plus by adding an equal amount of O. H minus two both sides of the equation when h plus in which minus are on the same side, they combined together to form water. If water is on both sides of the equation, then treat them as reaction intermediates. Alright, so let's see what that means. So we look and see how much. Eight plus we have left. We have 32 left as reactant. So we're gonna have 32 0 h minus to this side plus 32 0 H minus to this side. Remember, when you have HPE plus in O H minus together, they combined together to give me water, So we're gonna have 14 m n 04 minus plus five and two h four plus 32 h + gives me 14 m n o m n two plus plus 10 and all three minus plus 26 waters plus 32 0 h minus. Remember, we said if waters on both sides treat them as reaction intermediates, that means that all 26 of these cancel out and we have left here six. So that means that might balanced equation in basic solution would be This came and just bring this 32 oh h minus closer. So when we do that, this represents my balanced equation in basic solution. So again, you can see it's pretty intense with the amount of steps. But as long as you can understand these steps and then use them, you can balance a Redox reaction within any basic solution.

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Problem

Balance the following redox reaction in a basic solution

H_{2}O_{2} (aq) + ClO_{2 }(aq) → ClO_{2}^{- }(aq) + O_{2} (g)

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Problem

Balance the following redox reaction in a basic solution.

ClO_{2}^{-} (aq) → Cl^{- }(aq) + ClO_{4}^{- }(aq)

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