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Balancing redox reactions in base

How to balance a redox reaction in base. Created by Jay.
Video transcript
In the previous video, we saw how to balance redox reactions in acidic solution. In this video, we're going to balance a redox reaction in basic solution. And these are a little bit harder. But we're going to approach it the same way that we balanced the reactions in the acidic solution. So we're going to, once again, in step four, add some protons here. And we're going to go ahead and add the half reactions together. But after we do that, we're actually going to add some hydroxide anions in here as well. Since this takes place in basic solution, the hydroxide ions are necessary to neutralize the protons that we added in step four. So let's start off by starting with oxidation states, just like we did in the previous video. So if I start with the hypochlorite anion here, I know that oxygen has an oxidation state of negative 2. I know the total has to equal negative 1 here. So that means that chlorine must have an oxidation state of plus 1. Plus 1 and minus 2 give you negative 1. If I go to this chromium compound over here, the easiest way to approach it is to remember that the hydroxide anion has a charge of negative 1. And you have four of them, giving you a total of negative 4. In terms of oxidation states, a total has to add up to equal minus 1 here, the charge on the ion. Therefore, chromium must have an oxidation state of plus 3. Plus 3 and minus 4 give you negative 1. So chromium's oxidation state is plus 3 here. Move over here to the right to the chromate anion. So oxygen is negative 2. I have four of them for a total of minus 8. I need to get a total of minus 2, the charge on the ion. And so for an oxidation state for chromium, it would be plus 6 because plus 6 and minus 8 give you minus 2. So chromium's oxidation state is plus 6 here. And then finally, for the chloride anion negative 1 charge, therefore an oxidation state of negative 1. In terms of what is being oxidized and what is being reduced, you need to look for an increase in the oxidation state for oxidation. So going from an oxidation state of plus 3 to an oxidation state of plus 6 for chromium means chromium was oxidized. And if we look at the chlorine, it's going from an oxidation state of plus 1 all the way over here to an oxidation state of negative 1. That's a decrease in the oxidation state or a reduction in the oxidation state. So chlorine is being reduced. And so we know that something is being oxidized and something is being reduced. And so we can go ahead and do step one now. We're going to write our half reactions. So one oxidation half reaction and one reduction half reaction. This time, let's go ahead and write the reduction half reaction first. It doesn't really matter which one you do. So we'll go ahead and write the one using chlorine since chlorine was reduced here. So the hypochlorite anion going to the chloride anion here. So this is the reduction half reaction. The oxidation half reaction is the one that involved chromium. So we have this chromium compound, Cr(OH)4 with a negative charge, going to chromate, CrO42 minus. So this is our oxidation half reaction. So let's go to step two, balance the atoms other than oxygen and hydrogen. So the atoms other than oxygen and hydrogen. So for the reduction half reaction that applies to chlorine. We have one chlorine on the left and we have one on the right. So chlorine's balanced. In the oxidation half reaction, that applies to chromium. So one chromium on the left and one chromium on the right. So in this case, step two is already done for us. But always be careful because sometimes you actually have to balance these atoms and it will affect your final answer. So let's go to step three, balance the oxygens by adding water. So if we look at our reduction half reaction, we can see there's one oxygen on the left side and we have no oxygens on the right side. So we need to add one water molecule to the product side of our reduction half reaction. Now we have one oxygen on the right and one oxygen on the left and so oxygen is balanced for our reduction half reaction. Let's look at our oxidation half reaction. Remember, this four here applies to everything in the parentheses. So we have four oxygens on the left and we have four oxygens on the right. And so oxygen is already balanced. Next, we go to step four, balance the hydrogens by adding protons. So if I look at my reduction half reaction, we added a water molecule, and that introduced two hydrogens on the right side of my half reaction. So therefore, I need two hydrogens on the left side because right now I don't have any. And I'm going to balance them by adding them in the form of protons. I have to add two protons to the left side of my reduction half reaction. And now I have two hydrogens on the left and two hydrogens on the right. For my oxidation half reaction, remember, this four applies to the hydrogen as well. So I have four hydrogens on the left and none on the right side of my oxidation half reaction. So I need four on the right. I'm going to add those in the form of protons. So 4H plus on the right side. Next, we're going to add electrons to balance charge. So step five, we need to analyze the total charges on both sides of my half reactions. So let's look at the reduction half reaction first. I have two protons so that's two positive charges on the left side here. And I also have a negative charge on the hypochlorite anion here. So I have two positive charges and one negative charge. So, of course, that's a total of plus 1. And I have a habit of writing my charges like how I write oxidation states. So just remember that these are charges not oxidation states. And a total of plus 1 on the left side. And on the right side, I have about one negative charge on the chloride anion and that's it in terms of charges. And so I have a negative one charge on the right side here. So I need to get those charges to be equal. And I can only get that by adding electrons, which are, of course, negatively charged. So if I want to get both sides of my reduction half reaction equal, the only thing I could do is add two electrons to the left side right here because that gives me two more negative charges. So that takes us positive one to a negative one. So now I have a total of negative 1 charge on the left side of my half reaction. And that, of course, is now equal to the negative 1 on the right side of my half reaction. So I have balanced the charges by adding electrons. Let's do the oxidation half reaction now. So I have only one of these ions. So I have a negative 1 charge on the left side. On the right side, I have 1 chromate, but each chromate has two negative charges like that. And then I also have four positive charges from my protons. So negative 2 plus 4 gives me a total of plus 2. So I have plus 2 on the right side and negative 1 on the left side. So once again, my goal is to balance those charges by adding electrons only. And if I add three electrons to the right side of this half reaction, I would change this total charge from a plus 2 to a negative 1. And so now I have negative 1 on the right and I have negative 1 on the left, and so my charges are balanced. Another way to think about where those electrons go is to remember that Leo the lion goes "ger." So oxidation, Leo, loss of electrons is oxidation. So that's why you have to show your electrons and your oxidation half reaction being lost or given off on the product side, if you will. And then Leo the lion goes ger, so gain of electrons is reduction. So you have to show the electrons for your reduction half reaction being on your reactant side of your half reaction because they're being gained here. So that's another way just to double check yourself or think about it. So we're on to step six now. Make the number of electrons equal. So for step six, make the number of electrons equal. So the reason you have to do that is because the electrons that are being lost in the oxidation half reaction are the exact same electrons that are being gained in the reduction half reaction. So we had three electrons being lost, but only two electrons being gained. And that doesn't make any sense. Right? They're the same electrons, they should be the same number. So we need to make those number of electrons equal. And the way to do that, of course, is to think about lowest common denominator. So we have a two and the three. So we could make those six electrons. So we need to figure out a way to make both of our half reactions have six electrons. And we can do that by multiplying our first half reaction by a factor of 3. And multiplying our second half reaction by a factor of 2. And that will give us equal numbers of electrons. So let's go ahead and do that math here. So let's go ahead and take 3 times 2 electrons, that gives me 6 electrons. And then 3 times 2 protons gives me 6 protons. 3 times the 1, the coefficient in front of hypochlorite anion, gives me 3 CLO minus. The 3 times the 1 in front of the chloride anion gives me 3Cl minus. And then finally, there's a 1 in front of the water as a coefficient as well. So 3 times 1 gives me 3 waters. So let's go ahead and do the oxidation half reaction now. Let's get some more room here. So I take the 2, I multiply it through. There's a coefficient of 1 in front of here, in front of this ion. So now I would have to Cr(OH)4 minus. The 2 applies to the 1 coefficient in front of chromate as well. So I would have 2CrO42 Minus The 2 applies. 2 times 4 gives me 8 protons. So I have 8 protons here. And then, finally, 2 times the 3 electrons gives me those 6 electrons. And so now we have six electrons from a reduction half reaction and six electrons from an oxidation half reaction. Next, we're going to add our half reactions together. So once again, we're still pretending like we are in acid here. So let's go ahead and add our half reactions together. We're going to take all of our reactants and put them on one side. So let's go ahead and do that first. So we have 6 electrons plus 6 protons plus 3ClO minus plus 2Cr(OH)4 minus. And so on the right side, so we take all of this. And we add it all together for our product side. So let's see what we have. We have 3Cl minus plus 3H2O plus 2CRO42 minus plus 8H plus plus 6 electrons. So finally. All right. Now that we've written that out, I think it's easier to see that you have six electrons on the left and six electrons in the right. So you can cancel those out. You also have some protons on both sides. You have six protons here and eight protons here. So you could cancel out these six protons and you could make this two protons on the right side. So let's go ahead and rewrite what we have once again. So let's see what we have now, now that we've simplified it a little bit. We have 3ClO minus plus 2CR(OH)4 minus yields 3Cl minus plus 2H2O, sorry, 3H2O plus 2CrO42 minus plus 2H plus. All right. So now we have to remember that this reaction was actually done in base. So we're going to add in some hydroxide anions to neutralize the protons. So if we look at the number of protons we have, we have two protons over here on the right. So we need to add two hydroxide anions. So we're going to add two hydroxide anions to neutralize those protons. And since this is a balanced equation, what we do to the right side we have to the left side as well. So since I added two hydroxides to the right, I need to add in two hydroxides to the left. Now let's think about what adding those hydroxides to those protons will do. So they will neutralize each other. If you take H plus and OH minus, you will get H2O. And so we're going to get water molecules. And we're going to get two of them since we're neutralizing two protons with two hydroxide anions. And so actually on the right side, we're going to get two H2Os here, so two water molecules in the right side. So let's go ahead and write the final answer here including everything that we've done. So on the left side, we have 3ClO minus plus 2Cr(OH)4 minus. And now we added in those two hydroxides, so plus 2OH minus. And then on the right side, we had 3Cl minus plus-- well, over here, we have three waters. And we just made two more. And so we really have a total of five. So 5H2O. And then, once again, we have the 2 chromates over here. So 2CrO42 minus. And that should take care of everything. So we can go ahead and box our final answer. So we got there. It took us a long time, but we have our final answer. And remember, you can always check your final answer in terms of number of atoms and charge. So both atoms and charge should balance. And so you can go ahead and check this one on your own just like I showed you how to do it in the previous video.