In this example, it says, balance the following redox reaction if it is found to be in a basic solution. Alright. 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, I highly recommend you go and take a look at them again. So for step 1, we're going to break the full redox reaction into 2 half reactions. We do this by dealing with 2 half reactions involving nitrogen. So our 2 half reactions would be MnO4-→Mn2+ and N2H4→NO3-. Now let's look at the other steps.

Step 2, for each half reaction, balance elements that are not oxygen or hydrogen. So here we have 1 manganese, 1 manganese; they're balanced and fine. We have 2 nitrogens, 1 nitrogen, so throw a 2 here. For each half reaction, balance the number of oxygens by adding water. So come back up here. We have 4 oxygens here, but none on the product side, so we have to put 4 waters. Then we have 2 times 3, that's 6 oxygens here as products, so I have to add 6 waters here.

Next, for step 4, balance for each half reaction, you're going to balance the number of hydrogens by adding H^{+}. So if we come back up here we have 4 times 2, which is 8 hydrogens. So I have to add 8 H^{+} here. Now both sides have 8 hydrogens. Then we have 6 times 2, which is 12, plus 4, which is 16. So I need to add 16 H^{+} ions here. Now both sides have 16 hydrogens.

Then we go to step 5, we balance the overall charge by adding electrons to the more positively charged side of each half reaction. So here we're going to say we have minus 1, and then we have 8 times plus 1, so that's plus 8, which means the overall charge on this side is plus 7. For the product side, we have 2 manganese ions, so the overall charge here is plus 2 because water is neutral, it has no charge. So this side has a charge overall of plus 7, this side here has an overall charge of plus 2. I have to add enough electrons to the plus 7 side, so that it is also plus 2. So how many electrons do I need to add to go from plus 7 to plus 2? We would need to add 5 electrons.

Let's go to the other side. Both N2H4 and H2O are neutral; they have no charges that we can see. So the overall charge here is 0. Then we have 2 times minus 1, which is minus 2. 16 times plus 1 is plus 16, so this is plus 14. So we have plus 14 overall here. Now we have to add enough electrons to the plus 14 side so it has the same overall charge as the side with 0. 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'd 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 5. When I do that, I'll get the electrons for each half reaction.

Now at this point, I'm going to bring down everything, everything is getting multiplied by 14. So 14 MnO4- plus so we're going to do 8 times 14 here. So when we do 8 times 14 here that's going to give us 112 H^{+} plus 70 electrons. The product sign would be 14 Mn^{2+} plus 56 H2O.

Go to the other half reaction. Everything is getting multiplied by 5. So 5 N2H4 + 30 H2O gives me 10 NO3- plus we're going to have here 80 H^{+} plus 70 electrons. Here we're going to combine the half reactions and cross out reaction intermediates. So remember, reaction intermediates are things that look the same one's a product and one's a reactant. Electrons are always reaction intermediates, and they must always completely cancel out. We have all 30 of these H2Os cancelling out with 30 from here, so that gives us 26 left. All 80 of these H^{+}s 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 bring down all the reactants and you can see the process is pretty long. So 14Mn2+ + NO3- gives me 26 H2O. At this point, all we've done is balance it if it were in an acidic solution. Now with the basic solution, we do step 7, balance any remaining H^{+} by adding an equal amount of OH^{-} to both sides of the equation. When H^{+} and OH^{-} are on the same side, they combine 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 H^{+} we have left. We have 32 left as reactants, So we're going to add 32 OH^{-} to this side, plus 32 OH^{-} to this side. Remember, when you have H^{+}OH^{-} together, they combine to give me water. So we're going to have 14 MnO4- plus 5 N_{2}H_{4} plus 32H_{2}O gives me 14 Mn^{2+} + 10 NO3- plus 26 waters plus 32 OH^{-}. Remember, we said if waters are on both sides treat them as reaction intermediates, that means that all 26 of these cancel out, and we have left here 6. So that means that my balanced equation in a basic solution would be this. Okay. And then just bring this 32 OH^{-} closer. So when we do that, this represents my balanced equation in a 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.