Balancing Redox Reactions: Basic Solutions - Online Tutor, Practice Problems & Exam Prep

Balancing redox reactions in basic solutions builds upon the knowledge of balancing them in acidic solutions, with the addition of one extra step involving the presence of hydroxide ions (OH^-). Mastering the process in acidic environments is crucial as it forms the foundation for the additional step required to achieve a balanced redox equation in basic conditions. This final step is essential for accurately representing the chemical changes occurring within a basic medium, ensuring that the reaction adheres to the law of conservation of mass and charge. Understanding this process is vital for students delving into redox chemistry and its applications in various chemical reactions.

Balancing Redox Reactions: Basic Solutions

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 an 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, T7 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 OH-. All right.

So if you've watched my videos in terms of balancing in acidic solutions, you'll get most of this right off the bat. It's just that step 7 that's going to 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.

Balancing Redox Reactions: Basic Solutions Example 1

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, I highly recommend you go and take a look at them again. So for step one, we're going to 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 manganese with manganese, nitrogen with nitrogen. So our two half reactions would be

MnO 4 - → Mn 2 + N 2 H 4 → N 3

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 one manganese, 1 manganese. They're balanced and fine. We have two 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 what 2 * 3? That's six oxygens here as products. So I have to put 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 * 2 which is 8 hydrogens. So I have to put 8H⁺ here. Now both sides have 8 hydrogens. Then we have 6 * 2 which is 12 + 4 which is 16. I need to add 16 H⁺ ions here. Now both sides have 16 hydrogens. Then we go to step five. We balance the overall charge by adding electron well, the overall charge by adding electrons to the more positively charged side of each half reaction.

So how do we do that? Well, here we're going to say we have -1 and then we have 8 * + 1. So that's plus eight, which means the overall charge on this side is +7. For the product side, we have two manganese ions. So overall chart here is +2. Because water is neutral, it has no charge, so this side has a charge overall of +7. This side here has an overall charge of +2. I have to add enough electrons to the plus seven side so that it is also +2. So how many electrons do I need to add? To go from +7 to +2, you would need to add five 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 * -1 which is -2, 16 * + 1 is plus 16, so this is plus 14. So we have plus fourteen overall here. Now we have to add enough electrons to the plus fourteen side so it has the same overall charge as a side width 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 defer, so then you multiply to get the lowest common multiple between them.

O 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'd 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 going to bring down everything. Everything is getting multiplied by 14, so 14 MnO₄^- plus. So we're going to do 8 * 14 here. So when we do 8 * 14 here, that's going to give us a hundred and twelve H⁺ + 70 electrons. Product side would be 14 Mn²⁺ + 56 H₂O go to the other half reaction everything is getting multiplied by five.

So 5N₂H₄ plus 30H₂O gives me 10 NO₃^- plus. We're going to have here 80 H⁺ 70 electrons. Here we're going to combine the half reactions and crossover reaction intermediates. So remember, reaction intermediates are things that look the same except one's a product and one's a reactant. Electrons are always reaction intermediates and they must always completely cancel out. We have all thirty of these H₂O's canceling out with 30 from here. That gives us 26 left. All eighty of these H⁺es 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 reactants and you can see the process is pretty long. So 14 Mn²⁺ + 10 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.

All right, 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⁺ and OH^- together, they combine together to give me water. So we're going to have 14 MnO₄^- + 5 N₂H₄ + 32 H₂O gives me 14 Mn²⁺ + 10 NO₃^- + 26 H₂O + 32 OH^-. Remember we said if water is 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 basic solution would be this.

OK. And then just bring this 32 OH^- closer. So when we do that, this represents my balance 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.

Balance the following redox reaction in a basic solution

H₂O₂ (aq) + ClO₂ (aq) → ClO₂^- (aq) + O₂ (g)

• A. H₂O₂ (aq) + ClO₂ (aq) + 2 OH⁻ (aq) → ClO₂^- (aq) + O₂ (g) + 2 H₂O (l)
• B. H₂O₂ (aq) + 2 ClO₂ (aq) + 2 OH⁻ (aq) → 2 ClO₂^- (aq) + O₂ (g)
• C. H₂O₂ (aq) + 2 ClO₂ (aq) + 2 OH⁻ (aq) → 2 ClO₂^- (aq) + O₂ (g) + 2 H₂O (l)
• D. H₂O₂ (aq) + 2 ClO₂ (aq) + 2 OH⁻ (aq) → 2 ClO₂^- (aq) + O₂ (g) + H₂O (l)

Balance the following redox reaction in a basic solution.

ClO₂⁻ (aq) → Cl^- (aq) + ClO₄⁻ (aq)

• A. 3 ClO₂⁻ (aq) → 2 Cl⁻ (aq) + ClO₄⁻ (aq)
• B. 2 ClO₂⁻ (aq) → Cl⁻ (aq) + ClO₄⁻ (aq)
• C. 4 ClO₂⁻ (aq) → Cl⁻ (aq) + 3 ClO₄⁻ (aq)
• D. 4 ClO₂⁻ (aq) → 2 Cl⁻ (aq) + 2 ClO₄⁻ (aq)

Here’s what students ask on this topic:

How do you balance a redox reaction in a basic solution?

Balancing redox reactions in a basic solution involves a few steps. First, separate the reaction into two half-reactions: one for oxidation and one for reduction. Balance the atoms other than oxygen and hydrogen in each half-reaction. Then, for each half-reaction, balance the oxygen atoms by adding H₂O molecules to the side lacking oxygen. Next, balance the hydrogen atoms by adding OH^- ions to the side that needs hydrogen.

Once the oxygens and hydrogens are balanced, balance the charges by adding electrons to the more positive side of each half-reaction. Now, ensure that the electrons lost in the oxidation half-reaction are equal to the electrons gained in the reduction half-reaction by multiplying the half-reactions by appropriate coefficients.

After that, add the two half-reactions together and cancel out any species that appear on both sides of the resulting equation. Finally, verify that the atoms and charges are balanced. This will give you the balanced redox reaction in a basic solution.