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

Balancing redox reactions in basic solutions involves the same steps as in acidic solutions, with an additional step to account for hydroxide ions (OH^-). Mastery of acidic redox balancing is essential, as the process is similar, but the final step adjusts for the basic environment. Understanding oxidation and reduction, along with the roles of oxidizing and reducing agents, is crucial for successful balancing. This knowledge is foundational for grasping more complex chemical reactions and their applications in various fields.

Balancing Redox Reactions: Basic Solutions

Now before we start talking about balancing redox reactions in 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 an acidic solution, all this is is adding one last step, a step 7, 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^-. Alright. So if you watch my videos in terms of balancing in 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. 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 Mn²⁺ 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 N₂H₄ plus 32H₂O gives me 14 Mn²⁺ + 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.

Balance the following redox reaction in a basic solution

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

In this practice question, we're told to balance the following redox reaction in a basic solution. Alright. So following step 1, we know we look for elements that are not oxygen or hydrogen first. So, here we have chlorine with chlorine. So that's going to give us one half reaction which is ClO2 gives me ClO2−. Now all we have left is this compound and this compound. So by default, that's going to represent our other half reaction. So we have H2O2 gives us O2. Now that we have that, next, we balance elements that are not oxygen or hydrogen first. Here we have 1 chlorine and 1 chlorine, so that's already balanced. Now balance the number of oxygens by adding water. We have 2 oxygens on this side, 2 oxygens on this side. So it's already balanced. Let's go to the other side. We have 2 oxygens here and 2 oxygens here, still balanced. Alright. So move on to hydrogen. We balance hydrogens by adding H+. In the first half-reaction, neither side has hydrogen. In the other half-reaction, we have 2 hydrogens here, but none on this side. So, we put 2 H+. Now the hard part, figuring out the overall charge for each side of the two half-reactions. ClO2 here has no charge, so its overall charge on this side is 0. Here we have one negative charge, so the overall charge is minus 1. Remember, add electrons to the more positive side so that it has the same overall charge as this side. So, how do we go from 0 to minus 1? How many electrons do we need to add? We need to add 1 electron. Let's go to the other side. H2O2 which is hydrogen peroxide has no charge, so it's 0 for the overall side. Then here it's 2 times plus 1, so that's plus 2. Again, add electrons to the more positive side, so that equals this side. How do we go from a plus 2 charge to zero? We add 2 electrons. Notice your electrons do not match. When they don't match, we have to find the lowest common multiple. So, we'd have to multiply this equation by 2. Everything within it gets multiplied by 2. So, 2 ClO2 plus 2 electrons gives me 2 ClO2 minus. Then we're going to say here for our other half-reaction, we have H2O2 gives us O2 plus 2 H+ plus 2 electrons. Now, cancel out reaction intermediates. The electrons, remember, are always your reaction intermediates. They're always supposed to totally cancel out. Now, bring down everything because those are our only reaction intermediates in this question. So, we have 2 ClO2 + H2O2 gives me 2 ClO2 minus + O2 + 2 H+. At this point, we've done all the steps necessary to balance it in an acidic solution. But again, we're balancing it in a basic solution, so we add an additional step 7. We add OH− to both sides based on the amount of H+ remaining. And remember, when you have H+OH− together, they can combine to give us water. So, bring down everything. 2ClO2 + 2 OH− gives us 2 ClO2− + 2 H2O. So here, H2O is only found on one side, so we don't have to cancel those H2O molecules on the product side at all. So this would be our final balanced equation in a basic solution.

Balance the following redox reaction in a basic solution.

ClO₂^- (aq) → Cl^- (aq) + ClO₄^- (aq)

Here we're told to balance the following redox reaction in a basic solution. So you might notice that we have only 1 reactant but 2 products. We still follow the same rules. We're going to say here that this chlorine is attached to both this chlorine and this chlorine. So ClO2 gives us Cl-, and then ClO2- also gives us ClO4-. That's how we get our 2 half reactions. We balance elements that are not oxygen or hydrogen first, so look to see if the chlorines are the same on both sides. 1 and 1, 1 and 1. So we don't need to balance that. Next, balance oxygens by adding water. We have 2 oxygens here, but none on this side, so we have to put 2 waters. Here we have 2 oxygens, and here we have 4 oxygens, so we need to add 2 waters on this side. Next, we need to balance hydrogen by adding H+. So 2 times 2 is 4 H+ needed on this side. 2 times 2 is 4 H+ needed on this side. Next, balance the overall charge. So 4 times plus 1 is plus 4, minus 1 is plus 3 overall for this side. Then here, this side is just minus 1. Now remember, you add electrons to the more positive side in order for it to match the other side in terms of overall charge. So how do we go from plus plus 3 to minus 1? How many electrons do we need to add? We need to add 4 electrons to this side. Let's go to the other half reaction. Here we have minus one, and then here this is plus 4 because it's 4 times plus 1, minus 1. So this side here is plus 3. Again, add electrons to the more positive side. How many electrons do I need to add again to go from plus 3 to minus 1? Four electrons. So here the electrons are the same in both, so I don't need to multiply any of them to a lowest common multiple. So here we have ClO2+4H+4e-→ClO-+2H2O. Then we have here ClO2-+2H2O→ClO4-+4H++4e-. Cancel all reaction intermediates. So here, 4 electrons here, each cancel out completely. Then our 4 H+ here, cancel out completely as well. And what else canceled out completely? Our 2 waters cancel out completely. So if we look, we're not going to have any water left, any H+ left, any electrons left. Bring down everything. These 2 look the same, but they're on the same side. So they don't cancel out, they add up together. So we're going to have 2ClO2-→ClO4-+Cl-. Technically, we don't have any H+ left, so we don't have to do step 7. Because remember, step 7, we add OH- based on the amount of H+ remaining. Since there is none, this is our final balanced equation. So this here represents our balanced equation for both acidic solutions and basic solutions. Sometimes that will occur if there's no H+ remaining.

Here’s what students ask on this topic:

What are the steps to balance redox reactions in basic solutions?

Balancing redox reactions in basic solutions involves the following steps:

1. Write the unbalanced equation.
2. Separate the reaction into oxidation and reduction half-reactions.
3. Balance all elements except hydrogen and oxygen.
4. Balance oxygen atoms by adding H₂O.
5. Balance hydrogen atoms by adding H⁺.
6. Balance the charges by adding electrons (e^-).
7. Combine the half-reactions and cancel out electrons.
8. Add OH^- to both sides to neutralize H⁺ and form water (H₂O).
9. Simplify the equation by canceling out water molecules if necessary.

These steps ensure that the redox reaction is balanced in a basic solution.

Why is it important to understand balancing redox reactions in basic solutions?

Understanding how to balance redox reactions in basic solutions is crucial because many chemical reactions occur in basic environments, such as in biological systems and industrial processes. Mastery of this skill allows you to predict the behavior of substances in these environments, understand the mechanisms of reactions, and apply this knowledge to real-world problems. Additionally, it builds a strong foundation for more advanced topics in chemistry, such as electrochemistry and environmental chemistry.

How do you balance oxygen and hydrogen atoms in redox reactions in basic solutions?

To balance oxygen atoms in redox reactions in basic solutions, add H₂O to the side that needs oxygen. To balance hydrogen atoms, add H⁺ to the side that needs hydrogen. After balancing the half-reactions, add OH^- to both sides to neutralize the H⁺ ions, forming water (H₂O). Finally, simplify the equation by canceling out any water molecules that appear on both sides of the reaction.

What is the additional step in balancing redox reactions in basic solutions compared to acidic solutions?

The additional step in balancing redox reactions in basic solutions compared to acidic solutions is the addition of OH^- ions. After balancing the half-reactions and combining them, you add OH^- to both sides of the equation to neutralize any H⁺ ions present, forming water (H₂O). This step ensures that the reaction is balanced in a basic environment.

Can you provide an example of balancing a redox reaction in a basic solution?

Sure! Let's balance the redox reaction between MnO₄^- and I^- in a basic solution:

1. Write the unbalanced equation: MnO₄^- + I^- → MnO₂ + I₂
2. Separate into half-reactions:
• Oxidation: I^- → I₂
• Reduction: MnO₄^- → MnO₂
3. Balance elements other than H and O:
• Oxidation: 2I^- → I₂
• Reduction: MnO₄^- → MnO₂
4. Balance O by adding H₂O:
• Reduction: MnO₄^- → MnO₂ + 2H₂O
5. Balance H by adding H⁺:
• Reduction: MnO₄^- + 4H⁺ → MnO₂ + 2H₂O
6. Balance charges by adding e^-:
• Oxidation: 2I^- → I₂ + 2e^-
• Reduction: MnO₄^- + 4H⁺ + 3e^- → MnO₂ + 2H₂O
7. Combine half-reactions and cancel e^-:
• 2MnO₄^- + 8H⁺ + 6e^- + 6I^- → 2MnO₂ + 4H₂O + 3I₂
8. Add OH^- to neutralize H⁺:
• 2MnO₄^- + 8H⁺ + 8OH^- + 6I^- → 2MnO₂ + 4H₂O + 3I₂ + 8H₂O
9. Simplify:
• 2MnO₄^- + 6I^- + 4H₂O → 2MnO₂ + 3I₂ + 8OH^-