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. 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, 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^{-}. If you watch my videos on 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.

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# 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. The final step, which is unique to basic solutions, ensures the reaction is balanced correctly. Understanding these concepts is crucial for accurately determining oxidation states and identifying oxidizing and reducing agents in various chemical reactions.

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

## Balancing Basic Redox Reactions

### Balancing Redox Reactions: Basic Solutions

#### Video transcript

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

### Balancing Redox Reactions: Basic Solutions Example 1

#### Video transcript

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 focusing on the hydrogen with nitrogen. So our 2 half reactions would be:

M N O 4 − gives us M N 2 + ; N 2 H 4 gives NO 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 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 what? 2 times 3, that's 6 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 times 2 which is 8 hydrogens. So I have to put 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, 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 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 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? You 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 a 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 the 5. When I do that I'll get the electrons for each half reaction. Now at this point I'm gonna bring down everything. Everything is getting multiplied by 14. So 14MNO4− plus so we're gonna do 8 times 14 here. So when we do 8 times 14 here that's gonna give us a 112H+ plus 70 electrons. Product sign would be 14MN2+ plus 56H2O. Go to the other half reaction. Everything is getting multiplied by 5. So 5N2H4+30H2O gives me 10NO3− plus we're gonna have here 80 H+ plus 70 electrons. Here we're gonna 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 theseH2O cancelling out with 30 from here, so that gives us 26 left. All 80 of theseH+ 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 14MNO4−+N2H4 gives me 26H2O. 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 ofOH− to both sides of the equation. When H+ andOH− 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 muchH+ we have left. We have 32 left as reactants, So we're gonna add 32OH− to this side, plus 32OH− to this side. Remember, when you have H+OH−together, they combine together to give me water. So we're gonna have 14MNO4− plus 5N2H4 plus 30 2H2O gives me 14MNOH2++10NO3−+26H2O plus 32 OH−. 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 6. So that means that my balanced equation in 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 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

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

#### Problem Transcript

Balance the following redox reaction in a basic solution.

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

#### Problem Transcript

### 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 several steps. First, write the unbalanced equation. Second, separate the reaction into oxidation and reduction half-reactions. Third, balance all elements except hydrogen and oxygen. Fourth, balance oxygen atoms by adding H_{2}O. Fifth, balance hydrogen atoms by adding H^{+}. Sixth, balance the charges by adding electrons (e^{-}). Seventh, combine the half-reactions and cancel out electrons. Finally, add OH^{-} to both sides to neutralize H^{+} and form water, then simplify the equation by canceling out water molecules if necessary.

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

Understanding how to balance redox reactions in acidic solutions is crucial because the process in basic solutions builds upon it. The steps for balancing in acidic solutions form the foundation, and only one additional step is needed for basic solutions. This additional step involves adding OH^{-} ions to neutralize H^{+} ions, forming water. Mastery of acidic redox balancing ensures a smoother transition to balancing in basic solutions, making the learning process more efficient and less confusing.

How do you add the final step to balance redox reactions in basic solutions?

After balancing the redox reaction as if it were in an acidic solution, the final step for basic solutions is to add OH^{-} ions to both sides of the equation to neutralize any H^{+} ions present. This forms water (H_{2}O). For example, if you have H^{+} on one side, add the same number of OH^{-} ions to both sides. The H^{+} and OH^{-} will combine to form H_{2}O, which can then be simplified if water appears on both sides of the equation.

What role do hydroxide ions (OH^{-}) play in balancing redox reactions in basic solutions?

In basic solutions, hydroxide ions (OH^{-}) are crucial for neutralizing hydrogen ions (H^{+}) that may appear during the balancing process. When H^{+} ions are present, OH^{-} ions are added to both sides of the equation to form water (H_{2}O). This step ensures that the reaction remains balanced in a basic environment, where OH^{-} ions are naturally present. This additional step differentiates the balancing process in basic solutions from that in acidic solutions.

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

Sure! Let's balance the reaction between MnO_{4}^{-} and I^{-} in a basic solution. First, write the half-reactions: MnO_{4}^{-} → MnO_{2} and I^{-} → I_{2}. Balance the elements other than H and O. Add H_{2}O to balance O, and H^{+} to balance H. Add e^{-} to balance charges. Combine the half-reactions and add OH^{-} to neutralize H^{+}: MnO_{4}^{-} + 2H_{2}O + 3e^{-} → MnO_{2} + 4OH^{-} and 2I^{-} → I_{2} + 2e^{-}. Combine and simplify: MnO_{4}^{-} + 2H_{2}O + 2I^{-} → MnO_{2} + I_{2} + 4OH^{-}.