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

Balancing redox reactions involves understanding the transfer of electrons, where one reactant is oxidized (loses electrons) and another is reduced (gains electrons). This process can be approached under acidic or basic conditions. In acidic conditions, the presence of H⁺ ions is crucial. The balancing method utilizes half-reactions, separating oxidation and reduction processes. Identifying elements other than oxygen and hydrogen helps in forming these half-reactions, ensuring both mass and charge are conserved throughout the reaction.

Balancing Redox Reactions: Acidic Solutions

Up to this point, we've learned how to balance simple chemical reactions, but now we're going to learn the approach needed to balance redox reactions. Balancing these redox reactions requires a new approach that accounts for the transfer of electrons between reactants. Remember, one reactant loses electrons and therefore is oxidized, while the other one gains electrons and is reduced. When it comes to balancing redox reactions, we can either do it under acidic conditions or basic conditions. Here, we're going to take a look first at acidic conditions.

For acidic redox reactions, we generally have the presence of the H⁺ ion. In redox reactions, not only do we balance atoms and electrons, but also charge. In balancing redox reactions, a half-reaction is either an oxidation or a reduction reaction portion of a redox reaction. We're going to have two of them: one for oxidation and the other for reduction. Typically, a half-reaction is obtained by identifying the elements that are not oxygen or hydrogen. That's what we look for in our redox reaction to first determine how to break it up into its two half-reactions.

Now that we know that little piece of information, click on to the next video and let's take a look at an example question. We have to find the half-reactions of a redox reaction.

Balancing Redox Reactions: Acidic Solutions Example 1

Here in this example question, it says, identify the half reactions from the following redox reaction. So here we have permanganate + sulfurous acid producing manganese 2 ion + sulfide ion. Alright. Remember, to get the half reactions, we look for elements that are not oxygen or hydrogen. So we can see here we have manganese with manganese, so that would represent a half reaction. So we'd have MnO4- (aqueous) to Mn2+ (aqueous), given to us by the sulfite ion. So these represent our two half reactions.

Balancing Redox Reactions: Acidic Solutions Example 2

We're now going to take a look at how to balance an acidic redox reaction. Like I said before, it's going to require a brand new approach to balance these types of reactions. So follow the steps listed below and you'll be able to balance redox reactions. Now if we take a look here it says in this example, balance the following redox reaction if it is found to be in an acidic solution. Alright. So step 1 tells us that we have to break the full redox reaction into two half reactions. Now remember, we focus on the elements that are not oxygen or hydrogen to determine the two half reactions. So nitrogen is an element that is not oxygen or hydrogen and it's found here and here. That gives us one half reaction. Don't worry about the state that they're in, we'll put those in later after we're done balancing. And then what else? We see chromium here. Chromium is not oxygen or hydrogen so that's our other half reaction. So dichromate ion gives us chromium 3 ion. We've done step 1, so let's go to step 2. Step 2 tells us to for each half reaction, balance elements that are not oxygen or hydrogen. So we have here 1 nitrogen, 1 nitrogen, so they're balanced. Here we have 2 chromiums but only 1 chromium, so put a 2 there. Step 3. For each half reaction balance the number of oxygens by adding water. So here in the first one we have 2 oxygens and here we have 3 oxygens. So I'm going to put 1 water on this side so that this side has 3 oxygens as well. For the other half reaction, we have 7 oxygens here, but no oxygens on the product side. So we need to put 7 waters. So now we've done step 3. Step 4, balance each half reaction, when it comes to hydrogen by adding H⁺. So if we take a look here on the first half reaction we have 2 hydrogens within water. So this side on the left has 2 hydrogens. So to balance it out on the right so that it has 2 as well we put 2 H⁺. Now, the other half reaction. We have 7 waters. So that's 7 times 2, that's 14 hydrogens on the product side, which means I'd have to put 14 H⁺ here. So now both sides have the same number of hydrogens. Now things can get a little bit tricky here, so pay very close attention to step 5. Balance the overall charge by adding electrons to the more positively charged side of each half reaction. Alright. So let's look at how we determine their charges. The Nitrite ion has a charge of minus 1. Water is neutral. So the overall charge on this side is minus 1. Look up the other side. We have minus 1 from the nitrate ion, but then we have 2 times plus 1, that's plus 2 from the H⁺. So overall on this side, it is plus 1. Now we add electrons to the more positive side. We add the number of electrons necessary to make its charge on this right side, the same as on the left side. So since I'm starting out at plus 1, that would mean that I would need to add 2 electrons. 2 electrons make this side now equal to minus 1. Let's go to the other side. So if we look at the other side we have what? We have 2 minus 2 from the dichromide ion and then 14 times plus 1 is plus 14. So overall this side is plus 12. Then we have what? We have 2 times plus 3, this side is plus 6. Water is neutral so we don't have to account for a charge. Again, this side is plus 12, this side here is plus 6. I need to add enough electrons to the plus 12 side so that it comes out to be plus 6 as well because remember, add electrons to the more positive side. So for both sides to be plus 6, I would need to add 6 electrons to this side. So now doing that both sides will be plus 6. But here's the thing about step 5. If the number of electrons of both half reactions are different therefore, then multiply to get the lowest common multiple. So we have 2 electrons here and 6 electrons here. The lowest common multiple is 6, so I'd have to multiply this here by 3 so that this side also has 6 electrons just like the other half reaction. Now that we've done steps 1 through 5, step 6 is combine the half reactions and cross out reaction intermediates. Remember reaction intermediates are compounds that look the same with 1 look the same with 1 as a reactant and the other a product. So here, let me write them down now, this 3 gets distributed to everything inside of the, parentheses. So we're going to have 3 nitride ions plus 3 waters produce 3 nitrate ions plus 6 H⁺ plus 6 electrons. The other half reaction, I didn't have to multiply by anything so bring everything down. So dichromate + 14 H⁺ plus 6 electrons gives me 2 chromium 3 ions, plus 7 waters. Your electrons are always reaction intermediates because 1 will be a product, 1 will be a reactant. They must always completely cross out. Next, we see that we have water here and water here. All 3 of these waters cancel out with 3 from here. That leaves us 4 remaining. Then we have H⁺ here and H⁺ here. All 6 of these H⁺ cancel out with 6 from here which leaves us with 8. So at the end, our balanced redox reaction should come out as 3 NO₂⁻ + dichromate ion + 8 H⁺ gives me 3 nitrate ions + 2 dichromate ions or 2 chromium 3 ions + 4 H₂O. So this will be my balanced redox reaction within an acidic solution. I know the steps can be a little bit daunting, but again, if you want to be able to balance a redox reaction, these are the steps that you need to employ in order to answer the question correctly. So just remember keep practicing and you'll remember the order with enough of that.

Balance the following redox reaction in an acidic solution.

Cl₂ (g) + S₂O₃^2- (aq) → Cl^- (aq) + SO₄^2- (aq)

Here it says to balance the following redox reaction in an acidic solution. So remember, step 1: we look for elements that are not oxygen or hydrogen to determine our 2 half reactions. Here we have chlorine matched up with chlorine, so that's one half reaction. So Cl2 gives me Cl-1. Then we also have sulfur with sulfur. So S2O32- gives me SO42-. Step 2, we're going to balance elements that are not oxygen or hydrogen. We have here 2 chlorines, but then we only have 1 chlorine on the other side, so we're going to put a 2 here. Then we're going to say we have 2 sulfurs here, but only 1 over here, so we put a 2 here. Next, balance oxygens by adding water. We have no oxygens on the left half reaction so move on the to the one on the right. We have 3 oxygens here, but then we have 2 times 4 here so we have 8 oxygens on the product side. Alright? So 8 oxygens on this side, 3 oxygens on this side, which would mean I need to add 5 waters to this side so that both sides have 8. Next, balance hydrogens by adding H+. Again, the half reaction on the left has no hydrogens either, so move on to the one on the right. We have here 5 times 2, that's 10 hydrogens. So I have to put 10 H+ here on this side. Next, we have to balance the overall charge of each half reaction. Remember, add electrons to the more positively charged side of each half reaction. Here, Cl2 has no charge, so the overall charge here is 0. Then we have 2 times minus 1, so the overall charge here is minus 2. We add electrons to the more positive side, so the part the side with 0. We add enough electrons to it, so it has the same overall charge as this side here. So to do that I need to add 2 electrons. Alright. Next, go to the other half reaction. We're going to say we have, minus 2 here for our molecule or ion. Water has no charge so the overall charge here is minus 2. Now we have 2 times minus 2, that's minus 4. Mhmm. And then 10 times plus 1 is plus 10. So this side here is plus 6. We need to add electrons to this more positive side so that it has the same overall charge as this side here. So to go from plus 6 to minus 2 would require that we add 8 electrons. So we add 8 electrons to this side. Realize that the number of electrons do not match. Here we have, 2 electrons but here we have 8. They have to be the same number of electrons. Lowest common multiple between them is 8, so I multiply everything here by 4. So that’d be 4Cl2+8e-1→8Cl-1. Bring down the other half reaction, we didn’t have to multiply by anything so it just comes down. Reactants come down, so everything on this side, and then bring down the products. Remember to cross out your reaction intermediates, things that look the same except one's a reactant and one’s a product. Those will always be your electrons. They must always totally cancel out. There are no other reaction intermediates, so then bring down everything. 4Cl2+2S2O32-+5H2O→8Cl-1+2SO42-+10H+. So this here represents my balanced redox reaction in an acidic solution.

What is the coefficient of Fe³⁺ when the following reaction is balanced?

Bi³⁺ (aq) + Fe³⁺ (aq) + H₂O (l) → BiO₃^- (aq) + Fe²⁺ (aq) + H⁺ (aq)

• A. 1
• B. 2
• C. 3
• D. 4
• E. 6

Here’s what students ask on this topic:

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

To balance redox reactions in acidic solutions, follow these steps:

1. Write the unbalanced equation.
2. Separate the equation into two half-reactions: one for oxidation and one for reduction.
3. Balance all elements in each half-reaction except for oxygen and hydrogen.
4. Balance oxygen atoms by adding H₂O molecules.
5. Balance hydrogen atoms by adding H⁺ ions.
6. Balance the charge by adding electrons (e^-).
7. Combine the half-reactions, ensuring that the electrons cancel out.
8. Verify that both mass and charge are balanced in the final equation.

How do you identify the oxidation and reduction half-reactions in a redox reaction?

To identify the oxidation and reduction half-reactions in a redox reaction, follow these steps:

1. Determine the oxidation states of all elements in the reactants and products.
2. Identify which element's oxidation state increases; this element is oxidized, and its half-reaction is the oxidation half-reaction.
3. Identify which element's oxidation state decreases; this element is reduced, and its half-reaction is the reduction half-reaction.
4. Write the two half-reactions separately, showing the transfer of electrons.

Why is the presence of H⁺ ions crucial in balancing redox reactions in acidic solutions?

The presence of H⁺ ions is crucial in balancing redox reactions in acidic solutions because they help balance the hydrogen atoms in the half-reactions. When balancing redox reactions, you often need to add H⁺ ions to one side of the half-reaction to ensure that the number of hydrogen atoms is equal on both sides. This is essential for maintaining both mass and charge balance in the overall reaction.

What is the role of half-reactions in balancing redox reactions?

Half-reactions play a crucial role in balancing redox reactions by separating the oxidation and reduction processes. Each half-reaction focuses on either the loss of electrons (oxidation) or the gain of electrons (reduction). By balancing each half-reaction individually for mass and charge, it becomes easier to ensure that the overall redox reaction is balanced. Once both half-reactions are balanced, they are combined to form the complete balanced redox equation.

How do you balance oxygen and hydrogen atoms in redox reactions under acidic conditions?

To balance oxygen and hydrogen atoms in redox reactions under acidic conditions, follow these steps:

1. Balance oxygen atoms by adding H₂O molecules to the side that needs more oxygen.
2. Balance hydrogen atoms by adding H⁺ ions to the side that needs more hydrogen.

This ensures that both oxygen and hydrogen atoms are balanced in the half-reactions, which is essential for the overall balance of the redox reaction.