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

Balancing redox reactions is a crucial skill in chemistry that involves accounting for the transfer of electrons between reactants. In these reactions, one species is oxidized by losing electrons, while another is reduced by gaining electrons. Redox reactions can occur under acidic or basic conditions, with acidic conditions typically involving the presence of H⁺ ions. These reactions are unique in that they balance not only atoms and molecules but also charge and electrons. The process begins by identifying half-reactions, which are either the oxidation or reduction components of the overall redox reaction. To effectively balance redox reactions, one must first determine the elements involved, excluding oxygen and hydrogen, and then separate the reaction into its oxidation and reduction half-reactions. This approach ensures that 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 and with balancing these redox reactions it's going to be required a new approach that accounts for the transfer of electrons between reactants. Remember, 1 reactant loses electrons and therefore is oxidized while the other one gains electrons and is reduced.

We're going to say 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. We're going to say for acidic redox reactions we generally have the presence of the H⁺ ion. And we're going to say here, redox reactions not only balance the atoms of electrons, but also charge and electrons.

We're going to say here that when it comes to balancing redox reactions, we have what are called half reactions. We're going to say balancing a redox reaction begins with identifying its half reactions. 1/2 reaction is either the oxidation or reduction reaction portion of a redox reaction. So we're going to have two of them. One 1/2 reaction is going to be for oxidation, the other one's going to be for reduction.

Now we're going to say usually 1/2 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 plus sulforous acid producing manganese 2 ion plus bisulfite ion. All right.

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 1/2 reaction. So we'd have

MnO 4 - ( aq ) → Mn 2 ( aq )

Then we can see that the other element that is not oxygen or hydrogen is sulfur. So here we have sulfurous acid giving us by sulfate ion. So these will 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 I'll 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.

All right, so step one 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 1/2 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 one, 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 two chromiums but only one chromium. So put a two there, TE 3 for each half reaction, balance the number of oxygens by adding water. So here in the first one we have two oxygens and here we have 3 oxygens. So I'm going to put one water on this side so that this side has three 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 three.

Step 4. Balance each half reaction when it comes to hydrogen by adding H plus. So if we take a look here on the first half reaction we have two hydrogens within water. So this side on the left has two hydrogens. So to balance it out on the right so that it has two as well, we put 2H plus. Now the other half a reaction, we have 7 waters, so that's 7 * 2. That's 14 hydrogens on the product side, which means I'd have to put 14-H plus 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. All right, So let's look at how we determine their charges. The nitrite ion has a charge of -1. Water is neutral, so the overall charge on this side is -1. Look at the other side. We have one from the nitrate ion, but then we have 2 * + 1. That's +2 from the H pluses. So overall on this side it is +1. Now we add electrons to the more positive side. We add the number of electrons necessary to make its charge on this side, on this right side the same as on the left side. So since I'm starting out at one, that would mean that I would need to add 2 electrons. 2 electrons will make this side now equal to -1.

Let's go to the other side. So if we look at the other side, we have what we have 2 -, 2 from the dichromide ion and then 14 * + 1 is plus 14. So overall this side is plus 12. Then we have what we have 2 * + 3. This side is +6. Water is neutral, so we don't have to account for a charge Again. This side is plus 12. This side here is +6. I need to add enough electrons to the plus twelve side so that comes out to be +6 as well because remember add electrons but more positive side. So for both sides to be plus six, I would need to add 6 electrons to this side. So now doing that both sides will be +6.

But here's the thing about Step 5. If the number of electrons of both half reactions are different to fur, than multiply to get the lowest common multiple. So we have two electrons here and six electrons here. The lowest common multiple is 6, so I'd have to multiply this here by 3:00 so that this side also has six electrons just like the other half reaction. Now that we've done steps one through 5 steps 6 is combine the half reactions and cross out reaction intermediates. Remember, reaction intermediates are compounds that look the same with one look the same with one as a reactant and the other a product.

So here let me write them down. Now this three gets distributed to everything inside of the parentheses, so we're going to have three nitride ions plus 3 waters, roduce 3 nitrate ions plus six H + + 6 electrons. The other half reaction I didn't have to multiply by anything, so bring everything down. So dichromate plus fourteen H + + 6 electrons gives me two chromium 3 ions plus 7 waters. Your your electrons are always reaction intermediates because one will be a product, one will be a reactant. They must always completely cross out. Next we see that we have water here and water here. All three of these waters cancel out with three from here. That leaves us 4 remaining. Then we have H here and H here. All six of these H cancel out with six from here, which leaves us with eight O at the end.

Our balanced redox reaction should come out as three N2- plus dichromate ion. So let me write that down. Dichromate ion plus 8H plus gives ME3 nitrate ions +2 dichromate ions or two chromium 3 ions plus 4H2O. So this will be my balance 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)

• A. 2 Cl₂ (g) + S₂O₃^2- (aq) + 5 H₂O (l) → 4 Cl^- (aq) + 2 SO₄^2- (aq) + 10 H⁺ (aq)
• B. 4 Cl₂ (g) + S₂O₃^2- (aq) + 5 H₂O (l) → 8 Cl^- (aq) + SO₄^2- (aq) + 10 H⁺ (aq)
• C. 4 Cl₂ (g) + S₂O₃^2- (aq) + 5 H₂O (l) → 8 Cl^- (aq) + 2 SO₄^2- (aq) + 5 H⁺ (aq)
• D. 4 Cl₂ (g) + S₂O₃^2- (aq) + 5 H₂O (l) → 8 Cl^- (aq) + 2 SO₄^2- (aq) + 10 H⁺ (aq)

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:

How do you balance a redox reaction in an acidic solution?

Balancing a redox reaction in an acidic solution involves several steps:

1. Separate the reaction into half-reactions: Break down the equation into two separate half-reactions, one for oxidation and one for reduction.
2. Balance atoms other than H and O: For each half-reaction, balance all atoms except for hydrogen and oxygen.
3. Balance oxygen atoms with water: Add H₂O to the side that needs oxygen to balance the oxygen atoms in each half-reaction.
4. Balance hydrogen atoms with H⁺ ions: Add H⁺ ions to the side that needs hydrogen to balance the hydrogen atoms that were added with water in the previous step.
5. Balance the charges with electrons: Add electrons to one side of each half-reaction to balance the charges. The reduction half-reaction will gain electrons, and the oxidation half-reaction will lose electrons.
6. Make electron gain equal to electron loss: Multiply the half-reactions by appropriate coefficients so that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.
7. Combine the half-reactions: Add the two half-reactions together, canceling out the electrons, H⁺ ions, and water molecules that appear on both sides of the equation.
8. Verify: Check that both mass and charge are balanced.