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Worked example: Balancing a redox equation in acidic solution

AP.Chem:
TRA‑2 (EU)
,
TRA‑2.C (LO)
,
TRA‑2.C.1 (EK)
When balancing equations for redox reactions occurring in acidic solution, it is often necessary to add H⁺ ions or the H⁺/H₂O pair to fully balance the equation. In this video, we'll walk through this process for the reaction between dichromate (Cr₂O₇²⁻) and chloride (Cl⁻) ions in acidic solution.. Created by Jay.

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  • spunky sam blue style avatar for user Anshul Roy
    How do we balance the reactions if "Spectator ions" are also present in the redox reaction?
    (34 votes)
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  • female robot amelia style avatar for user Utkarsha_201
    I know this will sound stupid but please help me.
    How do I identify whether the reaction is in acid or base?
    (41 votes)
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    • ohnoes default style avatar for user ZekeEpling
      Look at the basic laws of the chemical reaction is it losing or gaining a hydrogen, if an atom is losing a hydrogen it is going to be an acid in which it becomes a conjugate base and if it is gaining a hydrogen atom it is a base in which it becomes a conjugate acid.
      (29 votes)
  • old spice man blue style avatar for user Lucian Rex
    Where are they getting the water (or the H+), where do the electrons go when something is oxidized, and what is the difference between oxidation states and charges?
    (8 votes)
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    • leafers ultimate style avatar for user JohnPeter
      The water is from the reaction taking place in an acidic solution - which basically (no pun intended) requires the reaction to occur in water. The H+ ions come from the acidic solution as well, as they are present in all acidic solutions.
      (14 votes)
  • old spice man green style avatar for user Shubham Paliwal
    How do we balance a reaction in which the same element disproportionates and gets reduced and oxidised ?
    (1 vote)
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    • spunky sam blue style avatar for user Ernest Zinck
      You write the same element in both half-reactions. If your reaction in basic solution is
      Cl₂ → Cl⁻ + ClO₃⁻
      you write the half-reactions as

      5×[Cl₂ +2e⁻ → 2Cl⁻]
      1×[Cl₂ + 6H₂O → 2ClO₃⁻ + 12 H⁺ + 10e⁻]

      6Cl₂ + 6H₂O→ 10Cl⁻ + 2ClO₃⁻ + 12H⁺
      12 H⁺ + 12OH⁻ →12 H₂O

      6Cl₂ + 12OH⁻ → 10Cl⁻ + 2ClO₃⁻ + 6H₂O
      (20 votes)
  • blobby green style avatar for user evan.guzman2017
    At why is it that the charge of the oxygen totals to 2-? I thought that because there are 7 that it would be -14? Could someone explain this to me please?
    (5 votes)
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    • leaf green style avatar for user Courtney Smith
      I had the same question when I first watched this video. However, I think it is because the molecule has a 2- charge, not necessarily each individual oxygen atom within that molecule. I do believe, however, that each individual oxygen atom does have a 2- oxidation state, which would total 14-, but then each chromium atom has a 6+ oxidation state, cancelling out 12 of those negative charges (because there are two chromium atoms), thus resulting in the 2- overall charge of the molecule.
      Someone please correct me if I'm wrong!

      Still confusing the way it was explained though, I agree.
      (10 votes)
  • female robot amelia style avatar for user naveed naiemi
    At why are we taking hole molecul of chromate for half reaction? Why not individually Cr2 and O7.
    Thanks.
    (6 votes)
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  • aqualine ultimate style avatar for user Bhavya Budhia
    Why we have to balance redox reaction this way why we don't just get it balanced like we balance other equations?
    (5 votes)
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    • blobby green style avatar for user James Li
      It is to remove the redundancies in an equation. In normal algebraic equations, redundancies like your typical leftover "x" or "y" are easy to remove because there are no electrons that change the quantifiable identity of each variable, if that makes sense. In order to keep track of electrons, we must be extra careful and account for charge of an atom. For example, Na+ is not the same as Na (the neutral one has an extra electron, which must be accounted for).
      (4 votes)
  • blobby green style avatar for user Claire Bell
    I thought that the redox half of the reaction was where the substance gained electrons and oxidation reaction was where the substance lost charge?
    (0 votes)
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  • leaf green style avatar for user Sarah Kristina
    Thank you very much for this clear, excellent explanation!

    I still have a question about a specific reaction. It is MnO4 - + H2C2O4 --> Mn2+ + CO2

    I can see that MnO4- --> Mn2+ is the oxidation, since electrons are lost. But what happens with the other reaction? I'm confused, since every oxidation requires a reduction, but there are no charges on the other parts.

    For the first reaction, I'd write the following: 8 H+ + MnO4- + 5e- --> Mn2+ + 4H2O

    But I have no idea how to continue. How can I proceed in such a case as it seems to differ from the equation you show in the video? Thanks for your help!
    (4 votes)
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    • spunky sam blue style avatar for user Manav Mishra
      You are wrong from the very beginning!
      Mn04- ----> Mn2+ is a reduction reaction (** Since oxidation number of Mn in MnO4- is +7 & in Mn2+ is +2). It can be seen that it is a reduction reaction since it has gained electron (+7 ---> +2)

      And H2C2O4 ----> CO2 is oxidation reaction (** Since oxidation number of C in H2C2O4 is +3 & in CO2 it is +4). So its oxidation as electron is lost. (+3 ---> +4)

      Basicallly the charges doesnt matter in the beginning...
      but as you go on you have to balance the charges.

      you determine whether it is oxidation or reduction by change in oxidation number not the charges
      (3 votes)
  • ohnoes default style avatar for user Marcus Fong
    At -
    Why is 12 added to the chromium? Where do you find the 12 being added?
    (3 votes)
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    • mr pants purple style avatar for user Ryan W
      Jay explains what he is doing right at the start of this video.

      He is working out the oxidation numbers of Cr and O in dichromate.

      Oxygen is -2, there are 7 of them, so that is a total of -14
      The sum of the oxidation numbers in a molecule equals the charge, so the oxidation numbers of each atom in dichromate need to sum to -2.
      So the 2 chromium atoms need to have oxidation numbers that equal +12 and there are 2 of them so each Cr has an oxidation number of +6
      (3 votes)

Video transcript

Our goal is to balance this redox reaction in acid. And before we get into the steps, let's talk about the fact that this is a redox reaction by assigning some oxidation states. And so we start over here with the dichromate anion. And we know that oxygen has an oxidation state of negative 2. We have seven oxygens. So negative 2 times 7 gives me negative 14 as our total here. We know that the total for the entire anion has to equal negative 2, which is the charge on the dichromate anion. Therefore, we must have plus 12 for all of our chromiums here. So plus 12 and minus 14 give us negative 2. Since we have two chromiums, each one must be plus 6. And so that's the oxidation state for chromium here. We go over here to the chloride anion. So the charge is negative 1, so our oxidation state is negative 1. Chromium ion over here. So plus 3. And then, finally, chlorine over here, so oxidation state of 0. So if we look at chlorine, chlorine went from an oxidation state of negative 1 to an oxidation state of 0. That's an increase in the oxidation state. Therefore, chlorine was oxidized here. Look at chromium. Chromium went from plus 6 to plus 3. That's a decrease in the oxidation state, or a reduction in the oxidation state. Therefore, chromium was reduced. And so this is a redox reaction because something is oxidized and something is reduced. In terms of balancing it, our first step is to write the different half reactions. And so we're going to break those into an oxidation half reaction and a reduction half reaction. So let's go ahead and get some space down here. And let's go ahead and write our half reactions. And so we had a chloride anion going to chlorine like that. And we said that this was our oxidation half reaction, so put that way over here on the right. So that's out oxidation. Our reduction half reaction involved chromium. So we had the chromate anion here, so Cr2O7 2 minus, going to chromium 3 plus like that. So this is our reduction half reaction. So that's step one. Write the different half reactions. Step two. Balance the atoms other than oxygen and hydrogen. And so, if you look at our first half reaction, we have one chlorine on the left and two chlorines on the right. So we need to balance it by putting a 2 over here on the left like that. We go down here to the reduction half reaction, and we have two chromiums on the left and only one on the right. And so we have to put a 2 right here to balance it. So step two is done. Step three. Balance the oxygens by adding water. So if I look at my oxidation half reaction, there are no oxygens so I don't need to worry about doing anything to this half reaction at the moment. I go down to the reduction half reaction, and I do have to balance my oxygens. So if I go over here and I can see that I have seven oxygens on the left side and none on the right. And so I need to do that by adding the water. And since I have seven oxygens on the left, I need seven oxygens on the right. So I'm going to go ahead and add seven water molecules. And that now gives me seven oxygens on the right of my half reaction. So that's this step, step three, right here. Step four. Balance the hydrogens by adding some protons. So let me go ahead and I'll use red for this. So step four. Balance the hydrogens by adding protons here. Once again, the oxidation half reaction, we don't have to do anything, because we don't have to balance oxygen or hydrogen here. But, again, we go down to our reduction half reaction and we have the oxygens balanced by adding water, but by adding water, now we have some hydrogens on the right side. So we can see we have a total of 14 hydrogens on the right side. So 7 times 2. And so we're going to balance that by adding protons. And so we need to add protons to the left side of our half reaction. So we need to add 14. So 7 times 2 is 14. So we go ahead and add 14 H plus to the left side of our half reaction. So step four is done. Step five. Balance the charges by adding electrons. So let's get some more space here, first of all. OK. So we're going to balance the charges by adding electrons. So first let's analyze what kinds of charges that we have here. So we'll start with the top oxidation half reaction. So we have the chloride anion, which is a negative 1 charge. And we have two of them. So we have two negative charges. Notice, these are not oxidation states. That's what gets people confused sometimes. These are charges. Over here on the right, we have no charges. So that's a neutral chlorine molecule here. So we have two negative charges on the left and 0 for a charge on the right. So we need to figure out how to balance those charges by adding electrons, and so it makes sense that we would have to add two electrons to the right over here, because that now gives us a total charge of negative 2 on the right. So that's one way to think about it-- just getting these numbers equal here. This negative 2 right here and this negative 2. Another way to do it would be, of course, you know the electrons have to go on the right side because this is the oxidation half reaction. And one of the ways to remember that-- LEO the lion. So Loss of Electrons is Oxidation. And so if you're losing electrons they must go on the product side of your half reaction. And so our top half reaction, our oxidation half reaction, is now balanced. Let's go down to our reduction half reaction. So LEO the lion goes GER. So Gain of Electrons is Reduction. So we already know we're going to have to add electrons to the reactant side of this half reaction. Let's see if we can figure out how many electrons we're going to need to use. So we have 14 positive charges from the protons. And then we have two negative charges from the dichromate anion here. So we have 14 positive charges and two negative charges which gives us a total of 12 positive charges on the left side. On the right side, we have a chromium ion. So this is a charge of 3 plus, and I have two of them. So 2 times positive 3 gives me positive 6. So I have positive 6 on the right side of my half reaction. So I have positive 12 on the left, positive 6 on the right. I need to add some electrons to balance out that charge. I already know I'm going to add them to the reactant side. I know that from LEO the lion goes GER. Or I could just think about the fact that, if I have 12 positive charges, I would need to add six negative charges to get me to a total charge of plus 6. So I need to add six electrons to the left side over here. So let me go plus six electrons like that. And now, we have the charges balanced. So this step is done. So step five is done. We move on to step six. So make the number of electrons equal. So what does that mean? Well, let's focus in here on the electrons that we just added to our half reactions. So if I go back up to here, we added two electrons to the oxidation half reaction, and we had six for the reduction half reaction. But we know that that number has to be the exact same number because the electrons that are lost from our oxidation half reaction are the exact same electrons that are gained in our reduction half reaction. So that is why we have to make these electrons equal in terms of the number. And so the way to do that would, of course, be to multiply my first half reaction by 3. Because if I multiply my first half reaction by 3, that would give me a total of six electrons, which is what we're looking for, making the number of electrons equal. So let's go ahead and do that. I'm going to rewrite our first half reaction. I'm going to multiply everything in our half reaction through by 3, so everything in parentheses. So if I take 3 and multiply that by 2 chloride anions, I would, of course, get 6. So we have six chloride anions like that. And then the three would go in front of the chlorine. So I have 3Cl2, and then 3 times two electrons gives me six electrons like that. OK. Let's go ahead and rewrite our reduction half reaction because we have a lot of stuff going on here. Some I'm just going to rewrite exactly what we have. Six electrons plus 14 protons plus the dichromate anion like that. And then I have two chromium ions and seven the waters. OK. So now I have my two half reactions. I've made the number of electrons equal. And we're ready for the last step. We just take our two half reactions and we're going to add them back together, and that's going to give us our overall balanced redox reaction. So let's go ahead say we did this. And let's go ahead and get some more room here so we can add those half reactions. So all I'm going to do is just take everything on the reactant side and add that together. So let's go ahead and just rewrite everything on our reactant side. So we have 6Cl minus and then six electrons, 14 protons. We have the dichromate anion here. And then I'm going to take everything on my product side. So I'm going to take all of this stuff right here and put that on my product side. So I have 3Cl2 plus six electrons plus 2 chromium ions plus 7 water molecules like that. And so now it's just a little bit easier, I find, for students to see that your electrons are going to cancel, right? You have these six electrons on the reactant side. You have these six electrons on the product side. So you can go ahead and take those out. And we're left with our final answer. So you can go ahead and rewrite it if you want to make it look better. We have 6Cl minus plus 14 H plus plus Cr2O7 2 minus yields 3Cl2 plus 2Cr 3 plus plus 7 H2O. And this should be our final answer. So I always like to box my final answer, so it just makes it easier for your instructor to grade. And the nice thing about redox reactions is, when you're finished, you can always check yourself. Because you know that you need to balance both the atoms and the charge. So let's go ahead and check that real fast. Let's first start with chlorine. So we have six chlorines on the left, then over here on the right we have 3 times 2, which is 6. So the chlorines balance. Hydrogen. We have 14 on the left, and then 7 times 2 gives us 14 on the right. The chromiums. We have two on the left and then over here we have two on the right. Oxygens. Seven oxygens on the left and then over here this seven applies to this oxygen. And so the atoms are balanced properly. Let's next check charge because you have to have the correct charge. Let's see, we have 6 negatives charges. I'll just put minus 6 here. And then I have 14 positives. And then I have a negative 2 right here. So when I add all those up, I get a total of plus 6 on the left side. So 14 minus 8. Over here on the right side, the only charge is this chromium ion. It's 3 plus, and I have two of them. So 2 times 3 gives me plus 6. And so the charge balance as well. And so this is the correct answer. It balances both in terms of atoms and in terms of charge.