If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Worked example: Balancing a redox equation in basic solution

AP.Chem:
TRA‑2 (EU)
,
TRA‑2.C (LO)
,
TRA‑2.C.1 (EK)
When balancing equations for redox reactions occurring in basic solution, it is often necessary to add OH⁻ ions or the OH⁻/H₂O pair to fully balance the equation. In this video, we'll walk through this process for the reaction between ClO⁻ and Cr(OH)₄⁻ in basic solution. Created by Jay.

Want to join the conversation?

  • leafers tree style avatar for user Roger Gerard
    Does someone know where I can find some good questions of balancing redox reactions to practice?
    (20 votes)
    Default Khan Academy avatar avatar for user
  • female robot amelia style avatar for user Ibraheem Rifthie
    Why do we use water to balance Oxygen, why not something else, like Oxygen itself?
    (18 votes)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user Nicki Seiden
    If the solution doesn't have a reactant with OH-, do we assume that the solution is acidic?

    Do we only use OH- to neutralize the H+ if its a basic solution?
    (10 votes)
    Default Khan Academy avatar avatar for user
  • aqualine ultimate style avatar for user manalinayak13
    At , Jay says that the oxidation state of oxygen is 2-, and since the net charge on the the molecule is 1-, the charge on chlorine has to be 1+. Can I not consider the charge on chlorine to be 1- as it usually is, and calculate the charge on oxygen accordingly?
    (7 votes)
    Default Khan Academy avatar avatar for user
    • leaf red style avatar for user Icedlatte
      When you see a molecule like H2O, ClO, etc, always do Oxygen first. Refer to the Electronegativity chart, and find the 2 most electronegative atoms you can find.

      https://en.wikipedia.org/wiki/Electronegativity#Electronegativities_of_the_elements

      If you look, it'll be first, Fluoride, then Oxygen. Since Oxygen is so electronegative, it will alway be considered first when calculating charge because it'll be more powerful at attracting electrons than the other molecule...

      ...Except in the case of Fluoride and Oxygen, such as chemical OF2 (Oxygen difluoride). The Fluorides will be more electronegative than oxygen and steal the electrons first. When OF2 is neutral, the charge will change for Oxygen to being 2+, and the Fluorides 1-.
      (14 votes)
  • old spice man blue style avatar for user Srujan Mhase
    How to find an oxidation number for a substance or an elemental entity in a chemical reaction?
    (6 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Just Keith
      You look at each chemical in the reaction individually, not the reaction as a whole, to assign oxidation states. It is quite likely, and in redox reactions a certainty, that some of the atoms will have have a change in their oxidation states as a consequence of the reaction.

      The oxidation state is an assigned value based upon am agreed-upon set of rules. You should be taught the basics of these rules in class (there are many details in the rules that we don't typically cover at the introductory level because some of these are for special situations). The oxidation state is a method of summarizing a large number of chemical properties.

      Here are some of the main rules, listed in order of importance from greatest to least. If there is a contradiction in the rules, the rule listed first should be followed:
      1. The oxidation of an element in its elemental form (that is, a neutral element that is bonded to nothing except other atoms of itself) is 0. Thus, H₂. F₂, Ne, S₈ all have the state 0. There are no exceptions to this rule.
      2. The oxidation state of a monatomic ion is the charge on that ion. There are no exceptions to this rule.
      3. Hydrogen has a state of +1 except in the hydrides of active metals in which case it has -1.
      4. Fluorine has a state of −1.
      5. Oxygen has a state of −2, except in peroxides in which case it has −1 or when directly bonded to fluorine in which case it may have a positive oxidation state.
      (There are some other exceptions, but these are obscure).
      6. Group 1 elements have a state of +1, except for hydrogen as mentioned above.
      7. Group 2 elements usually have a state of +2
      8. Group 17 (the halogens) usually have a state of −1 unless directly bonded to other halogens, in which case the lowest atomic number halogen has a state of −1. For example, in BrCl, Bromine has the +1 state and Cl has the −1 state. Also, halogens other than F may have different states when bonded directly to oxygen.
      9. The sum of oxidation states of all atoms in a neutral compound must be zero. For ions, the sum of the oxidation states of all atoms in the ion must equal the charge of that ion.
      10. Ag nearly always has the +1 state. Cd and Zn nearly always have a state of +2. Al nearly always has a state of +3.
      NOTE: not an actual rule, but the oxidation state is nearly always an integer.

      Atoms not explicitly mentioned in the above rules have their oxidation states determined by enforcing those rules. For example, in WCl₅, following the rules we assign −1 to each of the 5 Cl atoms. Thus, in order for the sum of the oxidation states to equal 0 (since this is a neutral compound) the state of W must be +5. However, in WCl₆, there are six atoms of Cl to assign the −1 charge to, thus W must have a state of +6.

      Here is a very difficult example: NaC₅H₅ (called Sodium cyclopentadienide).
      Following the rules: Na gets +1 and each of the five H gets +1
      There are six +1, so in order to sum to zero, the sum of the oxidation states of the five C atoms must sum to −6. Thus, carbon has a −⁶⁄₅ oxidation state. (Again, unusual because oxidation states are nearly always integers.)
      (9 votes)
  • piceratops ultimate style avatar for user Taylor Dover
    Does it matter when you add your OH-? In my AP Chemistry class, the standard is to add the OH- in the half reactions.
    (6 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user sriramklash
    In the beginning when writing the oxidation states, Jay wrote -2 for Oxygen in ClO- . But isn't the oxygen bonded to only one chlorine atom? How did it acquire a -2 charge?

    Thank You!
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user misha mathew
    in this video they add H+ IONS in both cases
    hows that possible
    its hould be OH-
    (4 votes)
    Default Khan Academy avatar avatar for user
    • female robot grace style avatar for user Rashmi Dahiya
      They're adding OH- in the end. Guess that doesn't make a difference though, 'cause I learnt another method at school in which you add twice as many OH- instead of H+. But as it turns out, that method sometimes makes things more complicated (well, dunno about others but at least for me, it most certainly does).
      (2 votes)
  • leaf green style avatar for user judyjaime
    Could you please explain at what time did you add the hydroxides?
    (4 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user andrea.s002
    How do you identify whether the redox reaction is in acid or a base when there are no hydrogens in the reaction? I am probably missing out on a key concept or something.
    (2 votes)
    Default Khan Academy avatar avatar for user

Video transcript

In the previous video, we saw how to balance redox reactions in acidic solution. In this video, we're going to balance a redox reaction in basic solution. And these are a little bit harder. But we're going to approach it the same way that we balanced the reactions in the acidic solution. So we're going to, once again, in step four, add some protons here. And we're going to go ahead and add the half reactions together. But after we do that, we're actually going to add some hydroxide anions in here as well. Since this takes place in basic solution, the hydroxide ions are necessary to neutralize the protons that we added in step four. So let's start off by starting with oxidation states, just like we did in the previous video. So if I start with the hypochlorite anion here, I know that oxygen has an oxidation state of negative 2. I know the total has to equal negative 1 here. So that means that chlorine must have an oxidation state of plus 1. Plus 1 and minus 2 give you negative 1. If I go to this chromium compound over here, the easiest way to approach it is to remember that the hydroxide anion has a charge of negative 1. And you have four of them, giving you a total of negative 4. In terms of oxidation states, a total has to add up to equal minus 1 here, the charge on the ion. Therefore, chromium must have an oxidation state of plus 3. Plus 3 and minus 4 give you negative 1. So chromium's oxidation state is plus 3 here. Move over here to the right to the chromate anion. So oxygen is negative 2. I have four of them for a total of minus 8. I need to get a total of minus 2, the charge on the ion. And so for an oxidation state for chromium, it would be plus 6 because plus 6 and minus 8 give you minus 2. So chromium's oxidation state is plus 6 here. And then finally, for the chloride anion negative 1 charge, therefore an oxidation state of negative 1. In terms of what is being oxidized and what is being reduced, you need to look for an increase in the oxidation state for oxidation. So going from an oxidation state of plus 3 to an oxidation state of plus 6 for chromium means chromium was oxidized. And if we look at the chlorine, it's going from an oxidation state of plus 1 all the way over here to an oxidation state of negative 1. That's a decrease in the oxidation state or a reduction in the oxidation state. So chlorine is being reduced. And so we know that something is being oxidized and something is being reduced. And so we can go ahead and do step one now. We're going to write our half reactions. So one oxidation half reaction and one reduction half reaction. This time, let's go ahead and write the reduction half reaction first. It doesn't really matter which one you do. So we'll go ahead and write the one using chlorine since chlorine was reduced here. So the hypochlorite anion going to the chloride anion here. So this is the reduction half reaction. The oxidation half reaction is the one that involved chromium. So we have this chromium compound, Cr(OH)4 with a negative charge, going to chromate, CrO42 minus. So this is our oxidation half reaction. So let's go to step two, balance the atoms other than oxygen and hydrogen. So the atoms other than oxygen and hydrogen. So for the reduction half reaction that applies to chlorine. We have one chlorine on the left and we have one on the right. So chlorine's balanced. In the oxidation half reaction, that applies to chromium. So one chromium on the left and one chromium on the right. So in this case, step two is already done for us. But always be careful because sometimes you actually have to balance these atoms and it will affect your final answer. So let's go to step three, balance the oxygens by adding water. So if we look at our reduction half reaction, we can see there's one oxygen on the left side and we have no oxygens on the right side. So we need to add one water molecule to the product side of our reduction half reaction. Now we have one oxygen on the right and one oxygen on the left and so oxygen is balanced for our reduction half reaction. Let's look at our oxidation half reaction. Remember, this four here applies to everything in the parentheses. So we have four oxygens on the left and we have four oxygens on the right. And so oxygen is already balanced. Next, we go to step four, balance the hydrogens by adding protons. So if I look at my reduction half reaction, we added a water molecule, and that introduced two hydrogens on the right side of my half reaction. So therefore, I need two hydrogens on the left side because right now I don't have any. And I'm going to balance them by adding them in the form of protons. I have to add two protons to the left side of my reduction half reaction. And now I have two hydrogens on the left and two hydrogens on the right. For my oxidation half reaction, remember, this four applies to the hydrogen as well. So I have four hydrogens on the left and none on the right side of my oxidation half reaction. So I need four on the right. I'm going to add those in the form of protons. So 4H plus on the right side. Next, we're going to add electrons to balance charge. So step five, we need to analyze the total charges on both sides of my half reactions. So let's look at the reduction half reaction first. I have two protons so that's two positive charges on the left side here. And I also have a negative charge on the hypochlorite anion here. So I have two positive charges and one negative charge. So, of course, that's a total of plus 1. And I have a habit of writing my charges like how I write oxidation states. So just remember that these are charges not oxidation states. And a total of plus 1 on the left side. And on the right side, I have about one negative charge on the chloride anion and that's it in terms of charges. And so I have a negative one charge on the right side here. So I need to get those charges to be equal. And I can only get that by adding electrons, which are, of course, negatively charged. So if I want to get both sides of my reduction half reaction equal, the only thing I could do is add two electrons to the left side right here because that gives me two more negative charges. So that takes us positive one to a negative one. So now I have a total of negative 1 charge on the left side of my half reaction. And that, of course, is now equal to the negative 1 on the right side of my half reaction. So I have balanced the charges by adding electrons. Let's do the oxidation half reaction now. So I have only one of these ions. So I have a negative 1 charge on the left side. On the right side, I have 1 chromate, but each chromate has two negative charges like that. And then I also have four positive charges from my protons. So negative 2 plus 4 gives me a total of plus 2. So I have plus 2 on the right side and negative 1 on the left side. So once again, my goal is to balance those charges by adding electrons only. And if I add three electrons to the right side of this half reaction, I would change this total charge from a plus 2 to a negative 1. And so now I have negative 1 on the right and I have negative 1 on the left, and so my charges are balanced. Another way to think about where those electrons go is to remember that Leo the lion goes "ger." So oxidation, Leo, loss of electrons is oxidation. So that's why you have to show your electrons and your oxidation half reaction being lost or given off on the product side, if you will. And then Leo the lion goes ger, so gain of electrons is reduction. So you have to show the electrons for your reduction half reaction being on your reactant side of your half reaction because they're being gained here. So that's another way just to double check yourself or think about it. So we're on to step six now. Make the number of electrons equal. So for step six, make the number of electrons equal. So the reason you have to do that is because the electrons that are being lost in the oxidation half reaction are the exact same electrons that are being gained in the reduction half reaction. So we had three electrons being lost, but only two electrons being gained. And that doesn't make any sense. Right? They're the same electrons, they should be the same number. So we need to make those number of electrons equal. And the way to do that, of course, is to think about lowest common denominator. So we have a two and the three. So we could make those six electrons. So we need to figure out a way to make both of our half reactions have six electrons. And we can do that by multiplying our first half reaction by a factor of 3. And multiplying our second half reaction by a factor of 2. And that will give us equal numbers of electrons. So let's go ahead and do that math here. So let's go ahead and take 3 times 2 electrons, that gives me 6 electrons. And then 3 times 2 protons gives me 6 protons. 3 times the 1, the coefficient in front of hypochlorite anion, gives me 3 CLO minus. The 3 times the 1 in front of the chloride anion gives me 3Cl minus. And then finally, there's a 1 in front of the water as a coefficient as well. So 3 times 1 gives me 3 waters. So let's go ahead and do the oxidation half reaction now. Let's get some more room here. So I take the 2, I multiply it through. There's a coefficient of 1 in front of here, in front of this ion. So now I would have to Cr(OH)4 minus. The 2 applies to the 1 coefficient in front of chromate as well. So I would have 2CrO42 Minus The 2 applies. 2 times 4 gives me 8 protons. So I have 8 protons here. And then, finally, 2 times the 3 electrons gives me those 6 electrons. And so now we have six electrons from a reduction half reaction and six electrons from an oxidation half reaction. Next, we're going to add our half reactions together. So once again, we're still pretending like we are in acid here. So let's go ahead and add our half reactions together. We're going to take all of our reactants and put them on one side. So let's go ahead and do that first. So we have 6 electrons plus 6 protons plus 3ClO minus plus 2Cr(OH)4 minus. And so on the right side, so we take all of this. And we add it all together for our product side. So let's see what we have. We have 3Cl minus plus 3H2O plus 2CRO42 minus plus 8H plus plus 6 electrons. So finally. All right. Now that we've written that out, I think it's easier to see that you have six electrons on the left and six electrons in the right. So you can cancel those out. You also have some protons on both sides. You have six protons here and eight protons here. So you could cancel out these six protons and you could make this two protons on the right side. So let's go ahead and rewrite what we have once again. So let's see what we have now, now that we've simplified it a little bit. We have 3ClO minus plus 2CR(OH)4 minus yields 3Cl minus plus 2H2O, sorry, 3H2O plus 2CrO42 minus plus 2H plus. All right. So now we have to remember that this reaction was actually done in base. So we're going to add in some hydroxide anions to neutralize the protons. So if we look at the number of protons we have, we have two protons over here on the right. So we need to add two hydroxide anions. So we're going to add two hydroxide anions to neutralize those protons. And since this is a balanced equation, what we do to the right side we have to the left side as well. So since I added two hydroxides to the right, I need to add in two hydroxides to the left. Now let's think about what adding those hydroxides to those protons will do. So they will neutralize each other. If you take H plus and OH minus, you will get H2O. And so we're going to get water molecules. And we're going to get two of them since we're neutralizing two protons with two hydroxide anions. And so actually on the right side, we're going to get two H2Os here, so two water molecules in the right side. So let's go ahead and write the final answer here including everything that we've done. So on the left side, we have 3ClO minus plus 2Cr(OH)4 minus. And now we added in those two hydroxides, so plus 2OH minus. And then on the right side, we had 3Cl minus plus-- well, over here, we have three waters. And we just made two more. And so we really have a total of five. So 5H2O. And then, once again, we have the 2 chromates over here. So 2CrO42 minus. And that should take care of everything. So we can go ahead and box our final answer. So we got there. It took us a long time, but we have our final answer. And remember, you can always check your final answer in terms of number of atoms and charge. So both atoms and charge should balance. And so you can go ahead and check this one on your own just like I showed you how to do it in the previous video.