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Disproportionation

Assigning oxidation states for the decomposition of hydrogen peroxide, a disproportionation reaction. . Created by Jay.

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  • old spice man green style avatar for user Shubham Paliwal
    What is the difference between a disproprtionation reaction and a decomposition reaction?
    (11 votes)
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    • spunky sam blue style avatar for user Ernest Zinck
      A disproportionation is a special case of a decomposition reaction.
      A decomposition reaction may or may not involve a redox reaction.
      The decomposition reaction H₂CO₃ → H₂O + CO₂ does not involve reduction or oxidation.
      But in the decomposition reaction 2H₂O → 2H₂ + O₂, H is reduced and O is oxidized.
      In a disproportionation reaction, the same element is simultaneously oxidized and reduced.
      The decomposition reaction 2H₂O₂ → 2H₂O + O₂ is also a disproportionation reaction because O is reduced in forming H₂O and oxidized in forming O₂.
      Some disproportionation reactions are not decompositions, but they all involve the simultaneous oxidation and reduction of the same element.
      (40 votes)
  • blobby green style avatar for user Juan Carlos Cedillo
    What i don't understand is why you didn't assign an oxidation state to the H in H2O on the right side of the equation. And why does H2O2 only have an oxidation state of -1 when it has two oxygens. Shouldn't it be -2?
    (3 votes)
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    • male robot johnny style avatar for user Dododeda
      An even simpler explanation is that, following the rules of assigning oxidation numbers, hydrogen takes precedence. So H is +1 (unless bonded to a metal, which H then becomes -1). Since H is +1 here, that gives +2 for hydrogen in H2O2. Therefore, the O must be a total of -2 to maintain neutrality. The only way it can do that is if O is -1 (-1 times 2 = -2). Bang.
      (24 votes)
  • blobby green style avatar for user Ahszha
    Is there a hint on how to know which ion to focus on in a molecule? I.e do you general analyze first the ion that has the most change on the product side? I can see why people don't understand why he only assigned the oxidation # to the Oxygen. In this case, it's clear to me but in some other videos where he was balancing redox reactions he only focused on one ion also and in those examples I was completely lost because there was reduction occurring too (as it appeared to me) for the other ion in the molecule.
    (1 vote)
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  • piceratops seed style avatar for user Hao Ni
    Is it ok for one molecule contains both reducing agent and oxidizing agent?
    (3 votes)
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  • female robot grace style avatar for user enlighteninglavender
    what about Hydrogen? How come its oxidaton state on the left isn't 1 (valence e-) -- 0 (assigned e-)
    (3 votes)
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  • duskpin tree style avatar for user Radoniser
    Could someone please explain the counting of only one of the electrons between the oxygen atoms in H2O2? Why is this happening? That way, shouldn't this happen in the H2O molecule as well?
    (2 votes)
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    • old spice man green style avatar for user Matt B
      Step 1: assign each H as +1.
      Step 2: typically you assign each oxygen with -2. However, H2O and H2O2 do not have any other atoms, therefore you start assigning the oxgens oxidation numbers. Since both are neutral molecules you know that for H2O the oxygen must be -2 because the two hydrogens are +1 each (+2 total). For H2O2, the two oxygens must combine to make an oxidation number of -1 each (-2 total) to cancel the hydrogens +1 each (+2 total)
      (2 votes)
  • piceratops seed style avatar for user Hao Ni
    Can this reaction also be called disproportionation " 2KMnO4--K2MnO4+MnO2+O2↑"? Since oxygen is been oxidized and manganese is been reduced.
    (2 votes)
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  • blobby green style avatar for user hiteshkumar.home
    I am confused bw disproportionation and intra molecular redox. Can you plase elaborate ?
    (1 vote)
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    • spunky sam blue style avatar for user Ernest Zinck
      Intra means "within"
      An intramolecular redox reaction is a reaction in which one element of a compound is oxidized while another element in the same compound is reduced.
      For example, 2KClO₃ → 2KCl + 3O₂.
      Cl(+5) in KClO₃ is reduced to Cl(-1) to in KCl, while O(-2) in KClO₃ is oxidized to O(0) in O₂.

      Disproportionation is like an intramolecular redox reaction, but here some atoms in the compound are oxidized while other atoms of the same element in the same compound are oxidized.
      For example, 2H₂O₂ → 2H₂O + O₂.
      Here O(-1) in H₂O₂ is reduced to O(-2) in H₂O, and other atoms in the same compound (H₂O₂) are being oxidized to O(0) in O₂.
      (4 votes)
  • leaf grey style avatar for user aditya.vasireddi
    Could this reaction be classified as a redox reaction? Since reduction and oxidation are still both occurring in the same reaction
    (2 votes)
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  • duskpin seedling style avatar for user Vaishnavi Muthu
    Could somebody give the exact difference between comproportionation and disproportionation reaction ?
    (1 vote)
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Video transcript

In most redox reactions something is oxidized and something else is being reduced. But in some redox reactions, one substance can be both oxidized and reduced in the same reaction. And those are called disproportionation reactions. And here we have one the most famous examples, the decomposition of hydrogen peroxide. And so over here on the left is hydrogen peroxide, H202. Which when you add a catalyst like potassium iodide will turn into water and oxygen. So let's go ahead and assign some oxidation states or oxidation numbers in this reaction. And we'll start with the oxygen and the water molecule. And there are a couple ways to assign oxidation states. The simplest is just, of course, to memorize the rules that you'll see in any general chemistry textbook. And when you have oxygen in water, the oxidation state is negative two for oxygen. When you have O2, so over here on the right, the oxidation state of oxygen is zero. And over here on the left, hydrogen peroxide is one of those weird examples where oxygen has an oxidation state of negative one. So you could, of course, memorize these. Or there's, of course, a different way of figuring out the oxidation state, and that involves drawing out your dot structures and thinking about electronegativity. And so let's start with, once again, the water molecule here. And we know that these bonds consists of two electrons, right? So I'm going to go ahead and draw in these electrons here. We know that oxygen is more electronegative than hydrogen. So when you're thinking about it that way, you could think about those electrons going towards the oxygen. So you treat it like an ionic bond even though it isn't an ionic bond. And now you can see the oxygen's going to get all of those electrons. Oxygen normally has six valence electrons. So oxygen normally has 6 valence electrons. Here it's being surrounded by eight. So 6 minus 8 gives us negative 2. And so that is, of course, the same number that we figured out we memorized it. But thinking about the electronegativity helps you understand oxidation states a little bit more. Let's next do oxygen in the O2 molecule here. And so once again, we think about these bonds. Each bond consists of two electrons, and so we have a double bond in this case. All right. So in this case, we have oxygen bonded directly to another oxygen atom, and of course they both have this exact same electronegativity. And so we have to share all those electrons. And so since we have four total electrons, each oxygen is going to get two. And so we can go and divide up those electrons that way. So once again, oxygen normally has six electrons around it. It is now surrounded by six. So 6 minus 6 gives you 0. So an oxidation state of 0, which is, of course, exactly what we got up here. All right, let's go ahead and do the hydrogen peroxide example over here. So once again, we draw in our electrons in these bonds. So we have it look like that. And we consider the oxygen. Let's go ahead and do the oxygen on the left here. So oxygen versus hydrogen, oxygen is more electronegative. Oxygen gets those electrons. And then over here, the two electrons between those oxygen atoms, once again, an equal electronegativity, and so therefore each oxygen gets one. And so you can see that is our situation. So this time we have 6 minus-- and if you count up those electrons in there that I've circled, you'll get seven. So 6 minus 7 gives you an oxidation state of negative 1 for oxygen in hydrogen peroxide. And so either way, the memorization way is, of course, faster, but sometimes the dot structure way is very useful. So now we have our oxidation states and we can analyze this a little bit better. And you can see that we have a case where oxygen is on the left, has an oxidation state of negative 1. And let's say this oxygen atom turned into one of the O2 oxygen atoms. So it's going from an oxidation state of negative 1 to an oxidation state of 0. So you have oxygen going from negative 1 to 0. And that is an increase in the oxidation state. That's an increase in the oxidation state, therefore by definition, we know that oxygen being oxidized here. So this is an example of an oxidation. Once again, just look at the numbers. Negative 1 is increasing to 0, so that's oxidation. All right, let's think about what else is happening, right? So we could start out with an oxygen state of negative one. And let's say that oxygen this time is going to an oxidation state of negative two, like the oxygen and water. And so that, of course, is a reduction or a decrease in the oxidation state. You're going from negative 1 to negative 2. So a decrease in the oxidation state is, of course, reduction. So in this case, oxygen is being reduced. And so you have a substance that is being both oxidized and reduced in the same reaction. And so once again, that's called a disproportionation reaction.