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# Disproportionation

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

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• What is the difference between a disproprtionation reaction and a decomposition reaction?
• 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.
• 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?
• 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.
• 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)
• You calculate the oxidation number of every atom in the molecule.
Then you ignore those that do not change oxidation number and focus on those that do change oxidation number.
• Is it ok for one molecule contains both reducing agent and oxidizing agent?
• Yes certainly possible. In fact, amino acids are an example molecule because they have an acid (carboxylic acid) on one end and a base (amine) on the other end. These are also known as zwitterions.
• what about Hydrogen? How come its oxidaton state on the left isn't 1 (valence e-) -- 0 (assigned e-)
• It is. Hydrgen's oxidation number in both H2O2 and H2O is +1
• 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?
• 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)
• Can this reaction also be called disproportionation " 2KMnO4--K2MnO4+MnO2+O2↑"? Since oxygen is been oxidized and manganese is been reduced.
• A Disproportionation reaction is a reaction in which a compound undergoes oxidation as well as reduction for example consider. 2H2O2=2H2O + O2. Here hydrogen peroxide is oxidized to oxygen and is reduced to water
• I am confused bw disproportionation and intra molecular redox. Can you plase elaborate ?
(1 vote)
• 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.
• Could this reaction be classified as a redox reaction? Since reduction and oxidation are still both occurring in the same reaction