What is the reaction between H2O2 and MnO2?

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    Manganese dioxide catalyzes the decomposition of hydrogen peroxide.


    Hydrogen peroxide, H2O2, decomposes naturally at a very slow rate to form oxygen gas and water.

    When manganese dioxide, MnO2, is added to a solution of hydrogen peroxide, the rate of the reaction increases significantly. Manganese dioxide acts as a catalyst for the decomposition of hydrogen peroxide, meaning that it is not consumed in the reaction.

    What the manganese dioxide does is it lowers the activation energy of the reaction from approximately 75 kJ/mol to a little under 60 kJ/mol.

    This allows more molecules of hydrogen peroxide to undergo decomposition in a shorter period of time. The balanced chemical equation for this reaction looks like this



    MnO2 + 2H2O2 –> MnO2 + O2 + 2H2O

    2H2O2 –> O2 + 2H2O

    Manganese dioxide is written above the arrow (you’ll sometimes see it written under the arrow) because it is not being consumed in the reaction.

    Hope this helps you.

    Correct me if I’m wrong.

    The MnO2 is not a reactant. It is a catalyst. The net reaction is 2H2O2 -> 2H2O + O2.


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    No reaction takes place between H2O2 and MnO2, but MnO2 is used in the preparation of Oxygen in laboratories.

    Hydrogen peroxide (H2O2) decomposes very slowly. But in presence of manganese dioxide (MnO2), the reaction speeds up. Thus, manganese dioxide acts as a catalyst.


    A small quantity of manganese dioxide is taken in flat bottom flask. Hydrogen peroxide is added drop by drop with the help of thistle funnel. Oxygen gas bubbles out into the water over the trough. Then the water level in the jar will keep decreasing and the space above the water is occupied by oxygen. Cover the jar with a greased lid and remove it from the water. This method is called downward displacement of water.

    Reaction: 2H2O2= 2H2O+O2

    Mark Fischer is correct. The reaction is 2H2O2 + 2MnO2 = 2H2O + 2MnO2 + O2. Delta G (at 20C) = -232.8kJ (reaction is spontaneous and exothermic).

    Monica Lopez is incorrect. There is no such compound as MnO3. Manganese oxides MnO2, Mn2O3, Mn3O4, yes. But MnO3 — no such thing. Sorry, Monica.


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    It’s interesting to see the catalytic decomposition of H2O2 by MnO2 being discussed, since the reaction of Mn+7 + H2O2→ O2 + OH- + MnO2 is used to determine the strength of H2O2 solutions.

    If MnO2 were able to react and not just catalyze, the titration would not work, and since I’ve done the titration, I had to look up the possible conflict.

    Turns out, the titration works because it’s done in acidic conditions, and MnO2 is not much of a catalyst in acid. Here, an article examines the catalytic activity in pH 3–14, and finds that the crystal structure is important as well. Studies on MnO2—III. The kinetics and the mechanism for the catalytic decomposition of H2O2 over different crystalline modifications of MnO2

    So there is no reaction between H2O2 and MnO2 per se, as others have noted, and at pH < 3 there’s probably little catalytic activity. Keeping the permanganate titration acidic is thus seen to be a vital step to avoid the error that would occur if pH were allowed to rise while MnO2 was present.

    Hope this is as useful for OP as it was for me.

    Pour some hydrogen peroxide into the cylinder and add a small spatula scoop of MnO2. Bubbles of O2 will form immediately (the reaction also produces heat).

    H2O2 + MnO2 = H2O + MnO3

    Reaction type: double replacement


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    If it’s not a radical(which means every atom is bonded to the amount of atoms you’d expect) H204 would take the shape of a chain of 4 oxyogen atoms with a hydrogen atom at each end.

    Its systematic name would be hydrogentetroxide, which means hydrogen with four oxygen. You could also call it hydroxyperoxide, which means that it’s two OH groups with two oxygen atoms inbetween.

    This molecule has been synthesized at very low temperatures and is quite unstable.

    I think you probably meant this as a joke tho, so it’s for drinking.


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    Well, RIP in advance.

    Symptoms of a hydrogen peroxide poisoning include:

    • Abdominal pain and cramping
    • Breathing difficulty (if large concentrations are swallowed)
    • Body aches
    • Burns in the mouth and throat (if swallowed)
    • Chest pain
    • Eye burns (if it gets in the eyes)
    • Seizures (rare)
    • Stomach swelling
    • Temporary white color to the skin
    • Vomiting (sometimes with blood)

    Drinking small dilution of hydrogen peroxide (3%) isn’t harmful. But, like mentioned above, it can cause irritation in the eyes and skin.

    Drinking the 35% solution can cause severe burns in the gastrointestinal tract, as far as even causing vomiting and death.

    So, if you want to drink hydrogen peroxide, drink it at your own risks.

    I know about two easy ways, how to prepare manganese dioxide in laboratory:

    1. Reduction of potassium permanganate in neutral medium: 2MnO4+3H2C2O4+2H+6CO2+2MnO2+4H2O

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      Instead of oxalic acid, we can use other reducing agents as well.

    2. Oxidation of manganese salts: Mn2++2Ag++4OHMnO2+2Ag+2H2O

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      , also other relativelly weak oxidating agents could be used, but not too strong, because then the product could be permanganate.


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    The Old Thermodynamist says:

    Manganese dioxide does not react with hydrogen peroxide, it acts as a catalyst in the decomposition of hydrogen peroxide into water + oxygen.

    The word equation is:

    Manganese dioxide + hydrogen peroxide = manganese dioxide + water + oxygen

    The real equation is:

    MnO2 + H2O2 = MnO2 + H2O + 0.5O2(g)

    Change in Free Energy: ΔG(20C) = -116.1kJ (negative, so the reaction runs)

    Change in Enthalpy: ΔH(20C) = -98.0kJ (negative, so the reaction is exothermic)


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    Guy Clentsmith’s answer is correct as far as it goes. However, the real question is whether the reaction occurs at all. If the desired reaction generates a positive voltage, it will occur spontaneously. That question can be answered by looking up the table of electrochemical series at, e.g., https://sites.chem.colostate.edu/diverdi/all_courses/CRC%20reference%20data/electrochemical%20series.pdf

    The two pertinent half reactions in that table are shown below. By adding them together to get the desired reaction, one finds that the driving force of the desired reaction is +1.176 volts, and so that reaction takes place. The half reactions and math are shown below (the ^ denotes a superscript):

    H2O2 + 2H+ + 2e- —> 2H2O +1.776,

    MnO4^-2 + 2H2O + 2e^- —> MnO2 + 4OH^-1 +0.60, and write the reverse of the first reaction and add to the second, getting:

    MnO4^-2 + 4H2O + 2e^- —> MnO2 + 4OH^- + H2O2 + 2H^ + 2e^- 0.60 – 1.776 = -1.176, and simplify by

    combining 2H^+ + 2OH^- → 2H2O, and dropping out identical amounts on each side, getting:

    MnO4^-2 + 2H2O. —>. MnO2 + 2OH^-1 + H2O2 -1.176, so the desired reverse reaction force is +1.176, meaning the desired reaction will occur spontaneously.

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