How is Mn2+ oxidised to MnO2 by KMnO4 in basic media?

2 Answers

  • 10 months ago

    Redox reaction ....

    In a basic solution, solid MnO2 is at a lower energy than either Mn^2+ or the permanganate ion.

    Mn^2+ + MnO4^- + OH- --> MnO2(s) + H2O ........ not balanced

    3(Mn^2+ + 4OH- --> MnO2 + 2H2O + 2e-) ........ oxidation half-reaction

    2(MnO4^- + 2H2O + 3e- --> MnO2 + 4OH-) ...... reduction half-reaction

    -------------------- --------------------- --------------------- -------------------- ---------

    3Mn^2+ + 2MnO4^- + 12OH- + 4H2O --> 5MnO2(s) + 6H2O + 8OH-


    3Mn^2+ + 2MnO4^- + 4OH-  --> 5MnO2(s) + 2H2O

    Start by understanding that aside from the Mn^2+ ion, Mn in MnO2 and Mn in MnO4^- don't actually have charges of +4 and +7, respectively.  When you think about it, that would be impossible to remove that many electrons from Mn atoms. It would require too much energy. The bonds in MnO2 and MnO4^- have significant covalent character.  The electrons are shared, not "lost" or "gained".  What we write in half-reactions, is an oversimplification, as if all the bonds were "ionic", which of course, they are not.  The half-reaction is merely a convenience in order to balance the equation.

    Like any chemical reaction, it's all about the energy of the species involved and the fact that the products are at a lower energy than the reactants.

  • Anonymous
    10 months ago

    Mn gives up two electrons, one to each of the two MnO4 groups.

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