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I'm having a little trouble determining oxidation numbers for elements in a redox reaction

KMnO4+HCl--> MnCl2+Cl2+H2O+KCl

So far what I have for the reactants side is: K= +1 Mn= +7 O4= -2 H= +1 Cl= -1

And for the products side I have:
Mn= +2 Cl2= -1 Cl2(diatomic)= 0 H2=+2 O= -2 K= +1 Cl= -1

Please tell me if these are right, and I wasn't sure about H2, but I'm guessing it could either be +2 or +1, since it was H20 I think it would be +1, but I am not entirely sure :(

  • Chemistry - ,

    You are right. It is always better to determine the oxidation state of EACH atom; therefore, H in H2O is +1 EACH and +2 total. O is -2 for the one. In that vein, Cl in MnCl2 is -1 each and -2 total.

  • Chemistry - ,

    Ah, ok, so another question I have is how would you determine which elements are oxidized and reduced?

  • Chemistry - ,

    You remember the definitions. Oxidation is the loss of electrons; reduction is the gain of electrons.
    So Mn changes from +7 to +2 which is a gain of electrons. That means MnO4^- is reduced to Mn^2+.
    Cl changes from -1 in HCl to zero in Cl2; that is a loss of electrons so Cl^- is oxidized (that's for Cl in HCl going to Cl2. The other Cl ions in HCl that go to MnCl2 are spectator ions since they don't change oxidation state.

  • Chemistry - ,

    Thanks for your help! Makes lots of sense :)

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