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 :(
4 answers
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.
Ah, ok, so another question I have is how would you determine which elements are oxidized and reduced?
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.
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.
Thanks for your help! Makes lots of sense :)