posted by Steve .
MnO4^-(aq) + H20(l) ==> MnO2 + OH^-
net charg is -1 +7 (-8) ==> 4(-4)
Manganese is reduced
MnO4^- +3e- ==> MnO2
H2) is the oxidizing agent in a basic solution
Mno4^- + H2O(l) --> MnO2(s) + OH^-
Add on OH^- to both sides of the equation for every H+ ion present
Cancel excess water
Check that atoms and charges balance
How did you get 2H2O(l) ==> 4 OH^-(aq)?
MnO4^- + H2O ==> MnO2 + OH^-
doesn't balance. Mn is OK. H doesn't balance and neither does O. Placing a 2 in front of H2O makes it balance.
I don't do it exactly as you did.
Here is how I do redox equations.
MnO4^- ==> MnO2 in basic soln.
a. oxidation state of Mn on left is +7 and on right is +4. So add 3 electrons to the left to balance the change in oxidation state.
MnO4^- + 3e ==> MnO2
b. Count the charges. On the left is -4 and on the right is zero.
b1. If acid solution, add H^+ to balance the charge. That doesn't apply in this case.
b2. If basic solution, add OH^- to balance the charge.
MnO4^- + 3e ==> MnO2 + 4OH^-
c. Now add water (not always needed) to balance the OH^- added.
MnO4^- + 3e + 2H2O ==>MnO2 + 4OH^-
d. check it.
Mn is ok.
6 O on left and right.
4 H on left and right.
4- charge on left and right.
Some profs use this method but delete part b2 and balance everything as if all equations are acidic. When everything is balanced, they add OH^- to each H^+ and re-balance. I don't like to do that because it's two balancing steps instead of one.
Complete and balance the reaction, Mno4-(aq) + C2O4- = MnO2(s) + Co3^2-(aq)