MnO4^-(aq) + H20(l) ==> MnO2 + OH^-

Question

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)?

Answer 1

The equation
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.
Everything OK.

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.

Answer 2

Complete and balance the reaction, Mno4-(aq) + C2O4- = MnO2(s) + Co3^2-(aq)