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)?
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.
Complete and balance the reaction, Mno4-(aq) + C2O4- = MnO2(s) + Co3^2-(aq)
To balance the equation and ensure that atoms and charges are balanced, we need to add OH^- ions to both sides of the equation for every H+ ion present. In this case, we see that there are 4 H+ ions on the left-hand side of the equation (from the MnO4^- and H2O), so we need to add 4 OH^- ions to the right-hand side to balance the equation.
Starting with 2 H2O(l), we can convert them to H+ and OH^- ions. Each water molecule dissociates into one H+ ion and one OH^- ion. So in 2 water molecules, we have a total of 2 H+ ions and 2 OH^- ions.
Now, we need to add an additional 2 OH^- ions to the right-hand side to account for the remaining 2 H+ ions. Therefore, we end up with a total of 4 OH^- ions on the right-hand side of the equation: 2 from the water molecules and an additional 2 to balance the equation.
So, 2H2O(l) would become 4OH^-(aq) in order to balance the equation.