posted by Dan on .
One more question- how do I balance this half-reaction in terms of atoms?
MnO4^-(aq) -> MnO2 (s)
i can't balance the oxygen
I have no idea!
Here is how to do it. It isn't the easiest method (there are easier ones) but this covers areas that you need to know. And this MUST be in basic or neutral solution. Either you didn't post that OR the person giving you the equation didn't provide it.
Step 1. MnO4^- ==> MnO2
Step 2. Determine the value of the oxidation state of Mn. That is +7 on the left and +4 on the right. If you don't understand how I obtained that number you need to repost another question.
Step 3. Add electrons to the appropriate side to balance the change of oxidation state.
MnO4^- +3e ==> MnO2
Step 4. Count up the charge on each side. The left side is -4; the right side is zero. Since this is a basic or neutral solution, add OH^- to the appropriate side to balance the charge.
MnO4^- + 3e ==> MnO2 + 4 OH^-
Step 4. Add water molecules to the appropriate side to balance the H atoms.
2H2O + MnO4^- + 3e ==> MnO2 + 4OH^-
Step 5. Check everything.
H atoms. 4 on left. 4 on right
O atoms. 6 on left . 6 on right.
Mn atoms. 1 on left. 1 on right.
-4 on left; -4 on right.
c. change of oxidation state:
Mn is +7 on left and +4 on right. On left, +7 +(-3) = +4.
Change in oxidation balances.
Repost if you don't understand but tell us exactly the trouble you are having.