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.
a. Atoms:
H atoms. 4 on left. 4 on right
O atoms. 6 on left . 6 on right.
Mn atoms. 1 on left. 1 on right.
atoms balance.

b. Charge:
-4 on left; -4 on right.
charge balances.

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.

To balance the given half-reaction in terms of atoms, follow these steps:

Step 1: Write the equation:

MnO4^-(aq) -> MnO2 (s)

Step 2: Determine the oxidation state of Mn on both sides of the equation.

On the left side, Mn is in the +7 oxidation state, and on the right side, Mn is in the +4 oxidation state.

Step 3: Balance the change in oxidation state by adding electrons to the appropriate side of the equation.

Since the oxidation state of Mn decreases from +7 to +4, you need to add 3e^- to the left side of the equation:

MnO4^- + 3e^- -> MnO2

Step 4: Check the overall charge on each side of the equation.

On the left side, the charge is -1 from the MnO4^- ion, and on the right side, the charge is neutral. Since we are assuming a neutral or basic solution, we need to balance the charge by adding OH^- ions to the right side of the equation:

MnO4^- + 3e^- -> MnO2 + 4OH^-

Step 5: Balance the hydrogen atoms by adding water molecules to the appropriate side of the equation.

Since there are no hydrogen atoms on either side, we do not need to add water molecules.

Step 6: Check that the final balanced equation satisfies the conditions.

a) Atoms: Count the number of atoms of each element on both sides of the equation.

H atoms: 0 on the left, 0 on the right (already balanced)
O atoms: 4 on the left (from MnO4^-), 7 on the right (from 4OH^-), so add 3H2O to the left side to balance the O atoms:

MnO4^- + 3e^- + 3H2O -> MnO2 + 4OH^-

Mn atoms: 1 on the left, 1 on the right (already balanced)

b) Charge: Check the total charge on both sides of the equation.

On the left side, the charge is -1 from MnO4^-. On the right side, the charge is -1 from the 4OH^- ions. Therefore, the charges balance.

c) Change in oxidation state: Verify that the oxidation state of Mn changes from +7 to +4.

On the left side, the oxidation state of Mn is +7, and on the right side, it is +4. The change in oxidation state is consistent with the balanced equation.

Therefore, the balanced half-reaction in terms of atoms is:

MnO4^- + 3e^- + 3H2O -> MnO2 + 4OH^-