Write a balanced equation for the following reduction-oxidation reaction.

SO3(2-) + MnO4(-) --- SO4(2–) + Mn(2+)

numbers in parenthesis means the charge

S goes from +4 on the left to +6 on the right.

Mn goes from +7 on the left to +2 on the right.

To write a balanced equation for a reduction-oxidation (redox) reaction, you need to balance both the elements and the charges on each side of the equation. Here's how to balance the given redox reaction:

1. Identify the elements present and their oxidation states:
- In SO3(2-), the oxidation state of sulfur is +6 and oxygen is -2.
- In MnO4(-), the oxidation state of manganese is +7 and oxygen is -2.
- In SO4(2-), the oxidation state of sulfur is +6 and oxygen is -2.
- In Mn(2+), the oxidation state of manganese is +2.

2. Divide the reaction into two half-reactions: reduction and oxidation.
- Reduction half-reaction: MnO4(-) + 8H+ + 5e- -> Mn(2+) + 4H2O
- Oxidation half-reaction: SO3(2-) + H2O -> SO4(2-) + 2H+ + 2e-

3. Balance the non-hydrogen/non-oxygen elements first.
- In the reduction half-reaction, the manganese is already balanced.
- In the oxidation half-reaction, the sulfur is already balanced.

4. Balance the oxygen atoms by adding water (H2O) molecules.
- In the reduction half-reaction, there are 4 oxygen atoms on the product side. Add 4 water molecules to balance them.
MnO4(-) + 8H+ + 5e- -> Mn(2+) + 4H2O
- In the oxidation half-reaction, there are 3 oxygen atoms on the product side. Add 3 water molecules to balance them.
SO3(2-) + H2O -> SO4(2-) + 2H+ + 2e-

5. Balance the hydrogen atoms by adding hydrogen ions (H+).
- In the reduction half-reaction, there are 8 hydrogen ions on the reactant side. Add 8 hydrogen ions to balance them.
MnO4(-) + 8H+ + 5e- -> Mn(2+) + 4H2O
- In the oxidation half-reaction, there are 2 hydrogen ions on the product side. Add 2 hydrogen ions to balance them.
SO3(2-) + H2O -> SO4(2-) + 2H+ + 2e-

6. Balance the charges by adding electrons.
- In the reduction half-reaction, there are 5 electrons on the reactant side. Add 5 electrons to balance the charge.
MnO4(-) + 8H+ + 5e- -> Mn(2+) + 4H2O
- In the oxidation half-reaction, there are 2 electrons on the product side. Add 2 electrons to balance the charge.
SO3(2-) + H2O -> SO4(2-) + 2H+ + 2e-

7. Finally, multiply the two half-reactions by appropriate coefficients, if needed, to balance the number of electrons transferred.
- Multiply the oxidation half-reaction by 5 to balance the electrons in both half-reactions:
5SO3(2-) + 5H2O -> 5SO4(2-) + 10H+ + 10e-
- The final balanced equation for the redox reaction is:
5SO3(2-) + MnO4(-) + 5H2O -> 5SO4(2-) + Mn(2+) + 10H+ + 10e-