MnO4 + SO3² MnO2 + SO4²

Answer 1

1: Separate into Half Equations:
MnO4{-} + _ = MnO2 + _
SO3{2-} + _ = SO4{2-} + _

2: Use water to balance O
MnO4{-} + _ = MnO2 + 2 H2O + _
SO3{2-} + H2O + _ = SO4{2-} + _

3: Use H{+} to balance H
MnO4{-} + 4 H{+} + _ = MnO2 + 2 H2O + _
SO3{2-} + H2O + _ = SO4{2-} + 2 H{+} + _

4: Neutralise H{+} with OH{-}
MnO4{-} + 4 H2O + _ = MnO2 + 2 H2O + 4 OH{-}
SO3{2-} + H2O + 2 OH{-} = SO4{2-} + 2 H2O + _

5: Cancel superfluous water
MnO4{-} + 2 H2O + _ = MnO2 + 4 OH{-}
SO3{2-} + 2 OH{-} = SO4{2-} + H2O + _

6: Balance charges with electrons
MnO4{-} + 2 H2O + 3 e = MnO2 + 4 OH{-}
SO3{2-} + 2 OH{-} = SO4{2-} + H2O + 2 e

7: Balance electron counts and combine
2 MnO4{-} + 3 SO3{2-} + 4 H2O + 6 OH{-} = 2 MnO2 + 3 SO4{2-} + 3 H2O + 8 OH{-}

8: Cancel superfluous molecules and ions
2 MnO4{-} + 3 SO3{2-} + H2O = 2 MnO2 + 3 SO4{2-} + 2 OH{-}

9: ... Eight. That's eight.
Balancing a red/ox reaction in basic solution in eight easy steps.