To complete and balance the reaction between $$Mn^{2+}$$ and $$PbO_2$$ that results in $$MnO_4^-$$ and $$H_2O$$, we need to take into account both the redox aspect of this reaction and the need for balancing atoms and charges. This reaction occurs in acidic solution, so we'll be using $$H^+$$ ions to balance hydrogen atoms and electrons to balance the charges.

The half-reactions are as follows:

Oxidation half-reaction: The manganese goes from a +2 oxidation state in $$Mn^{2+}$$ to a +7 oxidation state in $$MnO_4^-$$.

$$Mn^{2+} \to MnO_4^-$$

We add 4 $$H_2O$$ on the right side to balance the oxygen, giving 8 $$H^+$$ on the left side to balance the hydrogen. Then, we add electrons to balance the charge:

$$Mn^{2+} + 8H^+ + 4H_2O \to MnO_4^- + 5e^-$$

Reduction half-reaction: The lead goes from a +4 oxidation state in $$PbO_2$$ to a +2 oxidation state in the product not mentioned explicitly (since $$PbO_2$$ is being reduced, we can assume it becomes $$Pb^{2+}$$ in acidic solution).

$$PbO_2 + 4H^+ + 2e^- \to Pb^{2+} + 2H_2O$$

Now, to balance the electrons in both half-reactions, we need to equalize the number of electrons. Since the oxidation half-reaction produces 5 electrons and the reduction half-reaction consumes 2 electrons, we multiply the reduction half-reaction by 5 and the oxidation half-reaction by 2 to get 10 electrons in both:

$$2(Mn^{2+} + 8H^+ + 4H_2O \to MnO_4^- + 5e^-)$$

$$5(PbO_2 + 4H^+ + 2e^- \to Pb^{2+} + 2H_2O)$$

By multiplying, we have:

$$2Mn^{2+} + 16H^+ + 8H_2O \to 2MnO_4^- + 10e^-$$

$$5PbO_2 + 20H^+ + 10e^- \to 5Pb^{2+} + 10H_2O$$

Combining and simplifying these equations, while also canceling out electrons and what's common on both sides, we find:

$$2Mn^{2+} + 5PbO_2 + 16H^+ \to 2MnO_4^- + 5Pb^{2+} + 8H_2O$$

In this balanced equation, both the charges and the atoms are balanced, giving a clear representation of the chemical reaction that occurs.