Step 1: Write the skeletal ionic equation: $\text{MnO}_4^-(\text{aq}) + \text{I}^-(\text{aq}) \rightarrow \text{MnO}_2(\text{s}) + \text{I}_2(\text{s})$

Step 2: Write the half-reactions. Oxidation half: $\text{I}^-(\text{aq}) \rightarrow \text{I}_2(\text{s})$. Reduction half: $\text{MnO}_4^-(\text{aq}) \rightarrow \text{MnO}_2(\text{s})$

Step 3: Balance atoms in the oxidation half-reaction: $2\text{I}^-(\text{aq}) \rightarrow \text{I}_2(\text{s})$. For the reduction half, balance O with $\text{H}_2\text{O}$ and H with $\text{H}^+$: $\text{MnO}_4^-(\text{aq}) + 4\text{H}^+(\text{aq}) \rightarrow \text{MnO}_2(\text{s}) + 2\text{H}_2\text{O}(\text{l})$

Step 4: Convert to basic medium by adding $4\text{OH}^-$ to both sides of the reduction half-reaction: $\text{MnO}_4^-(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) \rightarrow \text{MnO}_2(\text{s}) + 2\text{H}_2\text{O}(\text{l}) + 4\text{OH}^-(\text{aq})$. Simplifying water gives: $\text{MnO}_4^-(\text{aq}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{MnO}_2(\text{s}) + 4\text{OH}^-(\text{aq})$

Step 5: Balance charges with electrons. Oxidation: $2\text{I}^-(\text{aq}) \rightarrow \text{I}_2(\text{s}) + 2\text{e}^-$. Reduction: $\text{MnO}_4^-(\text{aq}) + 2\text{H}_2\text{O}(\text{l}) + 3\text{e}^- \rightarrow \text{MnO}_2(\text{s}) + 4\text{OH}^-(\text{aq})$. Multiply oxidation by 3 and reduction by 2 to equalize electrons.

Step 6: Add the half-reactions and cancel electrons to get the final balanced equation: $6\text{I}^-(\text{aq}) + 2\text{MnO}_4^-(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) \rightarrow 3\text{I}_2(\text{s}) + 2\text{MnO}_2(\text{s}) + 8\text{OH}^-(\text{aq})$