Reduction Half Reaction:
\[ 2\text{MnO}_4^- \rightarrow 2\text{MnO}_2 \] \[ 2\text{MnO}_4^- + 4\text{H}_2\text{O} + 6e^- \rightarrow 2\text{MnO}_2 + 8\text{OH}^- \]
Oxidation Half Reaction:
\[ 2\text{I}^- \rightarrow \text{I}_2 + 2e^- \] \[ 6\text{I}^- \rightarrow 3\text{I}_2 + 6e^- \]
Adding the oxidation half and reduction half, we get the net reaction as:
\[ 2\text{MnO}_4^- + 6\text{I}^- + 4\text{H}_2\text{O} \rightarrow 3\text{I}_2 + 2\text{MnO}_2 + 8\text{OH}^- \]
Thus, \( z = 8 \).
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to: