Show that for a combination of half-cell reactions that produce a standard reduction potential for a half-cell that is not directly observable, the standard reduction potential is
$$E^{\circ}=\frac{\sum n_{i} E_{i}^{\circ}}{\sum n_{i}}$$
where $n_{i}$ is the number of electrons in each half-reaction of potential $E_{i}^{\circ} .$ Use the following half-reactions:
$$
\begin{array}{c}
\mathrm{H}_{5} \mathrm{IO}_{6}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{IO}_{3}^{-}(\mathrm{aq})+ \\
3 \mathrm{H}_{2} \mathrm{O}(1) \quad E^{\circ}=1.60 \mathrm{V} \\
\mathrm{IO}_{3}^{-}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq})+5 \mathrm{e}^{-} \longrightarrow \frac{1}{2} \mathrm{I}_{2}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{O}(1) \\
E^{\circ}=1.19 \mathrm{V} \\
2 \mathrm{HIO}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}_{2}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(1) \\
E^{\circ}=1.45 \mathrm{V} \\
\mathrm{I}_{2}(\mathrm{s})+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}^{-}(\mathrm{aq}) \quad \quad E^{\circ}=0.535 \mathrm{V}
\end{array}
$$
Calculate the standard reduction potential for
$$
\mathrm{H}_{6} \mathrm{IO}_{6}+5 \mathrm{H}^{+}+2 \mathrm{I}^{-}+3 \mathrm{e}^{-} \longrightarrow
$$
$$
\frac{1}{2} \mathrm{I}_{2}+4 \mathrm{H}_{2} \mathrm{O}=2 \mathrm{HIO}
$$