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 $$ \mathrm{H}_{6} \mathrm{IO}_{6}+5 \mathrm{H}^{+}+2 \mathrm{I}^{-}+3 \mathrm{e}^{-} \longrightarrow $$ $$ \frac{1}{2} \mathrm{I}_{2}+4 \mathrm{H}_{2} \mathrm{O}=2 \mathrm{HIO} $$

Step by step solution

01

Identify the reactions and the corresponding \(E_{i}^\circ\) and \(n_{i}\)

Here are four half-cell reactions: \[\mathrm{H}_{5} \mathrm{IO}_{6} + \mathrm{H}^{+} + 2\mathrm{e}^- \longrightarrow \mathrm{IO}_{3}^- + 3\mathrm{H}_{2} \mathrm{O} \quad (E_1^\circ = 1.60V, n_1 = 2)\] \[\mathrm{IO}_{3}^- + 6\mathrm{H}^{+} + 5\mathrm{e}^- \longrightarrow \frac{1}{2}\mathrm{I}_{2} + 3\mathrm{H}_{2}\mathrm{O} \quad (E_2^\circ = 1.19V, n_2 = 5)\] \[2\mathrm{HIO}+2\mathrm{H}^{+} + 2\mathrm{e}^- \longrightarrow \mathrm{I}_{2} + 2\mathrm{H}_{2}\mathrm{O} \quad (E_3^\circ = 1.45V, n_3 = 2)\] \[\mathrm{I}_{2}+2\mathrm{e}^- \longrightarrow 2\mathrm{I}^- \quad (E_4^\circ = 0.535V, n_4 = 2)\]

02

Perform the calculations

Plugging in the values into the equation: \[E^\circ = \frac{\sum n_{i}E_{i}^\circ}{\sum n_{i}} = \frac{n_1E_1^\circ + n_2E_2^\circ + n_3E_3^\circ + n_4E_4^\circ}{n_1 + n_2 + n_3 + n_4}\] \[E^\circ = \frac{(2)(1.60V) + (5)(1.19V) + (2)(1.45V) + (2)(0.535V)}{2 + 5 + 2 + 2}\] Calculating this gives the value of \(E^\circ\).

03

Simplify the result

Simplify the expression above to get the final answer.

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