The solubility product for \(\operatorname{CuI}(s)\) is \(1.1 \times 10^{-12}\) . Calculate the value of \(\mathscr{E}^{\circ}\) for the half-reaction $$\mathrm{CuI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)+\mathrm{I}^{-}(a q)$$

First, we need to write down the solubility product expression for CuI as per the given solubility constant. The dissociation of CuI in water can be represented as: $$\mathrm{CuI}(s) \rightleftharpoons \mathrm{Cu}^{+}(a q)+\mathrm{I}^{-}(a q)$$ The solubility product expression for this reaction is: $$K_{sp} = [\mathrm{Cu}^{+}][\mathrm{I}^{-}]$$ Given, \(K_{sp} = 1.1 \times 10^{-12}\).

The half-cell reaction given in the exercise is: $$\mathrm{CuI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)+\mathrm{I}^{-}(a q)$$ The cell potential of this half-cell reaction can be written as per the Nernst equation, modified for a half-reaction: $$\mathscr{E}=\mathscr{E}^{\circ}-\frac{RT}{nF} \ln Q$$ Here, \(\mathscr{E}\) is the cell potential, \(\mathscr{E}^{\circ}\) is the standard electrode potential, R is the gas constant, T is the temperature, n is the number of moles of electrons transferred, F is the Faraday constant, and Q is the reaction quotient.

In the given half-cell reaction, there is no \(\mathrm{Cu}^{+}\) ion, but we have the \(\mathrm{I}^{-}\) ion concentration. Our goal is to express \(\mathscr{E}^{\circ}\) in terms of given \(K_{sp}\). For the half-cell reaction, we can rewrite the cell potential expression like this: $$\mathscr{E}=\mathscr{E}^{\circ}-\frac{RT}{nF} \ln [\mathrm{I}^{-}]$$ At equilibrium, the cell potential \(\mathscr{E}\) is 0, and the reaction quotient Q becomes K (equilibrium constant). From the solubility product expression, we know \(K_{sp}=[\mathrm{Cu}^{+}][\mathrm{I}^{-}]\). Since in the half-cell reaction, we have only one mole of electrons transferred \((n=1)\), the modified Nernst equation becomes: $$0=\mathscr{E}^{\circ}-\frac{RT}{F} \ln [\mathrm{I}^{-}]$$

Let's rearrange the equation to solve for \(\mathscr{E}^{\circ}\): $$\mathscr{E}^{\circ}=\frac{RT}{F} \ln [\mathrm{I}^{-}]$$ Now, relieving the relation between \(K_{sp}\) and \([\mathrm{I}^{-}]\): $$[\mathrm{I}^{-}] = \frac{K_{sp}}{[\mathrm{Cu}^{+}]}$$ Since in the half-cell reaction, we have one mole of \(\mathrm{CuI}\) dissociating into one mole of \(\mathrm{I}^{-}\) and one mole of \(\mathrm{Cu}^{+}\), we can assume \([\mathrm{Cu}^{+}] = [\mathrm{I}^{-}]\). Therefore, $$[\mathrm{I}^{-}] = \sqrt{K_{sp}} = \sqrt{1.1 \times 10^{-12}}$$ Plugging the values of R (8.314 J/mol K), T (298 K, assuming room temperature), F (96485 C/mol), and \([\mathrm{I}^{-}]\) into the equation, we get: $$\mathscr{E}^{\circ}=\frac{8.314 \times 298}{96485} \ln \sqrt{1.1 \times 10^{-12}}$$

Finally, calculate the value of \(\mathscr{E}^{\circ}\): $$\mathscr{E}^{\circ} \approx -0.152\,\text{V}$$ Hence, the standard electrode potential \((\mathscr{E}^{\circ})\) for the given half-reaction is approximately -0.152 V.