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Chapter 18: Problem 92

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

Short Answer

Expert verified
The standard cell potential for the given half-reaction, \(\operatorname{CuI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)+\mathrm{I}^{-}(a q)\), is approximately 0.713 V.

Step by step solution

01

Identify the given information

We are given the following information: - The solubility product of CuI, \(K_{sp} = 1.1 \times 10^{-12}\) - The half-reaction: \(\operatorname{CuI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)+\mathrm{I}^{-}(a q)\) - Number of Moles of electrons transferred (n) = 1
02

Identify the constants

We need two constants to solve this problem. - Gas constant (R) = 8.314 J/(mol K) - Faraday's constant (F) = 96485 C/mol - Temperature at standard condition (T) = 298.15 K
03

Calculate the standard cell potential

Use the formula \(\mathscr{E}^{\circ}=-\frac{RT}{nF} \ln K_{sp}\) and substitute the given values and constants to calculate the standard cell potential. $$ \mathscr{E}^{\circ}=-\frac{8.314 \mathrm{J}/(\mathrm{mol} \cdot \mathrm{K})(298.15 \mathrm{K})}{(1)(96485 \mathrm{C}/\mathrm{mol})} \ln(1.1 \times 10^{-12}) $$ Now, go ahead and compute the standard cell potential. $$ \mathscr{E}^{\circ} \approx 0.713 \ \mathrm{V} $$ Therefore, the standard cell potential for the given half-reaction is 0.713 V.

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Most popular questions from this chapter

When balancing equations in Chapter 3, we did not mention that reactions must be charge balanced as well as mass

Consider the following half-reactions: $$ \begin{aligned} \mathrm{IrCl}_{6}{ }^{3-}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Ir}+6 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.77 \mathrm{~V} \\ \mathrm{PtCl}_{4}{ }^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.73 \mathrm{~V} \\ \mathrm{PdCl}_{4}{ }^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.62 \mathrm{~V} \end{aligned} $$ A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant \(1.0 M\) in chloride ion and \(0.020 M\) in each complex ion. Is it feasible to separate the three metals from this solution by electrolysis? (Assume that \(99 \%\) of a metal must be plated out before another metal begins to plate out.)

In the electrolysis of an aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\), what reactions occur at the anode and the cathode (assuming standard conditions)?

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 M\) separated by a porous disk from an aluminum metal electrode. a. What is the potential of this cell at \(25^{\circ} \mathrm{C}\) if the aluminum electrode is placed in a solution in which \(\left[\mathrm{Al}^{3+}\right]=7.2 \times\) \(10^{-3} M ?\) b. When the aluminum electrode is placed in a certain solution in which \(\left[\mathrm{Al}^{3+}\right]\) is unknown, the measured cell potential at \(25^{\circ} \mathrm{C}\) is \(1.62 \mathrm{~V}\). Calculate \(\left[\mathrm{Al}^{3+}\right]\) in the unknown solution. (Assume \(\mathrm{Al}\) is oxidized.)

Given the following two standard reduction potentials, $$ \begin{array}{ll} \mathrm{M}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{M} & \mathscr{E}^{\circ}=-0.10 \mathrm{~V} \\ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} & \mathscr{E}^{\circ}=-0.50 \mathrm{~V} \end{array} $$ solve for the standard reduction potential of the half-reaction $$ \mathrm{M}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{M}^{2+} $$ (Hint: You must use the extensive property \(\Delta G^{\circ}\) to determine the standard reduction potential.)
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