Question

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

Short answer

The standard electrode potential \((\mathscr{E}^{\circ})\) for the given half-reaction involving Copper Iodide (CuI) is approximately -0.152 V.

Step-by-step solution

Write the solubility product expression

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, \(K_{sp} = 1.1 \times 10^{-12}\).

Identify the half-cell reaction and write its cell potential expression

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.

Relate \(K_{sp}\) to the half-cell reaction

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: $$0=\mathscr{E}^{\circ}-\frac{RT}{F} \ln [\mathrm{I}^{-}]$$

Solve for \(\mathscr{E}^{\circ}\)

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: $$\mathscr{E}^{\circ}=\frac{8.314 \times 298}{96485} \ln \sqrt{1.1 \times 10^{-12}}$$

Calculate the value of \(\mathscr{E}^{\circ}\)

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.

Key concepts

Solubility Product (Ksp)

The solubility product constant, often denoted as \( K_{sp} \), is a special type of equilibrium constant that applies to the dissolution of sparingly soluble salts. When salts dissolve in water, they dissociate into their constituent ions. The \( K_{sp} \) specifically signifies the product of the concentrations of these ions each raised to the power of their stoichiometric coefficients.
For example, for the dissolution of copper iodide, \( \operatorname{CuI}(s) \), the equilibrium reaction is:

• \( \operatorname{CuI}(s) \rightleftharpoons \mathrm{Cu}^{+}(aq) + \mathrm{I}^{-}(aq) \)

Here, the solubility product expression is given by \( K_{sp} = [\mathrm{Cu}^{+}][\mathrm{I}^{-}] \).
In certain problems, like the one in the original exercise, you might need to solve for the concentration of one of the ions using the given \( K_{sp} \).
By acknowledging that in solution \( [\mathrm{Cu}^{+}] = [\mathrm{I}^{-}] \) due to the 1:1 stoichiometry, we can substitute and solve for the ion concentrations, frequently ending with \( [\mathrm{I}^{-}] = \sqrt{K_{sp}} \) because of the squared relationship.
This is crucial when combining knowledge of solubility with electrochemical potential tasks.

Standard Electrode Potential

The standard electrode potential, symbolized as \( \mathscr{E}^{\circ} \), is a measure of the tendency of a chemical species to be reduced, measured in volts under standard conditions. These conditions include solutions at 1 M concentration, gases at 1 atm pressure, and a temperature of 25 degrees Celsius (298 K).
When we talk about half-reactions, \( \mathscr{E}^{\circ} \) indicates the voltage difference between a given half-cell and the standard hydrogen electrode, which is defined at 0 V.
In the context of the CuI half-reaction:

• \( \mathrm{CuI}(s) + \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) + \mathrm{I}^{-}(aq) \)

The \( \mathscr{E}^{\circ} \) reflects how readily the copper(I) iodide can capture electrons to form solid copper and iodide ions.
Thus, a negative \( \mathscr{E}^{\circ} \) value, like the calculated -0.152 V, suggests that under standard conditions, the reaction is not spontaneous. This information, along with other electrochemical details, helps us determine conditions for practical applications, such as in galvanic cells.

Nernst Equation

The Nernst Equation provides a relationship between the standard electrode potential and the actual cell potential at any given conditions. This equation is crucial in electrochemistry, as it allows us to calculate the cell potential for conditions that are not standard.
It is given by:

• \( \mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF} \ln Q \)

Where:

• \( \mathscr{E} \) is the cell potential under non-standard conditions,
• \( R \) is the universal gas constant (8.314 J/mol K),
• \( T \) is the temperature in Kelvin,
• \( n \) is the number of moles of electrons transferred,
• \( F \) is the Faraday constant (96485 C/mol),
• \( Q \) is the reaction quotient, which is the ratio of product concentrations to reactant concentrations.

In equilibrium situations for half-cell reactions, the cell potential \( \mathscr{E} \) is zero, simplifying the Nernst equation for solving unknowns.
This equation was transformed during the step-by-step solution to incorporate the relationship between \( K_{sp} \) and the concentration of iodide ions, enabling the calculation of \( \mathscr{E}^{\circ} \) from equilibrium conditions. This all brings coherence between solubility products and electrochemical reactions.