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 $$ \operatorname{CuI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)+\mathrm{I}^{-}(a q) $$

Short answer

The standard cell potential \(\mathscr{E}^{\circ}\) for the given half-reaction is approximately 0.61 V.

Step-by-step solution

Write down the Nernst equation

The Nernst equation is given by: \[ \mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF} \ln{Q} \] where \(\mathscr{E}\) is the cell potential, \(\mathscr{E}^{\circ}\) is the standard cell potential, \(R\) is the gas constant, \(T\) is the temperature, \(n\) is the number of electrons transferred in the reaction, \(F\) is the Faraday constant, and \(Q\) is the reaction quotient.

Relate the solubility product to the ion product

The solubility product for CuI(s) is given as: \[ K_\text{sp} = [Cu^+][I^-] = 1.1 \times 10^{-12} \] Under standard conditions, the concentrations of the Cu⁺ and I⁻ ions are equal. The ion product is equal to the solubility product under these conditions: \[ Q = [Cu^+][I^-] = 1.1 \times 10^{-12} \]

Calculate the number of electrons transferred

In the reaction given, the number of electrons transferred is 1: \[ \operatorname{CuI}(s) + \mathrm{e}^− \longrightarrow \mathrm{Cu}(s) + \mathrm{I}^-(aq) \] So, \(n = 1\).

Set the temperature to 298 K and find the value of RT/F

Since we are not given a specific temperature, we assume standard temperature, which is 298 K. Therefore, \(T = 298\) K. The value of the gas constant \(R\) is \(8.314 \text{ J mol}^{-1}\text{ K}^{-1}\), and the value of the Faraday constant \(F\) is \(96485 \text{ C mol}^{-1}\). We can calculate the value of \(\frac{RT}{nF}\): \[ \frac{RT}{nF} = \frac{8.314 \cdot 298}{1 \cdot 96485} = 0.0257 \text{ V} \]

Calculate the standard cell potential

We can now calculate the value of \(\mathscr{E}^{\circ}\) using the Nernst equation, which relates the cell potential, the reaction quotient, and the standard cell potential. Since the cell potential \(\mathscr{E}\) is zero under standard conditions, we have: \[ 0 = \mathscr{E}^{\circ} - 0.0257 \cdot \ln{1.1 \times 10^{-12}} \] Solving for \(\mathscr{E}^{\circ}\), we get: \[ \mathscr{E}^{\circ} = 0.0257 \cdot \ln{1.1 \times 10^{-12}} = 0.61 \text{ V} \] So, the standard cell potential for the half-reaction is approximately 0.61 V.

Key concepts

Nernst Equation

Understanding the Nernst equation is pivotal when studying electrochemistry, as it allows us to calculate the electrical potential of a reaction under non-standard conditions. In its essence, the Nernst equation reflects how the potential will change in response to the concentration of the reacting species.

The equation is expressed as: \[\mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF} \ln{Q}\]Here, \(\mathscr{E}\) symbolizes the cell potential, while \(\mathscr{E}^{\circ}\) denotes the standard cell potential. The terms \(R\), \(T\), \(n\), and \(F\) represent the gas constant, temperature in Kelvin, the number of moles of electrons transferred in the reaction, and the Faraday constant respectively. The reaction quotient \(Q\) provides a snapshot of the reaction at a given point in time, comparing the concentration of products to reactants.

In the application for solubility product calculations, the Nernst equation elegantly marries electrochemical and solution equilibrium concepts. This marriage allows predictions on how the cell potential will adjust in the event of changes in solution concentration. By closely examining the steps given, it's clear that the educator has put significant effort into elucidating the relationship between solubility product and the Nernst equation to foster student understanding.

Standard Cell Potential

The standard cell potential \(\mathscr{E}^{\circ}\) is a key piece of information for any electrochemical cell, indicating the voltage generated under standard conditions. These conditions usually imply all solutes at 1 M concentration, all gases at 1 bar pressure, and all substances at 25°C or 298 K.

To calculate \(\mathscr{E}^{\circ}\) for a particular half-reaction, we rely on knowing the cell potential under standard conditions. The given exercise takes advantage of this concept, making use of the relation between solubility product and the standard cell potential to find its value for the dissolution of copper(I) iodide.

The solution walking through the computation for this half-reaction demonstrates how alterations in solubility inform the standard cell potential. This relationship has practical ramifications in various scientific fields, including material science, where the purity of the materials often hinges on their solubility characteristics.

Reaction Quotient

The reaction quotient, \(Q\), is a critical concept that quantifies the state of a reaction at a given moment. Unlike the equilibrium constant which applies when a reaction is at equilibrium, \(Q\) can be used to determine the direction in which the reaction will proceed to reach equilibrium.

It is calculated similarly to an equilibrium constant:\[ Q = \frac{[products]}{[reactants]} \]In the context of the exercise, the reaction quotient is equated to the solubility product (\(K_{sp}\)) of copper(I) iodide. This is because the exercise assumes that the system is in a state of dynamic equilibrium where the solid phase is dissolving while the dissolved ions are precipitating at the same rates.

By correlating \(Q\) with the solubility product, an implicit understanding is achieved that, under standard conditions, the concentrations of ions in the solution determine the forward drive or the 'push' of the reaction. The exercise uses this relationship to solve for the standard cell potential, providing a straightforward example of how solubility equilibrium and electrochemical potential intersect.