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 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

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

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

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.

Key concepts

Solubility Product

The solubility product, denoted as \( K_{sp} \), is a constant that helps us understand the solubility of slightly soluble ionic compounds in a solution. Specifically, it represents the maximum product of the molar concentrations of the ions involved, each raised to the power of its stoichiometric coefficient, that can exist in equilibrium with the solid at a given temperature.
For example, in the exercise, the solubility product for \( \operatorname{CuI}(s) \) is given as \( 1.1 \times 10^{-12} \).
This tiny value indicates that copper(I) iodide is only sparingly soluble in water.

• The smaller the \( K_{sp} \), the less soluble the compound is.
• The solubility product can be used to predict whether a precipitate will form under certain conditions.

To calculate the dissolution of a compound, use \( K_{sp} = [Cu^+][I^-] \), where \( [Cu^+] \) and \( [I^-] \) are the ionic concentrations in solution after dissolution.

Standard Cell Potential

The standard cell potential, denoted as \( \mathscr{E}^{\circ} \), is a measure of the electromotive force of a cell operating under standard conditions.
In electrochemistry, it's crucial for understanding the voltage produced by a galvanic cell and can also indicate the direction of electron flow.

• For standard conditions, the concentrations of any solutions are 1 mol/L, the pressure of any gases is 1 atm, and the temperature is 25°C.
• The standard cell potential is calculated using Nernst equation principles, as we did in the exercise with the formula \( \mathscr{E}^{\circ} = -\frac{RT}{nF} \ln K_{sp} \).

This formula links thermodynamics and electrochemistry by utilizing the gas constant \( R \), the Faraday constant \( F \), and the temperature \( T \), providing a deep understanding of how chemical equilibria and electric potential interrelate.

Half-Reaction

A half-reaction is a way of expressing the individual components of a redox (reduction-oxidation) reaction.
It splits the full reaction into two parts— one for oxidation and one for reduction.

• In the provided exercise, the half-reaction \( \operatorname{CuI}(s) + \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) + \mathrm{I}^{-}(aq) \) represents the reduction process.
• It shows the gain of an electron \( (\mathrm{e}^-)\) by \( \operatorname{CuI} \) forming copper \( (\mathrm{Cu}) \) and iodide ion \( (\mathrm{I}^{-}) \).

Analyzing half-reactions is essential as it helps in balancing redox reactions, understanding electron flow, and predicting the feasibility of redox processes under certain conditions.

Thermodynamics

Thermodynamics in electrochemistry involves studying the energy changes that accompany chemical reactions and the conversion of chemical energy into electrical energy.

• Key principles involve the concepts of energy conservation and entropy.
  
• In the context of electrochemistry, thermodynamics helps determine whether a process occurs spontaneously.
• The relationship between the Gibbs free energy \( \Delta G \) and the cell potential \( \mathscr{E} \) is given by the equation \( \Delta G = -nF\mathscr{E} \).

For the exercise, calculating the standard cell potential is a thermodynamic consideration, signaling the direction and magnitude of energy change, which in turn suggests whether the copper-iodine reaction is favorable or not under standard conditions. Studying such interactions provides essential insights into how cells generate electrical energy from spontaneous chemical changes.