Problem 101 - Chemistry Principles and Practice

Question

At \(298 \mathrm{~K}\), the solubility product constant \(\mathrm{for} \mathrm{PbC}_{2} \mathrm{O}_{4}\) is \(8.5 \times 10^{-10}\), and the standard reduction potential of the \(\mathrm{Pb}^{2+}(\) aq \()\) to \(\mathrm{Pb}(\mathrm{s})\) is \(-0.126 \mathrm{~V}\). (a) Find the standard potential of the half-reaction $$ \mathrm{PbC}_{2} \mathrm{O}_{4}(\mathrm{~s})+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}(\mathrm{s})+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(\mathrm{aq}) $$ (Hint: The desired half-reaction is the sum of the equations for the solubility product and the reduction of \(\mathrm{Pb}^{2+}\). Find \(\Delta G^{\circ}\) for these two reactions and add them to find \(\Delta G^{\circ}\) for their sum. Convert the \(\Delta G^{\infty}\) to the potential of the desired half-reaction.) (b) Calculate the potential of the \(\mathrm{Pb} / \mathrm{PbC}_{2} \mathrm{O}_{4}\) electrode in a \(0.025 M\) solution of \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\)

Step-by-Step Solution

Answer

(a) The standard potential is -0.414 V. (b) The electrode potential is -0.433 V.

Step 1: Calculate ΔG° for the Solubility Product

The reaction for the solubility product is: \( \mathrm{PbC}_{2} \mathrm{O}_{4}(\mathrm{s}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq}) + \mathrm{C}_{2} \mathrm{O}_{4}^{2-}(\mathrm{aq}) \), with \( K_{sp} = 8.5 \times 10^{-10} \). We use the formula \( \Delta G^{\circ} = -RT \ln K \): \[ \Delta G_1^{\circ} = - (8.314 \times 10^{-3} \ \mathrm{kJ/mol} \cdot \mathrm{K}) \times 298 \ \mathrm{K} \times \ln(8.5 \times 10^{-10}) \approx 55.5 \ \mathrm{kJ/mol} \]

Step 2: Calculate ΔG° for the Reduction Potential

The reduction half-reaction is \( \mathrm{Pb}^{2+}(\mathrm{aq}) + 2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}(\mathrm{s}) \), with potential \( E^{\circ} = -0.126 \ \mathrm{V} \). We use \( \Delta G^{\circ} = -nFE^{\circ} \), where \( n = 2 \), and \( F = 96485 \ \mathrm{C/mol} \): \[ \Delta G_2^{\circ} = -(2) \times 96485 \ \mathrm{C/mol} \times (-0.126) \ \mathrm{V} \approx 24.3 \ \mathrm{kJ/mol} \]

Step 3: Sum ΔG° Values

For the overall reaction \( \mathrm{PbC}_{2}\mathrm{O}_{4}(\mathrm{s}) + 2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}(\mathrm{s}) + \mathrm{C}_{2}\mathrm{O}_{4}^{2-}(\mathrm{aq}) \), we sum the \( \Delta G^{\circ} \) values: \[ \Delta G^{\circ}_{\text{overall}} = \Delta G_1^{\circ} + \Delta G_2^{\circ} = 55.5 \ \mathrm{kJ/mol} + 24.3 \ \mathrm{kJ/mol} = 79.8 \ \mathrm{kJ/mol} \]

Step 4: Convert ΔG° Overall to Standard Potential

Convert \( \Delta G^{\circ}_{\text{overall}} \) to \( E^{\circ} \) using \( E^{\circ} = -\frac{\Delta G^{\circ}}{nF} \): \[ E^{\circ}_{\text{overall}} = -\frac{79.8 \times 10^{3} \ \mathrm{J/mol}}{2 \times 96485 \ \mathrm{C/mol}} \approx -0.414 \ \mathrm{V} \]

Step 5: Calculate Potential Using the Nernst Equation

Given \( [\mathrm{C}_{2} \mathrm{O}_{4}^{2-}] = 0.025 \ \mathrm{M} \), use the Nernst equation for the potential of the electrode: \[ E = E^{\circ} - \frac{RT}{nF} \ln Q \]Since the overall reaction at equilibrium involves \( Q = [\mathrm{C}_{2} \mathrm{O}_{4}^{2-}] \), then:\[ E = -0.414 \ \mathrm{V} - \frac{8.314 \times 10^{-3} \ \mathrm{kJ/mol} \cdot \mathrm{K} \times 298 \ \mathrm{K}}{2 \times 96485 \ \mathrm{C/mol}} \ln(0.025) \approx -0.433 \ \mathrm{V} \]

Key Concepts

Standard Reduction Potential

The Standard Reduction Potential is a measure of a substance’s ability to gain electrons under standard conditions, and it is measured in volts. This potential tells us how readily a species will be reduced, transferring electrons to reduce ions into a solid or another form. The more positive the reduction potential, the greater the ability of the substance to act as an oxidizing agent. For \( \mathrm{Pb}^{2+}(\text{aq}) \), the standard reduction potential is \(-0.126\ \mathrm{V}\). This negative value implies it is not readily reduced compared to other substances with positive values. Knowing the reduction potential helps us evaluate and compare the behavior of reactions, especially in electrochemical cells.
Standard potentials are crucial when trying to determine the feasibility of a reaction. By comparing potential values, we understand how electrons might flow between various substances.

Half-Reaction

A half-reaction is one part of a redox process, where either oxidation or reduction occurs specifically. In this context, a half-reaction represents the transformation of one part of the overall cell reaction. In the exercise, the reaction \( \mathrm{Pb}^{2+} (\text{aq}) + 2 \mathrm{e}^{-} \rightarrow \mathrm{Pb} (\text{s}) \) represents the reduction half-reaction.
This transformation involves the gain of electrons, thereby prioritizing the reduction of the lead ions into solid lead. The inclusion of electrons is crucial for balancing the reaction in terms of charge, ensuring that all transferred electrons in a full redox reaction are accounted for. Understanding half-reactions is essential to writing balanced chemical equations, which is a fundamental concept in electrochemistry. They allow us to clearly depict the exact species that are gaining or losing electrons in a reaction.

Gibbs Free Energy

Gibbs Free Energy \((\Delta G^{\circ})\) represents the maximum reversible work that may be performed by a thermodynamic system at constant temperature and pressure. It's a crucial factor in predicting whether a reaction will be spontaneous. For a reaction to happen spontaneously, \( \Delta G^{\circ} \) should be negative.
In the exercise, the \( \Delta G^{\circ} \) for the solubility product and the reduction potential are calculated separately, then summed for the overall half-reaction. Calculation uses: \( \Delta G^{\circ} = -RT \ln K \) and \( \Delta G^{\circ} = -nFE^{\circ} \), showing the interplay between energy, equilibrium, and potential. These relationships help in determining the energy required or released, drawing connections between chemical equilibrium and electrochemical potential.

Nernst Equation

The Nernst Equation provides the relationship between cell potential, temperature, and concentrations of reactants and products. This equation helps in determining the potential of an electrochemical cell at non-standard conditions. Its form is:
\[ E = E^{\circ} - \frac{RT}{nF} \ln Q \]
Here, \( E \) is the actual potential of the cell, \( E^{\circ} \) is the standard electrode potential, \( R \) is the universal gas constant \((8.314 \ \mathrm{J/mol} \cdot \mathrm{K})\), \( T \) is the temperature in Kelvin, \( n \) is the number of moles of electrons transferred, \( F \) is Faraday's constant, and \( Q \) is the reaction quotient.
In our context, when solving for the electrode potential with a given concentration of \( \mathrm{C}_{2} \mathrm{O}_{4}^{2-} \), this equation allows for adjusting the standard cell potential to match the conditions of your specific system, emphasizing its versatility in real-world applications.