Problem 108 - Chemistry Principles and Practice

Question

A At \(298 \mathrm{~K},\) the solubility product constant for solid \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}\) is \(1.5 \times 10^{-9} .\) Use the standard reduction potential of \(\mathrm{Ba}^{2+}(\mathrm{aq})\) to find the standard potential for the half-reaction $$ \mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}(\mathrm{~s})+2 \mathrm{e}^{-} \rightarrow \mathrm{Ba}(\mathrm{s})+2 \mathrm{IO}_{3}^{-}(\mathrm{aq}) $$ (Hint: Find \(\Delta G^{\circ}\) for both the solubility equilibrium and the reduction half-reaction for \(\mathrm{Ba}^{2+},\) and add the reactions. Use the \(\Delta G^{\infty}\) for the sum reaction to find \(E^{a}\).)

Step-by-Step Solution

Answer

Use solubility and reduction Gibbs free energies to find overall ΔG°, then calculate standard potential.

Step 1: Calculate Gibbs Free Energy for Solubility Equilibrium

First, use the solubility product constant (Ksp) to find the Gibbs free energy change (ΔG°) for the solubility equilibrium of \(\mathrm{Ba(IO_3)_2}\) dissolving into its ions. The relationship is \(\Delta G^{\circ} = -RT \ln K_{sp}\), where \(R = 8.314 \; \text{J/mol·K}\) and \(T = 298 \; \text{K}\). Plug the values in to get:\[ \Delta G^{\circ}_{solubility} = - (8.314)(298) \ln(1.5 \times 10^{-9}) \].

Step 2: Calculate Gibbs Free Energy for the Reduction Half-Reaction

Now, write the Gibbs free energy in terms of the Nernst equation for the reduction half-reaction \(\mathrm{Ba}^{2+} + 2e^- \rightarrow \mathrm{Ba}\), using \(\Delta G^{\circ} = -nFE^{\circ}\). \(n = 2\) is the number of moles of electrons, \(F = 96485 \; \text{C/mol}\) is Faraday's constant, and \(E^{\circ}_{Ba^{2+}/Ba}\) is the standard reduction potential for this pair, which is given (or should be looked up) as \(-2.90 \; \text{V}\). Calculate \(\Delta G^{\circ}_{reduction}\) using:\[ \Delta G^{\circ}_{reduction} = -(2)(96485)(-2.90) \].

Step 3: Combine the Reactions and Calculate Overall ΔG°

For the overall reaction of the process described in the problem statement (\(\mathrm{Ba(IO_3)_2(s)} + 2e^- \rightarrow \mathrm{Ba(s)} + 2\mathrm{IO_3^-}\)), combine the two calculated \(\Delta G^{\circ}\) values:\[ \Delta G^{\circ}_{total} = \Delta G^{\circ}_{solubility} + \Delta G^{\circ}_{reduction} \].

Step 4: Calculate Standard Potential for the Overall Reaction

Using the result from Step 3, find the standard potential \(E^{\circ}_{a}\) for the overall reaction using the relation \(\Delta G^{\circ} = -nFE^{\circ}_{a}\). Rearrange to solve for \(E^{\circ}_{a}\):\[ E^{\circ}_{a} = -\frac{\Delta G^{\circ}_{total}}{nF} \]. This will give the standard potential for the given half-reaction.

Key Concepts

Solubility Product Constant

The solubility product constant, abbreviated as Ksp, is a valuable tool in determining how much of a solute can dissolve in a solvent until it reaches a point of saturation. When a compound dissolves, it breaks into its ions. For a hypothetical salt, given by the formula AB, that dissociates into A+ and B- ions, the expression for Ksp is given by \( K_{sp} = [A^+][B^-] \).Understanding Ksp helps in predicting which compounds can dissolve in water under given conditions, and how much of that compound can actually dissolve. In electrochemical reactions, knowing the Ksp is crucial when calculating the potential changes involved, as was seen in the standard potential half-reaction calculation for \( \mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2} \).By using this constant, we can predict the formation of precipitates and calculate how soluble a compound is. It's expressed in terms of molarity \( \text{mol/L} \), relating to the concentration of ions in the solution at equilibrium.

Gibbs Free Energy

Gibbs Free Energy, symbolized as \( \Delta G \), is a thermodynamic potential that can predict the direction of a chemical reaction as well as its feasibility under constant temperature and pressure. A negative change in Gibbs Free Energy \( (\Delta G < 0) \) indicates a spontaneous process. Conversely, a positive \( \Delta G \) suggests that the reaction is non-spontaneous.There are different contexts in which you might calculate \( \Delta G \):

• For solubility equilibrium: Here, \( \Delta G^{\circ} = -RT \ln K_{sp} \) is used to calculate how the solubility product relates to energy changes.
• For electrochemical reactions: In such cases, \( \Delta G^{\circ} = -nFE^{\circ} \) connects Gibbs Free Energy with electric potentials (E), electron transfer (n), and Faraday's constant (F).

In electrochemistry, understanding \( \Delta G \) allows students to bridge the gap between energy changes and electric work, providing insight into energy transformations in cells.

Nernst Equation

The Nernst Equation is a powerful equation used in electrochemistry to relate the concentration of reactants and products in a cell to its potential, especially away from standard conditions. The equation is expressed as:\[E = E^{\circ} - \frac{RT}{nF} \ln Q\]where:

• \( E \) is the cell potential under non-standard conditions.
• \( E^{\circ} \) is the standard cell potential.
• \( R \) is the gas constant \( 8.314 \text{ J/mol·K} \).
• \( T \) is the temperature in Kelvin.
• \( n \) is the number of moles of electrons transferred in the reaction.
• \( F \) is Faraday's constant \( 96485 \text{ C/mol} \).
• \( Q \) is the reaction quotient, representing the ratio of concentration of reaction products to reactants.

This equation allows chemists and students alike to predict how the potential of an electrochemical cell will shift when the concentrations of the ionic species involved change. It demonstrates how chemical equilibria are connected to electricity, providing critical insights into battery operations and redox reactions. By adjusting \( Q \), students can explore the impact of concentration changes on cell potential, making it a practical tool in both academic and real-world chemistry.