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Combine the equations $$\Delta G^{\circ}=-n F \mathscr{E}^{\circ} \text { and } \Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$$ to derive an expression for \(\mathscr{E}^{\circ}\) as a function of temperature. Describe how one can graphically determine \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) from measurements of \(\mathscr{E}^{\circ}\) at different temperatures, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. What property would you look for in designing a reference half-cell that would produce a potential relatively stable with respect to temperature?

Short Answer

Expert verified
To derive an expression for the standard cell potential, \(\mathscr{E}^{\circ}\), as a function of temperature, combine the given equations as follows: \( -n F \mathscr{E}^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \). Rearrange the terms to obtain, \( \mathscr{E}^{\circ} = \frac{\Delta H^{\circ}}{-n F} + \frac{T \Delta S^{\circ}}{-n F} \). To graphically determine the standard enthalpy change, \(\Delta H^{\circ}\), and standard entropy change, \(\Delta S^{\circ}\), plot \(\mathscr{E}^{\circ}\) against temperature, T. The plot will be linear, with \(\Delta H^{\circ} = -n F \times (T = 0 \text { intercept })\) and \(\Delta S^{\circ} = -n F \times (\text { slope })\). To design a reference half-cell with stable potential relative to temperature, look for a half-cell reaction with a low magnitude of the standard entropy change, \(\Delta S^{\circ}\). This minimizes the temperature dependence of the standard cell potential, \(\mathscr{E}^{\circ}\).

Step by step solution

01

Combine the given equations

First, let's rewrite both given equations to make it easier to combine them: \( \Delta G^{\circ} = -n F \mathscr{E}^{\circ}\) and \( \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \). Now, we can set both equations for \(\Delta G^{\circ}\) equal to each other: \( -n F \mathscr{E}^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \).
02

Solve for the standard cell potential \(\mathscr{E}^{\circ}\)

To get an expression for the standard cell potential, \(\mathscr{E}^{\circ}\), we need to solve the equation for it. We can do this by rearranging the terms in the equation as follows: \( \mathscr{E}^{\circ} = \frac{\Delta H^{\circ}}{-n F} + \frac{T \Delta S^{\circ}}{-n F} \) By looking at this expression, we can see that \(\mathscr{E}^{\circ}\) depends on temperature through the second term.
03

Graphically determine \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\)

To graphically determine the values for \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\), we can plot \(\mathscr{E}^{\circ}\) against temperature, T. From the expression we derived in Step 2, we can identify that the plot will be linear, with a slope of \(\frac{\Delta S^{\circ}}{-n F}\) and a \(T = 0\) intercept of \(\frac{\Delta H^{\circ}}{-n F}\). Once we have the slope and intercept of the plot, we can use these values to calculate the standard enthalpy change and standard entropy change using the following expressions: \(\Delta H^{\circ} = -n F \times (T = 0 \text { intercept })\) \(\Delta S^{\circ} = -n F \times (\text { slope })\)
04

Consider a property for a stable reference half-cell potential

If we want a reference half-cell potential that is relatively stable with respect to temperature, we would look for a half-cell reaction with a low magnitude of the standard entropy change, \(\Delta S^{\circ}\). As seen in the expression derived in Step 2, the temperature dependence of the standard cell potential, \(\mathscr{E}^{\circ}\), is proportional to the standard entropy change, \(\Delta S^{\circ}\). If \(\Delta S^{\circ}\) is small, the temperature dependence of the reference half-cell potential will also be small, resulting in more stable potential values with respect to temperature changes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Cell Potential
In electrochemistry, the standard cell potential, denoted as \( \mathscr{E}^{\circ} \), is a measure of the voltage, or electric potential, that can be generated by an electrochemical cell under standard conditions. These conditions typically include a temperature of 25°C (298 K), pressures of 1 bar for gaseous reactants, and a concentration of 1 M for any aqueous solutions involved.

The standard cell potential reflects the ability of the cell to do work under these defined conditions and is calculated from the potential difference between two electrodes. The equation \( \Delta G^{\circ} = -n F \mathscr{E}^{\circ} \) relates the standard cell potential directly to the Gibbs free energy change (\( \Delta G^{\circ} \)) of the cell's electrochemical reaction, where \( n \) is the number of moles of electrons transferred, and \( F \) is Faraday's constant.

By solving this equation for \( \mathscr{E}^{\circ} \), we can understand how factors like enthalpy and entropy influence the cell's capability to generate electric potential.
Temperature Dependence
Temperature can significantly affect the standard cell potential, \( \mathscr{E}^{\circ} \). As the temperature changes, so does the ability of an electrochemical cell to perform work. The equation \( \mathscr{E}^{\circ} = \frac{\Delta H^{\circ}}{-n F} + \frac{T \Delta S^{\circ}}{-n F} \) clearly illustrates this dependency.

In this expression:
  • \( \Delta H^{\circ} \) represents the standard enthalpy change of the reaction.
  • \( \Delta S^{\circ} \) stands for the standard entropy change.
The term \( \frac{T \Delta S^{\circ}}{-n F} \) shows the linear relationship between \( \mathscr{E}^{\circ} \) and temperature \( T \). This term suggests that the potential changes with temperature, depending on the entropy change. Therefore, in practice, establishing a reference electrode with minimal entropy change (\( \Delta S^{\circ} \)) would be advantageous, as this would result in minimal changes in \( \mathscr{E}^{\circ} \) with temperature shifts.
Standard Enthalpy Change
The standard enthalpy change, denoted \( \Delta H^{\circ} \), describes the heat absorbed or released during a reaction at constant pressure. It is a crucial concept when considering the energy changes occurring in reactions, especially in the context of electrochemical cells.

In electrochemical reactions, \( \Delta H^{\circ} \) indicates whether the reaction absorbs or releases heat (endothermic vs. exothermic processes). This value can be determined graphically by plotting \( \mathscr{E}^{\circ} \) versus temperature and determining the intercept of the plot at \( T = 0 \). The equation \( \Delta H^{\circ} = -n F \times (T = 0 \text{ intercept}) \) offers a straightforward way to find \( \Delta H^{\circ} \) and provides insight into how energy is transferred as heat during the reaction.

Understanding \( \Delta H^{\circ} \) helps predict the feasibility and spontaneity of reactions under different conditions, contributing to a more complete understanding of reaction thermodynamics.
Standard Entropy Change
Standard entropy change, \( \Delta S^{\circ} \), is a measure of the disorder or randomness associated with the reactants and products in a chemical reaction. In an electrochemical context, \( \Delta S^{\circ} \) helps predict how \( \mathscr{E}^{\circ} \) varies with temperature.

Entropy changes can be seen from the slope of the plot created by graphing \( \mathscr{E}^{\circ} \) against temperature. The slope given by \( \frac{\Delta S^{\circ}}{-n F} \) provides a way to calculate \( \Delta S^{\circ} \). If \( \Delta S^{\circ} \) is positive, it means the products have greater disorder compared to the reactants, often making the reaction favorable at higher temperatures. Conversely, a negative \( \Delta S^{\circ} \) indicates a decrease in disorder, potentially leading to unfavorability at elevated temperatures.

In designing reference electrodes with minimal temperature sensitivity, minimizing \( \Delta S^{\circ} \) could contribute to stability in \( \mathscr{E}^{\circ} \) by reducing temperature-induced variations.

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Most popular questions from this chapter

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

Hydrogen peroxide can function either as an oxidizing agent or as a reducing agent. At standard conditions, is \(\mathrm{H}_{2} \mathrm{O}_{2}\) a better oxidizing agent or reducing agent? Explain.

A copper penny can be dissolved in nitric acid but not in hydrochloric acid. Using reduction potentials from the book, show why this is so. What are the products of the reaction? Newer pennies contain a mixture of zinc and copper. What happens to the zinc in the penny when the coin is placed in nitric acid? Hydrochloric acid? Support your explanations with data from the book, and include balanced equations for all reactions.

The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at 298 \(\mathrm{K}\) is $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad K=1.28 \times 10^{83}$$ a. Calculate \(8^{\circ}\) and \(\Delta G^{\circ}\) at 298 \(\mathrm{K}\) for the fuel cell reaction. b. Predict the signs of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the fuel cell reaction. c. As temperature increases, does the maximum amount of work obtained from the fuel cell reaction increase, decrease, or remain the same? Explain.

Which of the following statement(s) is(are) true? a. Copper metal can be oxidized by \(\mathrm{Ag}^{+}\) (at standard conditions). b. In a galvanic cell the oxidizing agent in the cell reaction is present at the anode. c. In a cell using the half reactions \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow\) Al and \(\mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg}\) , aluminum functions as the anode. d. In a concentration cell electrons always flow from the compartment with the lower ion concentration to the compartment with the higher ion concentration. e. In a galvanic cell the negative ions in the salt bridge flow in the same direction as the electrons.

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