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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Most popular questions from this chapter

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Consider the following half-reactions: $$\begin{array}{ll}{\mathrm{IrCl}_{6}^{3-}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Ir}+6 \mathrm{Cl}^{-}} & {\mathscr{E}^{\circ}=0.77 \mathrm{V}} \\\ {\mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-}} & {\mathscr{E}^{\circ}=0.73 \mathrm{V}} \\\ {\mathrm{PdCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}+4 \mathrm{Cl}^{-}} & {\mathscr{E}^{\circ}=0.62 \mathrm{V}}\end{array}$$ A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant 1.0\(M\) in chloride ion and 0.020\(M\) in each complex ion. Is it feasible to separate the three metals from this solution by electrolysis? (Assume that 99\(\%\) of a metal must be plated out before another metal begins to plate out.)

Consider the following reduction potentials: $$\begin{array}{ll}{\mathrm{Co}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Co}} & {\mathscr{E}^{\circ}=1.26 \mathrm{V}} \\ {\mathrm{Co}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Co}} & {\mathscr{E}^{\circ}=-0.28 \mathrm{V}}\end{array}$$ a. When cobalt metal dissolves in 1.0\(M\) nitric acid, will \(\mathrm{Co}^{3+}\) or \(\mathrm{Co}^{2+}\) be the primary product (assuming standard conditions)? b. Is it possible to change the concentration of \(\mathrm{HNO}_{3}\) to get a different result in part a? Concentrated \(\mathrm{HNO}_{3}\) is about 16 \(M\) .

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