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Consider a galvanic cell based on the following theoretical half-reactions: $$\begin{array}{ll} & \mathscr{E}^{\circ}(\mathrm{V}) \\ \mathrm{M}^{4+}+4 \mathrm{e}^{-} \longrightarrow \mathrm{M} & 0.66 \\ \mathrm{N}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{N} & 0.39 \\ \hline \end{array}$$ What is the value of \(\Delta G^{\circ}\) and \(K\) for this cell?

Short Answer

Expert verified
The overall cell reaction is given by \(3\mathrm{M^{4+}}+ 12\mathrm{e^{-}} + 4\mathrm{N^{3+}} \longrightarrow 3\mathrm{M} + 4\mathrm{N}\) with a standard cell potential, \(\mathscr{E}^{\circ}_{cell}\), of \(0.54\ \mathrm{V}\). The standard change in Gibbs free energy, \(\Delta G^{\circ}\), is \(-7577.64\ \mathrm{J/mol}\), and the equilibrium constant, \(K\), is \(3.36 \times 10^{12}\).

Step by step solution

01

Determine the overall cell reaction

To do this, we need to combine the two given half-reactions and their standard reduction potentials (\(\mathscr{E}^{\circ}\)). We can combine them by multiplying the first equation by 3 and the second equation by 4 so that the number of electrons in both half-reactions is the same (12 electrons): \(3 (\mathrm{M^{4+} + 4 e^{-} \rightarrow M})\) with potential of \(3 \times 0.66\) \(4 (\mathrm{N^{3+} + 3 e^{-} \rightarrow N})\) with potential of \(4 \times 0.39\) Then, we can add these half-reactions to get the overall cell reaction: \(3\mathrm{M^{4+}}+ 12\mathrm{e^{-}} + 4\mathrm{N^{3+}} \longrightarrow 3\mathrm{M} + 4\mathrm{N}\) Now, subtract the two standard reduction potentials to find the standard cell potential (\(\mathscr{E}^{\circ}_{cell}\)): \(\mathscr{E}^{\circ}_{cell} = (3 \times 0.66) - (4 \times 0.39)\)
02

Calculate the standard change in Gibbs free energy, \(\Delta G^{\circ}\)

Use the Nernst equation and the standard cell potential to calculate the standard change in Gibbs free energy: \(\Delta G^{\circ} = -nFE^{\circ}_{cell}\) where \(n\) is the number of moles of electrons transferred (in this case, 12), \(F\) is the Faraday constant (equal to \(96485 \ \mathrm{C/mol}\)), and \(\mathscr{E}^{\circ}_{cell}\) is the standard cell potential (calculated in Step 1). Plug in the values and calculate \(\Delta G^{\circ}\):
03

Calculate the equilibrium constant, K

Use the relationship between Gibbs Free Energy and the equilibrium constant: \(\Delta G^{\circ} = -RT \ln K\) where \(R\) is the gas constant (equal to \(8.314\ \mathrm{J/(mol \cdot K)}\), \(T\) is the temperature (assume it is room temperature: \(298\ \mathrm{K}\)), and \(K\) is the equilibrium constant. Rearrange the equation for \(K\): \(K = e^{-(\Delta G^{\circ} / RT)}\) Plug in the values and calculate K:

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Key Concepts

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

Standard Reduction Potential
Understanding the standard reduction potential is crucial when studying galvanic cells, as it determines the voltage that a half-cell can produce during a reduction reaction under standard conditions. It is measured in volts and denoted by the symbol \( \mathscr{E}^{\circ} \). The higher the value of \( \mathscr{E}^{\circ} \), the greater the tendency of the species to gain electrons and, thus, be reduced.

For a galvanic cell consisting of two half-cells, the cell potential \( \mathscr{E}^{\circ}_{cell} \) is calculated by taking the difference between the standard reduction potentials of the cathode and the anode. It's essential to ensure that the reactions are correctly balanced concerning both the number of electrons and the stoichiometric coefficients to obtain the correct cell potential, as shown in the example exercise.
Gibbs Free Energy
Gibbs free energy, symbolized as \( \Delta G \), is a thermodynamic property that can be used to predict the direction of a chemical reaction and to determine whether a process is spontaneous at constant temperature and pressure. A negative \( \Delta G \) indicates that the reaction is spontaneous, while a positive value suggests a non-spontaneous reaction.

The relationship between Gibbs free energy and the standard cell potential is given by the equation \( \Delta G^{\circ} = -nFE^{\circ}_{cell} \), where \( n \) is the number of moles of electrons exchanged, \( F \) is the Faraday constant (around \( 96485 \ \mathrm{C/mol} \)), and \( \mathscr{E}^{\circ}_{cell} \) is the standard cell potential. This formula is applied to calculate \( \Delta G^{\circ} \) in the given exercise.
Nernst Equation
The Nernst equation is a foundational equation in electrochemistry that relates the cell potential to the concentration of reacting species. While the standard cell potential \( \mathscr{E}^{\circ} \) is measured under standard conditions, the Nernst equation allows us to adjust the cell potential for non-standard conditions. This is particularly important as most reactions occur under conditions that differ from standard ones.

The equation is given by: \(\mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF} \ln Q\), where \(\mathscr{E}\) is the cell potential at non-standard conditions, \(R\) is the universal gas constant, \(T\) is the temperature in kelvin, \(Q\) is the reaction quotient, and the other terms are as previously defined. In the example exercise, although we haven't used the full Nernst equation, the relation between Gibbs free energy and the standard cell potential stems from the same principles.
Equilibrium Constant
The equilibrium constant, represented as \( K \), is a dimensionless quantity that describes the ratio of concentrations of the products to reactants at equilibrium for a reversible reaction. It is related to Gibbs free energy through the equation \( \Delta G^{\circ} = -RT \ln K \). The equation indicates that if the standard free energy change for a reaction is known, the equilibrium constant can be calculated.

When \( \Delta G^{\circ} \) is negative, \( K \) will be greater than 1, suggesting that the reaction will favor the formation of products at equilibrium. Conversely, a positive \( \Delta G^{\circ} \) results in a \( K \) less than 1, indicating that reactants are favored. In the example exercise, by rearranging the equation to \( K = e^{-(\Delta G^{\circ} / RT)} \) and substituting the calculated value of \( \Delta G^{\circ} \) and known constants, we can determine the equilibrium constant for the galvanic cell.

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Most popular questions from this chapter

Consider the following half-reactions: $$\begin{array}{ll} \operatorname{IrCl}_{6}^{3-}+3 \mathrm{e}^{-} \longrightarrow \operatorname{Ir}+6 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.77 \mathrm{V} \\ \mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.73 \mathrm{V} \\ \mathrm{PdCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.62 \mathrm{V} \end{array}$$ 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 \(0.020 \mathrm{M}\) 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.)

Combine the equations $$ \Delta G^{\circ}=-n F \mathscr{E}^{\circ} \quad \text { and } \quad \Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} $$ 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?

The table below lists the cell potentials for the 10 possible galvanic cells assembled from the metals \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D},\) and \(\mathrm{E},\) and their respective \(1.00 \space \mathrm{M} \space 2+\) ions in solution. Using the data in the table, establish a standard reduction potential table similar to Table \(17-1\) in the text. Assign a reduction potential of \(0.00 \mathrm{V}\) to the half-reaction that falls in the middle of the series. You should get two different tables. Explain why, and discuss what you could do to determine which table is correct. $$\begin{array}{|lcccc|} \hline & \begin{array}{c} \mathrm{A}(s) \text { in } \\ \mathrm{A}^{2+}(a q) \end{array} & \begin{array}{c} \mathrm{B}(s) \text { in } \\ \mathrm{B}^{2+}(a q) \end{array} & \begin{array}{c} \mathrm{c}(s) \text { in } \\ \mathrm{c}^{2+}(a q) \end{array} & \begin{array}{c} \mathrm{D}(s) \text { in } \\ \mathrm{D}^{2+}(a q) \end{array} \\ \hline \mathrm{E}(s) \text { in } \mathrm{E}^{2+}(a q) & 0.28 \mathrm{V} & 0.81 \mathrm{V} & 0.13 \mathrm{V} & 1.00 \mathrm{V} \\ \mathrm{D}(s) \text { in } \mathrm{D}^{2+}(a q) & 0.72 \mathrm{V} & 0.19 \mathrm{V} & 1.13 \mathrm{V} & \- \\ \mathrm{C}(s) \text { in } \mathrm{C}^{2+}(a q) & 0.41 \mathrm{V} & 0.94 \mathrm{V} & \- & \- \\ \mathrm{B}(s) \text { in } \mathrm{B}^{2+}(a q) & 0.53 \mathrm{V} & \- & \- & \- \\ \hline \end{array}$$

What is the maximum work that can be obtained from a hydrogen-oxygen fuel cell at standard conditions that produces \(1.00 \mathrm{kg}\) water at \(25^{\circ} \mathrm{C} ?\) Why do we say that this is the maximum work that can be obtained? What are the advantages and disadvantages in using fuel cells rather than the corresponding combustion reactions to produce electricity?

a. Draw this cell under standard conditions, labeling the anode, the cathode, the direction of electron flow, and the concentrations, as appropriate. b. When enough \(\mathrm{NaCl}(s)\) is added to the compartment containing gold to make the \(\left[\mathrm{Cl}^{-}\right]=0.10 \mathrm{M},\) the cell potential is observed to be 0.31 V. Assume that \(\mathrm{Au}^{3+}\) is reduced and assume that the reaction in the compartment containing gold is $$ \mathrm{Au}^{3+}(a q)+4 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AuCl}_{4}^{-}(a q) $$ Calculate the value of \(K\) for this reaction at \(25^{\circ} \mathrm{C}\).

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