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

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 \(\mathscr{E}^{\circ}\) 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 \(A l^{3+}+3 e^{-} \longrightarrow A l\) 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.

An aqueous solution of \(\mathrm{PdCl}_{2}\) is electrolyzed for 48.6 seconds, and during this time 0.1064 g of Pd is deposited on the cathode. What is the average current used in the electrolysis?

Sketch the galvanic cells based on the following overall reactions. Show the direction of electron flow, and identify the cathode and anode. Give the overall balanced equation. Assume that all concentrations are \(1.0 M\) and that all partial pressures are 1.0 atm. a. \(\mathrm{Cr}^{3+}(a q)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cl}^{-}(a q)\) b. \(\mathrm{Cu}^{2+}(a q)+\mathrm{Mg}(s) \rightleftharpoons \mathrm{Mg}^{2+}(a q)+\mathrm{Cu}(s)\)

Explain the following relationships: \(\Delta G\) and \(w,\) cell potential and \(w,\) cell potential and \(\Delta G,\) cell potential and \(Q .\) Using these relationships, explain how you could make a cell in which both electrodes are the same metal and both solutions contain the same compound, but at different concentrations. Why does such a cell run spontaneously?

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