Question

Consider the galvanic cell based on the following half-reactions: $$\begin{array}{ll}\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{E}^{\circ}=1.50 \mathrm{~V} \\ \mathrm{Tl}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Tl} & \mathscr{E}^{\circ}=-0.34 \mathrm{~V} \end{array}$$ a. Determine the overall cell reaction and calculate \(\mathscr{E}_{\mathrm{ccll}}^{\circ}\) b. Calculate \(\Delta G^{\circ}\) and \(K\) for the cell reaction at \(25^{\circ} \mathrm{C}\). c. Calculate \(\mathscr{E}_{\text {cell }}\) at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{Au}^{3+}\right]=1.0 \times 10^{-2} M\) and \(\left[\mathrm{Tl}^{+}\right]=1.0 \times 10^{-4} \mathrm{M}\)

Short answer

a. The overall cell reaction is: \(\mathrm{Au^{3+}} + 3\mathrm{Tl} \longrightarrow \mathrm{Au} + 3\mathrm{Tl^+}\) and the standard cell potential is \(\mathscr{E}_{\mathrm{cell}}^{\circ}= 1.84 V\). b. The Gibbs free energy change at 25°C is \(\Delta G^{\circ}=-530.7 \, kJ/mol\) and the equilibrium constant is \(K \approx 3.68 \times 10^{69}\). c. The cell potential at 25°C and given concentrations is \(\mathscr{E}_{\mathrm{cell}} \approx 1.87 V\).

Step-by-step solution

Determine the overall cell reaction

To do this, we need to balance the oxidation and reduction half-reactions and add them together. Since there are 3 electrons in the reduction half-reaction, we will multiply the oxidation half-reaction by 3 to balance the electrons. \(3(\mathrm{Tl}^{+}+\mathrm{e}^{-}\longrightarrow \mathrm{Tl})\Rightarrow 3\mathrm{Tl}^{+}+3\mathrm{e}^{-} \longrightarrow 3\mathrm{Tl}\) Now, add the balanced reduction and oxidation half-reactions together to get the overall cell reaction: \(\mathrm{Au}^{3+}+3\mathrm{e}^{-}\longrightarrow \mathrm{Au} + 3\mathrm{Tl}^{+}+3\mathrm{e}^{-}\longrightarrow 3\mathrm{Tl}\) \(\Rightarrow \mathrm{Au}^{3+}+ 3\mathrm{Tl}\longrightarrow \mathrm{Au} + 3\mathrm{Tl}^{+} \)

Calculate the standard cell potential (\(\mathscr{E}_{\mathrm{cell}}^{\circ}\))

The standard cell potential is obtained by subtracting the standard potential of the anode reaction (\(\mathscr{E}^{\circ}_\mathrm{anode}\)) from the standard potential of the cathode reaction (\(\mathscr{E}^{\circ}_\mathrm{cathode}\)): \(\mathscr{E}_{\mathrm{cell}}^{\circ} = \mathscr{E}^{\circ}_\mathrm{cathode} - \mathscr{E}^{\circ}_\mathrm{anode}\) Since \(\mathscr{E}^{\circ}_\mathrm{cathode} = 1.50 V\) and \(\mathscr{E}^{\circ}_\mathrm{anode} = -0.34 V\): \(\mathscr{E}_{\mathrm{cell}}^{\circ}= 1.50 V- (-0.34 V) = 1.84 V\)

Calculate Gibbs free energy change (\(\Delta G^{\circ}\)) at 25°C

Use the formula: \(\Delta G^{\circ}=-nFE_{\text {cell}}^{\circ}\) where \(n\) is the number of moles of electrons transferred, \(F\) is the Faraday constant (96485 C/mol), and \(E_{\text {cell}}^{\circ}\) is the standard cell potential. In this case, \(n=3\) because 3 electrons are transferred. So, \(\Delta G^{\circ}=-(3)(96485)(1.84)=-530717 J/mol=-530.7 \, kJ/mol\)

Calculate the equilibrium constant (\(K\)) at 25°C

Use the formula: \(\Delta G^{\circ}=-RT\ln K\) where \(R\) is the universal gas constant (8.314 J/mol K), \(T\) is the temperature in Kelvin (298 K), and \(K\) is the equilibrium constant. Rearrange and solve for \(K\): \(\ln K = -\frac{\Delta G^{\circ}}{RT}\) \(\Rightarrow K = e^{-\frac{\Delta G^{\circ}}{RT}}\) Plug in the values: \(K = e^{\frac{530717}{(8.314)(298)}} \approx 3.68 \times 10^{69}\)

Calculate the cell potential (\(\mathscr{E}_{\mathrm{cell}}\)) at 25°C and given concentrations

First, use the Nernst equation: \(\mathscr{E}_{\mathrm{cell}} = \mathscr{E}_{\mathrm{cell}}^{\circ} - \frac{RT}{nF}\ln Q\) where \(Q\) is the reaction quotient and has the form: \(Q=\dfrac{[\mathrm{Tl}^{+}]^3}{[\mathrm{Au}^{3+}]}\) Plug in the values: \(Q = \dfrac{(1.0 \times 10^{-4})^3}{1.0 \times 10^{-2}}= 1.0 \times 10^{-10}\) And now calculate \(\mathscr{E}_{\mathrm{cell}}\): \(\mathscr{E}_{\mathrm{cell}} = 1.84 - \frac{(8.314)(298)}{(3)(96485)}\ln(1.0 \times 10^{-10}) \approx 1.87 V\)

Key concepts

Standard Cell Potential

Understanding the standard cell potential is essential when analyzing galvanic cells. Standard cell potential, denoted as \(\mathscr{E}_{\mathrm{cell}}^{\circ}\), is a measure of the driving force behind an electrochemical reaction. It represents the voltage between two half-cells at standard conditions: 25°C (or 298K), 1 atm of pressure, and 1M concentrations of all reactants and products.

In practice, it's obtained by referencing all potentials to a standard hydrogen electrode (SHE), which is set at 0 volts. The more positive the standard cell potential, the greater the cell's ability to drive electrons through the circuit, implying that the reaction is more spontaneous in the forward direction. Conversely, a negative standard cell potential means that the reaction is non-spontaneous and would require external energy to proceed.

To calculate the overall standard cell potential, the potential of the cathode (reduction half-cell) is subtracted from the potential of the anode (oxidation half-cell). The cell's ability to do work, power electronic devices, and the spontaneity of the reaction rely on a positive cell potential.

Gibbs Free Energy Change

The Gibbs free energy change, denoted \(\Delta G^{\circ}\), is a thermodynamic quantity that describes the maximum amount of work that can be performed by an electrochemical cell reaction at constant temperature and pressure when it reaches equilibrium. A negative \(\Delta G^{\circ}\) indicates a spontaneous reaction, whereas a positive value implies a non-spontaneous reaction.

It is directly related to the cell potential through the equation \(\Delta G^{\circ} = -nFE_{\text{cell}}^{\circ}\), where \(n\) is the number of moles of electrons transferred in the reaction, and \(F\) is Faraday's constant (approximately 96485 C/mol). This relationship signifies that the electrical work a cell can perform is equivalent to the decrease in free energy of the system. When discussing electrochemical cells, like the galvanic cell in the given problem, Gibbs free energy provides insight into the feasibility and efficiency of the cell's operation.

Equilibrium Constant

The equilibrium constant, represented by \(K\), quantifies the ratio of products to reactants at equilibrium for a given chemical reaction. It is a crucial concept in chemical thermodynamics and electrochemistry because it reflects the extent to which a reaction will proceed before reaching a state of balance.

The relationship between \(\Delta G^{\circ}\) and \(K\) is given by the equation \(\Delta G^{\circ}=-RT\ln K\), where \(R\) is the universal gas constant and \(T\) is the temperature in Kelvin. A higher value of \(K\) implies that, at equilibrium, the concentration of products is much larger than that of reactants, suggesting a greater tendency towards the products. For electrochemical cells, this equilibrium constant directly correlates with the standard cell potential and Gibbs free energy, providing another layer of understanding for the system's spontaneity and the position of equilibrium.

Nernst Equation

The Nernst equation is a fundamental equation in electrochemistry that adjusts the standard cell potential to account for non-standard conditions, such as different temperatures or reactant/product concentrations. The equation is \(\mathscr{E}_{\mathrm{cell}} = \mathscr{E}_{\mathrm{cell}}^{\circ} - \frac{RT}{nF}\ln Q\), where \(Q\) is the reaction quotient, which represents the immediate concentrations of the reactants and products.

The ability of the Nernst equation to predict cell potential under various conditions makes it an invaluable tool for scientists and engineers. It links thermodynamics with the practical behavior of cells under real-world conditions. For students, understanding how to apply the Nernst equation enables them to calculate the actual potential of an electrochemical cell at any point during its operation, not just under standard conditions.