Question

Consider the galvanic cell based on the following halfreactions: $$ \begin{array}{ll} \mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{E}^{\circ}=1.50 \mathrm{V} \\ \mathrm{Tl}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Tl} & \mathscr{E}^{\circ}=-0.34 \mathrm{V} \end{array} $$ a. Determine the overall cell reaction and calculate \(\mathscr{C}_{\text {cell. }}\) 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 \(\left[\mathrm{Au}^{3+}\right]=1.0 \times 10^{-2} \mathrm{M}\) and \(\left[\mathrm{Tl}^{+}\right]=1.0 \times 10^{-4} \mathrm{M}\)

Short answer

The overall cell reaction is Au³⁺ + 3Tl⁺ → Au + 3Tl, with a cell potential (E°cell) of 1.84 V. The standard Gibbs free energy change (ΔG°) for this reaction is -532,724 J/mol, and the equilibrium constant (K) at 25°C is 6.28 x 10^47. At the given concentrations of Au³⁺ and Tl⁺, the cell potential (E_cell) at 25°C is 1.79 V.

Step-by-step solution

Determine the overall cell reaction

First, we should balance the electrons in both half reactions. Since there is a difference in the number of electrons in both reactions (3e⁻ in the first reaction and 1e⁻ in the second), we should multiply the second reaction by 3 to balance the number of electrons. Then, we will add both reactions to obtain the overall cell reaction: Au³⁺ + 3e⁻ → Au (E° = 1.50 V) 3(Tl⁺ + e⁻ → Tl) (E° = -0.34 V) Now, add the reactions: Au³⁺ + 3Tl⁺ + 3e⁻ → Au + 3Tl

Calculate the cell potential

The cell potential (E°cell) is the difference between both half-reactions potentials. Since the Au³⁺/Au half-reaction has a higher potential value, it will act as the cathode, while the other half-reaction will act as the anode. Thus: E°cell = E°cathode - E°anode E°cell = (1.50 V) - (-0.34 V) E°cell = 1.84 V

Calculate the standard Gibbs free energy change

Now, we can calculate the standard Gibbs free energy change (ΔG°) using the formula: ΔG° = -nFE°cell Where n = number of moles of electrons transferred (in this case, 3), F = Faraday's constant (96,485 C/mol), and E°cell is the cell potential obtained in Step 2. ΔG° = -(3)(96485 C/mol)(1.84 V) ΔG° = -532,724 J/mol

Calculate the equilibrium constant

We can calculate the equilibrium constant (K) for the cell reaction at 25°C using the standard Gibbs free energy change and the equation: ΔG° = -RT ln K Where R = gas constant (8.314 J/mol K), T = temperature in Kelvin (25°C = 298.15 K), and ΔG° is the standard Gibbs free energy change calculated in Step 3. Rearranging the equation: ln K = -(ΔG°)/(RT) K = exp((532,724 J/mol)/((8.314 J/mol K)(298.15 K))) K = 6.28 x 10^47

Calculate the cell potential under given concentrations

We will now calculate the cell potential (E_cell) at 25°C, taking into account the given concentrations of both Au³⁺ ([Au³⁺] = 1.0 x 10⁻² M) and Tl⁺ ([Tl⁺] = 1.0 x 10⁻⁴ M) using the Nernst equation: E_cell = E°cell - (RT/nF) ln Q Where Q is the reaction quotient. For the balanced cell reaction, Q = [Au³⁺]/[Tl⁺]³. Now we can plug in the values: E_cell = 1.84 V - ((8.314 J/mol K)(298.15 K))/((3)(96485 C/mol)) ln (1.0 x 10⁻²)/(1.0 x 10⁻⁴)³ E_cell = 1.84 V - 0.0496 V E_cell = 1.79 V So, under the given concentration conditions, the cell potential at 25°C will be 1.79 V.

Key concepts

Electrochemical Cells

Electrochemical cells are devices that convert chemical energy into electrical energy. They function via redox reactions, where one substance is oxidized and another is reduced. In simpler terms, electrons are transferred between substances, generating electricity. There are two main types of electrochemical cells: galvanic (or voltaic) cells and electrolytic cells. The key difference is that galvanic cells generate electricity through spontaneous reactions, while electrolytic cells require an external power source to drive non-spontaneous reactions.

• **Galvanic Cells**: Utilize spontaneous reactions to produce electric current. The reaction takes place in two separate compartments, known as half-cells. Each half-cell contains an electrode and an electrolyte solution.
• **Cell Structure**: A typical galvanic cell consists of an anode, where oxidation occurs, and a cathode, where reduction happens. These are connected through a conductive wire, allowing electrons to flow between them, generating electricity.
• **Salt Bridge**: Often, a salt bridge or porous membrane is used to maintain electrical neutrality by allowing the flow of ions between the half-cells.

Through these components, electrochemical cells play a critical role in batteries and energy storage.

Nernst Equation

The Nernst equation allows for the calculation of cell potentials under non-standard conditions. This is essential because real-life applications rarely operate at standard state conditions, which involve 1 M concentrations and 1 atm pressure.
The equation is given by:
\[E_{ ext{cell}} = E^{ ext{0}}_{ ext{cell}} - \frac{RT}{nF} \ln Q\]
where:

• \( E_{\text{cell}} \) is the actual cell potential.
• \( E^{\text{0}}_{\text{cell}} \) is the standard cell potential.
• \( R \) is the ideal gas constant (8.314 J/mol K).
• \( T \) is temperature in Kelvin.
• \( n \) is the number of moles of electrons transferred.
• \( F \) is Faraday’s constant (96485 C/mol).
• \( Q \) is the reaction quotient, calculated similarly to the equilibrium constant but with non-equilibrium concentrations.

The Nernst equation helps us understand how concentration affects cell potential, making it invaluable for predicting battery performance in various conditions.

Gibbs Free Energy

Gibbs Free Energy (\( \Delta G \)) offers critical insight into the feasibility of a reaction. It's a measure of the maximum reversible work a thermodynamic system can do at constant temperature and pressure. For galvanic cells, \( \Delta G \) determines the spontaneity.

• **Negative \( \Delta G \)**: Indicates a spontaneous reaction, allowing the galvanic cell to generate electrical energy.
• **Positive \( \Delta G \)**: Implies a non-spontaneous reaction, as seen in electrolytic cells requiring energy input.
• **Zero \( \Delta G \)**: The system is at equilibrium and can no longer perform work.

For electrochemical cells, the relationship between Gibbs Free Energy and the cell potential is given by the equation:
\[\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}\]
Where \( n \) is the number of moles of electrons exchanged in the reaction and \( F \) is Faraday’s constant. Thus, a more negative \( \Delta G^{\circ} \) corresponds with a higher \( E^{\circ}_{\text{cell}} \), emphasizing the cell’s ability to do work.

Equilibrium Constant

The equilibrium constant (\( K \)) for a reaction provides insight into the concentrations of products and reactants at equilibrium. For electrochemical reactions, it connects to cell potentials and Gibbs Free Energy.

• For reactions that favor products at equilibrium, \( K \) is large.
• For reactions that favor reactants, \( K \) is small.

The equilibrium constant is linked to Gibbs Free Energy by the equation:
\[\Delta G^{\circ} = -RT \ln K\]
Where \( R \) is the gas constant and \( T \) is the temperature in Kelvin. At equilibrium, \( \Delta G \) is zero, and thus, the reaction's position can be assessed with \( K \). For galvanic cells, a large \( K \) indicates a higher tendency to proceed with the reaction, confirming a strong energy output. Understanding this relation helps in designing cells with optimized efficiencies.