Problem 78
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 \(\left[\mathrm{Tl}^{+}\right]=1.0 \times 10^{-4} \mathrm{M}\)
Step-by-Step Solution
Verified 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.
1Step 1: 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
2Step 2: 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
3Step 3: 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
4Step 4: 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
5Step 5: 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 CellsNernst EquationGibbs Free EnergyEquilibrium Constant
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.
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:
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.
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.
\[\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}\]
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.
- **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.
\[\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}\]
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.
\[\Delta G^{\circ} = -RT \ln K\]
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.
- For reactions that favor products at equilibrium, \( K \) is large.
- For reactions that favor reactants, \( K \) is small.
\[\Delta G^{\circ} = -RT \ln K\]
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.
Other exercises in this chapter
Problem 74
Consider the cell described below: $$ \mathrm{Al}\left|\mathrm{Al}^{3+}(1.00 M)\right|\left|\mathrm{Pb}^{2+}(1.00 M)\right| \mathrm{Pb} $$ Calculate the cell po
View solution Problem 77
Consider the galvanic cell based on the following halfreactions: $$ \begin{array}{ll} \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} & \mathscr{E
View solution Problem 79
An electrochemical cell consists of a standard hydrogen electrode and a copper metal electrode. a. What is the potential of the cell at \(25^{\circ} \mathrm{C}\
View solution Problem 80
An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 \mathrm{M}\) separated by a porous
View solution