Cell Potentials, Gibbs Free Energy, and Equilibrium in Electrochemistry

Cell Potentials and Equilibrium

Gibbs Free Energy, Standard Cell Potential, and Equilibrium Constant

The relationship between Gibbs Free Energy (ΔG), the standard cell potential (Ecell), and the equilibrium constant (K) is fundamental in electrochemistry. These quantities are interconnected and allow chemists to predict the spontaneity of reactions, the voltage produced by electrochemical cells, and the position of equilibrium.

• Gibbs Free Energy (ΔG): Indicates the spontaneity of a chemical reaction. A negative ΔG means the reaction is spontaneous.

• Standard Cell Potential (Ecell): The voltage produced by an electrochemical cell under standard conditions.

• Equilibrium Constant (K): Describes the ratio of product concentrations to reactant concentrations at equilibrium.

The connection between these quantities can be expressed by the following formulas:

• Relationship between ΔG and Ecell:

$\Delta G = -n F E_{cell}$

• Relationship between ΔG and K:

$\Delta G = -RT \ln K$

• Combined Formula for Ecell and K:

$E_{cell} = \frac{RT}{nF} \ln K$

Where:

• n = number of moles of electrons transferred

• F = Faraday's constant ($96,485$ C/mol e-)

• R = universal gas constant ($8.314$ J/mol·K)

• T = temperature in Kelvin

At standard temperature (25°C or 298 K), the formula simplifies to:

$E_{cell} = \frac{0.0257}{n} \ln K$

Or, using log base 10:

$E_{cell} = \frac{0.0592}{n} \log K$

Example Calculation

Example: A certain electrochemical reaction involves the transfer of 4 electrons. If the value of the equilibrium constant is $2.50 \times 10^{18}$ at 25°C, calculate the standard cell potential.

• Given: $n = 4$, $K = 2.50 \times 10^{18}$

• Use the formula: $E_{cell} = \frac{0.0592}{n} \log K$

• Calculation:

$E_{cell} = \frac{0.0592}{4} \log(2.50 \times 10^{18})$

$E_{cell} \approx 0.27$ V

Practice Problem

Practice: Calculate the equilibrium constant for the following reaction at 25°C.

Given the following reduction potentials:

• $Cu^{2+} + 2e^- \rightarrow Cu$    $E^\circ = +0.34$ V

• $Zn^{2+} + 2e^- \rightarrow Zn$    $E^\circ = -0.76$ V

Steps:

1. Write the cell reaction: $Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$

2. Calculate $E_{cell}^\circ$:

$E_{cell}^\circ = E_{cathode}^\circ - E_{anode}^\circ = 0.34 - (-0.76) = 1.10$ V

3. Calculate $K$ using the formula:

$E_{cell}^\circ = \frac{0.0592}{n} \log K$

$1.10 = \frac{0.0592}{2} \log K$

$\log K = \frac{1.10 \times 2}{0.0592} = 37.16$

$K = 10^{37.16}$

Conclusion: The equilibrium constant is extremely large, indicating the reaction proceeds nearly to completion.

Key Terms

• Electrochemical Cell: A device that generates electrical energy from chemical reactions.

• Standard Reduction Potential: The tendency of a chemical species to gain electrons under standard conditions.

• Faraday's Constant: The charge of one mole of electrons ($96,485$ C/mol).