Electrochemistry: Cell Potential, Gibbs Free Energy, and Faraday’s Constant

Cell Potential, Gibbs Free Energy, and Electrochemical Cells

Relationship Between Cell Potential and Gibbs Free Energy

Electrochemical cells convert chemical energy into electrical energy. The maximum electrical work that can be obtained from a cell is related to the change in Gibbs Free Energy (ΔG), which is a measure of the spontaneity of a reaction.

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

• Gibbs Free Energy (ΔG): Represents the maximum non-expansion work that can be extracted from a chemical reaction at constant temperature and pressure.

The relationship between cell potential and Gibbs Free Energy is given by:

$\Delta G = -nFE_{\text{cell}}$

• n: Number of moles of electrons transferred

• F: Faraday’s constant ($96,485\ \text{C/mol e}^-$)

• Ecell: Cell potential (V)

Example Calculation

Example: Calculate the maximum electrical work that can be produced by the cell:

• Given half-reactions:

  • Cu2+ + 2e- → Cu (E0 = +0.34 V)

  • Zn2+ + 2e- → Zn (E0 = -0.76 V)

• Overall cell reaction: Zn + Cu2+ → Zn2+ + Cu

• Cell potential: $E_{\text{cell}} = 0.34\ \text{V} - (-0.76\ \text{V}) = 1.10\ \text{V}$

• Number of electrons transferred: n = 2

• Gibbs Free Energy change:

$\Delta G = -nFE_{\text{cell}} = -(2)(96,485\ \text{C/mol})(1.10\ \text{V}) = -212,267\ \text{J}$

Faraday’s Constant and Electrochemical Cell Charge

Faraday’s constant (F) is the charge carried by one mole of electrons. It is a fundamental constant in electrochemistry, named after Michael Faraday.

• 1 Faraday (F) = 96,485 coulombs per mole of electrons

• Used to relate the amount of charge transferred in a reaction to the number of moles of electrons involved.

The conversion between coulombs and joules is:

$1\ \text{J} = 1\ \text{C} \cdot 1\ \text{V}$

Quantity	Symbol	Value
Faraday’s constant	F	96,485 C/mol e-
Elementary charge	e	1.602 x 10-19 C

Example Calculation

Example: Determine the overall charge (Q) when aluminum solid is oxidized as an anode.

• Number of moles of electrons transferred: 2.68 mol

• Charge: $Q = nF = (2.68\ \text{mol})(96,485\ \text{C/mol}) = 2.59 \times 10^5\ \text{C}$

Practice Problems and Applications

Calculating Gibbs Free Energy from Cell Potentials

Given a redox reaction and standard reduction potentials, you can calculate the Gibbs Free Energy change for the reaction.

• Write the half-reactions and identify the number of electrons transferred (n).

• Calculate the standard cell potential ($E_{\text{cell}}^0$) using the reduction potentials.

• Apply the formula $\Delta G = -nFE_{\text{cell}}$.

Example Practice

Given:

• Au3+ + 3e- → Au (E0 = +1.50 V)

• Li+ + e- → Li (E0 = -3.04 V)

Overall reaction: Au3+ + 3Li → Au + 3Li+

Cell potential: $E_{\text{cell}} = 1.50\ \text{V} - (-3.04\ \text{V}) = 4.54\ \text{V}$

Number of electrons transferred: n = 3

Gibbs Free Energy change:

$\Delta G = -3 \times 96,485 \times 4.54 = -1,315,212\ \text{J}$

Reduction of Chlorate Example

For the reduction of chlorate:

• Equation: ClO3- (aq) + 6H+ (aq) + 6e- → Cl- (aq) + 3H2O (l)

• If the standard cell potential is 1.37 V, and n = 6, then:

$\Delta G = -6 \times 96,485 \times 1.37 = -793,000\ \text{J}$

Summary Table: Key Electrochemical Quantities

Quantity	Symbol	Typical Value	Units
Cell Potential	Ecell	Varies	V (volts)
Faraday’s Constant	F	96,485	C/mol e-
Gibbs Free Energy	ΔG	Varies	J (joules)
Number of electrons	n	Varies	mol

Additional info: Practice problems and example calculations have been expanded for clarity and completeness. Tables have been reconstructed to summarize key constants and relationships.