Gibbs Free Energy from Cell Potential: 3 Examples

Understanding Gibbs Free Energy and Cell Potential

In electrochemistry, the Gibbs Free Energy (ΔG) is a crucial concept that indicates the spontaneity of a reaction. The relationship between Gibbs Free Energy and cell potential (E) is essential for predicting whether a reaction will occur spontaneously under standard conditions. The formula to calculate Gibbs Free Energy from cell potential is:

\[\Delta G = -nFE\]

where:

• ΔG = change in Gibbs Free Energy (in joules)
• n = number of moles of electrons transferred in the reaction
• F = Faraday’s constant (approximately 96485 C/mol)
• E = cell potential (in volts)

This formula allows us to assess the feasibility of redox reactions in electrochemical cells. Below are three diverse examples that illustrate how to calculate Gibbs Free Energy from cell potential.

Example 1: Standard Hydrogen Electrode Reaction

In a standard hydrogen electrode (SHE) setup, the half-cell reaction can be represented as:

\[\text{2H}^+ + 2e^- \rightleftharpoons \text{H}_2\]

This reaction serves as a reference for all electrochemical potentials. In this case, the standard cell potential (E) is 0.00 V since it is the baseline.

To calculate Gibbs Free Energy:

• The number of moles of electrons transferred (n) = 2 (for the 2 electrons involved in the half-cell)
• E = 0.00 V
• F = 96485 C/mol

Substituting into the formula:
\[\Delta G = -nFE = -2 \times 96485 \times 0.00 = 0\]

This indicates that the reaction at the standard hydrogen electrode is at equilibrium, with no Gibbs Free Energy change.

Notes:

• This example illustrates a baseline scenario. For reactions with different potentials, the values of E will change the Gibbs Free Energy accordingly.

Example 2: Zinc-Copper Galvanic Cell

Consider a galvanic cell involving a zinc electrode and a copper electrode:

\[\text{Zn}^{2+} + 2e^- \rightleftharpoons \text{Zn} \quad (E_{Zn} = -0.76 \, V)\] \[\text{Cu}^{2+} + 2e^- \rightleftharpoons \text{Cu} \quad (E_{Cu} = +0.34 \, V)\]

The overall cell reaction is:
\[\text{Zn} + \text{Cu}^{2+} \rightleftharpoons \text{Zn}^{2+} + \text{Cu}\]

To find the cell potential (E), we calculate:
\[E_{cell} = E_{cathode} - E_{anode} = 0.34 - (-0.76) = 1.10 \, V\]

Now, calculate Gibbs Free Energy:

• n = 2 (for the transfer of 2 electrons)
• E = 1.10 V

Substituting into the formula:
\[\Delta G = -nFE = -2 \times 96485 \times 1.10 = -212,130 \, J\]

This negative value indicates that the reaction is spontaneous under standard conditions.

Notes:

• The sign of ΔG informs us about spontaneity; a negative value means the reaction can occur naturally without external energy input.

Example 3: Electrolysis of Water

In the electrolysis of water, the half-reactions are:

\[\text{2H}_2O \rightleftharpoons \text{2H}_2 + \text{O}_2\]

This process requires energy input since it is non-spontaneous. The standard cell potential for the reaction can be estimated as:

• E = 1.23 V (based on thermodynamic data)

For the electrolysis of water, 4 moles of electrons are involved:

• n = 4

Now, calculate Gibbs Free Energy:
\[\Delta G = -nFE = -4 \times 96485 \times 1.23 = -474,440 \, J\]

However, since this process requires energy, we consider the absolute value:
\[\Delta G = +474,440 \, J\]

This positive value indicates that energy must be supplied for the reaction to proceed.

Notes:

• The electrolysis of water is a common application in renewable energy technology, highlighting the importance of Gibbs Free Energy calculations in practical scenarios.