Cell Potential and Equilibrium: Videos & Practice Problems

Gibbs free energy, denoted as ΔG°, represents the energy available to do work in a chemical reaction at standard conditions. It serves as a crucial link between the standard cell potential (E°_cell) and the equilibrium constant (K). The relationship can be expressed through two key equations.

The first equation connects standard Gibbs free energy to standard cell potential:

ΔG° = -nF E°_cell

In this equation, ΔG° is measured in kilojoules, n represents the number of moles of electrons transferred in the redox reaction, and F is Faraday's constant, approximately 96,485 coulombs per mole of electrons. The standard cell potential, E°_cell, is expressed in volts (V).

The second equation relates standard Gibbs free energy to the equilibrium constant:

ΔG° = -RT ln K

Here, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, and K is the equilibrium constant for the reaction.

Since both expressions equal ΔG°, we can set them equal to each other:

-nF E°_cell = -RT ln K

To isolate the standard cell potential, we can rearrange this equation. Dividing both sides by -nF gives:

E°_cell = (RT/nF) ln K

This final equation illustrates the relationship between the standard cell potential and the equilibrium constant, highlighting how thermodynamic properties influence electrochemical reactions.

In electrochemistry, the relationship between the standard cell potential and the equilibrium constant is crucial for understanding the feasibility of a reaction. The standard cell potential (\(E^\circ\)) can be calculated using the equation:

\(E^\circ = \frac{RT}{nF} \ln K\)

Where:

• R is the gas constant, approximately \(8.314 \, \text{J/(mol·K)}\).
• T is the temperature in Kelvin. For 25 degrees Celsius, this is calculated as \(25 + 273.15 = 298.15 \, \text{K}\).
• n is the number of moles of electrons transferred, which in this case is 4.
• F is Faraday's constant, approximately \(96,485 \, \text{C/mol}\).
• K is the equilibrium constant, given as \(2.50 \times 10^{12}\).

Substituting these values into the equation, we have:

\(E^\circ = \frac{(8.314 \, \text{J/(mol·K)})(298.15 \, \text{K})}{(4)(96,485 \, \text{C/mol})} \ln(2.50 \times 10^{12})\)

Calculating the natural logarithm of the equilibrium constant:

\(\ln(2.50 \times 10^{12}) \approx 27.631\)

Now, substituting this value back into the equation gives:

\(E^\circ = \frac{(8.314)(298.15)}{(4)(96,485)} \times 27.631\)

After performing the calculations, the result is:

\(E^\circ \approx 0.18335 \, \text{J/C}\)

Since \(1 \, \text{J/C} = 1 \, \text{V}\), we can express this as:

\(E^\circ \approx 0.183 \, \text{V}\)

This positive value indicates that the electrochemical reaction is spontaneous, as the equilibrium constant is greater than 1. Thus, the final answer for the standard cell potential is:

\(\mathbf{0.183 \, V}\)