Standard Cell Potential & the Equilibrium Constant: Videos & Practice Problems

Galvanic and voltaic cells generate electricity by remaining out of equilibrium, with cell potential (E_cell) decreasing to zero at equilibrium. The Nernst equation relates E_cell to the reaction quotient (q) and the equilibrium constant (K). The connection between Gibbs free energy (ΔG) and K is expressed as ΔG = -RT ln(K), while ΔG also relates to E_cell through ΔG = -nF E_cell. At standard conditions, E_cell is determined without the Nernst equation, emphasizing the importance of understanding these relationships in electrochemistry.

Cell Potential & the Equilibrium Constant

Galvanic and voltaic cells generate electricity by undergoing chemical reactions that are not yet at equilibrium. As these reactions progress, they will eventually reach equilibrium, at which point the reaction quotient (q) will equal the equilibrium constant (K), and the cell potential under non-standard conditions will become zero. The Nernst equation describes the relationship between cell potential and the reaction quotient:

E_{cell} = E^{\circ}_{cell} - \frac{0.05916 \, \text{V}}{n} \log(q)

In this equation, E_cell represents the cell potential under non-standard conditions, E^°_cell is the standard cell potential, n is the number of electrons transferred, and q is the reaction quotient. When equilibrium is reached, q becomes K, and the cell potential is zero. By rearranging the Nernst equation, we can express the equilibrium constant in relation to the standard cell potential:

K = 10^{\frac{n \cdot E^{\circ}_{cell}}{0.05916 \, \text{V}}}

This equation allows us to calculate the equilibrium constant K when the standard cell potential is known. Additionally, the relationship between the Gibbs free energy change (ΔG) and the equilibrium constant can be expressed as:

ΔG = -RT \ln(K)

Here, R is the universal gas constant (8.314 J/(mol·K)). The connection between ΔG and the standard cell potential is given by:

ΔG = -nF E^{\circ}_{cell}

where F is Faraday's constant (approximately 96485 C/mol). At standard conditions (1 M concentrations), the Nernst equation is not necessary; instead, the standard cell potential can be calculated simply by subtracting the anode potential from the cathode potential:

E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}

Understanding the transition from q to K is crucial as it highlights the relationship between the concentrations of reactants and products at equilibrium. The Nernst equation is particularly useful when dealing with concentrations that differ from 1 M, allowing for the calculation of cell potential under varying conditions. As we explore these concepts further, the interconnectedness of cell potential, Gibbs free energy, and equilibrium constants will become increasingly evident.

Calculations 1

To determine the equilibrium constant \( K \) for the Fong reaction, we start with the balanced chemical equation: 1 mole of gold ion (\( \text{Au}^+ \)) reacting with 1 mole of solid cerium (\( \text{Ce} \)) to produce 1 mole of cerium ion (\( \text{Ce}^{3+} \)) in aqueous form and 1 mole of solid gold (\( \text{Au} \)). The half-reactions and their corresponding cell potentials are essential for calculating the overall standard cell potential.

In this reaction, we identify the oxidation and reduction processes. The gold ion (\( \text{Au}^+ \)) is reduced, as its oxidation number decreases from +1 to 0, indicating it is the cathode. Conversely, cerium is oxidized, increasing its oxidation number from 0 to +3, making it the anode. The standard cell potential (\( E^\circ \)) can be calculated using the formula:

\( E^\circ = E_{\text{cathode}} - E_{\text{anode}} \)

Substituting the given values, we find:

\( E^\circ = 1.690 \, \text{V} - (-2.336 \, \text{V}) = 4.026 \, \text{V} \)

This substantial cell potential suggests a large equilibrium constant \( K \). The relationship between the equilibrium constant and the standard cell potential is given by the equation:

\( K = 10^{\frac{n \cdot E^\circ}{0.05916}} \)

Here, \( n \) represents the number of electrons transferred in the balanced reaction. Since the half-reactions involve different numbers of electrons, we multiply the overall reaction by 3 to ensure electron balance. Thus, \( n = 3 \).

Plugging in the values, we calculate:

\( K = 10^{\frac{3 \cdot 4.026}{0.05916}} \)

This results in:

\( K \approx 10^{204.158} \approx 1.44 \times 10^{204} \)

This extraordinarily high value for \( K \) indicates that the reaction is highly spontaneous, as evidenced by the positive standard cell potential and the equilibrium constant significantly greater than 1. Such large values may require advanced calculators to avoid computational errors.

Calculations 1

In electrochemistry, the relationship between the equilibrium constant (K) and the standard cell potential (E^°) is crucial for understanding redox reactions. Given an equilibrium constant of \( K = 6.79 \times 10^{30} \), we can determine the standard cell potential using the formula:

E^° = \(\frac{0.05916 \, \text{V}}{n} \log K\)

In this equation, \( n \) represents the number of electrons transferred in the reaction. For the half-reactions provided, one involves electrons as reactants (the reduction reaction at the cathode) and the other has electrons as products (the oxidation reaction at the anode). To ensure consistency, we need to balance the number of electrons transferred. If one half-reaction involves 3 electrons and the other involves 1, we multiply the latter by 3, maintaining the same cell potential value.

Substituting into the equation, we have:

E^° = \(\frac{0.05916 \, \text{V}}{3} \log(6.79 \times 10^{30})\)

Calculating this gives us a standard cell potential of 0.608 V. The overall standard cell potential is determined by the difference between the cathode and anode potentials:

0.608 V = E^°_cathode - E^°_anode

Given that the anode potential is 0.308 V, we can rearrange the equation to find the cathode potential:

E^°_cathode = 0.608 V + 0.308 V = 0.916 V

This calculation highlights the connection between the standard cell potential and the equilibrium constant, allowing us to isolate and solve for the missing variable, which in this case is the cell potential of the cathode compartment.

Here’s what students ask on this topic:

What is the Nernst equation and how is it used to calculate cell potential under non-standard conditions?

The Nernst equation is used to calculate the cell potential (E_cell) under non-standard conditions. It is given by:

$E_{\mathrm{cell}}=E_{\mathrm{cell}}-\frac{0.05916}{n}\times log\left(Q\right)$

Here, E_cell is the cell potential under non-standard conditions, E_cell° is the standard cell potential, n is the number of electrons transferred, and Q is the reaction quotient. This equation helps in determining the voltage of an electrochemical cell when the concentrations of the reactants and products are not at standard conditions (1 M concentration).

How do you calculate the equilibrium constant (K) from the standard cell potential (E_cell°)?

To calculate the equilibrium constant (K) from the standard cell potential (E_cell°), you can use the following equation:

$K={10}^{\frac{n\times E_{\mathrm{cell}}°}{0.05916}}$

Here, n is the number of electrons transferred in the reaction, and 0.05916 is a constant derived from the Nernst equation at 25°C. This equation shows the direct relationship between the standard cell potential and the equilibrium constant, allowing you to determine K when E_cell° is known.

What is the relationship between Gibbs free energy (ΔG) and the equilibrium constant (K)?

The relationship between Gibbs free energy (ΔG) and the equilibrium constant (K) is given by the equation:

$\mathrm{\Delta G}=-RT\times ln\left(K\right)$

Here, R is the universal gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and K is the equilibrium constant. This equation indicates that a negative ΔG corresponds to a spontaneous reaction and a large K value, while a positive ΔG corresponds to a non-spontaneous reaction and a small K value.

How is the standard cell potential (E_cell°) related to Gibbs free energy (ΔG)?

The standard cell potential (E_cell°) is related to Gibbs free energy (ΔG) by the equation:

$\mathrm{\Delta G}=-nF\times E_{\mathrm{cell}}°$

Here, n is the number of electrons transferred, F is Faraday's constant (96,485 C/mol), and E_cell° is the standard cell potential. This equation shows that a positive E_cell° corresponds to a negative ΔG, indicating a spontaneous reaction, while a negative E_cell° corresponds to a positive ΔG, indicating a non-spontaneous reaction.

Why is the Nernst equation not needed under standard conditions?

Under standard conditions, the concentrations of all reactants and products are 1 M, and the temperature is 25°C. In this scenario, the reaction quotient Q equals 1, making the logarithmic term in the Nernst equation zero. Therefore, the cell potential (E_cell) under standard conditions is simply the standard cell potential (E_cell°), and the Nernst equation simplifies to:

$E_{\mathrm{cell}}=E_{\mathrm{cell}}°$

Thus, the Nernst equation is not needed to calculate the cell potential under standard conditions.