Cell Potential: G and K: Videos & Practice Problems

Understanding the spontaneity of chemical reactions is pivotal in chemistry. Spontaneous reactions are characterized by a standard cell potential (E_cell°) greater than zero, a negative Gibbs free energy change (ΔG°), and an equilibrium constant (K) greater than one. These variables are interconnected through fundamental equations. The Nernst equation relates E_cell° to K, indicating:

$E_{\mathrm{cell}°}=\left(\frac{RT}{NF}\right)ln\left(K\right)$

where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, N is the number of moles of electrons transferred, and F is Faraday's constant (96,485 C/mol e-). The relationship between ΔG° and E_cell° is given by:

${\mathrm{\Delta G}}_{°}=-NF{E}_{\mathrm{cell}°}$

Lastly, the link between ΔG° and K is expressed by:

$K=e^{\frac{-{\mathrm{\Delta G}}_{°}}{RT}}$

These equations allow us to quantitatively assess reaction spontaneity and predict the direction of chemical processes.

Relationship between ∆E°, ∆G°, and K

In the study of spontaneous reactions, three key variables are essential: the standard cell potential (E_naught), the change in standard Gibbs free energy (ΔG_naught), and the equilibrium constant (K). A spontaneous reaction is characterized by a standard cell potential greater than 0, a Gibbs free energy change less than 0, and an equilibrium constant greater than 1. Understanding the relationships among these variables is crucial for determining the spontaneity of a reaction.

These relationships can be visualized through a triangle diagram that connects the three variables. Starting with the connection between the equilibrium constant (K) and the standard cell potential (E_naught), the equation is expressed as:

$$E_{naught} = \frac{RT}{nF} \ln(K)$$

In this equation, R represents the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, n is the number of moles of electrons transferred, and F is Faraday's constant (96,485 C/mol). This equation illustrates how the standard cell potential is influenced by the equilibrium constant.

Next, moving clockwise in the triangle, we examine the relationship between the standard cell potential and Gibbs free energy. The equation is given by:

$$\Delta G_{naught} = -nF E_{naught}$$

This equation indicates that the change in standard Gibbs free energy is directly related to the number of electrons transferred, Faraday's constant, and the standard cell potential. A negative ΔG_naught signifies a spontaneous reaction.

Finally, the connection between the equilibrium constant (K) and Gibbs free energy (ΔG_naught) is represented by the equation:

$$K = e^{-\frac{\Delta G_{naught}}{RT}}$$

This equation shows that the equilibrium constant can be derived from the Gibbs free energy change, reinforcing the interdependence of these three variables. By understanding these equations and their relationships, one can effectively analyze the spontaneity of chemical reactions.

Relationship between ∆E°, ∆G°, and K Example

To calculate the standard cell potential for the reaction involving solid phosphorus (P₄) and oxygen gas (O₂) producing tetraphosphorus decoxide (P₄O₁₀), we start by using the relationship between Gibbs free energy and cell potential. The equation governing this relationship is:

ΔG° = -nF E°

In this equation, ΔG° represents the change in standard Gibbs free energy, n is the number of moles of electrons transferred, F is Faraday's constant (approximately 96,485 coulombs per mole of electrons), and E° is the standard cell potential, which we are trying to find.

Given that 10 moles of electrons are transferred (n = 10), we need to determine ΔG° for the reaction. The change in standard Gibbs free energy can be calculated using the Gibbs free energies of formation for the products and reactants:

ΔG° = Σ(ΔG°_{f, products}) - Σ(ΔG°_{f, reactants})

For this reaction, the Gibbs free energy of formation for P₄O₁₀ is provided as -2984 kJ/mol. Since the reactants (P₄ and O₂) are in their standard states, their Gibbs free energies of formation are zero. Thus, we can simplify the calculation:

ΔG° = (-2984 kJ/mol) - (0) = -2984 kJ

To convert this value into joules, we use the conversion factor where 1 kJ = 1000 J:

ΔG° = -2984 kJ × 1000 J/kJ = -2984000 J

Now, substituting the values into the original equation:

-2984000 J = -10 moles × 96485 C/mol × E°

Rearranging to solve for E° gives:

E° = -2984000 J / (-10 moles × 96485 C/mol)

Calculating this yields:

E° = 3.09 V

Thus, the standard cell potential for the reaction is 3.09 volts, indicating the potential difference generated by the electrochemical reaction under standard conditions.

Given the following standard reduction potentials, determine Ksp for Hg₂Cl₂₍s) at 25 °C.
Hg₂²⁺ (aq) + 2 e^– → 2 Hg (l)                           E°red = + 0.789 V
Hg₂Cl₂ (s) + 2 e^– → 2 Hg (l) + 2 Cl ^– (aq)                E°red = + 0.271 V

• A. 4.93 x 10¹¹
• B. 2.18 x 10¹⁶
• C. 3.27 x 10¹⁷
• D. 6.74 x 10⁴

Answer: (C)

What is the value of the cell potential for the 4 electron transfer reaction below if the equilibrium mixture contains 0.255 M of CH₄, 1.10 M CO₂, 0.388 M CO and 0.250 M H₂ at 25ºC?
                                     CH₄ (g) + CO₂ (g)  ⇌. 2 CO (g) + 2 H₂ (g)

• A. +1.2720 V
• B. -0.2810 V
• C. +0.3021 V
• D. -0.0218 V

Answer: (D)

Given the reaction: 2 Cl2 (g) + 2 H2O (g) ⇌ 4 HCl (g) + O2 (g) Kp = 7.5x10-2, calculate the Gibbs Free Energy change for the reaction below at 30ºC.
8 HCl (g) + 2 O2 (g) ⇌ 4 Cl2 (g) + 4 H2O (g)

• A. -1.3 x 104 J/mol
• B. -2.9 x 103 J/mol
• C. +4.3 x 105 J/mol
• D. +7.7 x 107 J/mol

Answer: (A)

Here’s what students ask on this topic:

What is the relationship between standard cell potential (E°cell) and the equilibrium constant (K)?

The relationship between the standard cell potential (E°cell) and the equilibrium constant (K) is given by the Nernst equation:

$E_{\mathrm{cell}°}=\frac{RT}{NF}ln\left(K\right)$

Here, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, N is the number of moles of electrons transferred, and F is Faraday's constant (96,485 C/mol e-). This equation shows that a higher equilibrium constant (K) corresponds to a higher standard cell potential (E°cell), indicating a more spontaneous reaction.

How do you calculate the change in Gibbs free energy (ΔG°) from the standard cell potential (E°cell)?

The change in Gibbs free energy (ΔG°) can be calculated from the standard cell potential (E°cell) using the following equation:

${\mathrm{\Delta G}}_{°}=-NF{E}_{\mathrm{cell}°}$

In this equation, N is the number of moles of electrons transferred, and F is Faraday's constant (96,485 C/mol e-). This relationship indicates that a positive standard cell potential (E°cell) results in a negative ΔG°, signifying a spontaneous reaction.

What is the significance of a negative ΔG° in a chemical reaction?

A negative ΔG° (change in Gibbs free energy) in a chemical reaction signifies that the reaction is spontaneous under standard conditions. This means that the reaction can proceed without any external input of energy. A negative ΔG° is associated with a positive standard cell potential (E°cell) and an equilibrium constant (K) greater than one, indicating that the products are favored over the reactants at equilibrium.

How are the equilibrium constant (K) and Gibbs free energy (ΔG°) related?

The equilibrium constant (K) and Gibbs free energy (ΔG°) are related by the following equation:

$K=e^{\frac{-{\mathrm{\Delta G}}_{°}}{RT}}$

Here, R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin. This equation shows that a negative ΔG° results in a large equilibrium constant (K), indicating that the reaction favors the formation of products at equilibrium.

What are the conditions for a reaction to be spontaneous?

For a reaction to be spontaneous, it must meet the following conditions:

• The standard cell potential (E°cell) must be greater than zero.
• The change in Gibbs free energy (ΔG°) must be less than zero.
• The equilibrium constant (K) must be greater than one.

These conditions are interconnected through fundamental equations, allowing us to predict the spontaneity of a reaction based on these variables.