Question

If the equilibrium constant for a one-electron redox reaction at 298 \(\mathrm{K}\) is \(8.7 \times 10^{4}\) , calculate the corresponding \(\Delta G^{\circ}\) and \(E^{\circ} .\)

Short answer

The Gibbs free energy change (ΔG°) for this one-electron redox reaction is approximately -5.336 x 10⁴ J mol⁻¹, and the standard electrode potential (E°) is approximately 0.553 V.

Step-by-step solution

Use the relationship between ΔG° and K

We know that ΔG° and K are related by the following equation: \[\Delta G^{\circ} = -RT \ln K\] where R is the gas constant (8.314 J mol⁻¹ K⁻¹), T is the temperature (298 K), and K is the equilibrium constant (8.7 x 10⁴). First, we'll plug in the values to calculate ΔG°: \[\Delta G^{\circ} = -(8.314 \text{ J mol}^{-1} \text{K}^{-1})(298 \text{ K}) \ln (8.7 \times 10^{4})\]

Calculate ΔG°

Use a calculator to evaluate the equation and find ΔG°: \[\Delta G^{\circ} \approx -5.336 \times 10^{4} \text{ J mol}^{-1}\] The Gibbs free energy change for this one-electron redox reaction is approximately -5.336 x 10⁴ J mol⁻¹.

Use the relationship between ΔG° and E° for a one-electron redox reaction

Next, we will use the relationship between ΔG° and E°, which is given by the following equation: \[\Delta G^{\circ} = -nFE^{\circ}\] where n is the number of electrons transferred (which is 1 in a one-electron redox reaction), F is Faraday's constant (96485 C mol⁻¹), and E° is the standard electrode potential. For this reaction, we have: \[-5.336 \times 10^{4} \text{ J mol}^{-1} = -(1)(96485 \text{ C mol}^{-1})E^{\circ}\]

Calculate E°

To find E°, solve the equation: \[E^{\circ} = \frac{-5.336 \times 10^{4} \text{ J mol}^{-1}}{96485 \text{ C mol}^{-1}}\] \[E^{\circ} \approx 0.553 \text{ V}\] The standard electrode potential for this one-electron redox reaction is approximately 0.553 V.

Key concepts

Understanding Gibbs Free Energy (\(\Delta G^\circ\))

Gibbs free energy, typically represented as \(\Delta G^\circ\), plays a pivotal role in thermodynamics and particularly in the study of chemical reactions. It is used to determine the spontaneity of a chemical process at constant temperature and pressure. A negative value of \(\Delta G^\circ\) indicates a reaction that can occur spontaneously, while a positive value suggests a non-spontaneous reaction.

To calculate the change in Gibbs free energy for a reaction, we use the equation \(\Delta G^\circ = -RT \ln K\), where \( R \) is the universal gas constant, \( T \) is the absolute temperature, and \( K \) is the equilibrium constant. For the given redox reaction, plugging in the values yields a \(\Delta G^\circ\) of approximately -5.336 x 10^4 J/mol, suggesting that the reaction is spontaneous.

This calculation is crucial because it provides insight into the feasibility of chemical processes without needing to perform them. Therefore, grasping the concept of Gibbs free energy allows students and scientists alike to predict reaction behavior and drive the development of new reactions and materials.

The Role of Standard Electrode Potential (E°) in Redox Reactions

Standard electrode potential, designated as \(E^\circ\), is a measure of the individual potential of a reversible electrode at standard state conditions, which is a crucial concept in electrochemistry. It allows chemists to understand the tendency of a species to lose or gain electrons, known as the redox potential. The more positive the standard electrode potential, the greater the species' affinity for electrons; hence, it is a stronger oxidizing agent.

In a redox reaction, the difference in standard electrode potentials for the oxidation and reduction half-reactions can be used to calculate the cell potential for the overall reaction. Across a one-electron transfer reaction, the relationship \(\Delta G^\circ = -nFE^\circ\) equates Gibbs free energy change with standard electrode potential, where \( n \) represents the number of electrons transferred and \( F \) is Faraday's constant.

In our case, solving the equation with the given Gibbs free energy change results in a standard electrode potential of approximately 0.553 V for a one-electron redox reaction, which illustrates the potential's magnitude for driving the reaction.

The Equilibrium Constant (K) and Its Significance

The equilibrium constant, symbolized as \(K\), offers a quantitative measure of the position of equilibrium in a chemical reaction. In essence, it relates the concentrations of the products to the reactants at equilibrium for a reversible reaction. A larger \(K\) implies a reaction with a greater extent of product formation, while a smaller \(K\) indicates that the reactants are predominant at equilibrium.

The magnitude of the equilibrium constant for a given temperature is directly related to the standard Gibbs free energy change by the equation \(\Delta G^\circ = -RT \ln K\). Therefore, knowing \(K\) allows us to calculate \(\Delta G^\circ\), providing insight into the reaction's spontaneity. For the redox reaction in question, with \(K = 8.7 \times 10^4\), the substantial magnitude of \(K\) signifies a reaction that heavily favors the formation of products.

In educational contexts, understanding \(K\) is fundamental for students to predict the behavior and outcome of chemical reactions. It not only influences chemical manufacturing and experimental design but is also a key concept in many natural processes, such as in environmental, biological, and geophysical systems.