Question

Calculate \(\Delta G_{\mathrm{mn}}^{\mathrm{a}}\) and \(E_{\mathrm{cell}}^{\circ}\) for a redox reaction with \(n=2\) that has an equilibrium constant of \(K=\) 25 \(\left(\operatorname{at} 25^{\circ} \mathrm{C}\right)\)

Short answer

\(\Delta G_{\mathrm{mn}}^{\mathrm{a}} = -RT\ln K = -8.314 \text{ J/(mol*K)} \times 298.15 K \times \ln(25)\) and \(E_{\mathrm{cell}}^{\circ} = -\Delta G^{\circ}/(nF)\).

Step-by-step solution

Calculate the standard Gibbs free energy change \(\Delta G_{\mathrm{mn}}^{\mathrm{a}}\)

Use the equation \(\Delta G^{\circ} = -RT\ln K\). R is the universal gas constant \(8.314 \text{ J/(mol*K)}\) and T is the temperature in Kelvin. Since the temperature is given as \(25^\circ C\), convert it to Kelvin by adding 273.15 to get 298.15 K. Next, calculate \(\Delta G^{\circ}\) using the given equilibrium constant (K).

Insert values into the Gibbs free energy equation

Insert the values into the equation: \(\Delta G^{\circ} = -8.314 \text{ J/(mol*K)} \times 298.15 K \times \ln(25)\). Compute the value of \(\Delta G^{\circ}\).

Calculate the standard cell potential \(E_{\mathrm{cell}}^{\circ}\)

Use the equation relating Gibbs free energy change to cell potential: \(\Delta G^{\circ} = -nFE_{\mathrm{cell}}^{\circ}\) where \(n\) is the number of moles of electrons transferred and \(F\) is Faraday's constant \((96485 \text{ C/mol})\). Rearrange the equation to solve for \(E_{\mathrm{cell}}^{\circ}\) and insert the previously calculated \(\Delta G^{\circ}\) and the given \(n=2\) to find \(E_{\mathrm{cell}}^{\circ}\).

Key concepts

Equilibrium Constant

The equilibrium constant, represented as 'K', is central to understanding the behavior of a system at equilibrium. It quantifies the ratio of concentrations of products to reactants at equilibrium for a reversible chemical reaction, at a constant temperature. The higher the value of K, the more the reaction favors the formation of products.

In the context of a redox reaction, the equilibrium constant can reveal insights about the spontaneity and extent of the reaction. Given the equilibrium constant K=25, we can infer that the reaction is product-favored, meaning that the equilibrium lies to the right, resulting in more products than reactants at equilibrium.

To provide further clarity using the original exercise: When one calculates the Gibbs free energy change, \(\Delta G^{\circ} = -RT\ln K\), for K=25 at 298.15 K (25\(\degree\)C), it resonates with the product-favored nature of the reaction. The negative sign in front of RT in the equation implies that if K > 1 (which it is, since it's 25), the calculated Gibbs free energy change will also be negative, indicating a spontaneous reaction under standard conditions.

Standard Cell Potential

The standard cell potential, denoted as \(E_{\mathrm{cell}}^{\circ}\), is a measure of the electrical potential of a cell when all components are at their standard states, typically 1 bar for gases and 1 M for solutes, at a specified temperature (25\(\degree\)C unless otherwise noted). It is a thermodynamic property directly related to the Gibbs free energy change of a redox reaction via the equation \(\Delta G^{\circ} = -nFE_{\mathrm{cell}}^{\circ}\), where \(n\) is the number of moles of electrons transferred in the redox reaction, and \(F\) stands for Faraday's constant.

The standard cell potential provides an indication of how much voltage the cell can produce when operated in standard conditions. It is a crucial parameter for determining the efficiency of electrochemical cells and batteries. A positive standard cell potential suggests a spontaneous reaction, aligning with a negative Gibbs free energy. Using the example from the exercise, by solving the equation with the given \(n=2\) and the calculated Gibbs free energy, the standard cell potential can be determined, showcasing the linkage between the calculated physical properties and the electrochemical behavior of the reaction.

Redox Reaction

Redox reactions are a family of reactions where oxidation and reduction simultaneously occur. Oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. They are essential in numerous applications such as batteries, electrolysis, and biological systems.

Understanding the flow of electrons is key in analyzing redox reactions. The number of electrons involved in the reaction affects both the equilibrium constant through the relation with Gibbs free energy and the standard cell potential. In the given exercise, with \(n=2\) we understand that two moles of electrons are being transferred, impacting the calculation of the cell potential. This transfer of electrons is what drives the chemical energy to be converted into electrical energy, and vice versa, in electrochemical cells.

By knowing that the equilibrium constant \(K\) is related to the Gibbs free energy, and that the Gibbs free energy is also linked to the standard cell potential, there is a profound interconnection between these concepts that govern the behavior and quantitative assessment of redox reactions. This interplay is essential for predicting the directionality and spontaneity of the reactions under standard conditions.