Question

For a given overall cell reaction, \(\Delta S_{R}^{\circ}=16.5 \mathrm{Jmol}^{-1} \mathrm{K}^{-1}\) and \(\Delta H_{R}^{\circ}=-270.0 \mathrm{kJ} \mathrm{mol}^{-1}\) Calculate \(E^{\circ}\) and \(\left(\partial E^{\circ} / \partial T\right)_{P} .\) Assume that \(n=2\).

Short answer

The standard Gibbs free energy at \(T = 298.15 K\) is \(\Delta G_{R}^{\circ}=-270.0\,\text{kJ}\,\text{mol}^{-1} - \left(298.15\,\text{K}\right)(16.5\,\text{J}\,\text{mol}^{-1}\,\text{K}^{-1}) \approx -275.1\,\text{kJ}\,\text{mol}^{-1}\). The standard cell potential is \(E^{\circ} = \frac{-(-275.1\,\text{kJ}\,\text{mol}^{-1})}{(2)(96,485\,\text{C}\,\text{mol}^{-1})} \approx 1.426\,\text{V}\). The partial derivative of the standard cell potential with respect to the temperature at constant pressure is \(\left(\frac{\partial E^{\circ}}{\partial T}\right)_{P}=\frac{-16.5\,\text{J}\,\text{mol}^{-1}\,\text{K}^{-1}}{(2)(96,485\,\text{C}\,\text{mol}^{-1})} \approx -8.56 \times 10^{-5}\,\text{VK}^{-1}\).

Step-by-step solution

Determine standard Gibbs free energy using entropy and enthalpy

Using the given standard reaction entropy (\(\Delta S_{R}^{\circ}=16.5 Jmol^{-1}K^{-1}\)) and enthalpy (\(\Delta H_{R}^{\circ}=-270.0 kJmol^{-1}\)), we can determine the standard Gibbs free energy (\(\Delta G_{R}^{\circ}\)) at a specific temperature (T) using the equation: \[\Delta G_{R}^{\circ}=\Delta H_{R}^{\circ}-T\Delta S_{R}^{\circ}\] At standard conditions, the temperature is \(T = 298.15 K\). So, we can plug this value and the given data into the equation to get the Gibbs free energy.

Calculate the standard cell potential

To calculate the standard cell potential (\(E^{\circ}\)), we can use the relationship between standard Gibbs free energy and cell potential: \[\Delta G_{R}^{\circ} = -nFE^{\circ}\] Where n is the number of electrons transferred in the cell reaction (given as 2) and F is the Faraday constant (\(F \approx 96,485 Cmol^{-1}\)). We can rearrange the equation to solve for \(E^{\circ}\): \[E^{\circ} = \frac{-\Delta G_{R}^{\circ}}{nF}\] Substitute the values obtained in Step 1 and the given value of n to find \(E^{\circ}\).

Determine the partial derivative of cell potential with respect to temperature

We will now calculate the partial derivative of the standard cell potential with respect to temperature at constant pressure \(\left(\partial E^{\circ} / \partial T\right)_{P}\) using the Gibbs-Helmholtz equation: \[\left(\frac{\partial E^{\circ}}{\partial T}\right)_{P}=\frac{-\Delta S_{R}^{\circ}}{nF}\] In this equation, substitute the given value of entropy (\(\Delta S_{R}^{\circ}\)) and the number of electrons transferred in the cell reaction (n = 2) along with the Faraday constant to calculate the partial derivative of cell potential with respect to temperature at constant pressure.

Key concepts

Understanding Electrochemistry

Electrochemistry is the branch of chemistry that deals with the relationship between electricity and chemical reactions. It's particularly focused on the processes that involve electron transfer, known as redox reactions. In these reactions, oxidation, the loss of electrons, and reduction, the gain of electrons, occur simultaneously.

Devices like batteries, fuel cells, and electrochemical cells are practical applications of electrochemistry. These devices convert chemical energy into electrical energy or vice versa. The performance and efficiency of these cells are often characterized by their cell potential, which is a measure of the cell's ability to produce an electric current.

In the context of our exercise, the cell's overall energy change can be understood through the concepts of enthalpy, entropy, and Gibbs free energy, which helps determine the feasibility and spontaneity of the redox reactions that occur within the electrochemical cell.

Deciphering Cell Potential

Cell potential, often denoted as \(E^\circ\), is a key concept in electrochemistry that reflects the voltage, or electromotive force, produced by an electrochemical cell when no current is flowing. The standard cell potential is measured under standard conditions, which typically means a temperature of 298.15 K, 1 atm pressure, and 1 M concentrations for all solutions involved in the reaction.

The value of \(E^\circ\) is crucial because it indicates whether a cell will spontaneously react under standard conditions; a positive cell potential means that the reaction is spontaneous, while a negative value suggests non-spontaneity. Moreover, \(E^\circ\) can be related to the Gibbs free energy change and the number of electrons transferred in the reaction, showing that electrochemical reactions are fundamentally connected to thermodynamics.

Gibbs-Helmholtz Equation in Action

The Gibbs-Helmholtz equation is an essential relation in thermodynamics that describes the temperature dependence of the standard Gibbs free energy change (\(\Delta G_R^\circ\)). It links thermal dynamics with electrochemistry, providing a way to understand how temperature affects cell potential. The equation can be simplified in the electrochemical context to determine the effect of temperature on cell potential.

From our initial exercise, we used the partial derivative \(\left(\frac{\partial E^\circ}{\partial T}\right)_P\) to understand how the standard cell potential changes with temperature at constant pressure. The derivative is related directly to the reaction entropy, another key thermodynamic property. As temperature changes, the reaction entropy determines whether the cell potential will increase or decrease, further explaining the behavior of electrochemical systems under varying thermal conditions.