Question

The standard potential \(E^{\circ}\) for a given cell is \(1.135 \mathrm{V}\) at \(298.15 \mathrm{K}\) and \(\left(\partial E^{\circ} / \partial T\right)_{P}=-4.10 \times 10^{-5} \mathrm{V} \mathrm{K}^{-1}\). Calculate \(\Delta G_{R}^{\circ}, \Delta S_{R}^{\circ},\) and \(\Delta H_{R}^{\circ} .\) Assume that \(n=2\).

Short answer

The standard Gibbs free energy change, \(\Delta G_{R}^{\circ}\), is \(-218707.7 \mathrm{J/mol}\); the standard entropy change, \(\Delta S_{R}^{\circ}\), is \(7.9197 \mathrm{J/mol\cdot K}\); and the standard enthalpy change, \(\Delta H_{R}^{\circ}\), is \(-216586.5 \mathrm{J/mol}\).

Step-by-step solution

Calculate the Standard Gibbs Free Energy Change

We are given the standard potential, \(E^{\circ} = 1.135 \mathrm{V}\), and the number of electrons transferred in the redox reaction, \(n = 2\). The Faraday constant, \(F = 96485 \mathrm{C/mol}\), is needed to calculate the Gibbs free energy change using the equation \(\Delta G_{R}^{\circ} = -nFE^{\circ}\). \[\Delta G_{R}^{\circ} = -(2)(96485 \mathrm{C/mol})(1.135 \mathrm{V})\] \[\Delta G_{R}^{\circ} = -218707.7 \mathrm{J/mol}\]

Calculate the Standard Entropy Change

We are given the partial derivative of the standard potential with respect to temperature, \(\left(\partial E^{\circ} / \partial T\right)_{P} = -4.10 \times 10^{-5} \mathrm{V} \mathrm{K}^{-1}\). We will use the equation \(\left(\frac{\partial E^{\circ}}{\partial T}\right)_{P} = -\frac{\Delta S_{R}^{\circ}}{nF}\) in addition to the values of \(n\) and \(F\) to calculate \(\Delta S_{R}^{\circ}\). \[-4.10 \times 10^{-5} \mathrm{V} \mathrm{K}^{-1} = -\frac{\Delta S_{R}^{\circ}}{(2)(96485 \mathrm{C/mol})}\] \[\Delta S_{R}^{\circ} = -(-4.10 \times 10^{-5} \mathrm{V} \mathrm{K}^{-1})(2)(96485 \mathrm{C/mol})\] \[\Delta S_{R}^{\circ} = 7.9197 \mathrm{J/mol\cdot K}\]

Calculate the Standard Enthalpy Change

To calculate the standard enthalpy change, \(\Delta H_{R}^{\circ}\), we will use the equation \(\Delta H_{R}^{\circ} = \Delta G_{R}^{\circ} + T\Delta S_{R}^{\circ}\) and the given temperature, \(T = 298.15 \mathrm{K}\). \[\Delta H_{R}^{\circ} = -218707.7 \mathrm{J/mol} + (298.15 \mathrm{K})(7.9197 \mathrm{J/mol\cdot K})\] \[\Delta H_{R}^{\circ} = -216586.5 \mathrm{J/mol}\] So, the standard Gibbs free energy change, \(\Delta G_{R}^{\circ}\), is \(-218707.7 \mathrm{J/mol}\); the standard entropy change, \(\Delta S_{R}^{\circ}\), is \(7.9197 \mathrm{J/mol\cdot K}\); and the standard enthalpy change, \(\Delta H_{R}^{\circ}\), is \(-216586.5 \mathrm{J/mol}\).

Key concepts

Standard Electrode Potential

In electrochemistry, the standard electrode potential, denoted as \( E^{\circ} \), is a measure of the tendency of a chemical species to be reduced, and is under standard conditions: a 1 M concentration for each ion, a pressure of 1 atm for each gas, and a temperature of 25°C (298 K). It represents the voltage difference that drives electrons through the circuit in an electrochemical cell.

This potential is determined relative to the standard hydrogen electrode, which is assigned a potential of 0.00 V. If \( E^{\circ} \) is positive, the species is more likely to gain electrons (be reduced) than the hydrogen ion. Conversely, a negative \( E^{\circ} \) indicates a poorer ability to capture electrons compared to hydrogen ions.

In the context of this exercise, the standard electrode potential \( E^{\circ} = 1.135 \ \mathrm{V} \) suggests a strong ability of the cell, under standard conditions, to drive its redox reaction. The partial derivative of \( E^{\circ} \) with respect to temperature, \( \left( \partial E^{\circ} / \partial T \right)_{P} \), is also vital as it relates to the entropy change of the reaction.

Gibbs Free Energy

Gibbs Free Energy, denoted as \( \Delta G \), is a measure of the maximum usable energy that can be derived from a system, including chemical reactions, at constant temperature and pressure. It is defined as the energy associated with a chemical reaction that can be used to do work.

For spontaneous reactions (those that proceed without external input), \( \Delta G \) is negative. In electrochemical cells, the free energy change is connected to the number of moles of electrons transferred (\( n \)), the Faraday constant (\( F \)), and the standard cell potential (\( E^{\circ} \)): \[ \Delta G_{R}^{\circ} = -nFE^{\circ} \]
Using this formula, the standard Gibbs free energy change for the cell reaction is calculated to be \( -218707.7 \ \mathrm{J/mol} \). This value confirms the reaction's ability to perform work under standard conditions, indicating it can spontaneously drive the electron flow from anode to cathode at the given potential.

Entropy Change

Entropy change, represented as \( \Delta S \), indicates the degree of disorder or randomness in the system during a reaction. In simple terms, it refers to the dispersal of energy and matter. Entropy increase is common in natural processes and is often associated with state changes and energy transformations.

In electrochemical reactions, the entropy change can be calculated if the temperature dependence of the electrode potential is known. The equation used for this calculation is:
\[ \left( \frac{\partial E^{\circ}}{\partial T} \right)_{P} = -\frac{\Delta S_{R}^{\circ}}{nF} \]

From the exercise, the standard entropy change is found to be \( 7.9197 \ \mathrm{J/mol \cdot K} \), suggesting a mild increase in disorder during the reaction. This is typical because the electric energy is released and disorder slightly increases as electrons move from higher to lower potential.

Enthalpy Change

Enthalpy change, expressed as \( \Delta H \), embodies the total energy change in a reaction and is a sum of internal energy changes and work done due to volume changes at constant pressure. It provides insight into the energy required to break and form chemical bonds.

For electrochemical reactions, there's a direct relation linking enthalpy, Gibbs free energy, and entropy via the thermodynamic equation: \[ \Delta H = \Delta G + T\Delta S \]
In the task at hand, the enthalpy change calculated was \( -216586.5 \ \mathrm{J/mol} \). This negative value indicates that the reaction is exothermic, releasing heat to its surroundings. This makes sense since, in electrochemical cells, energy is usually released as electric work along with heat, confirming the likelihood of enthalpy helping drive the reaction.