Problem 37 - Chemistry The Molecular Science

Question

The standard cell potential for the oxidation of \(\mathrm{Mg}\) by $$ \begin{aligned} \mathrm{Br}_{2} \text { is } 3.42 \mathrm{~V} \\ & \mathrm{Br}_{2}(\ell)+\mathrm{Mg}(\mathrm{s}) \longrightarrow \mathrm{Mg}^{2+}(\mathrm{aq})+2 \mathrm{Br}^{-}(\mathrm{aq}) \end{aligned} $$ (a) Calculate \(\Delta_{\mathrm{r}} G^{\circ}\) for this reaction. (b) Suppose the equation is written with all coefficients doubled. Determine \(\Delta_{\mathrm{r}} G^{\circ}\) and \(E^{\circ}\) for this new equation.

Step-by-Step Solution

Answer

(a) ΔrG° = -660.56 kJ/mol; (b) ΔrG° = -1321.11 kJ/mol, E° = 3.42 V.

Step 1: Use the Nernst Equation for ΔrG°

The relation between Gibbs free energy change (ΔrG°) and standard cell potential (E°) is given by the equation:\[\Delta_{\text{r}} G^{\circ} = -nFE^{\circ}\]where \(n\) is the number of moles of electrons transferred, and \(F\) is the Faraday constant \((96485 \text{ C mol}^{-1})\).

Step 2: Determine the number of moles of electrons (n)

In the given reaction, \( \mathrm{Br}_2(\ell) + \mathrm{Mg}(\mathrm{s}) \rightarrow \mathrm{Mg}^{2+}(\mathrm{aq}) + 2\mathrm{Br}^{-}(\mathrm{aq}) \), two electrons are transferred, i.e., \( n = 2 \).

Step 3: Calculate ΔrG° for the original equation

Substitute \(n = 2\), \(E^{\circ} = 3.42 \text{ V}\), and \(F = 96485 \text{ C mol}^{-1}\) into the equation:\[\Delta_{\text{r}} G^{\circ} = -(2)(96485)(3.42) \text{ J mol}^{-1} = -660557.4 \text{ J mol}^{-1}\]Convert this value to kJ:\[-660.56 \text{ kJ mol}^{-1}\]

Step 4: Understand effect of doubling the equation

Doubling all the coefficients in a chemical equation means that the number of moles of electrons (n) will also double, becoming \(n = 4\).

Step 5: Recalculate ΔrG° for the doubled equation

Using \(n = 4\), substitute into the Gibbs energy equation:\[\Delta_{\text{r}} G^{\circ} = -(4)(96485)(3.42) \text{ J mol}^{-1} = -1321114.8 \text{ J mol}^{-1}\]Convert this to kJ:\[-1321.11 \text{ kJ mol}^{-1}\]

Step 6: Determine E° for the doubled equation

The standard cell potential \(E^{\circ}\) does not depend on the stoichiometry of the reaction; therefore, \(E^{\circ}\) remains the same at \(3.42 \text{ V}\) even if the equation is doubled.

Key Concepts

Standard Cell Potential

In electrochemistry, the Standard Cell Potential, denoted as \( E^{\circ} \), serves as a fundamental measure of a galvanic cell's ability to produce electrical energy. It is defined as the potential difference between the cathode and anode under standard conditions: 25°C, 1 atm pressure, and 1 M concentration for all aqueous solutions.
\( E^{\circ} \) is a critical value for predicting the direction of electron flow in a redox reaction. A positive \( E^{\circ} \) indicates a spontaneous reaction, suggesting the system can perform work without external energy input.
This concept can be visualized by comparing it to a battery: the higher the voltage, the more work the battery can perform. In our exercise example, the oxidation reaction of \( \mathrm{Mg} \) by \( \mathrm{Br}_2 \) has an \( E^{\circ} \) value of 3.42 V, indicating it's a strong reaction capable of transferring electrons effectively.
The use of \( E^{\circ} \) is abundant in calculating other important thermodynamic properties, allowing for a deeper understanding of the energy transformation processes within the cell.

Gibbs Free Energy Change

The Gibbs Free Energy Change, represented as \( \Delta_{\mathrm{r}} G^{\circ} \), is a key thermodynamic function that predicts the spontaneity of chemical reactions. A negative \( \Delta_{\mathrm{r}} G^{\circ} \) implies that the reaction occurs spontaneously under standard conditions. This relationship is explained through the equation:
\[\Delta_{\mathrm{r}} G^{\circ} = -nFE^{\circ} \]where \( n \) is the number of moles of electrons transferred, and \( F \) is Faraday's constant \((96485 \text{ C mol}^{-1})\).
In this context, the value of \( \Delta_{\mathrm{r}} G^{\circ} \) represents the maximum possible work in a chemical system other than pressure-volume work. It is a key determinant in industrial applications where minimizing energy consumption is essential.
For the given problem, we first calculate \( \Delta_{\mathrm{r}} G^{\circ} \) using these inputs and equations, with our reaction yielding a value of \(-660.56 \text{ kJ mol}^{-1}\). This assures that the reaction is not only possible but highly favorable. When the reaction equation is doubled, \( n \) doubles, just as our Gibbs Free Energy does, though \( E^{\circ} \) remains unchanged. This calculation helps to illustrate that while \( \Delta_{\mathrm{r}} G^{\circ} \) scales with the amount of substance, the potential (\( E^{\circ} \)) reflects reaction efficiency, not extent.

Nernst Equation

The Nernst Equation provides a powerful means to relate the concentrations of ions with the electric potential of a cell, thereby extending our understanding beyond standard conditions. The equation is expressed as:
\[ E = E^{\circ} - \frac{RT}{nF} \ln Q \]where \( E \) is the cell potential under non-standard conditions, \( R \) is the universal gas constant (8.314 J/mol·K), \( T \) is the temperature in Kelvin, \( n \) is the number of moles of electrons exchanged, \( F \) is Faraday's constant, and \( Q \) is the reaction quotient reflecting the concentrations of the reacting species.
By using the Nernst Equation, chemists can calculate cell potentials at any given concentration, bridging the gap between theory and real-world conditions where concentrations aren't always at 1 M. This adaptability is crucial for applications like batteries and electroplating, where efficiency must be optimized.
For the reaction we studied, by altering reactant or product concentrations, we could apply the Nernst Equation to determine new cell potentials, offering a predictive edge for understanding how changes in concentration impact cell behavior.