Question

Given the following two standard reduction potentials, $$\begin{array}{ll}{\mathrm{M}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{M}} & {\mathscr{E}^{\circ}=-0.10 \mathrm{V}} \\ {\mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M}} & {\mathscr{E}^{\circ}=-0.50 \mathrm{V}}\end{array}$$ solve for the standard reduction potential of the half-reaction $$\mathrm{M}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{M}^{2+}$$ (Hint: You must use the extensive property \(\Delta G^{\circ}\) to determine the standard reduction potential.)

Short answer

The standard reduction potential of the half-reaction M³⁺ + e⁻ → M²⁺ is -0.699 V.

Step-by-step solution

Find the ∆G° for M³⁺ → M and M²⁺ → M half-reactions

First, let's calculate the ∆G° values for both given reduction reactions using the equation: ∆G° = -nFE°. For M³⁺ → M: E° = -0.10 V; n = 3 ∆G° = -(3)(96,485 C/mol)(-0.10 V) ∆G° = 29,045.5 J/mol For M²⁺ → M: E° = -0.50 V; n = 2 ∆G° = -(2)(96,485 C/mol)(-0.50 V) ∆G° = 96,485 J/mol

Find the ∆G° for M³⁺ → M²⁺ half-reaction using given ∆G° values

We can relate the ∆G° of the reactions with the following equation: ∆G°(M³⁺ → M²⁺) = ∆G°(M²⁺ → M) - ∆G°(M³⁺ → M) ∆G°(M³⁺ → M²⁺) = 96,485 J/mol - 29,045.5 J/mol ∆G°(M³⁺ → M²⁺) = 67,439.5 J/mol

Find the standard reduction potential of M³⁺ → M²⁺ using the calculated ∆G°

Now we can find the standard reduction potential of the required half-reaction using the equation: E° = -∆G°/(nF) In this reaction (M³⁺ → M²⁺), only one electron is involved: n = 1. E°(M³⁺ → M²⁺) = -67,439.5 J/mol / (1 * 96,485 C/mol) E°(M³⁺ → M²⁺) = -0.699 V So, the standard reduction potential of the half-reaction M³⁺ + e⁻ → M²⁺ is -0.699 V.

Key concepts

Standard Reduction Potential

The standard reduction potential, often denoted as \( \mathscr{E}^{\circ} \), is a measure of the tendency of a chemical species to gain electrons and thereby be reduced. A more negative \( \mathscr{E}^{\circ} \) value indicates a lesser tendency to gain electrons. It is typically measured in volts (V), and each half-reaction has a unique value.
In our given problem, we are provided with the standard reduction potentials for two reactions involving the species \( M^{3+} \) and \( M^{2+} \). The values were \(-0.10\) V for \( M^{3+} + 3e^- \rightarrow M \) and \(-0.50\) V for \( M^{2+} + 2e^- \rightarrow M \).
These potentials are determined under standard conditions, which include 1 M concentration of ions, a pressure of 1 atm, and a temperature of 298 K, or 25 °C. It is crucial to note that these potentials allow us to predict the direction of electron flow in electrochemical reactions.

• Greater negative value: Less likely to be reduced.
• Greater positive value: More likely to be reduced.

Gibbs Free Energy

Gibbs free energy, represented by \( \Delta G^{\circ} \), is a thermodynamic property that indicates the maximum reversible work that can be performed by a thermodynamic system at constant temperature and pressure. In electrochemistry, it is directly related to the cell potential and the number of electrons transferred, following the equation \( \Delta G^{\circ} = -nFE^{\circ} \), where \( n \) is the number of moles of electrons, \( F \) is the Faraday constant (approximately 96,485 C/mol), and \( E^{\circ} \) is the standard cell potential.
In the solution of this exercise, we observed the calculation of \( \Delta G^{\circ} \) for the two half-reactions given:

• \( \Delta G^{\circ} \) for \( M^{3+} \rightarrow M \) was 29,045.5 J/mol.
• \( \Delta G^{\circ} \) for \( M^{2+} \rightarrow M \) was 96,485 J/mol.

These values were then used to find the \( \Delta G^{\circ} \) for the half-reaction \( M^{3+} \rightarrow M^{2+} \), giving us 67,439.5 J/mol. This indicates the energy change for the process of converting \( M^{3+} \) to \( M^{2+} \), guided by the principle that systems tend to move towards lower energy states.
This property is crucial for determining the feasibility of reactions, providing an understanding of the energy changes, and whether a process is spontaneous or non-spontaneous under standard conditions.

Half-Reaction

A half-reaction is a representation of either the oxidation or reduction process in a redox reaction. Each half-reaction shows the transfer of electrons, which are involved in the overall reaction, making it a cornerstone concept in electrochemistry.
In this exercise, the two half-reactions were:

• \( M^{3+} + 3e^- \rightarrow M \)
• \( M^{2+} + 2e^- \rightarrow M \)

These illustrate the reduction of \( M^{3+} \) and \( M^{2+} \) ions to \( M \). The step-by-step solution required determining the half-reaction \( M^{3+} + e^- \rightarrow M^{2+} \).
By analyzing half-reactions, we can discern the direction of electron flow and the specific elements involved in gaining or losing electrons. This is foundational for calculating properties such as reduction potentials and Gibbs free energy changes and is pivotal in determining the overall cell reactions and their feasibility. Understanding these half-reactions allows chemists and engineers to predict and manipulate the behavior of electrochemical systems, including batteries and corrosion processes.