Problem 43 - Chemistry The Science in Context

Question

The value of \(E_{\text {cell for the reaction below is } 0.500 \mathrm{V} . \text { What is }}\) the value of \(\Delta G_{\text {cell }}^{\text {? }}\) $$\mathrm{Mn}^{3+}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Mn}^{2+}+\mathrm{MnO}_{2}+4 \mathrm{H}^{+}$$

Step-by-Step Solution

Answer

Answer: The Gibbs free energy change (ΔG_cell) for the given reaction is -96,485 J/mol.

Step 1: Identify the half-reactions

The given reaction can be broken down into two half-reactions: 1. Oxidation half-reaction (Mn²⁺ being oxidized to Mn³⁺): Mn³⁺ → Mn²⁺ + e⁻ 2. Reduction half-reaction (MnO₂ being reduced to Mn²⁺): MnO₂ + 4H⁺ + 2e⁻ → Mn²⁺ + 2H₂O

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

The balanced equation for the given reaction is: Mn³⁺ + 2H₂O → Mn²⁺ + MnO₂ + 4H⁺ To balance the charges, we will need to transfer 2 moles of electrons (since the reduction half-reaction involves a two-electron transfer). So, n = 2.

Step 3: Calculate ΔG_cell using the equation

Now that we have the values for E_cell (0.500 V) and n (2), we can plug them into the equation: ΔG_cell = -nFE_cell ΔG_cell = -(2)(96,485 C/mol)(0.500 V) ΔG_cell = -96,485 J/mol (remember that 1 V = 1 J/C) The Gibbs free energy change (ΔG_cell) for the given reaction is -96,485 J/mol.

Key Concepts

Cell Potential and Its Significance

Cell potential, often represented as \(E_{\text{cell}}\), is a measure of the potential difference between two electrodes of an electrochemical cell. It serves as a measure of how much work a cell can do, reflecting the ability of the cell to drive an electric current through an external circuit.
The cell potential is expressed in volts (V) and is vital in determining the likelihood of a chemical reaction to occur spontaneously. A positive cell potential indicates that the reaction is thermodynamically favorable. In the given exercise, the \(E_{\text{cell}}\) is 0.500 V, signifying a reaction where products are more stable than reactants under standard conditions.

• This potential is affected by factors such as temperature, pressure, and concentration of solutions.
• It helps in calculating the Gibbs free energy, \(\Delta G\), which predicts the spontaneity of reactions.

Understanding cell potential also assists in studying battery functions where chemicals convert to electrical energy.

Understanding Half-Reactions

In electrochemistry, reactions are often split into two parts called half-reactions. These help in analyzing the process of electron transfer, highlighting each substance's role in the overall reaction.
For the equation given in the exercise, the breakdown results in:

• Oxidation half-reaction: where a substance loses electrons. In this case, \(\text{Mn}^{3+}\) changes to \(\text{Mn}^{2+}\).
• Reduction half-reaction: where a substance gains electrons, shown by \(\text{MnO}_2 + 4\text{H}^{+} + 2e^- \rightarrow \text{Mn}^{2+} + 2\text{H}_2\text{O}\).

The separation into half-reactions allows for a detailed view of oxidation and reduction processes, clarifying the transfer of electrons and balancing of charges. This is essential for calculating aspects such as \(n\), the number of moles of electrons transferred, as seen in the solution, where \(n = 2\).

The Role of Electron Transfer

Electron transfer is a core concept in understanding electrochemical reactions, where electrons shift from one reactant to another, changing their oxidation states.
This transfer is evident in the half-reactions where:

• Electrons are transferred from the MnO₂ molecule to the Mn³⁺ ions.
• The net effect results in a balanced flow of electrons, maintaining charge neutrality.

Such electron movements are depicted clearly from the balanced overall equation in the exercise where 2 electrons are transferred, correlating directly to the calculation of Gibbs free energy change using \(n\). This concept not only highlights the technical aspects of reactions but underscores fundamental properties of matter that drive circuit flows in batteries and other applications.

Oxidation and Reduction Reactions

Within any redox reaction, oxidation and reduction must occur simultaneously to conserve charge, adhering to the principle that electrons lost in oxidation are precisely gained in reduction.
This framework is pivotal in understanding the given reaction, where:

• Oxidation: \(\text{Mn}^{3+}\) is oxidized to \(\text{Mn}^{2+}\), illustrating the loss of an electron by the Mn ion.
• Reduction: MnO₂ is reduced to \(\text{Mn}^{2+}\), showing the gain of electrons, creating a reduced element.

Mastering these processes is essential because they are foundational to diverse applications such as energy storage and chemical synthesis and comprehend the calculation of Gibbs free energy change from cell potential, demonstrating the interplay of energy and electron flow in electrochemical systems.