Problem 44 - Chemistry The Science in Context

Question

What is the value of \(\Delta G_{\text {cell }}^{\circ}\) for an electrochemical cell based on a cell reaction described by the following net ionic equation? $$\mathrm{Mg}+2 \mathrm{Cu}^{+} \rightarrow \mathrm{Mg}^{2+}+2 \mathrm{Cu}$$

Step-by-Step Solution

Answer

Answer: The value of the standard change in Gibbs free energy (\(\Delta G_{\text{cell}}^{\circ}\)) for the given electrochemical cell is \(357290 \, J \cdot mol^{-1}\).

Step 1: Determine the Half-Reactions

Separate the given equation into two half-reactions in reduction form: 1) Mg reduction: \(\mathrm{Mg}^{2+} + 2e^- \rightarrow \mathrm{Mg}\) 2) Cu reduction: \(\mathrm{Cu}^+ + e^- \rightarrow \mathrm{Cu}\) Since we need the oxidation half-reaction for the cell, we will reverse the Mg reduction half-reaction to get the Mg oxidation half-reaction: Mg oxidation: \(\mathrm{Mg} \rightarrow \mathrm{Mg}^{2+} + 2e^-\) Now our half-reactions are: 1) Mg oxidation: \(\mathrm{Mg} \rightarrow \mathrm{Mg}^{2+} + 2e^-\) 2) Cu reduction: \(\mathrm{Cu}^+ + e^- \rightarrow \mathrm{Cu}\)

Step 2: Find Standard Reduction Potentials

Look up the standard reduction potentials for both half-reactions in a reference table: 1) Mg reduction: \(E_{\text{Mg}}^{\circ} = -2.37 \, V\) 2) Cu reduction: \(E_{\text{Cu}}^{\circ} = 0.52 \, V\) Since we have reversed the Mg reduction half-reaction to get the Mg oxidation half-reaction, the potential of the Mg half-reaction will be the opposite of the standard reduction potential: \(E_{\text{Mg}}^{\circ} = 2.37 \, V\)

Step 3: Calculate the Cell Potential

The standard cell potential can be found by subtracting the potential of the oxidation half-reaction from the potential of the reduction half-reaction: \(E_{\text{cell}}^{\circ} = E_{\text{Cu}}^{\circ} - E_{\text{Mg}}^{\circ} = 0.52 \, V - 2.37 \, V = -1.85 \, V\)

Step 4: Determine the Change in Gibbs Free Energy

Using the formula for the standard change in Gibbs free energy: $$\Delta G_{\text{cell}}^{\circ} = -nFE_{\text{cell}}^{\circ}$$ where \(n\) is the number of moles of electrons transferred, \(F\) is the Faraday constant (\(F = 96485 \, C \cdot mol^{-1}\)), and \(E_{\text{cell}}^{\circ}\) is the standard cell potential. From the half-reactions, we can see that the number of moles of electrons transferred (\(n\)) is 2. Substituting the values into the formula: $$\Delta G_{\text{cell}}^{\circ} = -2 \times 96485 \, C \cdot mol^{-1} \times -1.85 \, V = 357290 \, J \cdot mol^{-1}$$ The value of the standard change in Gibbs free energy (\(\Delta G_{\text{cell}}^{\circ}\)) for the given electrochemical cell is \(357290 \, J \cdot mol^{-1}\).

Key Concepts

Gibbs Free Energy

Gibbs Free Energy in the context of electrochemical cells is a measure of the maximum work that can be extracted from a chemical process at constant temperature and pressure. In simpler terms, it tells us how much usable energy is available from a reaction. In electrochemical cells, the change in Gibbs Free Energy (denoted as \( \Delta G \)) is directly related to the cell potential and the number of electrons involved in the reaction.The general equation to calculate \( \Delta G \) for electrochemical processes is given by:\[ \Delta G = -nFE_{\text{cell}} \]

• \( n \) is the number of moles of electrons transferred.
• \( F \) is the Faraday constant, which is approximately 96485 C/mol.
• \( E_{\text{cell}} \) is the cell potential.

In our problem, we see that \( \Delta G_{\text{cell}}^{\circ} = -2 \times 96485 \, C \cdot mol^{-1} \times -1.85 \, V = 357290 \, J \cdot mol^{-1} \), indicating the amount of energy available per mole of the reaction.

standard reduction potentials

Standard reduction potentials are essential in determining how much energy can be gained or lost during a chemical reaction in an electrochemical cell. They measure a tendency of a chemical species to gain electrons, hence being reduced. Each half-reaction in an electrochemical cell has a standard potential, denoted as \( E^{\circ} \), which is measured in volts against a standard hydrogen electrode.In the original exercise, the standard reduction potentials used are:

• For Mg: \( E_{\text{Mg}}^{\circ} = -2.37 \, V \)
• For Cu: \( E_{\text{Cu}}^{\circ} = 0.52 \, V \)

Since the Mg reaction was flipped to an oxidation reaction, its standard potential becomes positive \( 2.37 \, V \). The cell potential is then calculated by the difference:\[ E_{\text{cell}}^{\circ} = E_{\text{Cu}}^{\circ} - E_{\text{Mg}}^{\circ} = 0.52 \, V - (-2.37 \, V) = -1.85 \, V \]This negative potential indicates that the reaction, as written, is not spontaneous.

half-reactions

In electrochemical processes, reactions are often split into two half-reactions to simplify the analysis of electron transfers. Half-reactions show either the oxidation or the reduction process separately.For the given reaction, the half-reactions are:- **Oxidation:** \( \text{Mg} \rightarrow \text{Mg}^{2+} + 2e^- \)- **Reduction:** \( \text{Cu}^+ + e^- \rightarrow \text{Cu} \)Each half-reaction lets us track the electrons transferred more clearly:

• Oxidation involves the loss of electrons — here, magnesium loses 2 electrons.
• Reduction involves the gain of electrons — here, each copper ion gains 1 electron.

These half-reactions balance the number of electrons transferred, crucial for calculating the cell's overall potential and Gibbs Free Energy.

electron transfer

Electron transfer is at the heart of all electrochemical reactions. In these reactions, electrons are shifted from one reactant to another, often creating an electron flow that can be utilized as an electrical current in electrochemical cells.In our cell reaction:- Magnesium (Mg) undergoes oxidation, **losing electrons**: \( \text{Mg} \rightarrow \text{Mg}^{2+} + 2e^- \)- Copper ions (Cu⁺) undergo reduction, **gaining electrons**: \( \text{Cu}^+ + e^- \rightarrow \text{Cu} \)This electron transfer is balanced, meaning the number of electrons lost by Mg is equal to the number gained by Cu. For every 2 electrons released by one Mg atom, two Cu ions each gain one electron, resulting in two copper atoms.Understanding electron transfer allows us to calculate the cell potential and use the Faraday constant effectively to determine the Gibbs Free Energy of the system.

Faraday constant

The Faraday constant, denoted as \( F \), is a fundamental constant in electrochemistry, representing the charge of one mole of electrons. It quantifies the charge per mole of electrons, making it an essential factor in calculations involving electrochemical cells.The value of the Faraday constant:\[ F = 96485 \, C \cdot mol^{-1} \]This constancy is used in the Gibbs Free Energy equation:\[ \Delta G = -nFE_{\text{cell}} \]In the given exercise, knowing \( F \) aids in computing the change in Gibbs Free Energy (\( \Delta G = -2 \times 96485 \, C \cdot mol^{-1} \times -1.85 \, V \)). Thus, the whole reaction's energy yield is tied to this constant, highlighting its importance in determining efficiency and viability in delivering electricity from chemical sources.