Problem 35 - Chemistry: An Atoms-Focused Approach

Question

Starting with the appropriate standard free energies of formation from Appendix \(4,\) calculate the valucs of \(\Delta G^{\circ}\) and \(E_{\text {cell }}\) of the following reactions: a. \(2 \mathrm{Cu}^{+}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s)\) b. \(\mathrm{Ag}(s)+\mathrm{Fe}^{3+}(a q) \rightarrow \mathrm{Ag}^{+}(a q)+\mathrm{Fe}^{2+}(a q)\)

Step-by-Step Solution

Answer

For the two given reactions, calculate the standard Gibbs free energy change (\(\Delta G^{\circ}\)) and the cell potential (\(E_{\text{cell}}\)). a. \(2 \mathrm{Cu}^{+}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s)\) b. \(\mathrm{Ag}(s)+\mathrm{Fe}^{3+}(a q) \rightarrow \mathrm{Ag}^{+}(a q)+\mathrm{Fe}^{2+}(a q)\) Answer: For Reaction a: \(\Delta G^{\circ} = 223.8\,\mathrm{kJ/mol}\) and \(E_\text{cell} = -1.16\,\mathrm{V}\) For Reaction b: \(\Delta G^{\circ} = 291.8\,\mathrm{kJ/mol}\) and \(E_\text{cell} = 3.02\,\mathrm{V}\)

Step 1: Calculate the standard Gibbs free energy change

To calculate the \(\Delta G^{\circ}\), first, find the standard free energies of formation values in the given appendix or thermodynamics data table for each species in the reaction. Then use the equation mentioned in the analysis. \(\Delta G^{\circ} = (\Delta G_f^{\circ}(\text{Cu}^{2+}) + \Delta G_f^{\circ}(\text{Cu}(s))) - 2\Delta G_f^{\circ}(\text{Cu}^+) \) Insert the values from the data table: \(\Delta G^{\circ} = (65.0 + 0) - 2(-79.4) \mathrm{kJ/mol} = 223.8\,\mathrm{kJ/mol}\)

Step 2: Calculate the cell potential

To calculate the cell potential, use the Nernst equation and the calculated \(\Delta G^{\circ}\) value. Note that two moles of electrons are transferred in the reaction. \(\Delta G^{\circ} = -nFE_{\text{cell}}\) \(E_{\text{cell}} = -\frac{\Delta G^{\circ}}{nF} = -\frac{223.8\,\mathrm{kJ/mol}}{2 \cdot 96,485\,\mathrm{C/mol}} = -1.16\,\mathrm{V}\) The cell potential for Reaction a is -1.16 V. b. \(\mathrm{Ag}(s)+\mathrm{Fe}^{3+}(a q) \rightarrow \mathrm{Ag}^{+}(a q)+\mathrm{Fe}^{2+}(a q)\)

Step 3: Calculate the standard Gibbs free energy change

Use the same method as in Step 1 to calculate the \(\Delta G^{\circ}\) for this reaction: \(\Delta G^{\circ} = (\Delta G_f^{\circ}(\text{Ag}^+) + \Delta G_f^{\circ}(\text{Fe}^{2+})) - (\Delta G_f^{\circ}(\text{Ag}(s)) + \Delta G_f^{\circ}(\text{Fe}^{3+}))\) Insert the values from the data table: \(\Delta G^{\circ} = (-11.3 -78.4) - (0 + (-381.5))\,\mathrm{kJ/mol} = 291.8\,\mathrm{kJ/mol}\)

Step 4: Calculate the cell potential

Use the Nernst equation and the calculated \(\Delta G^{\circ}\) from Step 3. Note that only one mole of electrons is transferred in this case. \(\Delta G^{\circ} = -nFE_{\text{cell}}\) \(E_{\text{cell}} = -\frac{\Delta G^{\circ}}{nF} = -\frac{291.8\,\mathrm{kJ/mol}}{1 \cdot 96,485\,\mathrm{C/mol}} = 3.02\,\mathrm{V}\) The cell potential for Reaction b is 3.02 V. In summary: For Reaction a, \(\Delta G^{\circ} = 223.8\,\mathrm{kJ/mol}\) and \(E_\text{cell} = -1.16\,\mathrm{V}\) For Reaction b, \(\Delta G^{\circ} = 291.8\,\mathrm{kJ/mol}\) and \(E_\text{cell} = 3.02\,\mathrm{V}\)

Key Concepts

standard cell potential

The standard cell potential, represented as \( E^{\circ}_{\text{cell}} \), is a measure of the voltage difference between two half-cells in an electrochemical cell under standard conditions. These conditions are defined as the temperature at 298 K, pressure of 1 atm, and all solute concentrations at 1 M. The potential tells us how much voltage the electrochemical cell can produce and is an indication of its ability to perform work.

To calculate the standard cell potential, it's necessary to know the potentials of the individual half-reactions. These are often available in tables as standard electrode potentials. For the overall reaction, you calculate the cell potential by taking the difference between the reduction potential of the cathode and the oxidation potential of the anode:

• \( E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \)

By comparing \( E^{\circ}_{\text{cell}} \) for different reactions, we can determine which reactions are more favorable and strong under standard conditions.

Nernst equation

The Nernst equation is a fundamental formula in electrochemistry that allows us to calculate the cell potential at non-standard conditions. It connects the cell potential to the concentrations of the reactants and products involved in the electrochemical reaction.

Given by:\[E = E^{\circ} - \frac{RT}{nF} \ln Q\]

• \(E\) is the cell potential under non-standard conditions.
• \(E^{\circ}\) is the standard cell potential.
• \(R\) is the universal gas constant (8.314 J/mol·K).
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of moles of electrons transferred in the reaction.
• \(F\) is Faraday's constant (96485 C/mol).
• \(Q\) is the reaction quotient, a ratio of the concentrations of products over reactants.

The Nernst equation is crucial because it predicts how the cell potential changes as the concentrations of the ions in solution change. This is particularly useful in real-life applications where concentration is not always 1 M.

electrochemistry

Electrochemistry is the study of the relationships between chemical reactions and electricity. It encompasses a wide range of technologies, crucial for batteries, electrolysis, and more. The primary concept revolves around redox reactions that either produce or utilize electrical energy.

In an electrochemical cell, these redox reactions are divided into two half reactions: oxidation and reduction. Electrodes—an anode and a cathode—facilitate these reactions.

• The anode is where oxidation occurs, meaning it loses electrons.
• The cathode is where reduction takes place, it gains electrons.

Electrochemistry is essential because it provides insights into how these electron transfers can be harnessed for practical use, such as generating electricity in batteries or splitting water into hydrogen and oxygen gas. Understanding these basic principles allows for innovation in energy storage and conversion.