Problem 114 - Chemistry: The Central Science

Question

A voltaic cell is based on \(\mathrm{Ag}^{+}(a q) / \mathrm{Ag}(s)\) and \(\mathrm{Fe}^{3+}(a q) /\) \(\mathrm{Fe}^{2+}(\mathrm{aq})\) half-cells. (a) What is the standard emf of the cell? (b) Which reaction occurs at the cathode and which at the anode of the cell? (c) Use \(S^{\circ}\) values in Appendix \(\mathrm{C}\) and the relationship between cell potential and free-energy change to predict whether the standard cell potential increases or decreases when the temperature is raised above \(25^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

(a) The standard emf of the cell is 0.03 V. (b) The reaction occurring at the cathode is the silver reaction: \(\mathrm{Ag}^+ (aq) + e^- \rightleftharpoons \mathrm{Ag}(s)\). The reaction occurring at the anode is the iron reaction: \(\mathrm{Fe}^{3+} (aq) + e^- \rightleftharpoons \mathrm{Fe}^{2+} (aq)\). (c) To determine if the standard cell potential increases or decreases above \(25^\circ \mathrm{C}\), consult the provided ΔS˚ values for the reactants and products in Appendix C, calculate the ΔH˚ for the reaction, and follow the explanation given. If ΔH˚ > 0, the cell potential will increase with temperature; if ΔH˚ < 0, the cell potential will decrease with temperature.

Step 1: Identify the half-cell reactions

For this particular voltaic cell, the two half-cell reactions are the silver and iron reactions: 1. Silver reaction: \(\mathrm{Ag}^+ (aq) + e^- \rightleftharpoons \mathrm{Ag}(s)\) 2. Iron reaction: \(\mathrm{Fe}^{3+}(aq) + e^- \rightleftharpoons \mathrm{Fe}^{2+} (aq)\)

Step 2: Find the standard reduction potential (E°) of each half-cell reaction

Using a reduction potential table or, if provided, the appendix of a textbook, identify the standard reduction potentials for the two half-cell reactions: 1. Silver reaction: \(E_{\mathrm{Ag}^+/\mathrm{Ag}}^\circ = 0.80 \, \text{V}\) 2. Iron reaction: \(E_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}^\circ = 0.77 \, \text{V}\)

Step 3: Determine the overall electrochemical cell potential (E°)

Recognizing that anode reactions occur at lower, more negative potential compared to cathode reactions, we can assign the cell reactions to the appropriate location: 1. Oxidation (Anode): Iron reaction, as it has a lower standard reduction potential 2. Reduction (Cathode): Silver reaction, as it has a higher standard reduction potential Thus, the overall cell potential is given by the difference in their standard reduction potentials: \(E_{\text{cell}}^\circ = E_{\text{cathode}}^\circ - E_{\text{anode}}^\circ = E_{\mathrm{Ag}^+/\mathrm{Ag}}^\circ - E_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}^\circ\)

Step 4: Calculate the overall electrochemical cell potential (E°)

Using the values obtained in Step 2, we can calculate the standard cell potential: \(E_{\text{cell}}^\circ = 0.80 \, \text{V} - 0.77 \, \text{V} = 0.03 \, \text{V}\) (a) The standard emf of the cell is 0.03 V. (b) The reaction occurring at the cathode is the silver reaction: \(\mathrm{Ag}^+ (aq) + e^- \rightleftharpoons \mathrm{Ag}(s)\). The reaction occurring at the anode is the iron reaction: \(\mathrm{Fe}^{3+} (aq) + e^- \rightleftharpoons \mathrm{Fe}^{2+} (aq)\).

Step 5: Analyze the effect of temperature on cell potential

The relationship between cell potential and free-energy change is given by: \(\Delta G^\circ = -nFE_{\text{cell}}^\circ\) Also, using the Gibbs-Helmholtz equation: \( \frac{d(\Delta G^\circ/T)}{dT} = -\frac{\Delta H^\circ}{T^2} \) Substituting the relationship between cell potential and free-energy change into the Gibbs-Helmholtz equation: \(\frac{d(-nFE_{\text{cell}}^\circ/T)}{dT} = -\frac{\Delta H^\circ}{T^2}\) Rearranging for \(E_{\text{cell}}^\circ\): \(\frac{dE_{\text{cell}}^\circ}{dT} = \frac{\Delta H^\circ}{nFT^2}\) Thus, the change in cell potential with temperature is directly proportional to the change in enthalpy. By evaluating whether the ΔH˚ for the reaction is positive or negative, you can predict the response to a change in temperature. In this case, the ΔH˚ for redox reactions can be found from the entropy change, ΔS˚ (found in Appendix C). If ΔH˚ > 0, the cell potential will increase as temperature rises; if ΔH˚ < 0, the cell potential will decrease as temperature rises. To determine if the standard cell potential increases or decreases above \(25^\circ \mathrm{C}\), consult Appendix C and find the ΔS˚ values for the reactants and products, and then calculate the ΔH˚ for the reaction. (c) Without the actual Appendix C, the specific answer regarding temperature's effect on the standard cell potential cannot be provided. However, based on the explanation given above, the necessary calculations can be performed to arrive at the conclusion.

Key Concepts

Electrochemical Cell

An electrochemical cell is a fundamental device that converts chemical energy into electrical energy through redox reactions. It consists of two half-cells, each containing an electrode and an electrolyte solution. The electrodes are typically made of metals or conductive materials. In a voltaic cell, which is a type of electrochemical cell, the chemical reactions are spontaneous.

These half-cells are usually part of separate containers connected by a salt bridge or a porous diaphragm. The salt bridge allows ions to flow between the half-cells, maintaining electrical neutrality. This flow of ions, combined with the flow of electrons through a wire connecting the electrodes, results in a current that can perform electrical work.

In summary, electrochemical cells provide a mechanism to harness the energy of chemical reactions, which are at the heart of many batteries used in everyday devices.

Standard Reduction Potential

The standard reduction potential (E°) is a measure of the tendency of a chemical species to gain electrons and be reduced. It's determined under standard conditions of 1 M concentration for solutions, 1 atm pressure for gases, and at a temperature of 25°C (298 K).

Each half-reaction in an electrochemical cell has a specific standard reduction potential. These values are tabulated in reference tables. For a redox reaction, the half with the higher standard reduction potential acts as the cathode (reduction), while the other acts as the anode (oxidation).

Calculating the overall cell potential of an electrochemical cell involves subtracting the anode’s potential from the cathode’s potential: \( E_{cell}^ ext{°} = E_{cathode}^ ext{°} - E_{anode}^ ext{°} \). This cell potential indicates the maximum voltage the cell can provide under ideal conditions.

Understanding standard reduction potentials is crucial for predicting the feasibility of reactions and designing useful electrochemical cells.

Gibbs-Helmholtz Equation

The Gibbs-Helmholtz equation links the change in free energy \( \,\Delta G \) to the change in enthalpy \( \,\Delta H \) and entropy \( \,\Delta S^\circ \) of a system. It is specifically used to evaluate how cell potential might change with temperature:\[\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\]

In the context of electrochemical cells, it can also be expressed in terms of cell potential \( E_{cell}^\circ \):\[\frac{d(-nFE_{cell}^\circ/T)}{dT} = -\frac{\Delta H^\circ}{T^2}\]

This equation provides a way to understand how the cell potential varies with temperature. If the reaction's enthalpy change \( \,\Delta H^\circ \) is positive, an increase in temperature tends to increase the cell potential. Conversely, if \( \,\Delta H^\circ \) is negative, higher temperatures typically decrease the cell potential.

Using these insights, scientists and engineers can predict and manipulate the performance of electrochemical cells across different temperatures.