Problem 77 - Chemistry An Atoms First Approach

Question

Consider the galvanic cell based on the following halfreactions: $$ \begin{array}{ll} \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} & \mathscr{E}^{\circ}=-0.76 \mathrm{V} \\ \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} & \mathscr{E}^{\circ}=-0.44 \mathrm{V} \end{array} $$ a. Determine the overall cell reaction and calculate \(\mathscr{E}_{\text {cell. }}\) b. Calculate \(\Delta G^{\circ}\) and \(K\) for the cell reaction at \(25^{\circ} \mathrm{C}\). c. Calculate \(\mathscr{C}_{\text {cell }}\) at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{Zn}^{2+}\right]=0.10 \mathrm{M}\) and \(\left[\mathrm{Fe}^{2+}\right]=1.0 \times 10^{-5} \mathrm{M}\)

Step-by-Step Solution

Answer

The overall cell reaction is Zn + Fe^2+ → Zn^2+ + Fe with a standard electromotive force (E_cell) of 0.32 V. The standard change in Gibbs free energy (ΔG°) is -61752 J/mol, and the equilibrium constant (K) is approximately 1.14 × 10^17. At 25°C, with [Zn^2+] = 0.10 M and [Fe^2+] = 1.0 × 10^-5 M, the cell potential (E_cell) is approximately 0.08 V.

Step 1: Identify the reduction and oxidation half-reactions

We are given two half-reactions. The one with the more negative standard reduction potential E° will be the oxidation half-reaction and the one with the less negative E° will be the reduction half-reaction. Oxidation half-reaction: Zn^2+ + 2e^- → Zn, E° = -0.76 V Reduction half-reaction: Fe^2+ + 2e^- → Fe, E° = -0.44 V

Step 2: Determine the overall cell reaction

Combine the oxidation and reduction half-reactions and cancel the electrons. The oxidation half-reaction must be reversed to obtain the overall cell reaction: Zn → Zn^2+ + 2e^- Fe^2+ + 2e^- → Fe Overall cell reaction: Zn + Fe^2+ → Zn^2+ + Fe

Step 3: Calculate E_cell

E_cell is the difference between the two standard reduction potentials. Since the Zn^2+ half-reaction is the oxidation reaction, we take the difference between the reduction potential of Fe^2+ and Zn^2+: E_cell = E°_(Fe^2+) - E°_(Zn^2+) = (-0.44 V) - (-0.76 V) = 0.32 V b. Calculate ΔG° and K for the cell reaction at 25°C.

Step 4: Calculate ΔG° using E_cell

Use the formula: ΔG° = -nFE_cell, where n is the number of transferred electrons (2 in this case) and F is the Faraday constant (96485 C/mol). ΔG° = - (2 mol e^-) (96485 C/mol e^-) (0.32 V) = -61752 J/mol

Step 5: Calculate K using ΔG°

Use the formula: ΔG° = -RT ln(K), where R is the gas constant (8.314 J/mol K) and T is the temperature in Kelvin (25°C = 298 K). Rearrange this formula to solve for K: K = e^(-ΔG°/RT) K = e^(-(-61752 J/mol) / (8.314 J/mol K)(298 K)) ≈ 1.14 × 10^17 c. Calculate E_cell at 25°C when [Zn^2+] = 0.10 M and [Fe^2+] = 1.0 × 10^-5 M

Step 6: Use the Nernst equation to calculate E_cell

The Nernst equation: E = E° - (RT/nF) ln(Q), where Q is the reaction quotient. The reaction quotient for the overall cell reaction is Q = [Zn^2+]/[Fe^2+]. Using the given concentrations, we can plug in the values of R, T, n, F, E°, and Q: E = 0.32 V - (8.314 J/mol K × 298 K) / (2 mol e^- × 96485 C/mol e^-) × ln((0.10 M)/(1.0 × 10^-5 M)) E = 0.32 V - (0.0257 V) × ln(10^4) E = 0.32 V - (0.0257 V) × 9.210 ≈ 0.08 V

Key Concepts

Standard Reduction Potential

The concept of standard reduction potential, symbolized as \(\mathscr{E}^{\circ}\), plays a central role in the study of electrochemical reactions. It represents the tendency of a chemical species to acquire electrons and thereby be reduced. This value is measured under standard conditions, which include a solute concentration of 1 M, a pressure of 1 atmosphere, and a temperature of 25°C (298 K).

In the context of a galvanic cell, the standard reduction potential helps us predict the direction of an electron flow. The species with the higher (less negative) reduction potential will undergo reduction, while the species with the lower (more negative) reduction potential will be oxidized. Understanding this is crucial to determine the overall cell reaction and to predict which electrode (cathode or anode) will serve which purpose in the electrochemical cell.

For example, in the given exercise, the standard reduction potential of Fe(II) at \(\mathscr{E}^{\circ} = -0.44 V\) is higher than that of Zn(II) at \(\mathscr{E}^{\circ} = -0.76 V\). Therefore, Fe(II) will be reduced and Zn will be oxidized. Consequently, it's clear that Fe(II) acts as the cathode and Zn as the anode in the galvanic cell.

Electrochemical Cell Reaction

An electrochemical cell reaction involves an exchange of electrons between species at the electrodes, resulting in a flow of electric current through an external circuit as a chemical reaction takes place. When constructing an overall reaction for a galvanic cell, we need to carefully combine the half-reactions for the oxidation and the reduction processes, as seen in the given exercise.

To derive the overall reaction, two key steps are followed: (1) identify the half-reactions involved, and (2) balance and combine them. Once we determine the half-reactions based on standard reduction potentials, we reverse the oxidation half-reaction and then add it to the reduction half-reaction, while ensuring that the electrons cancel out. This ultimately gives us the overall electrochemical cell reaction, which, for the exercise, is: \(\text{Zn} + \text{Fe}^{2+} \rightarrow \text{Zn}^{2+} + \text{Fe}\).

The balanced overall reaction is fundamental for calculating other properties of the cell such as the cell potential, \(\mathscr{E}_{\text{cell}}\), Gibbs free energy change, \(\Delta G^\circ\), and the equilibrium constant, \(K\). This step-by-step understanding assures that students can tackle similar problems involving different redox pairs in electrochemical cells.

Gibbs Free Energy

Gibbs free energy, denoted by \(\Delta G\), is a thermodynamic quantity that predicts the direction of chemical reactions and whether they occur spontaneously. The change in Gibbs free energy, \(\Delta G^\circ\), during a reaction under standard conditions can be calculated from the electrochemical cell potential, \(\mathscr{E}_{\text{cell}}\), using the formula \(\Delta G^\circ = -nFE_{\text{cell}}\), where \(n\) is the number of moles of electrons exchanged in the reaction and \(F\) is the Faraday constant, approximately 96485 Coulombs per mole of electrons.

A negative value of \(\Delta G^\circ\) indicates a spontaneous reaction, whereas a positive value suggests a non-spontaneous process. Furthermore, \(\Delta G^\circ\) is intricately linked to the equilibrium constant, \(K\), of the reaction, with the relationship \(\Delta G^\circ = -RT \ln(K)\), where \(R\) is the universal gas constant and \(T\) the temperature in Kelvin. By rearranging this expression, we can find \(K\) if we know \(\Delta G^\circ\), as was performed in the exercise solution to calculate a very large equilibrium constant, indicating a reaction which lies far to the right (favoring the products).

This understanding of Gibbs free energy allows us to quantitatively analyze the spontaneity and equilibrium position of electrochemical reactions -- a critical factor in the world of chemistry, from batteries to biochemical reactions in living cells.