/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 Consider the galvanic cell based... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(\left[\mathrm{Fe}^{2+}\right]=1.0 \times 10^{-5} \mathrm{M}\)

Short Answer

Expert verified
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 by step solution

01

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
02

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
03

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.
04

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
05

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
06

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Consider a concentration cell that has both electrodes made of some metal M. Solution \(A\) in one compartment of the cell contains \(1.0 \mathrm{M} \mathrm{M}^{2+}\). Solution \(\mathrm{B}\) in the other cell compartment has a volume of 1.00 L. At the beginning of the experiment 0.0100 mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and 0.0100 mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$ \mathbf{M}^{2+}(a q)+\mathbf{S O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s) $$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be 0.44 V at 25 \(^{\circ}\) C. Assume that the process $$ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} $$ has a standard reduction potential of \(-0.31 \mathrm{V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C}\)

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(g)\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium $$2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ Consider the following standard reduction potentials at \(25^{\circ} \mathrm{C}:\) $$\begin{aligned} &3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\ & \mathscr{E}^{\circ}=0.957 \mathrm{V} \end{aligned}$$ $$\begin{aligned} &\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) &\mathscr{E}^{\circ}=0.775 \mathrm{V} \end{aligned}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and \(\mathrm{NO}_{2}\) mixture with only \(0.20 \% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and 1.00 atm? Assume that no other gases are present and that the change in acid concentration can be neglected.

A disproportionation reaction involves a substance that acts as both an oxidizing and a reducing agent, producing higher and lower oxidation states of the same element in the products. Which of the following disproportionation reactions are spontaneous under standard conditions? Calculate \(\Delta G^{\circ}\) and \(K\) at \(25^{\circ} \mathrm{C}\) for those reactions that are spontaneous under standard conditions. a. \(2 \mathrm{Cu}^{+}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s)\) b. \(3 \mathrm{Fe}^{2+}(a q) \rightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{Fe}(s)\) c. \(\mathrm{HClO}_{2}(a q) \rightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{HClO}(a q) \quad\) (unbalanced) Use the half-reactions: \(\mathrm{ClO}_{3}^{-}+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}_{2}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.21 \mathrm{V}\) \(\mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.65 \mathrm{V}\)

Consider the following galvanic cell at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Pt}\left|\mathrm{Cr}^{2+}(0.30 M), \mathrm{Cr}^{3+}(2.0 M)\right|\left|\mathrm{Co}^{2+}(0.20 M)\right| \mathrm{Co} $$ The overall reaction and equilibrium constant value are $$\begin{aligned} 2 \mathrm{Cr}^{2+}(a q)+\mathrm{Co}^{2+}(a q) & \longrightarrow \\ 2 \mathrm{Cr}^{3+}(a q)+\mathrm{Co}(s) & K=2.79 \times 10^{7} \end{aligned}$$ Calculate the cell potential, \(\mathscr{E}\), for this galvanic cell and \(\Delta G\) for the cell reaction at these conditions.

Consider a galvanic cell based on the following theoretical half-reactions: $$\begin{array}{lr} & \mathscr{E}^{\circ}(\mathrm{V}) \\ \mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & 1.50 \\ \mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg} & -2.37 \end{array}$$ a. What is the standard potential for this cell? b. A nonstandard cell is set up at \(25^{\circ} \mathrm{C}\) with \(\left[\mathrm{Mg}^{2+}\right]=\) \(1.00 \times 10^{-5} M .\) The cell potential is observed to be 4.01 V. Calculate \(\left[\mathrm{Au}^{3+}\right]\) in this cell. \mathscr{E}^{\circ}

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