/*! 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 the following electrochemical cell: a. If silver metal is a product of the reaction, is the cell a galvanic cell or electrolytic cell? Label the cathode and anode, and describe the direction of the electron flow. b. If copper metal is a product of the reaction, is the cell a galvanic cell or electrolytic cell? Label the cathode and anode, and describe the direction of the electron flow. c. If the above cell is a galvanic cell, determine the standard cell potential. d. If the above cell is an electrolytic cell, determine the minimum external potential that must be applied to cause the reaction to occur.

A chemist wishes to determine the concentration of \(\mathrm{CrO}_{4}^{2-}\) electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise 115 ) and a silver wire coated with \(\mathrm{Ag}_{2} \mathrm{CrO}_{4} .\) The \(8^{\circ}\) value for the following half-reaction is \(0.446 \mathrm{V}\) relative to the standard hydrogen electrode: $$\mathrm{Ag}_{2} \mathrm{CrO}_{4}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}+\mathrm{CrO}_{4}^{2-}$$ a. Calculate \(\mathscr{C}_{\text {cell }}\) and \(\Delta G\) at \(25^{\circ} \mathrm{C}\) for the cell reaction when \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \mathrm{mol} / \mathrm{L}\) b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant. c. If the coated silver wire is placed in a solution (at \(25^{\circ} \mathrm{C}\) ) in which \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \times 10^{-5} \mathrm{M},\) what is the expected cell potential? d. The measured cell potential at \(25^{\circ} \mathrm{C}\) is \(0.504 \mathrm{V}\) when the coated wire is dipped into a solution of unknown \(\left[\mathrm{CrO}_{4}^{2-}\right] .\) What is \(\left[\mathrm{CrO}_{4}^{2-}\right]\) for this solution? e. Using data from this problem and from Table \(17-1,\) calculate the solubility product \(\left(K_{\mathrm{sp}}\right)\) for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\)

Consider the galvanic cell based on the following halfreactions: $$ \begin{array}{ll} \mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{E}^{\circ}=1.50 \mathrm{V} \\ \mathrm{Tl}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Tl} & \mathscr{E}^{\circ}=-0.34 \mathrm{V} \end{array} $$ a. Determine the overall cell reaction and calculate \(\mathscr{C}_{\text {cell. }}\) b. Calculate \(\Delta G^{\circ}\) and \(K\) for the cell reaction at \(25^{\circ} \mathrm{C}\) c. Calculate \(\mathscr{E}_{\text {cell }}\) at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{Au}^{3+}\right]=1.0 \times 10^{-2} \mathrm{M}\) and \(\left[\mathrm{Tl}^{+}\right]=1.0 \times 10^{-4} \mathrm{M}\)

Aluminum is produced commercially by the electrolysis of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) in the presence of a molten salt. If a plant has a continuous capacity of 1.00 million \(A\), what mass of aluminum can be produced in \(2.00 \mathrm{h} ?\)

Balance the following oxidation-reduction reactions that occur in acidic solution using the half-reaction method. a. \(\mathrm{Cu}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{NO}(g)\) b. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{Cl}_{2}(g)\) c. \(\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)\) d. \(\mathrm{Mn}^{2+}(a q)+\mathrm{NaBiO}_{3}(s) \rightarrow \mathrm{Bi}^{3+}(a q)+\mathrm{MnO}_{4}^{-}(a q)\) e. \(\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{Zn}(s) \rightarrow \mathrm{AsH}_{3}(g)+\mathrm{Zn}^{2+}(a q)\)

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