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If the equilibrium constant for a one-electron redox reaction at 298 \(\mathrm{K}\) is \(8.7 \times 10^{4}\) , calculate the corresponding \(\Delta G^{\circ}\) and \(E^{\circ} .\)

Short Answer

Expert verified
The Gibbs free energy change (螖G掳) for this one-electron redox reaction is approximately -5.336 x 10鈦 J mol鈦宦, and the standard electrode potential (E掳) is approximately 0.553 V.

Step by step solution

01

Use the relationship between 螖G掳 and K

We know that 螖G掳 and K are related by the following equation: \[\Delta G^{\circ} = -RT \ln K\] where R is the gas constant (8.314 J mol鈦宦 K鈦宦), T is the temperature (298 K), and K is the equilibrium constant (8.7 x 10鈦). First, we'll plug in the values to calculate 螖G掳: \[\Delta G^{\circ} = -(8.314 \text{ J mol}^{-1} \text{K}^{-1})(298 \text{ K}) \ln (8.7 \times 10^{4})\]
02

Calculate 螖G掳

Use a calculator to evaluate the equation and find 螖G掳: \[\Delta G^{\circ} \approx -5.336 \times 10^{4} \text{ J mol}^{-1}\] The Gibbs free energy change for this one-electron redox reaction is approximately -5.336 x 10鈦 J mol鈦宦.
03

Use the relationship between 螖G掳 and E掳 for a one-electron redox reaction

Next, we will use the relationship between 螖G掳 and E掳, which is given by the following equation: \[\Delta G^{\circ} = -nFE^{\circ}\] where n is the number of electrons transferred (which is 1 in a one-electron redox reaction), F is Faraday's constant (96485 C mol鈦宦), and E掳 is the standard electrode potential. For this reaction, we have: \[-5.336 \times 10^{4} \text{ J mol}^{-1} = -(1)(96485 \text{ C mol}^{-1})E^{\circ}\]
04

Calculate E掳

To find E掳, solve the equation: \[E^{\circ} = \frac{-5.336 \times 10^{4} \text{ J mol}^{-1}}{96485 \text{ C mol}^{-1}}\] \[E^{\circ} \approx 0.553 \text{ V}\] The standard electrode potential for this one-electron redox reaction is approximately 0.553 V.

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Key Concepts

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

Understanding Gibbs Free Energy (\(\Delta G^\circ\))
Gibbs free energy, typically represented as \(\Delta G^\circ\), plays a pivotal role in thermodynamics and particularly in the study of chemical reactions. It is used to determine the spontaneity of a chemical process at constant temperature and pressure. A negative value of \(\Delta G^\circ\) indicates a reaction that can occur spontaneously, while a positive value suggests a non-spontaneous reaction.

To calculate the change in Gibbs free energy for a reaction, we use the equation \(\Delta G^\circ = -RT \ln K\), where \( R \) is the universal gas constant, \( T \) is the absolute temperature, and \( K \) is the equilibrium constant. For the given redox reaction, plugging in the values yields a \(\Delta G^\circ\) of approximately -5.336 x 10^4 J/mol, suggesting that the reaction is spontaneous.

This calculation is crucial because it provides insight into the feasibility of chemical processes without needing to perform them. Therefore, grasping the concept of Gibbs free energy allows students and scientists alike to predict reaction behavior and drive the development of new reactions and materials.
The Role of Standard Electrode Potential (E掳) in Redox Reactions
Standard electrode potential, designated as \(E^\circ\), is a measure of the individual potential of a reversible electrode at standard state conditions, which is a crucial concept in electrochemistry. It allows chemists to understand the tendency of a species to lose or gain electrons, known as the redox potential. The more positive the standard electrode potential, the greater the species' affinity for electrons; hence, it is a stronger oxidizing agent.

In a redox reaction, the difference in standard electrode potentials for the oxidation and reduction half-reactions can be used to calculate the cell potential for the overall reaction. Across a one-electron transfer reaction, the relationship \(\Delta G^\circ = -nFE^\circ\) equates Gibbs free energy change with standard electrode potential, where \( n \) represents the number of electrons transferred and \( F \) is Faraday's constant.

In our case, solving the equation with the given Gibbs free energy change results in a standard electrode potential of approximately 0.553 V for a one-electron redox reaction, which illustrates the potential's magnitude for driving the reaction.
The Equilibrium Constant (K) and Its Significance
The equilibrium constant, symbolized as \(K\), offers a quantitative measure of the position of equilibrium in a chemical reaction. In essence, it relates the concentrations of the products to the reactants at equilibrium for a reversible reaction. A larger \(K\) implies a reaction with a greater extent of product formation, while a smaller \(K\) indicates that the reactants are predominant at equilibrium.

The magnitude of the equilibrium constant for a given temperature is directly related to the standard Gibbs free energy change by the equation \(\Delta G^\circ = -RT \ln K\). Therefore, knowing \(K\) allows us to calculate \(\Delta G^\circ\), providing insight into the reaction's spontaneity. For the redox reaction in question, with \(K = 8.7 \times 10^4\), the substantial magnitude of \(K\) signifies a reaction that heavily favors the formation of products.

In educational contexts, understanding \(K\) is fundamental for students to predict the behavior and outcome of chemical reactions. It not only influences chemical manufacturing and experimental design but is also a key concept in many natural processes, such as in environmental, biological, and geophysical systems.

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Most popular questions from this chapter

Metallic magnesium can be made by the electrolysis of molten MgCl. (a) What mass of Mg is formed by passing a current of 4.55 A through molten MgCl \(_{2},\) for 4.50 days? (b) How many minutes are needed to plate out 25.00 \(\mathrm{g} \mathrm{Mg}\) from molten \(\mathrm{MgCl}_{2}\) using 3.50 \(\mathrm{A}\) of current?

(a) Assuming standard conditions, arrange the following in order of increasing strength as oxidizing agents in acidic solution: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Cu}^{2+}, \mathrm{Cl}_{2}, \mathrm{O}_{2} .\) (b) Arrange the following in order of increasing strength as reducing agents in acidic solution: \(\text {Zn,}\) \(\mathrm{I}^{-}, \mathrm{Sn}^{2+}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Al} .\)

(a) Based on standard reduction potentials, would you expect copper metal to oxidize under standard conditions in the presence of oxygen and hydrogen ions? (b) When the Statue of Liberty was refurbished, Teflon spacers were placed between the iron skeleton and the copper metal on the surface of the statue. What role do these spacers play?

Given the following half-reactions and associated standard reduction potentials: $$ \begin{array}{c}{\text { AuBr }_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q)} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {E_{\mathrm{red}}^{\circ}=-0.86 \mathrm{V}} \\ {\mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q)} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {E_{\mathrm{red}}^{\circ}=-0.43 \mathrm{V}}\end{array} $$ $$ \begin{array}{r}{\mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(a q)+2 \mathrm{OH}^{-}(a q)} \\\ {E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{V}}\end{array} $$ (a) Write the equation for the combination of these half-cell reactions that leads to the largest positive emf and calculate the value. (b) Write the equation for the combination of half-cell reactions that leads to the smallest positive emf and calculate that value.

Iron corrodes to produce rust, \(\mathrm{Fe}_{2} \mathrm{O}_{3},\) but other corrosion products that can form are \(\mathrm{Fe}(\mathrm{O})(\mathrm{OH}),\) iron oxyhydroxide, and magnetite, \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) . (a) What is the oxidation number of Fe in iron oxyhydroxide, assuming oxygen's oxidation number is \(-2 ?\) (b) The oxidation number for Fe in magnetite was controversial for a long time. If we assume that oxygen's oxidation number is \(-2,\) and Fe has a unique oxidation number, what is the oxidation number for Fe in magnetite? (c) It turns out that there are two different kinds of Fe in magnetite that have different oxidation numbers. Suggest what these oxidation numbers are and what their relative stoichiometry must be, assuming oxygen's oxidation number is - 2 .

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