Problem 53
Question
If the equilibrium constant for a two-electron redox reaction at \(298 \mathrm{~K}\) is \(1.5 \times 10^{-4}\), calculate the corresponding \(\Delta G^{\text {t }}\) and \(E_{\text {red }}\)
Step-by-Step Solution
Verified Answer
The corresponding change in Gibbs free energy (∆G) for this two-electron redox reaction at 298 K is 16386.6 J/mol, and the reduction potential (Ered) is -0.085 V.
1Step 1: Identify the known variables and required formula
We know the following variables:
- Equilibrium constant (K) = \(1.5 \times 10^{-4}\)
- Temperature (T) = 298 K
- The number of electrons transferred in the redox reaction (n) = 2
We need to find:
1. Change in Gibbs free energy (∆G)
2. Reduction potential (Ered)
For this, we'll use the following formulae:
1. Relationship between K and ΔG: \[\Delta G = -RT\ln K\]
Where R is the universal gas constant, 8.314 J/molK
2. Relationship between ΔG and E:
\[\Delta G = -nFE_{\text{red}}\]
Where F is the Faraday's constant, 96485 C/mol
2Step 2: Calculate the change in Gibbs free energy (∆G)
Now, we'll use the equation to find the change in Gibbs free energy (∆G):
\[\Delta G = -RT\ln K\]
\[\Delta G = - (8.314\, \text{J/molK})(298\, \text{K})\ln(1.5 \times 10^{-4})\]
Now, calculate the value:
\[\Delta G = 16386.6\, \text{J/mol}\]
3Step 3: Calculate the reduction potential (Ered)
Next, we'll find the reduction potential (Ered) using the relationship between ΔG and E:
\[\Delta G = -nFE_{\text{red}}\]
Rearrange the equation for Ered:
\[E_{\text{red}} = -\frac{\Delta G}{nF}\]
Now, plug in the values
\[E_{\text{red}} = -\frac{16386.6 \, \text{J/mol}}{(2)(96485\, \text{C/mol})}\]
Solve for Ered:
\[E_{\text{red}} = -0.085\, \text{V}\]
So, the corresponding change in Gibbs free energy (∆G) for this two-electron redox reaction at 298 K is 16386.6 J/mol, and the reduction potential (Ered) is -0.085 V.
Key Concepts
Gibbs Free EnergyReduction PotentialNernst Equation
Gibbs Free Energy
Gibbs free energy (G) is a thermodynamic quantity that serves as a measure of the maximum amount of work a system can perform at constant temperature and pressure. It is an important concept when studying chemical reactions, such as redox reactions, as it helps predict the spontaneity of the reaction.
In the context of redox reactions, the change in Gibbs free energy, denoted as \( \Delta G \), is particularly significant as it determines the feasibility of the reaction. A negative value of \( \Delta G \) generally indicates that the reaction is spontaneous, while a positive value suggests non-spontaneity.
The relationship between the equilibrium constant (K) and \( \Delta G \) can be identified through the formula:
\[\Delta G = -RT\ln K\]
Where:\
In the context of redox reactions, the change in Gibbs free energy, denoted as \( \Delta G \), is particularly significant as it determines the feasibility of the reaction. A negative value of \( \Delta G \) generally indicates that the reaction is spontaneous, while a positive value suggests non-spontaneity.
The relationship between the equilibrium constant (K) and \( \Delta G \) can be identified through the formula:
\[\Delta G = -RT\ln K\]
Where:\
- \( R \) is the universal gas constant
- \( T \) is the temperature in Kelvin
- \( K \) is the equilibrium constant
Reduction Potential
Reduction potential, expressed as \( E_{\text{red}} \), is a measure of the tendency of a chemical species to acquire electrons and be reduced. In electrochemistry, it provides a quantitative means of understanding how different substances donate or accept electrons. This concept is pivotal in determining the direction and flow of electrons in redox reactions.
\
\
Where:\
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Significance in Redox Reactions\
In any given redox reaction, the species with higher reduction potential will act as the oxidizing agent and gain electrons, while the one with lower reduction potential will lose electrons and act as the reducing agent.\
Calculating Reduction Potential from Gibbs Free Energy\
One can derive the reduction potential from the change in Gibbs free energy using the equation:\[\Delta G = -nFE_{\text{red}}\]Where:\
- \( n \) is the number of electrons transferred in the redox reaction
- \( F \) is the Faraday constant
Nernst Equation
The Nernst equation is a fundamental equation in electrochemistry that relates the reduction potential of a redox reaction to the standard electrode potential, temperature, and the activities (or concentrations) of the species involved in the reaction. It can be expressed as:
\[E = E^0 - \frac{RT}{nF} \ln Q\]
Where:\
\[E = E^0 - \frac{RT}{nF} \ln Q\]
Where:\
- \( E \) is the reduction potential
- \( E^0 \) is the standard electrode potential
- \( R \) is the universal gas constant
- \( T \) is the temperature in Kelvin
- \( n \) is the number of electrons involved in the reaction
- \( F \) is the Faraday constant
- \( Q \) is the reaction quotient, which is a measure of the relative amounts of reactants and products
Other exercises in this chapter
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The standard reduction potential for the reduction of \(\mathrm{RuO}_{4}^{-}(a q)\) to \(\mathrm{RuO}_{4}^{2-}(a q)\) is \(+0.59 \mathrm{~V}\). By using Appendi
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Given the following reduction half-reactions: $$ \begin{gathered} \mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) \quad E_{\text {red
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If the equilibrium censtant for a one-electron redox reaction at \(298 \mathrm{~K}\) is \(8.7 \times 10^{4}\), calculate the corresponding \(\Delta G^{\circ}\)
View solution Problem 56
Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at \(298 \mathrm{~K}\) - (a
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