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Magazine Chapter 20: Problem 53
If the equilibrium constant for a one-electron redox reaction at \(298 \mathrm{~K}\) is \(2.2 \times 10^{-5},\) calculate the corresponding \(\Delta G^{\circ}\) and \(E^{\circ}\).
Short Answer
Expert verified
The standard electrode potential (E掳) for the one-electron redox reaction at 298 K is -0.028 V, and the corresponding change in Gibbs free energy (螖G掳) is +2700 J/mol.
Step by step solution
01
Determine the relationship between K and Q
Since we are given K, which is the equilibrium constant, we need to express this in terms of the reaction quotient (Q). Whenever a reaction reaches equilibrium, the following relationship holds: \[K = Q\]
02
Determine the number of electrons
Since the reaction is a one-electron redox reaction, the number of electrons involved in the reaction is n = 1.
03
Use the Nernst Equation to find E掳
At equilibrium, the Nernst Equation becomes: \[\Delta E = E - E^\circ = 0\]
Now, plug in the values for n, R, T, and F, and rearrange the equation to solve for E掳: \[E^\circ = \frac{RT}{nF}\ln{K}\]
Where:
- R = the gas constant = 8.314 J/mol路K
- T = the temperature = 298 K
- n = the number of electrons transferred = 1
- F = the Faraday constant = 96,485 C/mol
- K = the equilibrium constant = 2.2 脳 10鈦烩伒
Now, substitute the given values into the equation: \[E^\circ = \frac{(8.314 \;\text{J/mol路K})(298 \;\text{K})}{(1)(96,485\; \text{C/mol})}\ln{(2.2 脳 10^{-5})}\]
Calculate E掳: \[E^\circ = -0.028 \;\text{V}\]
04
Calculate 螖G掳 using the Gibbs free energy equation
Now that we have the value of E掳, we can use the Gibbs free energy equation to find 螖G掳: \[\Delta G^\circ = -nFE^\circ\]
Plug in the values for n, F, and E掳: \[\Delta G^\circ = -(1)(96,485 \;\text{C/mol})(-0.028 \;\text{V})\]
Calculate 螖G掳: \[\Delta G^\circ = +2700 \;\text{J/mol}\]
05
State the final results
The standard electrode potential for the one-electron redox reaction at 298 K is E掳 = -0.028 V, and the corresponding change in Gibbs free energy is 螖G掳 = +2700 J/mol.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equilibrium Constant
The equilibrium constant, denoted as \( K \), is a fundamental concept in chemistry that measures the ratio of products to reactants at equilibrium for a given reaction. It is calculated from the concentrations or partial pressures of the chemical species involved.
When the reaction reaches equilibrium, the state where the rate of the forward reaction equals the rate of the backward reaction, the value of \( K \) remains constant.
When the reaction reaches equilibrium, the state where the rate of the forward reaction equals the rate of the backward reaction, the value of \( K \) remains constant.
- In the case of redox reactions, the equilibrium constant can often be related to the standard electrode potential \( E^\circ \) and the Gibbs free energy change \( \Delta G^\circ \).
- A small \( K \) indicates a reaction that favors reactants, as seen here with \( 2.2 \times 10^{-5} \), telling us that under standard conditions, the reaction does not proceed significantly towards products.
- Conversely, a large \( K \) indicates that the reaction strongly favors the formation of products.
Nernst Equation
The Nernst Equation is crucial in electrochemistry as it relates the cell potential to the standard electrode potential, temperature, and activities or concentrations of reacting species. The form of the Nernst Equation used at equilibrium is especially simple, because at this stage, \( \Delta E \), the difference between the actual cell potential \( E \) and the standard potential \( E^\circ \), is zero.
- The equation becomes \( E^\circ = \frac{RT}{nF}\ln{K} \), notice how it connects \( E^\circ \) with \( K \), temperature \( T \), and the gas constant \( R \).
- Inserting the given values, compute \( E^\circ = \frac{(8.314 \, \text{J/mol路K})(298 \, \text{K})}{(1)(96,485\, \text{C/mol})}\ln{(2.2 \times 10^{-5})} \), resulting in \( E^\circ = -0.028 \text{ V} \).
- This expression helps us understand how cell potentials change with concentration or pressure differences.
Gibbs Free Energy
Gibbs free energy \( \Delta G \) and its standard state \( \Delta G^\circ \) reflect the maximum reversible work performed by a thermodynamic system at constant temperature and pressure. The relation between Gibbs free energy change and the equilibrium constant at standard conditions is given by \( \Delta G^\circ = -RT\ln{K} \).
Moreover, in electrochemical reactions, \( \Delta G^\circ \) is related to the cell potential by this expression:
\( \Delta G^\circ = -nFE^\circ \).
Moreover, in electrochemical reactions, \( \Delta G^\circ \) is related to the cell potential by this expression:
\( \Delta G^\circ = -nFE^\circ \).
- Substituting the calculated \( E^\circ = -0.028 \text{ V} \) into this equation gives \( \Delta G^\circ = -(1)(96,485 \, \text{C/mol})(-0.028 \, \text{V}) \).
- The result is \( \Delta G^\circ = +2700 \, \text{J/mol} \).
- This positive value indicates non-spontaneity under standard conditions, with reactants being favored.
Standard Electrode Potential
The standard electrode potential \( E^\circ \) is a measure of the individual potential of a reversible electrode at standard state, green light to the extent of spontaneous electron flow direction in an electrochemical cell.
- In redox reactions, it provides insight into the tendency of the reaction to proceed. A negative \( E^\circ \) like -0.028 V suggests the reverse reaction is favored under standard conditions.
- This information is pivotal when predicting spontaneity and calculating related thermodynamic parameters, such as \( \Delta G^\circ \).
- The calculated standard electrode potential tells us about the equilibrium position of electrons between two redox species, guiding the pathway of reactions under fixed conditions.
