Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Problem:**
Determine \( E^\circ \) for a galvanic (voltaic) cell if \( \Delta G^\circ = -4.9 \, \text{kJ/mol} \) and \( n = 3 \).
(Faraday's constant, \( F = 96,500 \, \text{J/(V·mol)} \))
**Solution:**
To find \( E^\circ \) (the standard cell potential), you can use the relationship between Gibbs free energy change (\( \Delta G^\circ \)) and cell potential:
\[
\Delta G^\circ = -nFE^\circ
\]
Where:
- \( \Delta G^\circ = -4.9 \times 10^3 \, \text{J/mol} \) (since 1 kJ = 1000 J)
- \( n = 3 \) (number of moles of electrons transferred)
- \( F = 96,500 \, \text{J/(V·mol)} \)
Rearrange the formula to solve for \( E^\circ \):
\[
E^\circ = -\frac{\Delta G^\circ}{nF}
\]
Substitute the values:
\[
E^\circ = -\frac{-4.9 \times 10^3 \, \text{J/mol}}{3 \times 96,500 \, \text{J/(V·mol)}}
\]
Calculate:
\[
E^\circ = \frac{4.9 \times 10^3}{289,500} \, \text{V}
\]
\[
E^\circ \approx 0.0169 \, \text{V}
\]
Thus, the standard cell potential \( E^\circ \) is approximately 0.0169 volts.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fec3b779b-d0d6-4768-9dcd-21f6961c75e0%2F86f04988-9406-4408-92d1-5573303bd9f7%2Fiqxklu_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem:**
Determine \( E^\circ \) for a galvanic (voltaic) cell if \( \Delta G^\circ = -4.9 \, \text{kJ/mol} \) and \( n = 3 \).
(Faraday's constant, \( F = 96,500 \, \text{J/(V·mol)} \))
**Solution:**
To find \( E^\circ \) (the standard cell potential), you can use the relationship between Gibbs free energy change (\( \Delta G^\circ \)) and cell potential:
\[
\Delta G^\circ = -nFE^\circ
\]
Where:
- \( \Delta G^\circ = -4.9 \times 10^3 \, \text{J/mol} \) (since 1 kJ = 1000 J)
- \( n = 3 \) (number of moles of electrons transferred)
- \( F = 96,500 \, \text{J/(V·mol)} \)
Rearrange the formula to solve for \( E^\circ \):
\[
E^\circ = -\frac{\Delta G^\circ}{nF}
\]
Substitute the values:
\[
E^\circ = -\frac{-4.9 \times 10^3 \, \text{J/mol}}{3 \times 96,500 \, \text{J/(V·mol)}}
\]
Calculate:
\[
E^\circ = \frac{4.9 \times 10^3}{289,500} \, \text{V}
\]
\[
E^\circ \approx 0.0169 \, \text{V}
\]
Thus, the standard cell potential \( E^\circ \) is approximately 0.0169 volts.
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