7. Determine the cell potential (in V to two decimal places) for the following electrochemica cell: 2+ Cu(s) | Cu²+ (aq, 0.02 M) || Cu²+ (aq, 0.50 M) | Cu(s) A: 0.04
7. Determine the cell potential (in V to two decimal places) for the following electrochemica cell: 2+ Cu(s) | Cu²+ (aq, 0.02 M) || Cu²+ (aq, 0.50 M) | Cu(s) A: 0.04
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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![**Exercise Title: Determining Cell Potential**
**Objective:**
Calculate the cell potential (in volts, accurate to two decimal places) for the specified electrochemical cell.
**Electrochemical Cell Description:**
- Cell Notation: Cu(s) | Cu²⁺(aq, 0.002 M) || Cu²⁺(aq, 0.50 M) | Cu(s)
**Calculation Outcome:**
- Answer: 0.04 V
### Explanation:
This electrochemical cell consists of two copper electrodes immersed in copper ion solutions of different concentrations. The task involves calculating the potential difference arising due to the concentration gradient between the two half-cells.
**Note:** The calculations likely involve using the Nernst equation:
\[ E = E^\circ - \left( \frac{0.0592}{n} \right) \log \frac{[\text{Cu}^{2+}]_{\text{anode}}}{[\text{Cu}^{2+}]_{\text{cathode}}} \]
Where:
- \( E^\circ \) is the standard electrode potential for copper.
- \( n \) is the number of electrons transferred (for Cu, \( n = 2 \)).
- The concentration gradient involves 0.002 M and 0.50 M solutions.
This exercise provides insight into the principles of electrochemistry and demonstrates the effect of concentration differences on cell potential.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8e77b376-6a5c-4963-870c-f49a6e984799%2F09f8b165-0fcb-45ec-bed5-eaa25d35f566%2Fbhdy9ll_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Exercise Title: Determining Cell Potential**
**Objective:**
Calculate the cell potential (in volts, accurate to two decimal places) for the specified electrochemical cell.
**Electrochemical Cell Description:**
- Cell Notation: Cu(s) | Cu²⁺(aq, 0.002 M) || Cu²⁺(aq, 0.50 M) | Cu(s)
**Calculation Outcome:**
- Answer: 0.04 V
### Explanation:
This electrochemical cell consists of two copper electrodes immersed in copper ion solutions of different concentrations. The task involves calculating the potential difference arising due to the concentration gradient between the two half-cells.
**Note:** The calculations likely involve using the Nernst equation:
\[ E = E^\circ - \left( \frac{0.0592}{n} \right) \log \frac{[\text{Cu}^{2+}]_{\text{anode}}}{[\text{Cu}^{2+}]_{\text{cathode}}} \]
Where:
- \( E^\circ \) is the standard electrode potential for copper.
- \( n \) is the number of electrons transferred (for Cu, \( n = 2 \)).
- The concentration gradient involves 0.002 M and 0.50 M solutions.
This exercise provides insight into the principles of electrochemistry and demonstrates the effect of concentration differences on cell potential.
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