**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.

Given that an electrochemical cell is Cu(s)|Cu2+(aq, 0.02 M)||Cu2+(aq, 0.50 M)|Cu(s). We have to…

Answered: 7. Determine the cell potential (in V…

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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Concept explainers

Electrolysis

Electrolysis is defined as the process of breaking down ionic compounds into their constituent elements by passing a direct electric current through the compound in a fluid state. The cathode reduces cations, while the anode oxidizes anions. An electrolyte, electrodes, and some form of external power source are the main components required for conducting electrolysis. A partition, such as an ion-exchange membrane or a salt bridge, may also be used, but this is optional. These are primarily used to prevent the products from diffusing near the opposing electrode.

Question

**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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