One half-cell in a voltaic cell is constructed from a copper wire electrode in a 3.9 × 10-³ M solution of Cu(NO3)2. The other half-cell consists of a zinc electrode in a 0.10 M solution of Zn(NO3)2 . Calculate the cell potential. You may need to use the following data: Zn2+ (aq) + 2 e- + Zn(s) E' = -0.763 V Cu?* (aq) + 2 e¯ → Cu(s) E' = +0.337 V Potential II

One half-cell in a voltaic cell is constructed from a copper wire electrode in a 3.9 × 10-³ M solution of Cu(NO3)2. The other half-cell consists of a zinc electrode in a 0.10 M solution of Zn(NO3)2 . Calculate the cell potential. You may need to use the following data: Zn2+ (aq) + 2 e- + Zn(s) E' = -0.763 V Cu?* (aq) + 2 e¯ → Cu(s) E' = +0.337 V Potential II

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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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

One half-cell in a voltaic cell is constructed from a copper
wire electrode in a 3.9 × 10-8 M solution of Cu(NO3)2 .
The other half-cell consists of a zinc electrode in a 0.10 M
solution of Zn(NO3)2 - Calculate the cell potential.
You may need to use the following data:
Zn2+ (aq) + 2 e- → Zn(s) E° = -0.763 V
Cu?+ (aq) + 2 e¯ → Cu(s) E° = +0.337 V
Potential =
|v

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