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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
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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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
Transcribed Image Text: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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