-3M solution of Cu(NO3)2. The other half-cell consists of a zinc electrode inOne half-cell in a voltaic cell is constructed from a copper wire electrode in a 3.9 x 10a 0.45 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 VCu²+ (aq) + 2 e¯ →Cu(s) E° = +0.337 VPotential =V

Points to note: 1) Electrode with more reduction potential forms the cathode 2) Reduction takes…

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 0.45 M solution of Zn(NO3)2. Calculate the cell potential. You may need to use the following data: Zn²+ (aq) + 2 e¯ → Zn(s) Cu²+ (aq) + 2 e¯ → Cu(s) E° =+0.337 V E° = -0.763 V

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 0.45 M solution of Zn(NO3)2. Calculate the cell potential. You may need to use the following data: Zn²+ (aq) + 2 e¯ → Zn(s) Cu²+ (aq) + 2 e¯ → Cu(s) E° =+0.337 V E° = -0.763 V

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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