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One half-cell in a voltaic cell is constructed from a copper wire electrode in a 4.3 x 10-8 M solution of Cu(NO3)2 . The other half-cell consists of a zinc electrode in a 0.60 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 Cu²+ (aq) + 2 e¯→ Cu(s) E' +0.337 V Potential V

One half-cell in a voltaic cell is constructed from a copper wire electrode in a 4.3 x 10-8 M solution of Cu(NO3)2 . The other half-cell consists of a zinc electrode in a 0.60 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 Cu²+ (aq) + 2 e¯→ Cu(s) E' +0.337 V Potential V

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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One half-cell in a voltaic cell is constructed from a copper wire electrode in a 4.3 x 10-8 M solution of Cu(NO3)2. The other half-cell consists of a zinc electrode in a 0.60 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 Cu²+ (aq) + 2 e-+ Cu(s) E° ,2+ :+0.337 V Potential = V %3D
Transcribed Image Text:One half-cell in a voltaic cell is constructed from a copper wire electrode in a 4.3 x 10-8 M solution of Cu(NO3)2. The other half-cell consists of a zinc electrode in a 0.60 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 Cu²+ (aq) + 2 e-+ Cu(s) E° ,2+ :+0.337 V Potential = V %3D
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