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Item 19: Question Calculate the voltage of the cell. Ag(s)|AgBr(s), NaBr(aq, 1.0 M)||CdCl₂(aq, 0.050 M)| Cd(s) AgBr(s) + e = Ag(s) + Br |E° = 0.071 V Ag+ + e Ag(s) E° = 0.799 V Cd²++ 2e Cd(s) | Ksp (AgBr(s)) = 5.0 × 10-13 -0.678 V -0.511 V -0.423 V 0.511 V E° = -0.402 V

Item 19: Question Calculate the voltage of the cell. Ag(s)|AgBr(s), NaBr(aq, 1.0 M)||CdCl₂(aq, 0.050 M)| Cd(s) AgBr(s) + e = Ag(s) + Br |E° = 0.071 V Ag+ + e Ag(s) E° = 0.799 V Cd²++ 2e Cd(s) | Ksp (AgBr(s)) = 5.0 × 10-13 -0.678 V -0.511 V -0.423 V 0.511 V E° = -0.402 V

Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter18: Electrochemistry
Section: Chapter Questions
Problem 18.95QE
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19. How do you set up and solve voltage problems like the one below?

Item 19: Question Calculate the voltage of the cell. Ag(s)|AgBr(s), NaBr(aq, 1.0 M)||CdCl₂(aq, 0.050 M)| Cd(s) AgBr(s) + e = Ag(s) + Br |E° = 0.071 V Ag+ + e Ag(s) E° = 0.799 V Cd²++ 2e Cd(s) | Ksp (AgBr(s)) = 5.0 × 10-13 -0.678 V -0.511 V -0.423 V 0.511 V E° = -0.402 V
Transcribed Image Text:Item 19: Question Calculate the voltage of the cell. Ag(s)|AgBr(s), NaBr(aq, 1.0 M)||CdCl₂(aq, 0.050 M)| Cd(s) AgBr(s) + e = Ag(s) + Br |E° = 0.071 V Ag+ + e Ag(s) E° = 0.799 V Cd²++ 2e Cd(s) | Ksp (AgBr(s)) = 5.0 × 10-13 -0.678 V -0.511 V -0.423 V 0.511 V E° = -0.402 V
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