A voltaic cell consists of a Pb/Pb?+ half-cell and a Cu/Cu2+ half-cell at 25 °C.The initial concentrations of Pb?+ and Cu?+ are 0.0530 mol L1 and 1.60mol L-', respectively.Part BWhat is the cell potential when the concentration of Cu+ has fallen to 0.240 mol L?Pb?+ (ag) +2 e → Pb(s) E° =0.13 VExpress your answer using two decimal places.Cu (ag) + 2 e → Cu(s) E = 0.34 VEcal = 0.489VSubmitPrevious Answers Request AnswerX Incorrect; Try AgainPart CWhat are the concentrations of Pb?+ and Cu?+ when the cell potential falls to 0.36 V?Express your answers using two significant figures separated by a comma.?[Pb* ]- [Cu*] =mol LSubmitRequest Answer

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A voltaic cell consists of a Pb/Pb?+ half-cell and a Cu/Cu2+ half-cell at 25 °C. The initial concentrations of Pb?+ and Cu?+ are 0.0530 mol L-1 and 1.80 mol L-1. respectively. Pb?+ (ag) +2 e → Pb(s) E = Cu?+ (ag) + 2 e- → Cu(s) E =0.34 V 0.13 V

A voltaic cell consists of a Pb/Pb?+ half-cell and a Cu/Cu2+ half-cell at 25 °C. The initial concentrations of Pb?+ and Cu?+ are 0.0530 mol L-1 and 1.80 mol L-1. respectively. Pb?+ (ag) +2 e → Pb(s) E = Cu?+ (ag) + 2 e- → Cu(s) E =0.34 V 0.13 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

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Question

A voltaic cell consists of a Pb/Pb?+ half-cell and a Cu/Cu2+ half-cell at 25 °C.
The initial concentrations of Pb?+ and Cu?+ are 0.0530 mol L1 and 1.60
mol L-', respectively.
Part B
What is the cell potential when the concentration of Cu+ has fallen to 0.240 mol L?
Pb?+ (ag) +2 e → Pb(s) E° =
0.13 V
Express your answer using two decimal places.
Cu (ag) + 2 e → Cu(s) E = 0.34 V
Ecal = 0.489
V
Submit
Previous Answers Request Answer
X Incorrect; Try Again
Part C
What are the concentrations of Pb?+ and Cu?+ when the cell potential falls to 0.36 V?
Express your answers using two significant figures separated by a comma.
?
[Pb* ]- [Cu*] =
mol L
Submit
Request Answer

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