4QEWhen thePb²+ concentration is 1.47 M, the observed cell potential at 298 K for an electrochemical cell with the following reaction is 0.732 V.What is theZn²+ concentration?DPb²+ (aq) +Zn(s) → Pb(s) + Zn²+ (aq)Submit Answer$Pb2+ (aq) + 2eZn²+ (aq) + 2e4RF%5[Review Topics][References]Use the References to access important values if needed for this question.M→ Pb(s) Ee = -0.126 Vred→ Zn(s) E = -0.763 VredRetry Entire GroupTCengage Learning | Cengage Technical SupportG^9 more group attempts remaining6&7HU*8J(9K0)OPPreviousEmail InstructorNext>Save and Exit

Answer: Nernst equation will be used to solve this problem which is shown below:…

When the Pb²+ concentration is 1.47 M, the observed cell potential at 298 K for an electrochemical cell with the following reaction is 0.732 V. What is the Zn²+ concentration? 2+ Pb²+ (aq) +Zn(s)→ Pb(s) + Zn²+ (aq) Pb2+ (aq) + 2e Zn²+ (aq) + 2e [Zn²+] = M → Pb(s) E → Zn(s) E red = -0.126 V -0.763 V

When the Pb²+ concentration is 1.47 M, the observed cell potential at 298 K for an electrochemical cell with the following reaction is 0.732 V. What is the Zn²+ concentration? 2+ Pb²+ (aq) +Zn(s)→ Pb(s) + Zn²+ (aq) Pb2+ (aq) + 2e Zn²+ (aq) + 2e [Zn²+] = M → Pb(s) E → Zn(s) E red = -0.126 V -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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