The free energy change for the following reaction at 25 °C, when [Pb²*] = 1.15 M and [Fe²*] = 5.40×103 M, is -73.9 kJ: Pb"(1.15 M) + Fe(s)→ Pb(s) + Fe²*(5.40×10³ M) AG=-73.9 kJ What is the cell potential for the reaction as written under these conditions? Answer: V Would this reaction be spontaneous in the forward or the reverse direction?

The free energy change for the following reaction at 25 °C, when [Pb²] = 1.15 M and [Fe²] = 5.40×103 M, is -73.9 kJ: Pb"(1.15 M) + Fe(s)→ Pb(s) + Fe²*(5.40×10³ M) AG=-73.9 kJ What is the cell potential for the reaction as written under these conditions? Answer: V Would this reaction be spontaneous in the forward or the reverse direction?

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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