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The free energy change for the following reaction at 25 °C, when Sn2+ = 0.00776 M, is -133 kJ: Sn²+ = 1.11 M and 2+ Sn²+ (1.11 M) + Zn(s) → Sn(s) + Zn²+ (0.00776 M) = AG = -133 kJ What is the cell potential for the reaction as written under these conditions? Ecell V Would this reaction be spontaneous in the forward or the reverse direction? O forward direction O reverse direction

The free energy change for the following reaction at 25 °C, when Sn2+ = 0.00776 M, is -133 kJ: Sn²+ = 1.11 M and 2+ Sn²+ (1.11 M) + Zn(s) → Sn(s) + Zn²+ (0.00776 M) = AG = -133 kJ What is the cell potential for the reaction as written under these conditions? Ecell V Would this reaction be spontaneous in the forward or the reverse direction? O forward direction O reverse direction

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 18.44QE: For each of the reactions, calculate E from the table of standard potentials, and state whether the...
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The free energy change for the following reaction at 25 °C, when Sn²+ 2+ [Zn²+] = 0.00776 M, is -133 kJ: [Sn²+] = 1.11 M and Sn²+ (1.11 M) + Zn(s) → Sn(s) + Zn²+ AG = -133 kJ What is the cell potential for the reaction as written under these conditions? Ecell V Sn(s) + Zn²+ (0.00776 M) Would this reaction be spontaneous in the forward or the reverse direction? O forward direction O reverse direction
Transcribed Image Text:The free energy change for the following reaction at 25 °C, when Sn²+ 2+ [Zn²+] = 0.00776 M, is -133 kJ: [Sn²+] = 1.11 M and Sn²+ (1.11 M) + Zn(s) → Sn(s) + Zn²+ AG = -133 kJ What is the cell potential for the reaction as written under these conditions? Ecell V Sn(s) + Zn²+ (0.00776 M) Would this reaction be spontaneous in the forward or the reverse direction? O forward direction O reverse direction
2+ The free energy change for the following reaction at 25 °C, when Hg² = 1.14 M and [Pb²+] = 0.00903 M, is -201 kJ: 2+ 2+ Hg²+ (1.14 M) + Pb(s) → Hg(1) + Pb²+ (0.00903 M) AG: = What is the cell potential for the reaction as written under these conditions? Ecell Would this reaction be spontaneous in the forward or the reverse direction? forward direction O reverse direction -201 kJ
Transcribed Image Text:2+ The free energy change for the following reaction at 25 °C, when Hg² = 1.14 M and [Pb²+] = 0.00903 M, is -201 kJ: 2+ 2+ Hg²+ (1.14 M) + Pb(s) → Hg(1) + Pb²+ (0.00903 M) AG: = What is the cell potential for the reaction as written under these conditions? Ecell Would this reaction be spontaneous in the forward or the reverse direction? forward direction O reverse direction -201 kJ
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