The free energy change for the following reaction at 25 °C, when [Zn“]= 6.68×10³ M and [H"]= 1.17 M, is 160. kJ: Zn2*(6.68x103 M) + H,(g) Zn(s) + 2H*(1.17 M) AG = 160. kJ What is the cell potential for the reaction as written under these conditions? Answer: Would this reaction be spontaneous in the forward or the reverse direction?
The free energy change for the following reaction at 25 °C, when [Zn“]= 6.68×10³ M and [H"]= 1.17 M, is 160. kJ: Zn2*(6.68x103 M) + H,(g) Zn(s) + 2H*(1.17 M) AG = 160. kJ What is the cell potential for the reaction as written under these conditions? Answer: Would this reaction be spontaneous in the forward or the reverse direction?
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter17: Electrochemistry
Section: Chapter Questions
Problem 73QAP: Consider the reaction below at 25°C: 2MnO4(aq)+16H+(aq)+10Br(aq)2Mn2+(aq)+5Br2(l)+8H2O Use Table...
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![The free energy change for the following reaction at 25 °C, when [Zn“]= 6.68×10³ M and [H"]= 1.17 M, is 160. kJ:
Zn2*(6.68x103 M) + H,(g) Zn(s) + 2H*(1.17 M) AG = 160. kJ
What is the cell potential for the reaction as written under these conditions?
Answer:
Would this reaction be spontaneous in the forward or the reverse direction?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F75937fc1-2331-4616-9cdd-3b5cf4cb7dc9%2F1dd4958f-d966-4c15-9795-f655b43991aa%2Fg0h7a7.jpeg&w=3840&q=75)
Transcribed Image Text:The free energy change for the following reaction at 25 °C, when [Zn“]= 6.68×10³ M and [H"]= 1.17 M, is 160. kJ:
Zn2*(6.68x103 M) + H,(g) Zn(s) + 2H*(1.17 M) AG = 160. kJ
What is the cell potential for the reaction as written under these conditions?
Answer:
Would this reaction be spontaneous in the forward or the reverse direction?
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