Calculate AG° at 298K for the following reaction:Cu(s) + 2Ag*(aq)2Ag(s) + Cu²+(aq)Cu2+(aq) + 2e →Cu(s)E° = 0.24 VAg*(aq) + e →Ag(s)0.957 VAG = –nFEF=96485 C/mol eO +69.2 kJO +257 kJ-69.2 kJ+138 kJ-138 kJ

We have a reaction whose half cell potential is given , we've to calculate the∆G° value.

Calculate AG° at 298K for the following reaction: Cu(s) + 2Ag*(aq) + 2Ag(s) + Cu2+(aq) Cu2+(aq) + 2e¯ → Cu(s) Ag*(aq) + e E° = 0.24 V %3D Ag(s) E° = 0.957 V AG = -nFE F=96485 C/mol e +69.2 kJ +257 kJ -69.2 kJ +138 kJ -138 kJ

Calculate AG° at 298K for the following reaction: Cu(s) + 2Ag(aq) + 2Ag(s) + Cu2+(aq) Cu2+(aq) + 2e¯ → Cu(s) Ag(aq) + e E° = 0.24 V %3D Ag(s) E° = 0.957 V AG = -nFE F=96485 C/mol e +69.2 kJ +257 kJ -69.2 kJ +138 kJ -138 kJ

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter19: Applications Of Standard Electrode Potentials

Section: Chapter Questions

Problem 19.8QAP

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Question

Calculate AG° at 298K for the following reaction:
Cu(s) + 2Ag*(aq)
2Ag(s) + Cu²+(aq)
Cu2+(aq) + 2e →
Cu(s)
E° = 0.24 V
Ag*(aq) + e →
Ag(s)
0.957 V
AG = –nFE
F=96485 C/mol e
O +69.2 kJ
O +257 kJ
-69.2 kJ
+138 kJ
-138 kJ

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