5. Followed from the previous question, it was determined that K for the reaction at25°C between CIO and H2O2 was 1.2 x 10. Calculate AG° and E celi from thegiven K. What is the implication of K having such an enormous value? Use Faradayconstant, F=96485 C/mol.6. Calculate the Ecel value (non-standard state) at 298 K for the cell based on the reaction,Fe"(ag) + Cu'(ag)– Fe"(aq) + Cu (ag),when [Fe"]=[Cu'] - 2.56 x 10-3 M and [Fe2*]= [Cu*]-8.25 x 10 M.Fe"(aq) + e - Fe2"(aq) E = 0.770 VCu"(aq) + e - Cu (ag) E = 0.153 VRT-In Q and Q=nF[FeTCuFe* ][Cu*]Note: Use E= E-

Answered: Image /qna-images/answer/7c19a040-2d7b-4eff-a4c6-9cc58223a500.jpg

5. Followed from the previous question, it was determined that K for the reaction at 25°C between CIO and H;Oz was 1.2 x 107. Calculate AG and Ecel from the given K. What is the implication of K having such an enormous value? Use Faraday constant, F=96485 C/mol.

5. Followed from the previous question, it was determined that K for the reaction at 25°C between CIO and H;Oz was 1.2 x 107. Calculate AG and Ecel from the given K. What is the implication of K having such an enormous value? Use Faraday constant, F=96485 C/mol.

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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Electrolysis

Electrolysis is defined as the process of breaking down ionic compounds into their constituent elements by passing a direct electric current through the compound in a fluid state. The cathode reduces cations, while the anode oxidizes anions. An electrolyte, electrodes, and some form of external power source are the main components required for conducting electrolysis. A partition, such as an ion-exchange membrane or a salt bridge, may also be used, but this is optional. These are primarily used to prevent the products from diffusing near the opposing electrode.

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