Solve all parts otherwise I will downvote Question 27Calculate AGel in kJ/mol at 655 K givencell -0.830 voltsn= 6F= 96500 C/mole electronsAGnFEelcellcellO 0.00498721O 2131,600481 Question 28Calculate E in volts at 533 K givencellAGOcell 222 kJ/moln= 4F= 96500 C/mole electronsAGOcell– nFEcellO-1.73O-0.575O-0.288-0.383O-1.15

Formula, ∆G° = -nFE°cell then,

Answered: Question 27 Calculate AGel in kJ/mol at…

Question 27 Calculate AGel in kJ/mol at 655 K given cell -0.830 volts n= 6 F= 96500 C/mole electrons AG nFE, cell= cell O 0.00498 O721 O213 1,600 481

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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Concept explainers

Potentiometric Titrations

Potentiometric titration is one of the important titration methods and is mainly carried out to characterize acid solutions. It follows a method similar to that of the normal titration technique, but the main difference is that there is no indicator used in the potentiometric titrations. Instead of using an indicator, electrodes are employed for monitoring the whole titration process. A scientist named Robert Behrend did the potentiometric titration for the first time, and it was done using the mercurous solution and potassium chloride.

Question

Solve all parts otherwise I will downvote

Question 28
Calculate E in volts at 533 K given
cell
AGO
cell 222 kJ/mol
n= 4
F= 96500 C/mole electrons
AGO
cell
– nFE
cell
O-1.73
O-0.575
O-0.288
-0.383
O-1.15

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