9. 10. E for the following cell at 298 K is Ag(s) AgCl(s) | Cl-(aq.) || Br-(aq.) | AgBr(s) | Ag(s) (Given: K of AgCl = 2 x 10-10, K of AgBr = 4 x 10-13) (A) 0.0 V (B) 0.16 V (C) 0.16 V Calculate the emf of the call 8500 (D)-0.32 V

9. 10. E for the following cell at 298 K is Ag(s) AgCl(s) | Cl-(aq.) || Br-(aq.) | AgBr(s) | Ag(s) (Given: K of AgCl = 2 x 10-10, K of AgBr = 4 x 10-13) (A) 0.0 V (B) 0.16 V (C) 0.16 V Calculate the emf of the call 8500 (D)-0.32 V

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

pls give detailed answer

9.
10.
for the following cell at 298 K is
Ag(s) | AgCl(s) | Cl-(aq.) || Br (aq.) | AgBr(s) | Ag(s)
(Given: K of AgCl = 2 x 10-10, K of AgBr = 4 × 10-¹³)
(A) 0.0 V
(B) 0.16 V
(C)-0.16 V
Calculate the emf of the cell at 25°C
Pt III (1.0
(D)-0.32 V

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