Q2. Find the cell potential of a galvanic cell based on the reaction below at 25 c°?CdPb*2 + 2e-Where {Cd+2}=0.015M , {Pb*2}= 0.35MCd+2 + 2e-E = - 0.514 v- PbE°= - 0.25 vB. A constant current was passed though solution of AuCl- ions between gold electron , aftera period of 15 min the cathode increase in mass by 2.613 grams . How much was passed ? whatwas the current. (At.wt. Au= 197)?AuCla- + 3e → Au + 4CI

According to Nernst equation- Ecell = Estd cell - 0.0591/n * log [Product/Reactant] Estd cell =…

Answered: Q2. Find the cell potential of a…

Q2. Find the cell potential of a galvanic cell based on the reaction below at 25 c? Cd Pb+2 + 2e- Where (Cd+)=0.015M, {Pb*}= 0.35M Cd+2 + 2e- E =- 0.514 v + Pb E = - 0.25 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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Question

Q2. Find the cell potential of a galvanic cell based on the reaction below at 25 c°?
Cd
Pb*2 + 2e-
Where {Cd+2}=0.015M , {Pb*2}= 0.35M
Cd+2 + 2e-
E = - 0.514 v
- Pb
E°= - 0.25 v
B. A constant current was passed though solution of AuCl- ions between gold electron , after
a period of 15 min the cathode increase in mass by 2.613 grams . How much was passed ? what
was the current. (At.wt. Au= 197)?
AuCla- + 3e → Au + 4CI

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