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
![Mg(s) | Mg*(0.025 M) || Mg**(1.5 M) | Mg(s)
Use the Nernst equation to calculate the emf of this cell.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff57c5c9b-acb3-4a3d-a39b-55eeff7ee4ad%2F279228d1-a1b3-41b3-aac5-01ee0ac74f07%2Fb02ycqdh_processed.png&w=3840&q=75)
Transcribed Image Text:Mg(s) | Mg*(0.025 M) || Mg**(1.5 M) | Mg(s)
Use the Nernst equation to calculate the emf of this cell.
Expert Solution
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Step 1
The given cell is a concentration cell.
The electrodes are same in both the half cells only the difference is the concentration of electrolytes.
An equilibrium is established due to the transfer of electrons from low concentration to high concentration.
Due to this Mg2+ will be formed in anode and Mg(s) will be formed in cathode.
The cell potential can be calculated by using Nernst equation:
Anode is the cell where oxidation takes place
and cathode is the cell where reduction takes place.
n= number of electrons transferred.
Eocell = 0 (since both the electrodes are same in concentration cell).
Step 2
Given cell is:
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