Mg(s) | Mg*(0.025 M) || Mg**(1.5 M) | Mg(s)Use the Nernst equation to calculate the emf of this cell.

Answered: Mg(s) | Mg²*(0.025 M) || Mg²*(1.5 M) |…

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

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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

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:

$E_{c e l l} = {E^{o}}_{c e l l} - \frac{0.0592}{n} \log \frac{[a n o d e]}{[c a t h o d e]}$

Anode is the cell where oxidation takes place

and cathode is the cell where reduction takes place.

n= number of electrons transferred.

E^o_cell= 0 (since both the electrodes are same in concentration cell).

Step 2

Given cell is:

$E_{c e l l} = {E^{o}}_{c e l l} - \frac{0.0592}{n} \log \frac{[a n o d e]}{[c a t h o d e]} F o r t h e g i v e n c e l l, E_{c e l l} = 0 - \frac{0.0592}{2} \log \frac{0.025 M}{1.5 M} = 0.0526 V$

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