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Electrochemistry and the Nernst Equation

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1.Consider the following cell at equilibrium:
Mg(s) / Mg+2 (0.01 M) // OH- (?) / Mg(OH)2 (s) / Mg (s)
Given: Mg+2 + 2e- ---> Mg(s) Eo = -2.360 V
Mg(OH)2 (s) + 2e- ---> Mg(s) + 2 OH- Eo = -2.690 V
a) Write each half-reaction and the net cell reaction. Find the standard net cell potential.
b) Find the pH of the right cell when the cell is at equilibrium.
c) Draw the cell, clearly mark all the components, and show the anode and the cathode, and the direction of the flow of electrons. Which side of the cell is the reduction taking place?

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Solution Summary

Cell pH is found using the standard cell potential and the Nernst equation. Full working is shown and a digaram is included.

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1.Consider the following cell at equilibrium:
Mg(s) / Mg+2 (0.01 M) // OH- (?) / Mg(OH)2 (s) / Mg (s)
Given: Mg+2 + 2e- ↔ Mg(s) Eo = -2.360 V
Mg(OH)2 (s) + 2e- ↔ Mg(s) + 2 OH- Eo = -2.690 V
a) Write each half-reaction and the net cell reaction. Find the standard net cell potential.
Mg ↔ Mg2+ + 2e E0 = +2.360 V
Mg(OH)2 (s) + 2e- ↔ Mg(s) + 2 OH- Eo = -2.690 V
Mg(OH)2(s) ---> Mg2+ + 2 OH- E0cell = -0.330 V

When interpreting a cell diagram, the anode is always listed ...

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