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Example! Effect of Concentration Changes on Cell Potentials

posted Dec 17, 2016, 1:06 AM by Grace Ong   [ updated Dec 17, 2016, 1:06 AM ]
The following post was first posted on Blogger on Monday, 11 February 2013.



Happy Lunar New Year! This is my excuse for my tardiness in posting... I've been busy preparing for the festivities!

How do changes in reactant or product concentrations affect cell potential? An example of such a question is the 2011 National Junior College (NJC) preliminary examination paper:

2011 NJC/Paper 1/Q33

2011 NJC Preliminary Examination Paper 1 Question 33

A typical solution goes like this:


E / V
I Cl2 + 2e ⇌ 2Cl   +1.36
II   S4O62− + 2e ⇌ 2S2O32−
+0.09

In this cell, the Cl2/Cl half cell is the cathode, while the S4O62−/S2O32− half cell is the anode.
EcellE(Cl2/Cl) − E(S4O62−/S2O32−) = (+1.36)  − (+0.09) = +1.27 V

Statement 1:
When pressure of Cl2 increases, by Le Chatelier's Principle, equilibrium I shifts right to reduce the pressure. E(Cl2/Cl) becomes more positive than +1.36, and Ecell increases above +1.27 V.

Statement 2:
When water is added to the S4O62−/S2O32− half cell, the number of ions per unit volume decreases. By Le Chatelier's Principle, equilibrium II shifts right to increase the number of ions per unit volume. E(S4O62−/S2O32−) becomes more positive than +0.09 V, and Ecell decreases below +1.27 V.

Statement 3:
E(I2/I) = +0.54 V is more positive than E(S4O62−/S2O32−). When iodine crystals are added to the anode, i.e. the Cl2/Cl half cell, both Cl2 and I2 are reduced by S2O32−. This decreases [S2O32−], and by Le Chatelier's Principle, equilibrium II shifts right to increase [S2O32−]. E(S4O62−/S2O32−) becomes more positive than +0.09 V, and Ecell decreases below +1.27 V.

Thus Statements 2 and 3 will reduce the cell potential.



I shall now attempt to use a more concretised approach, by using the Nernst equation, to explain how concentration changes affect cell potential. The Nernst equation is not in the H2 Chemistry Syllabus, but the equation has appeared frequently enough in examination questions to warrant a short discussion.

The following file contains a brief introduction of the Nernst equation, followed by a simple example on how dilution affects the value of Ecell. The Nernst equation is used to calculate the new Ecell after dilution is made to one or both half cells.

(To view an expandable version, click here to open slides on SlideShare.)

Note:
In my calculations, I have adopted a simplified approach by substituting molar concentrations directly into the Nernst equation, although I really should be using dimensionless activities.
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