2013‎ > ‎

02

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.

The Hybridisation Theory... as applied to simple inorganic molecules

posted Dec 16, 2016, 7:21 AM by Grace Ong   [ updated Dec 16, 2016, 7:22 AM ]

The following post was first posted on Blogger on Monday, 18 February 2013.



I shall use the Hybridisation Theory to explain the shapes of the ammonia and beryllium chloride molecules.

The Ammonia Molecule

Electronic configuration of nitrogen:

electron-in-box diagram for nitrogen

If the atomic orbitals of the central nitrogen atom are not hybridised, the N−H bonds will be oriented at angles of the px, py and pz orbitals.

hypothetical unhybridised orbitals of nitrogen in ammonia

This does not explain why the ammonia molecule is trigonal pyramidal.

If the 2s and 2p orbitals of nitrogen are mixed to form four sp3 hybrid orbitals,

hybridisation of nitrogen

three of the sp3 orbitals will each form an end-on overlap with the 1s orbital of a hydrogen atom (to form three N−H σ-bonds), while the fourth sp3 orbital will house the unbonded pair of electrons:

sp3 hybrid orbitals of nitrogen in ammonia and shape of ammonia molecule

Note: In a typical situation,

No. of hybrid orbitals
formed
 = 
No. of atoms the central atom
is bonded to
 + 
No. of non-bonded electron pairs
on the central atom

The electron geometry with respect to the central nitrogen atom is tetrahedral, and the resulting molecular geometry of the molecule is trigonal pyramidal. However the observed H−N−H bond angle is 107.5 °, less than the 109 ° of a tetrahedral angle. This is because the unbonded electron pair is closer to the nitrogen atom and occupies a greater volume than a bond pair, resulting in a lone pair-bond pair repulsion being greater than a bond pair-bond pair repulsion.

The Beryllium Chloride Molecule

Electronic configuration of beryllium:

electron-in-box diagram of beryllium
Since there are only two chlorine atoms bonded to beryllium, and no unbonded electron pairs on beryllium, only one 2p orbital is used to combine with the 2s orbital of beryllium to form two sp hybrid orbitals:

hybridisation of beryllium

Each of the two sp hybrid orbitals then forms an end-on overlap with a 3p orbital of a chlorine atom (to form two Be−Cl σ-bonds):

sp hybrid orbitals of beryllium in beryllium chloride

The resulting molecular geometry of the molecule is linear, with a bond angle of 180 °.

shape of beryllium chloride

Hybridisation of Carbon Atomic Orbitals

posted Dec 15, 2016, 8:10 AM by Grace Ong   [ updated Dec 17, 2016, 1:20 AM ]

The following post was first posted on Blogger on Monday, 18 February 2013.



The following slides describe the three types of hybidisation that carbon atomic orbitals can undergo
  • sp3 hybridisation, as in the methane molecule
  • sp2 hybridisation, as in the ethene molecule
  • sp hybridisation, as in the ethyne molecule

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

1-3 of 3