Friday, February 1, 2008

Electrochemical Potentials, Membrane Potentials

From last quarter in thermodynamics and the first half of this quarter, we have underscored the great importance of the chemical potential, noting that matter flows spontaneously from high to low μ. When these particles are charged, however, electrical work is performed in addition to chemical work. To take this into account, we transmogrify the chemical potential into the new-and-improved function, the electrochemical potential μ~ = μ + zFφ or

μi~ = μi° + RTlnai + zi

From this thermodynamic function, we can derive most of the basics of electrochemistry:
  • voltaic/galvanic cells use spontaneous redox reactions to generate voltage
  • the cathode is the site of reduction and the anode is the site of oxidation
  • electrons flow towards cathode making it the positive terminal
  • standard reduction potentials use the conventional standard [E°(H+=0]
  • cell potentials can be calculated by E°=E°cathode-anode
  • cell notation is in form anode half cell || cathode half cell
  • nonstandard potentials are calculated using Nernst eqn: E=E°-(RT/νF)lnQ
  • pH is defined as -log(aH+)
  • concentration cells generate voltage using same half-cell at different concentrations
When a membrane separates two ionic environments, we again see that balancing the electrochemical potential leads to an electrical potential difference [the membrane potential]. Note again that, unfortunately, we are employing the term potential twice, the first being a form of Gibbs energy and the latter being a type of voltage. The membrane potential can be found from:

∆φ = (RT/ziF)ln(aαi/aβi)

A more realistic scenario, called the Donnan effect, is scene we consider a membrane in which everything is permeable [ions, solvent] except for a macro-ion M. In addition to the electrochemical potential, the macro-ion will generate an osmotic pressure force on the α side, although this term is typically small enough to be neglected if the macro-ion is sufficiently dilute. With a half-page of work, we find the following relationships:

∆φ = (RT/z+F)ln(Y)

where Y = mα+/mβ+ (the concentration imbalance)
and, when M is dilute, Y ≅ 1 - zMmM/2mβ


Anonymous said...

Which problems from HW #3 are we responsible for knowing how to do for the quiz tomorrow? Are the solutions going to be up today?

rod said...

You'll notice that there are little breaks in the homework, in anticipation of questions like this. So, 1 thru 6 are fair game for the quiz, the rest possibly before the exam.

Anonymous said...

i was wondering how you know when to include the value of the counter ion.

Anonymous said...

How does the natural log on prob. 5 part (a) hw. 2 turn into a log function?

rod said...

a) You use the counter-ion when calculating the ionic strength (which is a function of all ions present in solution).

b)ln x = 2.303 log x

Anonymous said...

i understand that it's associated with the ionic strength, but, for instance in number 5 in the solutions, the counter ion wasn't calculated into the ionic strength. is this because it was small enough to be negligible?

rod said...

For hw.2 problem 5, the ionic strength does in fact take the counter-ion into account. Had we ignored the chloride (which is *equal* to sodium ion, and therefore not negligible) we would've gotten I=0.0050. Don't forget that 1/2 factor in the ionic strength formula.