September 20, 2024, 09:00:13 PM
Forum Rules: Read This Before Posting


Topic: Electrode potentials  (Read 1301 times)

0 Members and 1 Guest are viewing this topic.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Electrode potentials
« on: June 09, 2013, 01:34:12 PM »
My electrode potentials book says that the potential difference between an electrode and the solution in which it is dipped is given by:

Φ[electrode]-Φ[solution]=ΔΦ°+RT/F·ln(Concentration of Reactants/Concentration of Products)

Where reactants and products each refer to the equilibrium Reactants + e-  ::equil:: Products, established for the half-cell. ΔΦ° is a constant for the electrode.

The book seems to make use of this frequently, and in all given examples we've got the reactants on top and products on the bottom of the ln() expression.

Then we move to a 2 electrode situation, let's say Electrode 1 / Solution / Electrode 2. Now

E=(Φ[electrode 2]-Φ[solution])-(Φ[electrode 1]-Φ[solution])
E=ΔΦ°[electrode 2]-ΔΦ°[electrode 1]+RT/F·ln{(Concentration of Reactants in 2 + Products in 1)/(Concentration of Reactants in 1 + Products in 2}

Where ΔΦ°[electrode 2]-ΔΦ°[electrode 1] is often written E°.

But then we get to the common form of the Nernst equation, E=E°-RT/F·ln(Q). Q clearly corresponds to the reaction Reactants in 2 + Products in 1  ::equil:: Reactants in 1 + Products in 2. My question, though, is then wouldn't Q (and by extension E) depend on which way around our circuit is connected? Whereas, being a quantity of the solution, it doesn't make sense to me that Q would depend on which electrode is Electrode 2 and which electrode is Electrode 1.

Sponsored Links