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What prevents infinite potential with Nernst Equation

  1. Jun 4, 2009 #1
    The nernst equation has a log Q term. The denominator of that could be very close to zero. That would make Q close to infinity and log Q close to infinity. How high can the EMF be due to concentration in the real world and why?
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  3. Jun 4, 2009 #2

    Andy Resnick

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    The Nernst equation is [itex] E = \frac{R T}{z F} \ln\frac{[\text{ion outside cell}]}{[\text{ion inside cell}]}[/itex].

    As the concentration of ions on one side of the membrane goes down (towards zero), the driving force on ions on the other side of the membrane goes up- that's simple diffusion. Since the log function is sublinear, I'ts not clear what you are having problems with.
  4. Jun 4, 2009 #3
    I wasn't thinking of membranes in particular. The log of infinity is infinity. My guess is the limiting factor is some other half reaction or diffusion of ions from one side of a battery to the other. Even if this is the case the Nernst equation makes it appear that a much higher voltage can be obtained from a concentration gradient than I think can occur.
  5. Jun 6, 2009 #4
    I would imagine the Nernst equation loses accuracy at low ion concentrations. The physics would change and the equation would no longer be correct.

    Wiki (not the most authoritative source, I realize) seems to agree:

    At very low concentrations of the potential determining ions, the potential predicted by Nernst equation tends to ┬▒infinity. This is physically meaningless because, under such conditions, the exchange current density becomes very low, and then other effects tend to take control of the electrochemical behavior of the system.
  6. Jun 13, 2009 #5
    How would you explain a 0.1V current when the concentration of copper ions in solution is 0M ?

    This is in a zinc || copper galvanic cell (with salt bridge) and the [Zn2+] is 1M?

    I thought that the copper ions attract the electrons? Am I mistaken?
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