I am very new to electrochemistry and ion selective electrodes. The membrane potential of an ion selective electrode is given by $$E=\frac{RT}{nF}\ln{\frac{c_2}{c_1}}$$ Where R is universal gas constant, T temperature in kelvin, n the number of electrons transfered, F Faraday of electricity, $c_2$ the concentration of internal standard solution and $c_1$ is the concentration of the ion to be calculated.

But according to the Nernst equation $$E= E^\circ -\frac{RT}{nF}\ln k$$ Can someone explain how to come up with the membrane potential equation using the Nernst equation?


1 Answer 1


The Nernst Equation is $E = E^\circ - \frac{RT}{nF} \ln{Q_c}$. From that you can derive the approximation you gave. $Q_c$ is the product of the activities of the products over those of the reactants. This can be, for dilute solutions where activity approximates concentration, the typical [B]/[A] for the reaction A→B. So, the only thing you need to "explain" is why E° would be zero. Review the definition of E°. hint:What is ln(1/1)? See the discussion on concentration cells in: http://www.chem1.com/acad/webtext/elchem/ec4.html

  • $\begingroup$ If the concentration of the solutions is different, then their activities are also different (since concentration is a measure of activity). Why E° value is zero? (please correct me if I am wrong) $\endgroup$
    – Aurges
    Sep 5, 2016 at 3:48
  • $\begingroup$ $E^\circ$ only depends on the standard states, not on actual states. It thus doesn't really depend on the activities of anything. For most ion-sensitive electrode applications, there is no electrochemical cell across the ion-selective membrane. $\endgroup$
    – Curt F.
    Dec 3, 2016 at 21:41
  • $\begingroup$ Also there is a misplaced $nF$ in this answer. I'll edit to move it, but make sure you agree with this change please. $\endgroup$
    – Curt F.
    Jan 3, 2017 at 4:41

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