Effects of concentration on \(E^o\) (Nernst equation)

What is the effect of concentration on \(E^o\) ?

Consider the following standard electrochemical cell potential

Cu(s)+2Ag^+(aq)\rightleftharpoons Cu^{2+}(aq)+2Ag(s)\; \; \; \; \; \; E^o=+0.46\; V

where [Cu2+] = [Ag+] = 1 M.

If [Ag+] > 1 M, the equilibrium will shift to the right according to Le Chatelier’s principle. This implies that the reaction has become more spontaneous from left to right, resulting in a more positive Eo. Conversely, if [Ag+] < 1 M, the equilibrium will shift to the left, giving a less positive Eo. In general, the effects of concentration of chemical species on the electrochemical cell potential is given by the Nernst equation:



    • E is the potential of the cell under non-standard conditions, e.g. when T is not 298 K or concentrations of species are not at 1 M and 1 atm.
    • Eo is the overall standard electrode potential of the cell.
    • R is the gas constant
    • T is temperature
    • F is the Faraday constant
    • Q is the reaction quotient, which for the above reaction is:


For example, if [Cu2+] = 1.8 M and [Ag+] = 0.02 M at rtp,

E=+0.46-\frac{(8.31)(298)}{(2)(96485)}ln\frac{1.8}{(0.02)^2}=+0.35\; V

It is important to note that the Nernst equation applies to situations where no current flows, i.e. it allows the potential of a cell to be calculated under open circuit situations. This means that the calculated value, E, for the above example, is the predicted potential of the cell (containing[ Cu2+] = 1.8 M and [Ag+] = 0.02 M at rtp), which is connected to an external emf that exactly balances this calculated potential.



What If we want to calculate the potential of a cell where a current flows, e.g. an electrolytic cell ?


We will have to account for changes in electrode potentials due to a phenomenon called polarisation. Please see this article in the advanced section for details.



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