My questions relates to the fundamental concept of electrochemistry, more specifically the electrode potentials.

1) First, why is there a potential difference at the interface of two phases? Considering the simplest case of a metal rod dipped in the solution of its ions, why should there be a potential difference. Out of many ways, one of the ways of understanding this provided by this site here,is in terms of equilibrium between the metal atoms and ions. But then, why is the presence of the other phase (ions for the metal piece) required to effect this equilibrium? For example, for the equilibria of ammonia and nitrogen/hydrogen, even if we take pure ammonia, the equilibrium is set up according to the surrounding conditions. Next,then can this equilibrium be described in terms of the equation $\Delta G=-RT\ln(K)$. Lastly, if this is an equilibrium, then there must also be an equilibrium between the metal and the hydrogen or the hydroxyl ions in solution as these might also take up the electrons from the metals?

2) When we measure the difference in electrode potentials of two different half cells, we are just measuring the potential difference between the two metal electrodes. How can we get an idea of the potential difference of the metal piece and its solution as the absolute value of potential of the two solutions might also differ. That is, if the two solutions are at different electrostatic potentials ( though not measurable), the potential difference between the two electrodes will not be the difference of their electrode potentials(which are the potential difference between the electrode and the electrolyte in each case). So, in order for the potential difference of the electrodes be equal to the difference in electrode potentials of the two half-cells, the solutions should be at the same potential?

  • $\begingroup$ I'm sorry, I'm having trouble making heads or tails of your question. For Question 1, I don't see where hydroxyl ions come into play with the metals. For Question 2, I'm not sure what you mean by absolute potential. Potentials are always relative to each other. The closest thing to an "absolute" potential is the vacuum, and if you want to know the potential relative to the vacuum, you just add 4.44V en.wikipedia.org/wiki/Absolute_electrode_potential $\endgroup$
    – chipbuster
    Aug 11, 2013 at 17:31
  • $\begingroup$ @chipbuster In question 1, i am asking that in the presence of any other ionic species apart from the metal, that species should also be free to take the electrons accumulated over the electrode and hence enter an equilibrium with the metal electrode. In second part, i am just asking, unless the absolute electrostatic potentials of the two solutions are same, how can the voltage difference between the electrodes be the difference between their electrode potential. $\endgroup$ Aug 12, 2013 at 1:42

1 Answer 1


There are a bunch of of sub parts to these questions; let me see if I can tease some of them out.

Why does a potential difference occur at an interface?

The easiest way to answer this question is through thermodynamics. If we consider the situation of a reactive metal rod (e.g. magnesium) being exposed to water, the reaction $$\ce{Mg -> Mg^{2+} + 2 e-}$$ is favorable. In other words, the $\Delta G$ for this reaction will be less than zero. I've done nothing more than summarize the website you reference in your question, which does a nice job at describing what happens when a piece of metal comes in contact with water, so I don't feel the need to repeat that work. We then get to your next question:

Why is a second phase required to generate a potential difference?

If we have a universe that consists just of magnesium, then there would be no place for the above reaction to occur. In this trivialized case, there is no driving force to make magnesium give up its electrons. That doesn't mean it would not happen, only that any dynamic equilibrium that is established would highly favor the pure metal. The meat of question one, then, is (may be?):

Why doesn't a potential difference between a metal and gas occur?

Actually it does. There's an interesting paper (which is behind a paywall if you are not at a university that subscribes to this journal) that describes the measurement of potentials across metal/gas interfaces. The challenge in studying metal/gas potentials is the ability of controlling this potential externally. With the metal/liquid scenario, an electrochemist (or budding electrochemist) can add another electrode to the solution and apply a potential between the electrodes. It's much harder to do the same thing with a gas. (Well, technically it's *not harder to connect two electrodes to one another and hold them in air, but the redox work functions, resistance and capacitance issues that arise in gas-phase electrochemistry make this a really challenging feat.)

The paper I reference above uses chemisorbed species that have active IR or Raman vibration modes to measure the surface potential using spectroscopy and something called Stark tuning (not the best reference, but it's also not behind a paywall) which states that the vibrational frequency of an harmonic oscillator is proportional to an applied electric field. Not many metal/gas interfaces have been explored using this method, but those that have display surface potentials fairly close to 0 vs. SCE or 5 eV vs. vacuum.

What about the rest of the stuff in solution?

Back to the metal/liquid scenario, the last part of your first question asks about other ions in solution. They most definitely play a role in defining the potential of a metal in solution. In order to accurately predict electrode potentials, one must know concentrations and possible reactions (including states of matter) for all species present.

What about question 2? I'm afraid question 2 might be based on the false premise that we can measure the potential of an electrode in the absence of anything else (liquid, gas, ions or molecules). When electrochemists refer to a 'half cell', they have in mind both the reduced (in this case, metal electrode) and oxidized (metal ions) forms of a redox couple contained within the same system. (Think about a beaker with a copper wire placed in it and the beaker is filled with a solution containing 0.1 M $\ce{CuCl2}$. That half-cell or system has a given potential based upon (primarily) the standard reduction potential of $\ce{Cu^{2+}}$ to $\ce{Cu}$ and the concentration of $\ce{CuCl2}$. If I have misread your second question, please edit and I will update my answer as needed.

  • $\begingroup$ That explained quite a bit. I have edited my second question. I require a couple of clarifications. Is the requirement of the second phase just another way to tell that the ions would constitute a very small concentration in case of only one metal? Next, is it the standard equilibrium we are talking about,( standard as in describable by $\Delta G=-RT\ln(K)$, or a mixture of adsorption effects into the equilibrium? $\endgroup$ Aug 12, 2013 at 2:32
  • $\begingroup$ I also read about the double layer but couldn't fit it in the equilibrium concept. Can you help? $\endgroup$ Aug 12, 2013 at 2:46
  • 2
    $\begingroup$ @SatwikPasani The edit to your 2nd question still refers to electrostatic potentials of metals and solutions individually, which is similar to asking for free energy for the reaction $\ce{H2O ->}$. Without knowing the products and reactants one cannot answer the question. The electrode double layer arises from aligning opposite charges at the interface. Lastly, interfacial potentials can be described using the relationship between $\Delta G$ and $K$ as you mention, although adsorption does play a role in advanced treatments. $\endgroup$ Aug 12, 2013 at 11:25

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