My background: I've read several books and articles on electrochemistry and even do photoelectrochemical research in a lab, so I'm somewhat familiar with the concepts and techniques.

In texts on electrochemistry, the standard reduction potential and its meaning is usually axiomatically presented early on and everything follows from there. What I would like to see in a treatment of the subject is an explanation of how the chemical equations of the reduction half reactions are actually determined. For example, let's take the equation $\ce{Cu^{2+} + 2e- <-> Cu}$. How did they figure out that it is in fact two electrons reducing the $\ce{Cu^{2+}}$ ion?

One could do open circuit potential measurements to determine the standard reduction potential, but it would seem to me that you would have to know the ionic state of the element a priori to use this technique. One could also apply a voltage until the reaction "turns on" then approximate that onset voltage as the standard reduction potential, but this wouldn't be strictly true as it would be a non-equilibrium measurement and would also be subject to the kinetics of the surface. Does anyone know of a book or other treatment showing how the chemical equations are determined or better yet has experience doing this?


The standard reduction potentials were mostly discovered by carefully measuring the reduction and oxidation of solid metals (or dissolved ions) in electrochemical cells, and monitoring the electrical current through the wire connecting the two electrodes. (The image below should have an ammeter instead of of a voltmeter, but you get the idea)

Take, for example, an electrochemical cell of zinc and copper.

enter image description here

At the anode, the solid zinc will dissolve to become Zn2+. At the cathode, the Cu2+ ions will be reduced, and become solid copper. During the redox reaction, the scientist measures the current across the two electrodes.

After some time, the scientist could pull the two electrodes out of solution, dry them off and mass them. They could then calculate the moles of zinc that dissolved from the anode, and the moles of copper added to the cathode. In addition, by measuring the current over the period of the experiment, they would also be able to calculate the number of electrons that passed from the zinc anode to the copper cathode.

In this instance, they would calculate that for every mole of zinc that dissolved, 2 moles of electrons were produced. Concurrently, 2 electrons were consumed for each mole of copper reduced! These measurements, over time, made it possible to determine the oxidation states of many metals and soluble ions.

(image from moodle2.rockyview.ab.ca/)

  • $\begingroup$ That seems pretty reasonable and probably historically accurate. Would you mind giving a citation supporting why you believe this is true? $\endgroup$
    – Zeruff
    Jul 28 '15 at 1:34
  • $\begingroup$ No problem - the modern table of standard reduction potentials was first developed early on using Michael Faraday's research (published in early 1800's). His works culminated in a set of rules called Faraday's laws of electrolysis. There's a nice treatment in the J. of Chem. Ed. (dx.doi.org/10.1021%2Fed031p226), and a quick search for laws of electrolysis produce a summary in wikipedia. $\endgroup$
    – Dan Burden
    Jul 28 '15 at 12:15
  • $\begingroup$ Re-read this answer after four months and realized that it doesn't fully answer the question. First of all, if you had a voltmeter in series with two half cells as drawn above, there would be negligible electrolysis due to its high resistance. Did you mean to put it in parallel? In any case, even if it were in parallel if you had the reaction running I don't think you'd quite measure the equilibrium voltage (i.e. the standard reduction potential). How would this be corrected for? $\endgroup$
    – Zeruff
    Dec 2 '15 at 3:18
  • $\begingroup$ Yeah, it should be an ammeter to measure current instead of a voltmeter. I didn't create the image, but should have made that significant change more clear in the answer (edited for clarity) $\endgroup$
    – Dan Burden
    Aug 3 '16 at 0:21

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