What? Chemistry not fun?? Next thing you know you’ll be telling me physics is hard. I always preferred chemistry to things like psychology or even straight biology, though I admit that Gould could get me interested even in snails and pandas.

I don’t know if this will be helpful or not, but I thought it might be fun to try and correlate two of the questions that have come up recently, namely the idea of redox potential, and predicting or explaining what happens to methylene blue in the yeast experiment that seems to be giving people fits. Hopefully, this won’t lead to even more confusion, but it seems to me to be a decent object lesson in basic chemistry as pertains to biology.

The standard definition of a “redox” potential is usually given as a “reduction” potential. You could do it either way, as reduction or oxidation potentials, but as a reduction potential is historically how it was defined. Compounds or elements that have a high affinity for electrons have a greater tendency to accept electrons (that is, to be reduced) compared to those with lower affinity for electrons. One of the more confusing points sometimes is the nomenclature: an oxidizing agent is itself reduced in a spontaneous redox reaction, while a reducing agent is itself oxidized in the same reaction. The trick is to figure out which in any particular redox reaction is which. To put the potentials on a scale, a reference is chosen as a convenient zero point and the potentials of everything else are measured relative to the standard. Something with a greater affinity for electrons than the reference will have a positive redox potential, while things with less affinity for electrons than the reference will have a negative potential.

Another possible source of confusion is the way the standard half-cells are presented. Voet, for example, lists them in decreasing order from the most positive to the most negative. Wood, Wilson, Benbow & Hood list them in the opposite order. It doesn’t really matter how it is tabulated, but electrons will flow in the direction from the most negative-potential to the most positive-potential reactant for any two half-cells being considered.

So what about methylene blue in the presence of oxidative metabolism—what are the half-cells and potentials involved? You might pick two of several, but perhaps the most instructive are the methylene blue ox/red half-cell and the oxygen/H2O half-cell. They are:

MB(ox) + 2H+ + 2e- –> MBH2(red) potential = +0.1 V
½ O2 + 2H+ + 2e- –> H2O potential = +0.815 V

Which will be the oxidizing agent (which is the same as asking which reduction reaction will go as written)? The oxygen/H2O half-cell has the greater positive reduction potential, so this half-cell will go as written; oxygen will preferrentially be reduced to water. What about MB? The reduction potential of MB ox/red is lower than that of oxygen/water so this MB will preferrentially be oxidized when coupled with the oxygen/water half-cell; the MB reaction must go the other way. So when the half-cells are combined to make the full-cell, they are:

MBH2(red) –> MB(ox) + 2H+ + 2e- potential = -0.1 V (i.e., reversed)
½ O2 + 2H+ + 2e- –> H2O potential = +0.815 V
MBH2(red) + ½ O2 –> MB(ox) + H2O potential = +0.715 V (full-cell reaction)

So what happens to methylene blue in the presence of active oxidative metabolism, then? Well, on first mixing MB is likely to be mostly oxidized (and blue) because any colorless MBH2(red) that might be present should be oxidized by any oxygen in the system. But what happens if oxidative metabolism persists? The full-cell is written as a one-way reaction, but in reality it is an equilibrium (as are most reactions) even though its equilibrium point may lay far to the right. But as respiration consumes oxygen, by LeChatlier’s Principle the full-cell reaction will shift to the left in an attempt to maintain the equilibrium. So long as oxygen continues to be consumed via respiration, eventually the solution will run out of MB(ox) and the solution will be colorless. On the other hand, if respiration is no longer happening (in the case of the yeast experiment, because the important enzymes have been irreversibly denatured by heat treatment) then there is no driving force to reverse the redox reaction, and MB will remain oxidized (and blue).

This little analysis demonstrates how the redox potential is used to predict the “direction” of a redox reaction as well as the application of a concept taught in general chemistry (LeChatlier’s Principle). It also suggests a cautionary note in designing the yeast experiment. It is possible to add so much methylene blue that you exceed the capacity of the system. Under such a condition you will never see the decolorization even though it is happening because, in essence, you’ve given the system an unlimited supply of MB(ox). To exhaust the supply of MB(ox) you will have to let the metabolism continue for a very long time or else somehow step up the rate of respiration.