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The standard reduction potential of diatomic oxygen in acidic conditions is +1.23 volts. However, the standard reduction potential of diatomic oxygen in basic conditions is only +0.40 volts.

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Why is the reduction of diatomic oxygen by acidic media more favored than the reduction of oxygen by water that is basic?

I can think of a few reasons:

1) We are likely going to be forming negative two oxidation state oxygen anions when we oxidize diatomic oxygen. Oxygen adopts an oxidation state of negative two except in peroxides, which aren't very stable, and I don't think the oxygen radical will be very stable in solution.

Can someone provide a less fluffy reason that we're going to be forming negative two oxidation state oxygen atoms? I'd imagine that the electronegativity and electron affinity of oxygen are two reasons it might favor the negative two oxidation state over the negative one oxidation state in solution. I.e. why doesn't this occur to a significant extent?

$\ce{O2 + 2e^- -> 2O^-}$

Or this?

$\ce{O2 + 2e^- -> HOOH}$

2) From the above we can deduce that water is the product; water is pretty stable in either acidic or basic medium - i.e. it won't react to a large extent with either hydronium or hydroxide ion. Plus the oxygen in water has a -2 oxidation state. So if we look at our skeleton half reaction:

$\ce{O2 + 4e^- -> 2H2O}$

We realize that we need a source of protons. And the proton source in acidic medium is likely hydronium ion.

However, in basic medium, the proton source is water. I'd imagine that water is a poorer proton source than hydronium ion.

So would this account for the difference in standard reduction potential of oxygen in media that vary in pH?

I understand that voltage is an extension of K through the Nernst equation; is what I'm seeing here basically that the protonation of a base ($\ce{O^2-}$) is more favored in acidic rather than basic media? Could there also be kinetic factors? In other words, why is this statement true?

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Also:

1) What's the mechanism for the reduction of diatomic oxygen?

2) Why is the reduction of protons at high pH less favored than the reduction of protons at low pH? My first guess: you don't find as many protons in basic solutions. But then, protons don't exist by themselves anyway. So what can the proton source be? I'm guessing that in basic pH, there doesn't exist much hydronium ion. So the only two sources of protons are water and hydroxide anion. Heterolytically cleaving the hydroxide anion to form the highly basic oxide anion doesn't seem thermodynamically favorable. Cleaving water to form hydroxide anions ... are we running into a Le Chatlier's Principle effect?

ETA: Or should I be considering proton activities? Is the activity of the proton less in basic solution than in acidic solution?

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  • $\begingroup$ 1+2) standard acidic conditions: pH=0, standard basic conditions: pH=14, 1.23-0.40=0.83, 0.059*14=0.826, suspicious. Applying Nernst equation correctly may give an answer 1) It depends on the material of the electrode, but many metals and metal oxides can (reversibly) adsorb/lose oxygen on the surface. For details you'll have to study surface chemistry, which is relatively unexplored. 2)Nernst equation properly applied and learning what's behind it should give the answer $\endgroup$
    – permeakra
    Jun 27, 2014 at 4:44
  • $\begingroup$ Thank you. I obtained the numbers from some presentation I found on Google. The exact numbers might be off but I'm interested in why they might differ in the way they do. What exactly about the Nernst equation should I be studying? $\endgroup$
    – Dissenter
    Jun 27, 2014 at 4:47
  • $\begingroup$ Derivation process. Basically, Nernst equation is an equation for equilibrium constant including electron in external electric chain. $\endgroup$
    – permeakra
    Jun 27, 2014 at 5:23
  • $\begingroup$ May I know which book that information is from? $\endgroup$
    – Newton
    Feb 23, 2017 at 13:52

1 Answer 1

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If you look at the half equations, you will see when you add them together, you get 1.23 V, 1.22 V, and 1.23 V.

As the mixture becomes more basic, it is more difficult for the oxygen to grab $H^+$ ions and easier for $H_2$ to lose electrons to generate $H^+$ ions.

The mechanism probably involves these combinations: $O_2, H_2O_2, H^+, HO_2^-, H_3O_2^-, H_2O, and\ HO^-$.

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  • $\begingroup$ Which half reactions are you referring to? $\endgroup$
    – Dissenter
    Jun 27, 2014 at 5:23
  • $\begingroup$ Also why would H+ be formed in basic solution? That seems counterproductive. $\endgroup$
    – Dissenter
    Jun 27, 2014 at 5:23
  • $\begingroup$ @Dissenter 1. You have 3 pairs of half reactions in the table. 2. $H^+$ would immediately combine with $OH^-$ in basic solutions. $\endgroup$
    – LDC3
    Jun 27, 2014 at 13:37
  • $\begingroup$ How does that affect the voltage? $\endgroup$
    – Dissenter
    Jun 27, 2014 at 17:14

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