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In biological systems, energy is often extracted from molecules with reduced carbon backbones — the oxidation of those molecules has $\Delta G < 0$ and that energy can be captured, often in the reduction of other molecules. I'm wondering if it is always the case that oxidation is thermodynamically favorable and reduction thermodynamically unfavorable.

I suppose I'm not asking if the oxidation of any molecule is favorable (I'm pretty sure that's not the case), but for redox reactions that do occur, is it ever the case the oxidation half reaction is thermodynamically unfavorable and the reduction half reaction thermodynamically favorable?

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    $\begingroup$ You can't measure the thermodynamics of a half-reaction. $\endgroup$ – Ivan Neretin Dec 8 '16 at 15:06
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    $\begingroup$ You need two half reactions and the value of $\Delta G$ will depend where they are situated relative to one another in redox potential. Wikipedia has a large table of redox potentials, you can choose some reactions for yourself. $\endgroup$ – porphyrin Dec 8 '16 at 15:11
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    $\begingroup$ @IvanNeretin - Is it therefore meaningless to talk about the ΔG for the oxidation of a particular molecule? The free energy of succinate in cells is higher than the free energy of fumarate, which is a result of the oxidation of succinate. Should I not say that the oxidation of succinate has a ΔG<0? $\endgroup$ – kevbonham Dec 8 '16 at 15:25
  • $\begingroup$ Why, that's a full reaction, which is another story. It certainly has some $\Delta G$ and everything. $\endgroup$ – Ivan Neretin Dec 8 '16 at 15:37
  • $\begingroup$ That is, if we write it as a full reaction: succinate plus oxygen $\to$ fumarate plus water. $\endgroup$ – Ivan Neretin Dec 8 '16 at 15:39
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As the term redox implies you have an oxidation and a reduction in every redox reaction. It makes no sense to just talk about a half reaction, but you have to look at the cumulated energy difference of educts and products. Now you have to remember two things:

  1. You can easily convert between the energy and potential difference of a redox reaction by multiplying with the exchanged charges: $$\Delta G= Uzq$$ $U$ is the potential (usually measured in Volt), $z$ the number of exchanged elementary charges and $q$ the elementary charge (usually measured in Coulomb).
  2. Energy differences are transitive. That means: $$\Delta G (A\rightarrow C) = \Delta G (A \rightarrow B) + \Delta G (B \rightarrow C)$$
  3. There is no absolute zero point for the energy in chemical reactions. You can only talk about differences.

What chemists did, was to define a Standard electrode potential. With this it is possible to measure half reactions compared to this standard. Because of sloppy naming, people call this the potential of the half reaction. But strictly speaking it is just the difference to the standard electrode potential. Since differences are transitive that is not a problem when working with potentials.

This means: The oxidation can be unfavoured compared to the standard electrode potential if the reduction is even more favoured.

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