Can someone explain (on an advanced level) what phys-chem property of compounds/elements determines what their half-cell redox potential will be? For example, Zn and Cu are structurally very similar and their ionization energies, electronegativities, etc. won't be all that different. So, why is the 2-electron redox potential for each so different? ie Zn is -0.76 V (SHE) and Cu is +0.34 V (SHE), which is a significant 1.1 V different.

Isn't electric potential the amount of "stored" energy? So, wouldn't removing or adding 2 electrons to Zn or Cu be very similar since they are next to each other on the periodic table?

Edit: I've added some sources and clarity to the discussion in the answer below.

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    $\begingroup$ I wonder, what you mean by "advanced level"? Potentials are connected with Gibbs' enthalpy in simple way, but don't arise from electron configuration in any simple way, neither do other common properties. $\endgroup$ – Mithoron Jan 11 '17 at 20:14
  • $\begingroup$ I meant at a graduate level, preferably from a theoretical standpoint. For example, if I were going to computationally derive the redox potential, what properties would go into that model? Can you elaborate or provide a link that would help? $\endgroup$ – prof.kvothe Jan 12 '17 at 15:11
  • $\begingroup$ Also, I found NIST enthalpy for ionization energies and Cu and Zn 2 electron ionizations were very similar, but their redox is not? $\endgroup$ – prof.kvothe Jan 12 '17 at 15:18
  • $\begingroup$ From a theoretical standpoint you will have to account for solvation energies. $\endgroup$ – porphyrin Jan 12 '17 at 21:53

As I understand it, the potential is related to the difference in energy between the oxidized and reduced state (via the Gibb's energy as Mithoron noted). While their individual electron orbital energies are probably similar given the similar nuclear charges, the oxidized and reduced states are differently occupied and so have different energies.

In zinc for example, you have the reduced state of 4s2, 3d10 and oxidized state of 3d10. Copper on the other hand goes from 4s1, 3d10 to 3d9. $\ce{Zn^2+}$ has a fully occupied d-shell while $\ce{Cu^2+}$ does not, meaning that the reduced state for zinc is lower energy and more stable. As a result, zinc has a stronger tendency to oxidize and consequently a lower reduction potential.

  • $\begingroup$ Okay, that makes sense about not having a complete shell, but 1.1 V of a difference? I still can't wrap my head around going from a full d shell to 1 e- off equating to that huge of potential difference. Why isn't iron or cobalt or silver similar in difference? Why doesn't redox potential follow a periodic trend like other phenomena (assuming they are all eg 1 electron reaction). The explanation is probably complex, but perhaps someone can point me toward a good source? I couldn't find anything in my chemistry and EC texts. $\endgroup$ – prof.kvothe Jan 12 '17 at 15:17

I was able to get some clarity on my own question, which may lead to further discussion. From "Inorganic Chemistry in Aqueous Solution by Jack Barrett," Section 7.4.2. The portion below is paraphrased for brevity.

The two-electron reduction potential related to the enthalpy of oxidation of half a mole of dihydrogen to give one mole of hydrated protons and ionization energies is as follows: $\ce{E(M^2+/M)=-[-\Delta _{hyd}H(M^2+) + 840] + [I_2 + I_1 + \Delta _aH(M)]/2F}$

The term [$\ce{-\Delta_{hyd}H(M^2+) + 840}$] tends to make the value negative, while the latter term has the opposite effect (paraphrased). For Ca, V, Cr, Mn, Fe, Co, Ni, and Zn, these two terms combine to give a negative SRP, while Cu has a positive value.

The unique behavior of Cu is attributed to its inability to liberate $\ce{H_2}$ from dilute acids (i.e., those with molar concentrations of 1 M). The reason for this behaviour is the high energy needed to transform Cu(s) to $\ce{Cu^2+ (g)}$, which is not exceeded by the hydration enthalpy of the ion. (as @porphyrin suggested with regards to solvation energies)


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