5
$\begingroup$

The reduction potential is the electrical potential of the species to be reduced. It's basically how positive it is (compared to a reference, since voltages are relative).

Tables list the following

$$\begin{align}\ce{Cu^{2+} + 2e- &-> Cu}\quad &&E^\circ=+0.340\ \mathrm V\\ \ce{Cu^{2+} + e- &-> Cu+}\quad &&E^\circ=+0.15\ \mathrm V\\ \ce{Cu+ + e- &-> Cu}\quad &&E^\circ=+0.522\ \mathrm V\end{align}$$

Why is an electron more attracted to the less positive charge? And why is the two electron reduction potential not the sum of the individual steps? In fact, if both steps are exothermic, why is the both steps done sequentially, less exothermic than just the second step?

I am aware that $\ce{Cu^2+}$ has a higher enthalpy of hydration, and so $\ce{Cu+}$ will disproportionate in (aqueous) solution, but as I understand, that is not what is being measured here. These potentials should apply in the gas phase or in hexane.

____ Edit____

I realised that the two electron reduction is the average energy for both reductions, so it does indeed sum approximately to the other two. But my underlying question of why the second step is more exothermic remains. I'll leave the second part here in case anyone else with this question is following the same line of logic.

$\endgroup$

2 Answers 2

3
$\begingroup$

We should qualify this properly. Water-solvated copper(II) has a lower reduction potential than water-solvated copper(I). As will be mentioned below, copper -- alone among fourth-period transition elements -- has the property that this difference is small and readily reversible with suitable other ligands.

More is involved here than just ionization energy. It takes more energy to ionize one copper atom twice than two copper atoms once apiece, but the greater electrostatic attraction of copper(II) to hard-base ligands such as fluoride ions or the oxygen ends of water molecules overcomes this difference and makes the +2 oxidation state more favored in such environments.

However, copper(I) has greater polarizability than copper(II), which favors it when more covalent bonding with softer, more polariz-able bases is introduced. This answer discusses the effect of chloride as a softer, more polarizable and more covalently-bonding ligand on copper redox thermodynamics. By contrast, with all preceding transition metals in the fourth period (and most of those in later periods), the electrostatically-driven preference for a +2 or higher oxidation state is too large to be as easily reversed, and the +1 state is realized only is specialized nonaqueous environments.

$\endgroup$
1
$\begingroup$

TL;DR: Standard electrode potentials are indicative of the thermodynamics involved in the half-cell redox reactions. Electrode potentials cannot be directly added since these are intensive quantities and only extensive quantities can be added.

The Problem

The standard reduction potentials do not (directly) reflect the energy changes or heats involved; therefore, we cannot deduce which step is more or less exothermic from the reduction potentials.

The Solution

The following shows calculations of Gibbs free energy for each step:

Gibbs free energy for one-electron and two-electron reductions of copprous and cupric ions

It is clear that reduction of $\ce{Cu(II)}$ to $\ce{Cu}$ is more exothermic than that of $\ce{Cu(I)}$ to $\ce{Cu}$. The reduction of $\ce{Cu(I)}$ to $\ce{Cu}$ is more exothermic than $\ce{Cu(II)}$ to $\ce{Cu(I)}$ because we are going from a charged (unstable) moiety to a neutral (stable) moiety.

This diagram should be useful:

Gibbs energy changes in serial reduction of cupric ion

Caution

Note: Gibbs Potential is not an accurate predictor of the heat involved in a reaction. However, since pressure and volume change only negligibly during the reduction, heat of the reaction may be approximated by change in Gibbs potential.

$\endgroup$
5
  • $\begingroup$ Thanks for the reply. I do understand that the energy for the two electron reaction is indeed the sum of the individual steps because the energy for that step is the $\Delta V$ multiplied by 2 electrons. I'm sorry if that wasn't clear from the edit at the bottom. My question is why is Cu$^+$ at a more positive potential than Cu$^{2+}$? These voltages are relative to the same SHE reference potential. The diagram suggests that the $\Delta V$ is between Cu$^+$ and Cu$^{2+}$. $\endgroup$ May 7, 2023 at 2:11
  • $\begingroup$ @FurrierTransform as stated in the answer, since in going from $\ce{Cu+}$ to $\ce{Cu}$, we are making a neutral moiety from a charged one, and this is a favorable reaction. $\endgroup$
    – ananta
    May 7, 2023 at 2:54
  • $\begingroup$ @FurrierTransform I may be misunderstanding your question. My point is that standard reduction potentials should not be compared with each other; a more accurate understanding of the reactions is obtained from comparing the corresponding Gibbs potentials. However, I have saved this question and will be reading up more about the absolute values of the reduction potential and what can be inferred from these. $\endgroup$
    – ananta
    May 7, 2023 at 3:25
  • $\begingroup$ @ananta: In solution section does not make sense because there might be at least one typo in the section. Would you please go throuth paragraph and correct it? $\endgroup$ May 7, 2023 at 21:21
  • 1
    $\begingroup$ @MathewMahindaratne thank you for pointing it out, I have edited the answer for typos. Hope this addresses your comment. $\endgroup$
    – ananta
    May 8, 2023 at 2:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.