Origin of electrode potential and its connection with electric double layer

Question

Recently, I was learning about electrode processes and I came across the topic of "electrode potentials". I understood that at equilibrium, a potential difference was generated at the interface between the electrode and the electrolyte because of the formation of an electric double layer.

I was reading this article which says that on dipping metals into their respective salt solutions, positive ions readily form and move into the solution leaving behind electrons in the electrode and a dynamic equilibrium is achieved soon.

Across different metals, only the position of the equilibrium (magnitude of negative charges on the metal electrode) differs due to their differing tendencies to oxidize (or reduce).

(I understood this part)

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On the other hand, I’ve also seen in some other texts that at equilibrium, the charge acquired by the metal might be either positive or negative depending on its tendency to oxidize (or reduce) (unlike the $1$st source where they say the charge will be always negative).

Examples of

$\ce{Zn}$ ($\ce{Zn}$ has greater tendency to form $\ce{Zn^2+}$ ions hence metal acquires negative charge) $$\ce{Zn<=>>Zn^2+ + 2e^-}$$

and

$\ce{Cu}$ ($\ce{Cu^2+}$ ions have greater tendency to get reduced hence $\ce{Cu}$ metal acquires positive charge) $$\ce{Cu^2+ + 2e^-<=>>Cu}$$ are usually given.

(I also understood this explanation)

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But what is confusing me is which of the above explanations is correct (or more accurate)?

Because if we go by the $2$nd view, we have that $\ce{Cu}$ ions are more likely to get reduced so let’s say the $\ce{Cu}$ electrode becomes positively charged when ions from the solution stick to the metal, but at the same time few other $\ce{Cu}$ atoms in the metal will also get oxidized (because there are limited electrons in the $\ce{Cu}$ electrode) and dissolve so there’s no net oxidation or reduction. If this indeed happens then at equilibrium the $\ce{Cu}$ electrode will have no net charge (which isn't possible).

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Note-I've seen a few other answers here on this site like this, this and this but none of these seem to address the confusion I'm having.

Any suggestion will be of great importance.

Answer 1

Let's see if we can tease out the main points here. The images below are taken from Lance and Cole's Analytical Chemistry POGIL workbook.

Microscopic processes at a metal surface exposed to an aqueous solution containing the corresponding metal ion(1) will result in a slight negative charge on the metal surface due to oxidation of the metal and a slight positive charge in the region of the solution close to the metal surface owing to the sudden generation (and dissolution) of positively charged metal ions.

The amount of negative charge generated is metal specific and as of now, unknowable in an absolute sense. Therefore, we are required to compare the extent of this process at one metal electrode to that of another. If we compare the Zn electrode above to a Pb electrode, we would find that the Pb electrode has less negative charge on its surface relative to Zn. Both of the electrodes, however, have a slightly negative charge at their surfaces. Because the charge is greater at the Zn electrode, we say Zn is easier to oxidize than Pb.

To answer your question which of the above explanations is correct, the answer is both (although the clarity of the explanations is up for debate). A negative charge is generated at both electrodes, but the the amount relative to one another differs, and that gives rise to the notions of easier/harder to oxidize/reduce.

(1) In my experience, microscopic theory of electrochemical processes typically assume the presence of the metal and metal ion in part to avoid issues which identifying what in the system is being oxidized and reduced.

Answer 2

Theoretically, we could do any reaction. Practically, some are only possible in one direction; others are under special conditions.

For the case of copper: the standard potential of the pair $ \ce {Cu^{2 +} / Cu} $ being positive, this means that naturally the reaction is done in the direction $ \ce {Cu^{2+} -> Cu} $. But we can still force the reaction in the other direction by imposing a potential.

For the case of zinc ($ \ce {Zn^{2+} / Zn} $), the natural reaction is that of oxidation so the reaction is $ \ce {Zn} -> \ce{Zn^{2+}} $.

So if we put copper ions $ \ce {Cu ^{2 +}} $ and zinc together, a reaction takes place between these species; if, on the other hand, we place $\ce{Cu}$ and $ \ce {Zn^{2 +}} $ ions, nothing happens naturally.

To know if a reaction is possible or not, we must look at the standard potentials of the couples (for similar concentrations otherwise we must calculate the potential using the Nernst relation): the reaction is naturally possible then between the oxidant of the highest potential torque and the lower potential torque reducer.