0
$\begingroup$

The electrical double layer, is commonly used to describe the origins of electric potentials that is fundamental to electrochemistry. This model is quite outdated from my understanding, but it is still used inside my textbook.

Right now, my understanding is that when a metal rod is placed within a solution of its respective ions, some of those metals atoms change to metal ions and go into the solution leaving a negative charge at the electrode surface. Hence, the metal cations inside the solution move towards the electrode surface. At this electrolyte - electrode interface there is now a seperation of charge, that gives rise to the phenomenon of electric potential. At least this how my cambridge textbook explains it.

Some answers on stackechange reinforce the theory as stated above. However, my problem lies in understanding why a negative layer of surface charge builds up in the first place. How can metal atoms just convert to metal cations and move into the solution? This doesn't make sense to me. The surface of the metal rod comes into contact with the electrolyte solution, so what is causing the surface to gain this initial negative charge?

The rest seems to make sense, but this part specifically doesn't.

$\endgroup$
3
  • $\begingroup$ Consider also inert electrodes, where only exchange of electrons happen, like $\ce{Pt|V^2+/V^3+}$ // Double layers are rather the consequence than cause. $\endgroup$
    – Poutnik
    Commented Mar 31 at 11:53
  • $\begingroup$ Very similar to my question a while ago: chemistry.stackexchange.com/questions/181451/… $\endgroup$
    – Buck Thorn
    Commented Mar 31 at 12:08
  • $\begingroup$ "processes which can lead to a surface being charged, including adsorption of ions, protonation or deprotonation" $\endgroup$
    – Stark
    Commented Mar 31 at 21:14

2 Answers 2

3
$\begingroup$

Consider an electrode immersed into an electrolyte, two things can occur.

  1. Metal ion $M^{n+}$ from the solution may collide with the electrode, gaining $n$ electrons from it, and convert to metal atoms. This means that the ions are reduced.
  2. Metal atom on the surface may lose $n$ electrons to the electrode and enter the solution as the ion $M^{n+}$. This means that the metal atoms are oxidized

Notice that for different elements, the redox equilibrium established is different, either further to the right or further to the left. Hence, depending on the electrode, one of the processes above is going to be more favourable then the other, creating either a positive or negative surface charge.

$\endgroup$
7
  • $\begingroup$ The question is about a metal bar (a single electrode or cell in an open circuit), not a galvanic cell (closed circuit). $\endgroup$
    – Buck Thorn
    Commented Apr 1 at 9:27
  • $\begingroup$ wqeq's answer is related to a metal bar, not to a galvanic cell. $\endgroup$
    – Maurice
    Commented Apr 1 at 11:43
  • $\begingroup$ @BuckThorn Does the answer given by wqeq apply for galvanic cells? $\endgroup$
    – Stark
    Commented Apr 1 at 19:49
  • $\begingroup$ @Maurice the answer above relates to a galvanic cell, but is the content of the answer all correct? Is the validity of the answer 100% but in the context of galvanic cells? $\endgroup$
    – Stark
    Commented Apr 1 at 19:51
  • $\begingroup$ How does a galvanic cell differ from a single metal bar dipped into an electrolyte, the former has 2 metals bars dipped. I understand the former is a closed circuit, however initially it should behave as the latter right? $\endgroup$
    – Stark
    Commented Apr 1 at 20:06
0
$\begingroup$

Negative charge is kept near surface by attraction of excess positive ions in solution. The reason for ions going into solution and electrons staying in metal is electron's De broglie wavelength strongly depending on environment. Creating a bubble larger than intermolecular distance in a liquid made of polar molecules needs energy. That does not happen in the situation stated in the question.

For a nice example of a bubble filled with electron, see https://en.wikipedia.org/wiki/Electron_bubble

There are counterexamples, where both cations and electrons go into solution, such as alkali metals in liquid ammonia. They can do this because in these cases surface electron affinities of the alkali metals are lower.

$\endgroup$

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.