3
$\begingroup$

Example electrolysis setting:
Cathode: copper
Anode: zinc
Electrolyte: dilute aqueous sodium chloride [$\ce{NaCl(aq)}$]


During electrolysis, zinc is more reactive than copper, hence electron flows from zinc anode to copper electrode.

At the zinc anode: $\ce{Zn(s)->Zn^2+ (aq) + 2e-}$
Q: What is this process known as?
("Dissociation" is usually applied for salts that dissolve to form mobile ions, and "discharge" is usually used for ions in the electrolyte itself that gain/lose electrons. Hence, what would a proper term be?)

When the zinc anode "dissociates" in the electrolyte, $\ce{Zn^2+}$ ions are formed.
Q: Do the $\ce{Zn^2+}$ ions remain "isolated"?
(Because $\ce{H+}$ions are selectively discharged at the copper cathode, this leaves the $\ce{Zn^2+}$ions $\ce{Na^+}$ions are bonded to $\ce{Cl-}$ions, so does the $\ce{Zn^2+}$ions form just move into the electrolyte just like that, with an unstable electronic configuration?)

$\endgroup$
0

1 Answer 1

2
$\begingroup$

Q: What is this process known as?

This is a particular case of dissolution. Call it electrolytic dissolution, if you want to be more specific. The word "dissociation" certainly does not apply here. It is used when talking about salts and other ionic compounds.

Q: Do the $\ce{Zn^2+}$ ions remain "isolated"?

Well, $\ce{Zn^2+}$ just goes into the solution and gets solvated (hydrated), that is, surrounded by interacting water molecules, pretty much like any other ion. Some people in some contexts would probably call these ions "free", though they are surely not as free as in vacuum. The same applies to the $\ce{Na+}$ and $\ce{Cl-}$ ions out there, and no, they are not linked to each other in any way except for electrostatic interaction. Have you heard of dissociation? That's what it is about. A $\ce{Cl-}$ ion does not "know" that it "belongs" to $\ce{Na+}$ and not $\ce{Zn^2+}$. They all just coexist there together, probably with some other ions like $\ce{OH-}$. As the process goes on, I expect some $\ce{Zn(OH)2}$ precipitate to appear.

$\endgroup$
1
  • $\begingroup$ The process in the particular case of dissolution is also known as oxidation. $\endgroup$ Nov 9, 2015 at 5:18

Your Answer

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

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