Virtually every source I've read pertaining to zinc-air batteries explain the chemistry of it either using these formulae or a simplified version of them:

$$ \begin{align} &\text{Anode:} &\quad \ce{Zn + 4 K(OH) &-> [Zn(OH)4]^{2-} + 4 K+ + 2 e-} \\ &\text{Electrolyte:} &\quad \ce{[Zn(OH)4]^{2-} + 2 K+ &-> ZnO + 2 K(OH) + H2O} \\ &\text{Cathode:} &\quad \ce{1/2 O2 + 2 K+ + H2O + 2 e- &-> 2 K(OH)} \end{align} $$

For all intents and purposes, I believe it is acceptable to simplify the formulae (correct me if I'm wrong).

How does the anode reaction occur in the first place without some sort of catalyst? It momentarily pushes the system out of equilibrium, at least until the cathode reaction can occur, so I would think that the activation barrier would be some significant obstacle. What am I missing?

Apologies if this is a secretly a simple question, I'm not very adept in chemistry.

  • $\begingroup$ The first equation as written is a spontaneous reaction. Zinc readily dissolves in KOH soln. $\endgroup$ – M. Farooq Jun 16 at 4:30
  • $\begingroup$ A couple of questions : 1) Why don't you erase the potassium atoms and ions in all these equations. They are always spectator. 2) What the use of the so-called Electrolyte reaction. 3) Why do you think the first equation needs a catalyst ? 4) Where do you see an activation barrier ? 5) Which equilibrium are you speaking of ? $\endgroup$ – Maurice Jun 16 at 9:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.