Electrochemistry

[Anjaline]

Hi
Can anyone explain the below-mentioned reaction? Here Co₃O₄ is used as electrode material and KOH is used as electrolyte. Redox peaks were observed in CV graph. But couldn't understand the actual reaction mechanism taking place in the electrochemical cell. I need help......
Co₃O₄+OH+H₂O = 3CoOOH+e
CoOOH+OH = CoO₂+H₂O+e

[ChenBeier]

Co₃O₄ is Co₂O₃ * CoO Cobalt- II/III-Oxide
In first step CoO will be oxidised to CoOOH
CoO + OH- => CoOOH + e-
The Co₂O₃ reacts with water to the same product.
Co₂O₃ + H₂O => 2 CoOOH
In Addition we have
Co₃O₄+ OH- + H₂O = 3 CoOOH+e-

The next Oxidation goes to Co-IV

CoOOH + OH- => CoO₂ + H₂O+ e-

[Anjaline]

Thank you so much for your reply.
I have few more questions.
1. what happened to K+ ion. (because KOH is an electrolyte)
2. Is all mentioned reactions reversible?

[ChenBeier]

For this have to know the built up of the cell.Because all reaction were oxidations. Where is the reduction. Probably 2H+ + 2 e- => H₂
K+ will be not touched.

If it is a rechable cell then the reactions must be reversible.