Chemistry Forum

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

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-

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?

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.