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Anjaline
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Electrochemistry

Post by 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
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ChenBeier
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Re: Electrochemistry

Post by 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
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Re: Electrochemistry

Post by 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?
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ChenBeier
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Re: Electrochemistry

Post by 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.
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