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Topic: Water electrolysis  (Read 1041 times)

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Offline captaincook

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Water electrolysis
« on: March 02, 2013, 09:02:02 PM »
Hi

I am very bad at electrochemistry please help me with this one:

What are the electrode reactions for water electrolysis in 5M KOH?

How can I determine the cell potential and the energy required to produce 1 kg hydrogen?

This is what i think but I don't understand what to do with KOH concentration.

Anode (oxidation): 2 H2O(l) → O2(g) + 4 H+(aq) + 4e−
Cathode (reduction): 2 H2O(l) + 2e− → H2(g) + 2 OH-(aq)

Offline Borek

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Re: Water electrolysis
« Reply #1 on: March 03, 2013, 04:15:02 AM »
Anode (oxidation): 2 H2O(l) → O2(g) + 4 H+(aq) + 4e−
Cathode (reduction): 2 H2O(l) + 2e− → H2(g) + 2 OH-(aq)

These are OK. Alternatively the first reaction could start with OH-, or with H2O + OH-. Not that it would change much, it would just reflect the fact that H+ - even if they are produced - can't survive long is highly concentrated alkali solution.

but I don't understand what to do with KOH concentration.

Neither do I. Perhaps it was intended to be plugged into the Nernst equation, but at such a high concentration ionic strength of the solution can't be ignored and in practice it is not possible to calculate the reliable value of the potential, the only viable approach is to determine it experimentally (which was probably already done somewhere by someone).
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