3 replies to this topic

[heisenberg]

Dear all,

I have to performe a mass and energy balance on the electrolysis of water. I know this might be obvious for most of you but I am confused when dealing with the energy part:

First that all, I am not really sure about the equations I need to use, I found different half equations depending on the type of electrolyte is used. I would run my calculation on a conventional (KOH) electrolyte.

I understand that in order to produce 1 kg of Hydrogen (H₂) I need between (50-65 kWh of energy) depending on the electrolyser efficiency.

I also found that the oxidation taking place on the anode has the follow equation:

2H₂O_(l)------> O_2(g) + 4H⁺_(aq) + 4e^- E^o_ox = 1.23 V

What is the relationship between the 1.23 V and the 50-65 kWh of energy?

For the mass part can I say that if I use 18 kg of H₂O I get 4 kg of H₂?

Could you recommend any text book or website that can help me undestand this topic.

Thank you in advance for your suggestions and help.

Kind Regards,

Kevin

Edited by heisenberg, 11 September 2012 - 10:47 AM.

[francisco.angel]

Hello, I m not specialist too, but I have to say a couple of words, the [V] referred in the semireaction, once corrected (to reflect no standard conditions), reflect the energy by charge unit that is required/release for the reaction procedd, so if you know the total charge (mol electron) required to produce a certain amount of product, can multiply by potential to obtain energy.
On the second question if you refer to a simple stochiometric relation in terms of mass for the semireaction my first glance will be, if you use 18[kg] of water obtain 2 [kg] of hydrogen.

[heisenberg]

Dear Francisco Angel,

Thank you for your response.

Is it possible for you to expand a little on your answer, perhaps referring to the stoichemetric equation.

Regaring the second question I made a typing mastake and therefore you are right by stating that using 18 kg of water will give me 2 kg of hydrogen.

Thank you again for your help.

[kkala]

Having seen this topic just now, following could be useful.

1. Schaum's ''Theory and Problems of College Chemistry" by D Schaum and J Rosenberg (present publisher: McGraw-Hill) contains useful Chapter 19, titled "Electrochemistry".

- Mantel's "Electrochemical Engineering" ('Industrial Electrochemistry' in older editions), McGraw-Hill, is no longer in circulation though you could get a used book. It was classical at 1960s per my information, but I have never had the chance to read it.

2. Reported equation of H2O electrolysis can be seen in Schaum's book, table 19-1. However look at < http://en.wikipedia....olysis_of_water>, paras: equations - thermodynamics of the process - efficiency, for a probably more complete explanation. Water electrolysis in theory requires a Voltage of 1.23 V (*) at standard conditions (including pH=0). Besides F=96500 Cb are needed for 1 greq of H2, thus 2*96500 Cb per gmol H2 (2H⁺+2e-->H2). So theoretical energy required 2*96500*1.23 CbV / 2 g H2 = 118695 VAs/g H2 = 118695 kJ/kg H2 = 33 kWh/kg H2, versus 50-65 kWh/kg H2 in practice. 

(*) other views support higher than 1.23 V as theoretical minimum, as explained in wikipedia.

Edited by kkala, 08 February 2013 - 10:33 AM.