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I'm having some problems understanding Gibbs energy of formation, and how it's applied to electrolysis. So, I'm hoping someone can explain what I'm doing wrong..

According to his site: http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/electrol.html (image)

The change in the Gibbs energy of formation is the same as the electrical input energy required to disassociate one mole of water.. Which I've been told isn't correct. But it seemed to work with the following calculations/assumptions

Here's what I did...

According to Faraday law:

107.205 Amps in a cell over one hour should generate 73.338 Liters of (2 moles H2 and 1 mole O2) at 100 percent efficiency (at 25C 101.325 kPa)

Then I tried using the "Gibbs Energy of Formation" to double check the Faraday efficiency. With information from: http://hyperphysics.phy-astr.gsu.edu.../electrol.html [Broken]

At 25C and 101.325 kPa the change in Gibbs Energy of formation is 237.18 kilojoules / mol ...

Which I assumed meant 237.18 (kJ / mol) of electrical input energy is required to convert 1 mole of H2O into 1 mole of H2 gas and a 1/2 mole of O2 gas (at 25C and 101.325 kPa) at 100 percent efficiency.

So with the above Faraday calculations: 107.205 Amps continuous for 1 Hour will create 2 moles of H2 and 1 mole of O2 gas (3 moles total), which has a volume of 73.338 Liters.

Then I assumed if I multiply the Gibbs Free Energy of formation (energy used to create 1.5 moles of gas) by 2, I should have the actual energy required for 3 moles of gas (at 100 percent efficiency, in the above conditions).

237.18 kJ * 2 = 474.36 kJ

Convert 474.36 kJ to Watts:

474360 Joules / 3600 seconds = 131.7666 Watts

Then I put Faraday and "Gibbs" efficiency together..

131.7666 Watts = 107.204 Amp * Volts

So, V = (131.7666 W) / (107.204 A)

V = 1.23 Volts

Which seems to imply deltaG is related to the electrical input energy required to disassociate 1 mole of water (at 25C 101.325 kPa)r. Is this right? If not, can you explain what I'm doing?

Thanks in advance.

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# Gibbs energy of formation and electrolysis efficiency

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