[gre]

Hello,

Is the Change in Gibbs free energy of formation in electrolysis the same as  "the minimum electrical input energy" required to disassociate 1 mole of H2O?

Thanks.

[gre]

At 100 percent efficiency, I guess I should add.


Thanks in advance.

[Acid_Guru]

No it is not, voltage is what you are thinking of, delta G is simply to determine whether the reaction is spontaneous or not.

[gre]

I was going by this reference at hyperphysics:  http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/electrol.html

Where it seems to imply that delta G  is the "Electrical Input Energy" in the diagram and is 237.1 kJ, in the diagram.

What is that about, exactly?

Thanks.

[Acid_Guru]

Delta G is the amount of free energy lost or gained during a reaction, if free energy is lost and delta g is negative that means a reaction happens. E total on the other hand is the amount of voltage needed to make an electrolysis reaction happen(if E tot is negative) or the amount of voltage made when the reaction happens(if the E tot is positive).

[gre]

Delta G is the amount of free energy lost or gained during a reaction, if free energy is lost and delta g is negative that means a reaction happens. E total on the other hand is the amount of voltage needed to make an electrolysis reaction happen(if E tot is negative) or the amount of voltage made when the reaction happens(if the E tot is positive).

Thanks for the response.  Maybe you can explain what I'm doing here..   

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


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.

[Acid_Guru]

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.

Wrong assumption, change in Delta G is inversely related to equilibrium(Q, and K). E tot is what you are talking about, and 1 amp = 1 coulomb per second = 1 joule

[gre]

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.

Wrong assumption, change in Delta G is inversely related to equilibrium(Q, and K). E tot is what you are talking about, and 1 amp = 1 coulomb per second = 1 joule

http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/electrol.html

Is misleading,  since it clearly states Delta G is "electrical input energy".   

Is there another name for "E tot" or is it just "Total energy"? Is this always the absolute value of Gibbs energy of formation?

[Borek]

Is the Change in Gibbs free energy of formation in electrolysis the same as  "the minimum electrical input energy" required to disassociate 1 mole of H2O?

I can be utterly wrong, but it doesn't sound wrong to me.

Gibbs free energy measures energy that can be obtained from the system. Shouldn't matter in what form the energy is obtained (assuming it is isobaric and isothermic). Could be electric energy. And "input" vs "output" are just a matter of sign.

Note, that  :delta: G = -nF  :delta: E

[Acid_Guru]

Yes delta G is related to E total, but it is not "the minimum electrical input energy required for a reaction", take for example the formation of calcium metal and fluorine gas from CaF₂ you need 5.74v to make the reaction happen because when delta G is at zero the E tot is at -5.74 volts.

[Borek]

when delta G is at zero the E tot is at -5.74 volts.

Assuming

:delta: G = -nF  :delta: E

and using numbers from your example

0 = -2 * 96500 * -5.74

Something is wrong.

[Acid_Guru]

I was always taught that at Delta G of zero, the delta E is equal to the change of voltage. Guess I was taught wrong.