change in gibbs free energy of formation

[gre]

Hello,

 

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

 

Thanks.

[hermanntrude]

is this a homework question?

 

what does water dissociate into during electrolysis?

 

what does the Gibbs free energy of formation represent?

[gre]

is this a homework question?

 

This isn't a homework question, I'm just curious.

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I was going by this information:

 

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

 

Which seems to imply the deltaG is "Electrical Energy Input"

 

Is that right?

[hermanntrude]

not really, no. There is a relationship between the cell potential and the gibbs free energy, though:

 

[math]\Delta G = -nFE[/math]

 

where [math\Delta G[math] is the gibbs free energy, n is the number of moles of electrons involved in the process, F is the Faraday constant (the charge on a mole of electrons, so that "nF" is equal to the charge used), and E is the voltage required or provided (depending on whether the process is spontaneous or not). In reality, though, the calculated value of E and the real value of E are usually different, due to a phenomenon known as overpotential, which is more prevalent when gases are involved.

 

In general, if [math]\Delta G[/math] is negative, E will be positive, meaning the process is spontaneous, and that electricity will be generated, and if the reverse is true, the process will require energy input for it to occur. The latter situation is often an electrolysis.

[gre]

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/Hbase/thermo/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

 

Is this right?

[hermanntrude]

I don't think so. Gibb's free energy is related to the energy required but it's not the same. Gibb's free energy is actually the negative of the change in the entropy of the universe for the process multiplied by the temperature in kelvins ([math]\Delta G = -T \Delta S_{universe}[/math])

[gre]

Which of my assumptions and calculations are wrong? I figgured it made sense because the end result 1.23 V.

 

 

Thanks in advance.

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I'm lost. Can you explain what I did, in the previous calculation?

[hermanntrude]

i'm not sure about your calculation, but electrical input energy and Gibb's free energy aren't the same thing, as far as I know. I may be wrong, however