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Gibbs energy of formation and electrolysis efficiency

  1. May 21, 2009 #1
    Hello -

    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.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 25, 2009 #2
    Anyone?
     
  4. May 26, 2009 #3
    Thats because it is. You can think of the gibbs free energy as the energy required to split the water molecules apart or the energy released when they are combined (enthalpy of formation). To understand this more clearly with electrical work we can apply the first law of thermodynamics and Faraday's Law and take 1 mole of H2O as our system.

    Energy into H2O = Energy used by breaking H2O bonds

    Voltage x Current = Gibbs Free Energy (dG)

    Since we know that the current is proportional to the number of electrons being redoxed in the reaction and that electrical charge is quantized, we can state that I = n x F. F is Faraday's constant (#moles x electron charge) and n is the number of electrons being transferred in the reaction (1 electron from each H). So,

    V x n x F = dG

    Solving for Voltage,

    V = dG/nF = 1.23V (at STP)

    From this relationship you can see that the electrical input is directly proportional to the gibbs free energy. More importantly, the current is directly proportional to the production rate of the reaction. In reality, the voltage will always be higher than 1.23V do to losses.
     
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