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Homework Help: The electrical work needed to produce a mole of lead in a chemical reaction

  1. Jun 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider the charging of a lead-acid batter at atmospheric pressure and room room temperature.

    2PbSO4 +2H2O -> Pb + PbO2 + 4H+ + 2SO42-

    How much electrical work must be provided to produce one mole of lead?

    3. The attempt at a solution

    We're given the values of Gibbs Free Energy and Enthalpy for each of the reactants.

    I was doing a past paper for a uni exam and I came across that question. Now correct me if I'm wrong but isn't the electric work equal to the [tex]\Delta[/tex]G?

    So Welec = Gfinal - Ginitial?

    I'm fairly sure that's the case but the answer shows it calculated as
    W = Hfinal - Hinitial

    Again, correct me if I'm wrong but wouldn't that give compression/expansion work?
  2. jcsd
  3. Jun 9, 2009 #2
    Hello there,

    I am unsure about how your solution addressed this question, but this is what I would do.

    1) You need to understand this equation: ∆G = nFE0cell.

    2) You can calculate ∆G for the entire reaction using tabulated values of ∆G for each reactant. That is, ∆G for the reaction = ∆G products - ∆G reactants.

    3) The chemical equation is in fact a redox equation. Find the number of electrons transferred in that equation to determine the value of n for the above equation.

    4) Now that you know ∆G, n, and F (This is Faraday's constant, which can be found online or in your textbook), you can find E0cell by rearranging the above equation.

    I hope that this helps! Feel free to write back!
  4. Jun 10, 2009 #3
    Thanks for the reply.

    Not sure why my topic was moved here though because the question came up in a Thermal Physics course, so we're not particularly interested in electron transfer or what have you.

    We were given the values of Gibbs Free Energy and Enthalpy - the change was easily calculated. My question was more a conceptual one - namely, what exactly a change in the Gibbs Free Energy means physically. I'm now confident that it is the change in electrical work (or any work that isn't compression/expansion) and that the solution to the exam must have been incorrect.
  5. Jun 11, 2009 #4
    Hello there,

    Unfortunately, I'm not an expert in the area of thermal physics. I have always interpreted Gibbs Free Energy as the amount of energy available to do work in a system.

    Does that help?
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