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General Chemistry - gibbs free energy

  1. May 1, 2006 #1
    I don't know where to start with this problem. I must be missing an equation or something....

    Q: [tex] Pb(s)+2H^{+}(aq)\rightarrow Pb^{2+}(aq)+H_2(g) \,\,\,\,\,\,\,\,E^{\degree}_{cell}=+0.126V [/tex]

    What is the [tex] \Delta G^{\degree} [/tex] in [itex] \frac{kJ}{mol} [/itex] for this reaction.

    a) -24
    b) 24
    c) -12
    d) 12
    e) 50

    a shove in the right direction would be awesome
     
  2. jcsd
  3. May 1, 2006 #2

    siddharth

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    FrogPad, you are missing the equation, which relates the potential of a cell to the change in Gibbs free energy. Try searching your text again.
     
  4. May 1, 2006 #3
    Woops I think I missed reading like 3 pages out of the chapter :)

    So, does this look ok?
    [tex] \Delta G^\circ = -nFE^\circ [/tex]

    [tex]Pb(s)+2H^{+}(aq)\rightarrow Pb^{2+}(aq)+H_2(g)\,\,\,\,E^\circ = 0.126 [/tex]

    [tex] Pb(s)\rightarrow Pb^{2+}(aq)=2e^{-} \,\,\,\,E^{\circ}=-0.126[/tex]
    [tex] 2H^{+}(aq)+2e^{-} \righarrow H_2 (g) \,\,\,\,E^\circ = 0 [/tex]
    [tex] \Delta G^\circ = -nFE^\circ = -2\left( \frac{96485 J}{Vmol}\right) (0.126V)=-243142\frac{J}{mol}=-24\frac{kJ}{mol}[/tex]
     
    Last edited: May 1, 2006
  5. May 1, 2006 #4

    siddharth

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    Yes, it looks fine.
     
  6. May 1, 2006 #5
    Thanks man. :)

    I appreciate it!
     
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