General Chemistry - gibbs free energy

[FrogPad]

I don't know where to start with this problem. I must be missing an equation or something...

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

What is the \Delta G^{\degree} in \frac{kJ}{mol} for this reaction.

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

a shove in the right direction would be awesome

 

[siddharth]

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.

 

[FrogPad]

Woops I think I missed reading like 3 pages out of the chapter :)

So, does this look ok?
\Delta G^\circ = -nFE^\circ

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

Pb(s)\rightarrow Pb^{2+}(aq)=2e^{-} \,\,\,\,E^{\circ}=-0.126
2H^{+}(aq)+2e^{-} \righarrow H_2 (g) \,\,\,\,E^\circ = 0
\Delta G^\circ = -nFE^\circ = -2\left( \frac{96485 J}{Vmol}\right) (0.126V)=-243142\frac{J}{mol}=-24\frac{kJ}{mol}

 

[siddharth]

Yes, it looks fine.

 

[FrogPad]

Thanks man. :)

I appreciate it!