# General Chemistry - gibbs free energy

1. May 1, 2006

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

2. May 1, 2006

### 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.

3. May 1, 2006

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

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

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

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

Last edited: May 1, 2006
4. May 1, 2006

### siddharth

Yes, it looks fine.

5. May 1, 2006