8.6 Cell Potential Relationships
A spontaneous redox reaction will have a positive cell potential. Therefore, we can relate E°cell to ΔG° and K.
| E°cell | ΔG° | K | Summary |
|---|---|---|---|
> 0 |
< 0 |
> 1 |
Spontaneous; product favored |
0 |
0 |
1 |
Equilibrium; reactants and products equally present |
< 0 |
> 0 |
< 1 |
Non-spontaneous; reactant favored |
8.6.3 E°cell and K
\[E_{\mathrm{cell}}^{\circ} = \left (\dfrac{RT}{nF}\right )\ln K\]
Example
Determine the standard free energy (in kJ mol–1) and equilibrium constant for the following galvanic cell at 25 °C.
\[\mathrm{Cr}(s) ~\rvert ~\mathrm{Cr^{2+}}(aq) ~\rvert\lvert~ \mathrm{Pb}^{2+}(aq) ~ \lvert ~\mathrm{Pb}(s)\]
Find the standard cell potential.
\[\begin{align*} E_{\mathrm{cell}}^{\circ} &= E_{\mathrm{cathode}}^{\circ} - E_{\mathrm{anode}}^{\circ}\\ &= E_{\mathrm{Pb^{2+}/Pb}}^{\circ} - E_{\mathrm{Cr^{2+}/Cr}}^{\circ}\\ &= (-0.13~\mathrm{V}) - (-0.913~\mathrm{V})\\ &= 0.78~\mathrm{V}\\ &= 0.78~\mathrm{J~C^{-1}} \end{align*}\]
Convert the standard cell potential into a standard free energy.
\[\begin{align*} \Delta G^{\circ} &= -nFE^{\circ}_{\mathrm{cell}}\\ &= -(2~\mathrm{mol}~e^-)(96,485~\mathrm{C~mol^{-1}}~e^-)(0.78~\mathrm{J~C^{-1}})\left( \dfrac{1~\mathrm{kJ}}{\mathrm{10^3~J}} \right ) \\ &= -150.5~\mathrm{kJ~mol^{-1}} \end{align*}\]
Determine K from the standard free energy.
\[\begin{align*} \Delta G^{\circ} &= -RT\ln K \\ K &= e^{-\dfrac{\Delta G^{\circ}}{RT}}\\ &= e^{-\dfrac{-150.5~\mathrm{kJ~mol^{-1}}}{(0.008315~\mathrm{kJ~mol^{-1}~K^{-1}})(298.15~\mathrm{K})}} \\ &= 2.32\times 10^{26} \end{align*}\]
Concept Check: We expect a large K given the very negative ΔG° value. A very spontaneous reaction is very product favored.
Alternative: You could also determine K from the standard cell potential. Be sure to convert E°cell from J C–1 (which is V) into kJ C–1.
\[\begin{align*} E_{\mathrm{cell}}^{\circ} &= \left (\dfrac{RT}{nF}\right )\ln K \\ K &= e^{\dfrac{nFE_{\mathrm{cell}}^{\circ}}{RT}}\\ &= e^{\dfrac{(2~\mathrm{mol}~e^-)(96,485~\mathrm{C~mol^{-1}}~e^-)(0.00078~\mathrm{kJ~C^{-1}})}{(0.008315~\mathrm{kJ~mol^{-1}~K^{-1}})(298.15~\mathrm{K})}}\\ &= 2.33\times 10^{26} \end{align*}\]
Despite which way you solve for K, you will notice that the answers match to one decimal place.