What if I'm not at standard conditions?
Imagine you have a voltaic cell at standard conditions to start. Now, you let the chemistry start by wiring up the cell and allowing the electrons to flow from anode to cathode. As the chemistry proceeds, the concentrations are changing and the potential is changing. The reaction will "go" until the system reaches equilibrium. At this point, the potential will be zero. Clearly the potential must change as the concentrations are changing. How can we determine the potential under any concentration conditions not just standard conditions?
We can use our free energy relations. Before, we could figure out if a reaction was at equilibrium by comparing the ratios of the concentrations "now" to what they would be at equilibrium. This was a comparison between the mass action expression now Q and what it is at equilibrium K. Thinking back to free energy and equilibrium we can remember that
\[\Delta G = \Delta G^{\circ} +RT\ln Q\]
If we put in our relationships for ΔG and E we get
\[E = E^{\circ} - {RT \over nF}\ln Q\]
This equation is called the Nernst Equation. The last term (\(RT/nF \ln Q\))can be converted to a more "user friendly" form by calculating for the constants R, F, and T (assuming 298.15K) and then converting from natural log (ln) to log base 10 (log). That means using the mathematical relation: ln x = (ln 10)(log x) = (2.303)(log x)
\[E = E^{\circ} - {(8.314)(298.15) \over n (96485)}(2.303)\log Q\]
Which gives our "nicer" version of the Nernst Equation:
\[E = E^{\circ} - {0.05916 \over n}\log Q\]
Why do we need or even want to switch to "log" instead of "ln"? There is no other reason that for our human preferences of thinking better and faster in multiples of 10 as opposed to multiples of e which is 2.7182818... (not exactly the multiple most of us humans think of).
The Nernst Equation© 2013 mccord/vandenbout/labrake