The Nernst equation can be used to calculate the potential of an electrochemical cell under any conditions. Let's look at an example of this using the very simple cell
\[\rm{ Zn(s) \;\bigl| \;Zn^{2+}(aq) \;\bigl|\bigl| \;Cu^{2+}(aq) \;\bigl| \;Cu(s) }\]
The balanced equation for this cell is
\[\rm{Zn(s)\; + \;Cu^{2+}(aq)\; \rightarrow\; Zn^{2+}(aq)\; + \;Cu(s)}\]
For this reaction, we can look up the standard half reduction potentials and calculate the standard potential E°. In this case, you find that E° = 1.100 V.
Now the question is, what is the potential under other conditions. For this let's look at an example where we [Zn2+] = 0.1 M and [Cu2+] = 0.001 M.
To find the potential, we need to use the Nernst equation
\[\mathcal{E} = \mathcal{E}^{\circ} - {0.05916 \over n}\log Q\]
Here n=2 since the equation as written has 2 electrons per mole reaction. Now we need Q.
\[Q = {\rm [Zn^{2+}] \over [Cu^{2+}]} = {0.1 \over 0.001} = 100\]
Plugging this into the Nerst Equation with the standard potential, we get
\[\eqalign{ \mathcal{E} &= +1.100 - {0.05916 \over 2}\log 100 \cr &= +1.100 - 0.05916 \cr &= +1.041\;{\rm V} \cr }\]
Since the conditions are closer to equilibrium (more Zn(II) ion than Cu(II) ion) than the standard conditions the potential is lower.
You should be able to manipulate the Nernst equation in any way. Given a potential and standard potential, you should be able to determine ratios of concentrations or an unknown concentration. Given concentration and standard potential, you should be able to find the potential, etc...
Application of the Nernst Equation