We can calculate the standard potential for any electrochemical cell from the standard potentials of the two half reactions.
For example, imagine we had the following cell.
\[\rm{Ni\;|\;Ni^{2+} \;\; || \;\; Fe^{3+} , \;\;Fe^{2+} \;| \;Pt}\]
We have at the anode (left) side the reaction
\[\rm{Ni \rightarrow Ni^{2+} + 2e^-}\]
and at the cathode (right) side the reaction
\[\rm{Fe^{3+} + 1 e^- \rightarrow Fe^{2+}}\]
The standard potential for this cell is simply
\[\rm{\mathcal{E}^{\circ} = \mathcal{E}^{\circ}_{cathode} - \mathcal{E}^{\circ}_{anode}}\]
It's even OK to be very literal about the two sides and you can get the standard cell potential with this equation as well
\[\rm{\mathcal{E}^{\circ} = \mathcal{E}^{\circ}_{right} - \mathcal{E}^{\circ}_{left}}\]
Where \(\mathcal{E}^{\circ}\) for each electrode is the standard reduction potential for the half-reaction. Using our table of standard reduction potentials we find for
\[\rm{Ni^{2+} + 2e^- \rightarrow Ni(s) \qquad \mathcal{E}^{\circ} = -0.25 V}\]
\[\rm{Fe^{3+} + e^- \rightarrow Fe^{2+} \qquad \mathcal{E}^{\circ} = +0.77 V}\]
So the standard potential is
\[\rm{\mathcal{E}^{\circ} = \mathcal{E}^{\circ}_{cathode} - \mathcal{E}^{\circ}_{anode} = 0.77 - (-0.25) = +1.02 V }\]
VERY IMPORTANT. Note: the potential is simply the energy difference between the two half reactions. Do NOT try to multiply the potentials by the number of electrons! The number of electrons simply relates how many electrons there are per reaction. How many Fe3+ will be reduced per Ni atom that is oxidized. The potential difference (free energy difference) between the two half-reactions is not dependent on the number of electrons.
© 2013 mccord/vandenbout/labrake