Relation of E o and \(\Delta G^o\) E o is the cell potential under standard conditions. Applying this potential to the galvanic cell under standard conditions will bring the reaction to a halt as it will exactly oppose any flow of electrons. This suggests a relation between E o and \(\Delta G^o\). We can derive the relation by considering the definition of electrical work from physics: \[w_{elec} = CV \label{EQ:welec}\] where C is the Coulombs of charge moved through the voltage potential and V is the magnitude of the potential change. When this work is done reversibly, we are balancing the potential exactly and can get the maximum work, which is just \(-\Delta G\) under the current conditions. The (-) sign comes from the convention that released energy is negative. If we use E o for the voltage we are under standard and reversible conditions so equation \(\ref{EQ:welec}\) becomes: \[-\Delta G^o = CE^o \label{EQ:Nern1}\] We usually calculate the free energy change with respect to a moles of a reactant. If we use n e to represent the number of moles of electrons per unit of reaction and utilize the fact that there are F = 96485.33212 C/mol of electrons (F is the Faraday constant ), we have C = n e F. Substituting into equation \(\ref{EQ:Nern1}\): \[-\Delta G^o = n_e FE^o \quad\text{or}\quad \Delta G^o = -n_e FE^o\label{EQ:Nern2}\] This subscript 'e' is often dropped and other symbols are often used for n e (most common is v ). The Nernst equation The connection between \(\Delta G\) and potential allows us to utilize our knowledge of concentration effects to develop an equation for how \(E_{cell}\) varies with concentrations (really the reactions quotient, \(Q\)). The resulting relation is the Nernst equation . First we remember that at nonstandard conditions: \[\Delta G = \Delta G^o + RT\ln Q\nonumber\] then by analogy with equation \(\ref{EQ:Nern2}\): \[\Delta G = -n_e FE = \Delta G^o + RT\ln Q\nonumber\] where we can substitute in for \(\Delta G^o\) with equation \(\ref{EQ:Nern2}\) to get: \[ -n_e FE = -n_e FE^o + RT\ln Q\label{EQ:Nern3}\] Dividing through by -n e F produces the standard form of the Nernst equation showing the relationship of the cell potential to the reaction quotient: \[ E = E^o - \frac{RT}{n_e F}\ln Q\label{EQ:Nernst}\] Since we know at equilibrium \(\Delta G= -n_e FE = 0\) this makes it clear that when a galvanic cell (battery) is dead (the voltage, \(E = 0\)) it is at chemical equilibrium. Concentration cells Equation \(\ref{EQ:Nernst}\) has a very important implication. If you connect two half-cells of the same material but at different concentrations you will see a potential difference because of the concentration difference. If the two cells are the same \(E^o = E^o_2-E^o_1 = 0\) so the equation reduces to: \[ E = - \frac{RT}{n_e F}lnQ\label{EQ:Conc_cell}\] This will give a non-zero voltage as long as Q ≠ 1. It also means that any ion that is at a different concentration in the two half-cells will generate a potential difference. Although spectator ions (the counter ions) are often ignored in calculations using the Nernst equation, that is not completely correct as they do contribute to a concentration difference induced cell potential and impact the activity of all the ions through the ionic strength. This idea is used in biochemical applications to assign contributions of specific ions to cell membrane potentials to differences in concentrations of the ion on each side of the membrane. Calculating Cell Potentials Using values measured relative to the SHE, it is fairly easy to calculate the standard cell potential of a given reaction. For example, consider the reaction \[\ce{ 2 Ag^{+}(aq) + Cu(s) \rightarrow 2 Ag(s) + Cu^{2+}(aq)} \nonumber \] Before calculating the cell potential, we should review a few definitions. The anode half reaction, which is defined by the half-reaction in which oxidation occurs, is \[\ce{Cu(s) \rightarrow Cu^{2+}(aq) + 2 e^{-}} \nonumber \] And the cathode half-reaction, defined as the half-reaction in which reduction takes place, is \[\ce{Ag^+(aq) + e- \rightarrow Ag(s)}\nonumber \] Using standard cell notation , the conditions (such as the concentrations of the ions in solution) can be represented. In the standard cell notation, the anode is on the left-hand side, and the cathode on the right. The two are typically separated by a salt bridge , which is designated by a double vertical line. A single vertical line indicates a phase boundary. Hence for the reaction above, if the silver ions are at a concentration of 0.500 M, and the copper (II) ions are at a concentration of 0.100 M, the standard cell notation would be Example \(\PageIndex{1}\): Cell Potential Under nonstandard Conditions (ignoring the counter ions) Calculate the cell potential at 25 °C for the cell indicated by \[Cu(s) | Cu^{2+}(aq, \, 0.100 \, M) || Ag^+ (aq,\, 0.500 \, M) | Ag(s) \nonumber \] Solution In order to calculate the cell potential (\(E\)), the standard cell potential must first be obtained. The standard cell potential at 25 °C is given by \[\begin{align*} E_{cell} &= E^o_{cathode} -E^o_{anode} \\[4pt] &= 0.799 \,V - 0.337\,V \\[4pt] &=0.462\,V \end{align*} \] And for a cell at non-standard conditions, such as those indicated above, the Nernst equation can be used to calculate the cell potential. At 25 °C, The cell potential is given by \[ \begin{align*} E_{cell} &= E^o_{cell} - \dfrac{RT}{nF} \ln \left( \dfrac{[Cu^{2+}]}{[Ag^+]^2} \right) \\[4pt] &= 0.462\,V - \dfrac{(8.314 \,J/(mol\,K) (298\,K) }{2(96485\,C/mol)} \ln \left( \dfrac{0.100}{0.500^2} \right) \end{align*} \] Noting that \(1\, J/C = 1\, V\), \[E = 0.480\,V \nonumber \] Example \(\PageIndex{2}\): Cell Potential Under nonstandard Conditions (with the counter ions) Calculate the cell potential at 25 °C for the cell indicated by \[Cu(s) | Cu^{2+}(aq, \, 0.100 \, M) || Ag^+ (aq,\, 0.500 \, M) | Ag(s) \nonumber \] Solution If the other ions in solution are just the counter ions that make the overall charge in each half-cell neutral they can just be included in Q using the reaction stoichiometry. Because the concentrations on the two sides are different, they do not cancel out. Assuming a singly charged counter ion (eg. NO 3 - ). We have at 25 °C: \[ \begin{align*} E_{cell} &= E^o_{cell} - \dfrac{RT}{nF} \ln \left( \dfrac{[Cu^{2+}][NO_3^- ]_{Cu}^2}{[Ag^+]^2[NO_3^- ]_{Ag}^2} \right) \\[4pt] &= 0.462\,V - \dfrac{(8.314 \,J/(mol\,K) (298\,K) }{2(96485\,C/mol)} \ln \left( \dfrac{(0.100)(0.200)^2}{0.500^4} \right) \end{align*} \] \[E = 0.468\,V \nonumber \] Example \(\PageIndex{3}\): Cell Potential under Non-Standard Conditions Calculate the cell potential at 25 °C for the cell defined by \[Ni(s) | Ni^{2+}\, (aq, \,0.500\, M) || Cu(s) | Cu^{2+}(aq, \,0.100\, M) \nonumber \] Solution We will use the Nernst equation. First, we need to determine \(E^o\). Using Table P1, it is apparent that \[ \ce{Cu^{2 }+ 2 e^{-} \rightarrow Cu } \nonumber \] \(E^o = 0.337 \,V\) \[\ce{ Ni^{2+} + 2 e^{-} \rightarrow Ni} \nonumber \] with \(E^o = -0.250\, V\) So copper, having the larger reduction potential will be the cathode half-reaction while forcing nickel to oxidize, making it the anode. So E o for the cell will be given by \[ \begin{align*} E_{cell} &= E^o_{cathode} -E^o_{anode} \\[4pt] &= 0.337 \,V -(-0.250\,V) \\[4pt] = 0.587\,V \end{align*} \] And the cell potential is then given by the Nernst Equation \[ \begin{align*} E_{cell} &= E^o_{cell} - \dfrac{RT}{nF} \ln Q \\[4pt] &= 0.587 - \dfrac{(8.314 \,J/(mol\,K) (298\,K) }{2(96485\,C/mol)} \ln \left( \dfrac{0.500}{0.100} \right) \\[4pt] &= 0.566\,V \end{align*} \]