The standard cell potential,
\[ E^\circ \]describes an electrochemical cell only when all reactants and products are in their standard states. In practice, however, concentrations and partial pressures rarely have these standard values.
As the composition of an electrochemical cell changes, so does the chemical potential of the reacting species. Consequently, the cell potential also changes.
Beginning with the thermodynamic relationship
\[ \Delta G = \Delta G^\circ + RT\ln Q \]and substituting
\[ \Delta G=-nFE \]and
\[ \Delta G^\circ=-nFE^\circ \]gives the Nernst equation:
\[ E = E^\circ - \frac{RT}{nF} \ln Q \]This equation relates the cell potential to the reaction quotient and therefore to the composition of the electrochemical cell.
At
\[ 298\ \text{K} \]the Nernst equation becomes
\[ E = E^\circ - \frac{0.02569}{n} \ln Q \]or, using common logarithms,
\[ E = E^\circ - \frac{0.05916}{n} \log Q \]These forms are commonly used in electrochemical calculations performed at room temperature.
The reaction quotient has exactly the same form as in chemical equilibrium:
\[ Q = \frac{\text{activities of products}} {\text{activities of reactants}} \]with pure solids and pure liquids omitted because their activities are unity.
The Nernst equation shows how the cell potential changes as the reaction proceeds.
At equilibrium,
\[ Q = K \]and no useful electrical work can be extracted from the cell.
Consequently,
\[ E=0 \]This is entirely consistent with the thermodynamic condition
\[ \Delta G=0 \]for equilibrium.
One of the most striking applications of the Nernst equation is the concentration cell. In a concentration cell, both electrodes involve the same half-reaction, so
\[ E^\circ=0 \]The only driving force is the difference in concentration between the two half-cells.
As ions move and the concentrations become more nearly equal, the reaction quotient approaches unity and the cell potential gradually decreases toward zero.
Big picture: The Nernst equation extends electrochemistry beyond standard conditions by relating the cell potential to the composition of the reaction mixture. It is the electrochemical analogue of the Gibbs energy equation, \[ \Delta G = \Delta G^\circ + RT\ln Q, \] showing that cell potentials, like Gibbs energies, depend on the state of the system rather than simply on the chemical reaction itself.
Calculate the cell potential at 25 °C for the electrochemical cell
\[ \mathrm{Cd(s)\,|\,Cd^{2+}(aq,\ 0.100\ M)\,||\,Cu^{2+}(aq,\ 0.400\ M)\,|\,Cu(s)} \]Use the following standard reduction potentials:
| Reduction half-reaction | \(E^\circ_{\text{red}}\) |
|---|---|
| \(\mathrm{Cu^{2+}(aq)+2e^- \rightarrow Cu(s)}\) | \(+0.337\ \text{V}\) |
| \(\mathrm{Cd^{2+}(aq)+2e^- \rightarrow Cd(s)}\) | \(-0.403\ \text{V}\) |
Copper has the larger reduction potential, so copper is reduced at the cathode:
\[ \mathrm{Cu^{2+}(aq)+2e^- \rightarrow Cu(s)} \]Cadmium is oxidized at the anode:
\[ \mathrm{Cd(s)\rightarrow Cd^{2+}(aq)+2e^-} \]The overall cell reaction is
\[ \mathrm{Cd(s)+Cu^{2+}(aq)\rightarrow Cd^{2+}(aq)+Cu(s)} \]The standard cell potential is
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] \[ E^\circ_{\text{cell}} = 0.337-(-0.403) = 0.740\ \text{V} \]For this reaction,
\[ n=2 \]and the reaction quotient is
\[ Q= \frac{[\mathrm{Cd^{2+}}]} {[\mathrm{Cu^{2+}}]} = \frac{0.100}{0.400} = 0.250 \]At 25 °C, the Nernst equation is
\[ E = E^\circ - \frac{0.05916}{n} \log Q \]Substituting the values:
\[ E = 0.740 - \frac{0.05916}{2} \log(0.250) \] \[ E = 0.740 - (0.02958)(-0.602) \] \[ E = 0.758\ \text{V} \]Therefore,
\[ \boxed{E=0.758\ \text{V}} \]Physical interpretation: Since \(Q < 1\), the reaction mixture contains relatively more reactant \(\mathrm{Cu^{2+}}\) than product \(\mathrm{Cd^{2+}}\), compared with standard conditions. This makes the cell potential slightly larger than \(E^\circ\), increasing the driving force for the spontaneous cell reaction.
Calculate the cell potential at 25 °C for the concentration cell
\[ \mathrm{Cu(s)\,|\,Cu^{2+}(aq,\ 0.100\ M)\,||\,Cu^{2+}(aq,\ 2.00\ M)\,|\,Cu(s)} \]The half-reaction in both compartments is
\[ \mathrm{Cu^{2+}(aq)+2e^- \rightarrow Cu(s)} \]In a concentration cell, both half-cells use the same redox couple. Therefore,
\[ E^\circ=0 \]The cell potential arises entirely from the concentration difference between the two \(\mathrm{Cu^{2+}}\) solutions.
Reduction occurs in the compartment with the larger \(\mathrm{Cu^{2+}}\) concentration:
\[ \mathrm{Cu^{2+}(aq,\ 2.00\ M)+2e^- \rightarrow Cu(s)} \]Oxidation occurs in the compartment with the smaller \(\mathrm{Cu^{2+}}\) concentration:
\[ \mathrm{Cu(s)\rightarrow Cu^{2+}(aq,\ 0.100\ M)+2e^-} \]The net effect is to reduce the concentration difference between the two half-cells.
For this concentration cell,
\[ Q = \frac{[\mathrm{Cu^{2+}}]_{\text{dilute}}} {[\mathrm{Cu^{2+}}]_{\text{concentrated}}} = \frac{0.100}{2.00} = 0.0500 \]At 25 °C, the Nernst equation is
\[ E = E^\circ - \frac{0.05916}{n} \log Q \]Since \(E^\circ=0\) and \(n=2\),
\[ E = - \frac{0.05916}{2} \log(0.0500) \] \[ E = 0.0385\ \text{V} \]Therefore,
\[ \boxed{E=0.0385\ \text{V}} \]Physical interpretation: The cell potential is small because the only driving force is the concentration difference. Electrons flow in the direction that causes copper metal to dissolve in the dilute half-cell and copper ions to plate out in the concentrated half-cell, reducing the difference between the two \(\mathrm{Cu^{2+}}\) concentrations.
Two half-reactions are selected randomly. Use the concentrations provided to calculate the cell potential at 25 °C.
\(E =\) V
A concentration cell uses the same metal/ion couple in both half-cells. Calculate the cell potential at 25 °C.
\(E =\) V
Big picture: The Nernst equation extends electrochemistry beyond standard-state conditions by relating the cell potential to the reaction quotient. It shows that the voltage produced by an electrochemical cell depends not only on the identity of the chemical reaction, but also on the composition of the reaction mixture. As equilibrium is approached, the cell potential decreases and ultimately becomes zero.