Potentials at Non-Standard Conditions: The Nernst Equation

Most of the redox processes that interest science and society do not occur under standard state conditions, and so the potentials of these systems under nonstandard conditions are a property worthy of attention. Having established the relationship between potential and free energy change in this section, the previously discussed relation between free energy change and reaction mixture composition can be used for this purpose.

$$ΔG=ΔG°+RT\cdot ln\, Q$$

Notice the reaction quotient, Q, appears in this equation, making the free energy change dependent upon the composition of the reaction mixture. Substituting the equation relating free energy change to cell potential yields the Nernst equation:

$$-nFE_{cell}=-nFE^o _{cell}+RT\cdot ln\, Q \\ E_{cell}=E^°_{cell}-\frac{RT}{nF}ln\, Q$$

This equation describes how the potential of a redox system (such as a galvanic cell) varies from its standard state value, specifically, showing it to be a function of the number of electrons transferred, n, the temperature, T, and the reaction mixture composition as reflected in Q. A convenient form of the Nernst equation for most work is one in which values for the fundamental constants (R and F) and a factor converting from natural to base-10 logarithms have been included:

$$E_{cell}=E^°_{cell}-\frac{0.0592\;V}{n}log\, Q$$

Predicting Redox Spontaneity Under Nonstandard Conditions
Use the Nernst equation to predict the spontaneity of the redox reaction shown below.

$$Co(s)+Fe^{2+}(aq,\;1.94\;M)⟶Co^{2+}(aq,\;0.15\;M)+Fe(s)$$

Solution
Collecting information from the Appendix and the problem,

Anode (oxidation): $Co(s)⟶Co^{2+}(aq)+2e^-\qquad E^°_{Co^{2+}/Co}=-0.28\;V$
Cathode (reduction): $Fe^{2+}(aq)+2e^-⟶Fe(s)\qquad E^°_{Fe^{2+}/Fe}=-0.447\;V$
$E^°_{cell}=E^°_{cathode}-E^°_{anode}=-0.447\;V-(-0.28\;V)=-0.17\;V$

Notice the negative value of the standard cell potential indicates the process is not spontaneous under standard conditions. Substitution of the Nernst equation terms for the nonstandard conditions yields:

$$Q=\frac{[Co^{2+}]}{[Fe^{2+}]}=\frac{0.15\;M}{1.94\;M}=0.077$$ $$E_{cell}=E^°_{cell}-\frac{0.0592\;V}{n}logQ$$ $$E_{cell}=-0.17\;V-0.033\;V=-0.20\;V$$

The cell potential remains negative (slightly) under the specified conditions, and so the reaction remains nonspontaneous.

Check Your Learning
For the cell schematic below, identify values for n and Q, and calculate the cell potential, Ecell.

$$Al(s)|Al^{3+}(aq,\;0.15\;M)||Cu^{2+}(aq,\;0.025\;M)\;|\;Cu(s)$$
Answer:

n = 6; Q = 1440; Ecell = +1.97 V, spontaneous.

A concentration cell is constructed by connecting two nearly identical half-cells, each based on the same half-reaction and using the same electrode, varying only in the concentration of one redox species. The potential of a concentration cell, therefore, is determined only by the difference in concentration of the chosen redox species. The example problem below illustrates the use of the Nernst equation in calculations involving concentration cells.

Concentration Cells
What is the cell potential of the concentration cell described by

$$Zn(s)|Zn^{2+}(aq,\;0.10\;M)||Zn^{2+}(aq,\;0.50\;M)|Zn(s)$$

Solution
From the information given:

Anode: $Zn(s)⟶Zn^{2+}(aq,\;0.10\;M)+2e^-$ $E^°_{anode}=-0.7618\;V$
Cathode: $Zn^{2+}(aq,\;0.50\;M)+2e^-⟶Zn(s)$ $E^°_{cathode}=-0.7618\;V$
Overall: $Zn^{2+}(aq,\;0.50\;M)⟶Zn^{2+}(aq,\;0.10\;M)$ $E^°_{cell}=0.000\;V$

Substituting into the Nernst equation,

$$E_{cell}=0.000\;V-\frac{0.0592\;V}{2}log\frac{0.10}{0.50}=+0.021\;V$$

The positive value for cell potential indicates the overall cell reaction (see above) is spontaneous. This spontaneous reaction is one in which the zinc ion concentration in the cathode falls (it is reduced to elemental zinc) while that in the anode rises (it is produced by oxidation of the zinc anode). A greater driving force for zinc reduction is present in the cathode, where the zinc(II) ion concentration is greater (Ecathode > Eanode).

Check Your Learning
The concentration cell above was allowed to operate until the cell reaction reached equilibrium. What are the cell potential and the concentrations of zinc(II) in each half-cell for the cell now?

Answer:

Ecell = 0.000 V; [Zn2+]cathode = [Zn2+]anode = 0.30 M