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:

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.

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, *E*_{cell}.

## Answer:

*n* = 6; *Q* = 1440; *E*_{cell} = +1.97 V, spontaneous.