Interpreting Electrode and Cell Potentials

Thinking carefully about the definitions of cell and electrode potentials and the observations of spontaneous redox change presented thus far, a significant relation is noted. The previous section described the spontaneous oxidation of copper by aqueous silver(I) ions, but no observed reaction with aqueous lead(II) ions. Results of the calculations on the previous page have shown the spontaneous process is described by a positive cell potential while the nonspontaneous process exhibits a negative cell potential. And so, with regard to the relative effectiveness (“strength”) with which aqueous Ag+ and Pb2+ ions oxidize Cu under standard conditions, the stronger oxidant is the one exhibiting the greater standard electrode potential, E°. Since by convention electrode potentials are for reduction processes, an increased value of corresponds to an increased driving force behind the reduction of the species (hence increased effectiveness of its action as an oxidizing agent on some other species). Negative values for electrode potentials are simply a consequence of assigning a value of 0 V to the SHE, indicating the reactant of the half-reaction is a weaker oxidant than aqueous hydrogen ions.

Applying this logic to the numerically ordered listing of standard electrode potentials in the Appendix shows this listing to be likewise in order of the oxidizing strength of the half-reaction’s reactant species, decreasing from strongest oxidant (most positive E°) to weakest oxidant (most negative E°). Predictions regarding the spontaneity of redox reactions under standard state conditions can then be easily made by simply comparing the relative positions of their table entries. By definition, E°cell is positive when E°cathode > E°anode, and so any redox reaction in which the oxidant’s entry is above the reductant’s entry is predicted to be spontaneous.

Reconsideration of the two redox reactions in the Cu-Ag cell seen previously provides support for this fact. The entry for the silver(I)/silver(0) half-reaction is above that for the copper(II)/copper(0) half-reaction, and so the oxidation of Cu by Ag+ is predicted to be spontaneous (E°cathode > E°anode and so E°cell > 0). Conversely, the entry for the lead(II)/lead(0) half-cell is beneath that for copper(II)/copper(0), and the oxidation of Cu by Pb2+ is nonspontaneous (E°cathode < E°anode and so E°cell < 0).

Recalling the chapter on thermodynamics, the spontaneities of the forward and reverse reactions of a reversible process show a reciprocal relationship: if a process is spontaneous in one direction, it is non-spontaneous in the opposite direction. As an indicator of spontaneity for redox reactions, the potential of a cell reaction shows a consequential relationship in its arithmetic sign. The spontaneous oxidation of copper by lead(II) ions is not observed,

$$Cu(s)+Pb^{2+}(aq)⟶Cu^{2+}(aq)+Pb(s)\qquad E^°_{forward}=-0.47\;V\;(\text{negative, non-spontaneous})$$

and so the reverse reaction, the oxidation of lead by copper(II) ions, is predicted to occur spontaneously:

$$Pb(s)+Cu^{2+}(aq)⟶Pb^{2+}(aq)+Cu(s)\qquad E^°_{forward}=+0.47\;V\;(\text{positive, spontaneous})$$

Note that reversing the direction of a redox reaction effectively interchanges the identities of the cathode and anode half-reactions, and so the cell potential is calculated from electrode potentials in the reverse subtraction order than that for the forward reaction. In practice, a voltmeter would report a potential of −0.47 V with its red and black inputs connected to the Pb and Cu electrodes, respectively. If the inputs were swapped, the reported voltage would be +0.47 V.

Predicting Redox Spontaneity
Are aqueous iron(II) ions predicted to spontaneously oxidize elemental chromium under standard state conditions? Assume the half-reactions to be those available in the Appendix.

Referring to the tabulated half-reactions, the redox reaction in question can be represented by the equations below:


The entry for the putative oxidant, Fe2+, appears above the entry for the reductant, Cr, and so a spontaneous reaction is predicted per the quick approach described above. Supporting this predication by calculating the standard cell potential for this reaction gives

$$E^°_{cell}=E^°_{cathode}-E^°_{anode}$$ $$E^°_{cell}=E^°_{Fe(II)}-E^°_{Cr}$$ $$=-0.447\;V–0.744\;V=+0.297\;V$$

The positive value for the standard cell potential indicates the process is spontaneous under standard state conditions.

Check Your Learning
Use the data in the Appendix to predict the spontaneity of the oxidation of bromide ion by molecular iodine under standard state conditions, supporting the prediction by calculating the standard cell potential for the reaction. Repeat for the oxidation of iodide ion by molecular bromine.


$$I_2(s)+2Br^(aq)⟶2I^-(aq)+Br_2(l)\qquad E^°_{cell}=-0.5518\;V\;(\text{nonspontaneous})$$ $$Br_2(s)+2I^(aq)⟶2Br^-(aq)+I_2(s)\qquad E^°_{cell}=+0.5518\;V\;(\text{spontaneous})$$