Analysing reactions using standard electrode potentials

Standard electrode potentials are used to predict reactions.

Under standard conditions, a reaction is thermodynamically feasible or spontaneous if the standard electrode potential of the electrochemical cell, Ecell, is positive. Let’s look at the following example:

Cr_2O_7^{\: 2-}(aq)+14H^+(aq)+6e^-\rightleftharpoons 2Cr^{3+}(aq)+7H_2O(l)\; \; \; \; \; E^o=+1.33\; V

Sn^{4+}(aq)+2e^-\rightleftharpoons Sn^{2+}(aq)\; \; \; \; \; E^o=+0.15\; V

The more positive the potential of a half-cell is, the more spontaneous its reaction from left to right. Hence, the first half-cell equation occurs at the cathode and the second one at the anode. Multiplying the second equation by a factor of three and combining it with the first,

Cr_2O_7^{\: 2-}(aq)+14H^+(aq)+3Sn^{2+}(aq)\rightleftharpoons 2Cr^{3+}(aq)+3Sn^{4+}(aq)+7H_2O(l)

E_{cell}=E_{red,cathode}-E_{red,anode}=+1.33\; V-(+0.15\; V)=+1.18\; V

Note that the \(E^o\) value of the second half-cell equation is unchanged when the equation is multiplied by a factor of three. This is because the unit of standard electrode potential is joules per coulomb. When we multiply the equation by three, we triple the number of moles of electrons (coulomb) as well as the amount of energy (joules) conveyed by the electrons. Thus, we say that the value of \(E^o\) is intensive.

In general, a reaction with a positive Ecell value is thermodynamically feasible. There are, however, exceptions. One such example is

Cr_2O_7^{\: 2-}(aq)+14H^+(aq)+6e^-\rightleftharpoons 2Cr^{3+}(aq)+7H_2O(l)\; \; \; \; \; E^o=+1.33\; V

O_2(g)+4H^+(aq)+4e^-\rightleftharpoons 2H_2O(l)\; \; \; \; \; E^o=+1.23\; V

The combined reaction has a positive value of +0.10 V but it does not proceed. This is because the combined reaction, even though thermodynamically feasible, is not kinetically viable, which implies that the activation energy for the reaction is high.


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