Recall that we could set up tables of ΔfGo and ΔfHo for ions in solution by referencing to a standard, say ΔfGo for H+ equals zero by definition. For example, consider the Na+(aq) ion. From the tables of thermodynamics properties we find,
(1)What is the reaction?
Na(s) → Na+(aq) + e− (2)This is a "half-reaction, a hypothetical reaction, written as an oxidation. By international convention, electrochemists have agreed to write half-reactions as reductions. Rewrite our half-reaction as a reduction,
Na+(aq) + e− → Na(s). (3)Then the Gibbs free energy change for the reduction reaction is just the negative of the Gibbs free energy of formation of the positive ion,
(4)We can calculate an emf associated with this half reaction from,
(5a,b)This E o is called the "half-cell" potential for the Na+(aq)|Na half cell.
Recall that was not measured in an absolute sense, but was measured relative to an arbitrary standard which we (by convention) pick to be the formation of the H+(aq) ion. That is, we measure the ΔfG o of all other ions relative to ΔG o of the reaction,
1/2 H2(g) → H+(aq) + e− ,for which (by convention) we set,
(6)Then is relative to
. (7)This is easy to show. Write the half-reaction for the reduction of hydrogen ion,
H+(aq) + e− → ½ H2(g), (8)for which
(10)There are tables of half-cell potentials or you can make your own from tables of ΔfGoion(aq). (Modern half-cell tables will always be written for half-reactions written as reductions. Tables from older American books have the half-reactions written as oxidations and the tables are oxidation potentials.)
Conventions and Usage
Given a cell, A,B|C,D,...|...|Z, write the reaction for the Right-hand-side as a Reduction, and left-hand-side as an oxidation.
Right = Reduction --- (call the value from table EoR.)Then, for the complete reaction,
Left = Oxidation --- (call the table value, which will be a reduction potential, EoL.)
Eo = EoR − EoL. (11)I emphasize that both values are table REDUCTION potentials. You do not change the sign of either one to compensate for anything. All the signs have been taken care of in the above equation. Then,
If Eo > 0 then ΔGo < 0, reaction proceeds spontaneously as written,
If Eo < 0 then ΔGo > 0, reaction does not proceed spontaneously as written.
Zn(s)|ZnSO4(aq)||CuSO4(aq)|Cu(s) (13a)From the half-reactions we see that n = 2.
R Cu2+ + 2 e− → Cu
L Zn → Zn2+ + 2 e−
Cu2+ +Zn → Cu − Zn2+ (13b)
From tables, both as reductions. (14a, b)Then
(16)Recall, that if Eo > 0 then ΔGo < 0 so the reaction is spontaneous.
Let the reaction run free in the spontaneous direction, which side is positive?
Notice that the half-reaction,
Zn → Zn2+ + 2 e− , (17)is running forward. Zn is "giving off" electrons, Zn is negative, therefore Cu is positive.
We can get the equilibrium constant from Eo:
(18a, b, c)
Other Thermodynamic Functions From Electrochemical Cell Data
We have seen that we can get ΔrGo for a reaction from the emf of its electrochemical cell, can we get other thermodynamic functions? Yes, we can get other thermodynamic functions, but it requires additional data.
(20)So we can get ΔSo if we know Eo as a function of T. Then we can also get ΔHo from,
ΔHo = ΔGo + TΔ So. (21)Sometimes we want to do this graphically. Plot Eo vs T.
(23)into Equation 22 to get,
(24)Then we plot Eo/T vs 1/T. We should get a line much closer to a straight line, with slope equal to
Copyright 2004, W. R. Salzman
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