CHM 501 Lecture

Thermodynamic Considerations

The tendency for an electron transfer reaction to occur is embodied in the potential, E, which is directly related to the Gibb’s free energy, G

G = –nFE

F is Faraday’s constant = 96485 Coulombs/mole

n = the number of moles of electrons transferred in the reaction

E is typically more convenient because it can be measured readily using electrochemical techniques.

At standard conditions (1 bar pressure, 25 ^oC, 1 mole reactant, pH = 0), potentials are denoted E^o and are further partitioned (arbitrarily) into reduction potentials for each reacting species:

E^o = E^o_red(reduced species) – E^o_red(oxidized species)

Reduction potentials are tabulated in a variety of formats.

At nonstandard conditions, the Nernst equation is used to find a reaction potential:

Q is the reaction quotient (mass action expression)

The Nernst equation can also be applied to half-reactions:

Q_red is the mass action expression of the reduction half-reaction, ignoring the electrons.

Q_red is only a relative value since it is based on the arbitrary partitioning done to obtain E^o_red

This explains the strong pH dependence of many redox reactions:

O₂(g) + 4H⁺(aq) + 4e^– 2H₂O(l)

at pH = 0 (standard conditions) E^o_red = +1.23 V

at other pH values:

assuming an oxygen pressure of 1 bar

Latimer Diagrams

This is a method of conveying large amounts of reduction potential information about a series of reductions for a given element.

An example using the Latimer Diagram for chlorine under acidic conditions:

Simple redox reactions:

ClO₄^–(aq) + 2H⁺(aq) + 2e^– ClO₃^–(aq) + H₂O(l)

E^o_red = 1.20 V

More complex redox reactions (one of many possible examples)

ClO₄^–(aq) + 8H⁺(aq) + 8e^– Cl^–(aq) + 4H₂O(l)

Using Gibb's Free Energies gives the same result:

ClO4–(aq) + 2H+(aq) + 2e– ClO3–(aq) + H2O(l)	Eo = 1.20	Go = –(2)(96485)(1.20) = –232000 J
ClO3–(aq) + 2H+(aq) + e– ClO2(aq) + H2O(l)	Eo = 1.18	Go = –(1)(96485)(1.18) = –114000 J
ClO2(aq) + H+(aq) + e– HClO2(aq)	Eo = 1.19	Go = –(1)(96485)(1.19) = –115000 J
HClO2(aq) + 2H+(aq) + 2e– HClO(aq) + H2O(l)	Eo = 1.70	Go = –(2)(96485)(1.70) = –328000 J
HClO(aq) + H+(aq) + e– ½Cl2(aq) + H2O(l)	Eo = 1.63	Go = –(1)(96485)(1.63) = –157000 J
½Cl2(aq) + e– Cl–(aq)	Eo = 1.36	Go = –(1)(96485)(1.36) = –131000 J
ClO4–(aq) + 8H+(aq) + 8e– Cl–(aq) + 4H2O(l)	Go = [–232000 – 114000 – 115000 – 328000 – 157000 – 131000] = –1077000 J

Disproportionation (one of many possible examples)

Cl₂(aq) + H₂O(l) HClO(aq) + Cl^–(aq) + H⁺(aq)

E^o = 1.36 – 1.63 = –0.27 V

This is unfavored at standard conditions.

However, at pH = 7:

At equilibrium E = 0 and

At slightly acidic pH values (e.g., 5 - 6) the amount of hypochlorous acid in solution can be quite high relative to the amount of chlorine, i.e., the conversion is substantial.