Organometallic Solutions: #5

5.* (1995 3 2) Below are electrode potentials for conventional half-cell reactions (reduction potentials) versus the "normal hydrogen electrode."

Fe(III) + e- --> Fe(II)	0.77V
O₂ + 2H⁺ + 2e^- --> H₂O₂	0.7V
H₂O₂ + 2H⁺ + 2e^- --> 2H₂O	1.4V

A. Which substance among all of these components is the strongest reducing agent?

The strongest reducing agent is the compound that most wants to lose electrons (thereby becoming oxidized itself). All of the half reactions shown are not oxidations but reductions; it is necessary to write them backwards (flipping the sign of E^o) to obtain the corresponding oxidation half reaction. Since all of the E^o's are negative when we write the half-reactions backwards, we want the least negative E^o, which would signal the least non-spontaneous reaction. This would come from writing the second reaction backwards. Thus, the strongest oxidizing agent is H₂O₂.

B. Which substance is the strongest oxidizing agent?

The strongest oxidizing agent is the one that most wants to gain electrons (thereby becoming reduced itself). Since all of the half reactions are already written as reductions and all of the E^o's are already positive, we are here looking for the largest E^o. Thus, H₂O₂ is the strongest oxidizing agent.

C. Is it feasible to find a catalyst to make H₂O₂ by bubbling O₂ into water according to the following equation?

O₂ + 2H₂O --> 2H₂O₂

O₂ + 2H⁺ + 2e^- --> 2H₂O₂	E^o = 0.70V
2H⁺ + 2e^- + H₂O₂ --> 2H₂O	E^o = 1.40V

Write the second reaction backwards and add the two equations to obtain:

O₂ + 2H₂O --> 2H₂O₂ E^o = -0.70V

Therefore since the reaction has negative E^o, it has positive DG, and is therefore nonspontaneous. Because this reaction is not thermodynamically favorable, it is not feasible to find a catalyst to make it go (catalysts can't change thermodynamics, only kinetics).