Section: Chemical and Physical Foundations of Biological Systems

24) Given the reduction potentials for the following half-reactions,

Br₂ (l) + 2e^– -> 2 Br^–(aq) E^o_red= + 1.07 V

Au³⁺ (aq) + 3 e^– -> Au (s) E^o_red= + 1.55 V

What is the electrochemical potential for the following chemical reaction?

2 Au³⁺ (aq) + 6 Br^–(aq) -> 3 Br₂ + 2 Au (s) E^ocell = ?

1. E^o = – 0.48 V
2. E^o  = – 0.11 V
3. E^o  = + 0.11 V
4. E^o  = + 0.48 V

This question is testing your understanding of electrochemical cells and using half-reactions to determine the cell potential. To answer this question, you must first know that the electrochemical potential for a cell is equal to the reduction potential plus the oxidation potential:

E^ocell = E^ored + E^o_ox

In this chemical reaction, Au³⁺ is reduced to Au making this the reduction half-reaction (E^ored= + 1.55 V). Oppositely, Br^– is oxidized to Br₂ so this would be the oxidation half-reaction.

Therefore, you must reverse the reduction potential given for Br₂ so that it matches the chemical reaction given, as follows:

2 Br^–(aq) → Br2 (l) + 2e^– E^o_ox= – E^o_red= -(+1.07 V) = -1.07 V

Therefore, the E^ocell = E^ored + E^oox = (+1.55 V) + (-1.07 V) = + 0.48 V making D the correct answer.

Note, when calculating the cell potentials, do not multiply the reduction potential or oxidation potential by the coefficients.