Given the reduction potentials for the following half-reactions,
Br2 (l) + 2e– → 2 Br–(aq) Eored= + 1.07 V
Au3+ (aq) + 3 e– → Au (s) Eored= + 1.55 V
What is the electrochemical potential for the following chemical reaction?
2 Au3+ (aq) + 6 Br–(aq) → 3 Br2 + 2 Au (s) Eocell = ?
A) Eo = – 0.48 V
B) Eo = – 0.11 V
C) Eo = + 0.11 V
D) Eo = + 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:
Eocell = Eored + Eoox
In this chemical reaction, Au3+ is reduced to Au making this the reduction half-reaction (Eored= + 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– Eoox= – Eored= -(+1.07 V) = -1.07 V
Therefore, the Eocell = Eored + Eoox = (+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.