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MCAT Chemistry Question — Electrochemical Potential

  • by John
  • May 28, 2018
  • MCAT Question of the Day

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

Click for Explanation
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, Bris 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) + 2eEoox= – 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.

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