Return to MCAT® Blog Homepage

MCAT Chemistry Question — Electrochemical Potential

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.

Want more MCAT practice?

We’ve got options for every schedule and learning style!

From the best online MCAT course created by top instructors with 524+ MCAT scores to the most representative full-length practice exams and private tutoring, we can custom tailor your MCAT prep to your goals!

Not sure which option is right for you? Schedule a free MCAT consultation with an MCAT expert using the form below. No obligation, just expert advice.

Create your Free Account to access our MCAT Flashcards SIGN UP NOW
MCAT is a registered trademark of the Association of American Medical Colleges (AAMC), which is not affiliated with Blueprint.