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MCAT Chemistry Question — Electrochemical Potential

by Allison Chae
May 28, 2018
MCAT Question of the Day
Reviewed By: Liz Flagge

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^o_cell = ?

A. E^o= – 0.48 V
B. E^o= – 0.11 V
C. E^o= + 0.11 V
D. E^o= + 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:

E^o_cell= E^o_red+ E^o_ox

In this chemical reaction, Au³⁺is reduced to Au making this the reduction half-reaction (E^o_red= + 1.55 V). Oppositely, Br^–is oxidized to Br₂so this would be the oxidation half-reaction. Therefore, you must reverse the reduction potential given for Br₂so 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^o_cell= E^o_red+ E^o_ox= (+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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