Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 17 - Electrochemistry and Its Applications - Questions for Review and Thought - Topical Questions - Page 782b: 31d


$E^{o}_{cell}=1.164\,V$ The reaction is product-favored.

Work Step by Step

Anode: $2Zn(s)\rightarrow 2Zn^{2+}(aq)+4e^{-}$ Cathode: $O_{2}(g)+2H_{2}O(l)+4e^{-}\rightarrow 4OH^{-}$ $E^{o}_{cell}=E^{o}_{cathode}-E^{o}_{anode}$ $= (+0.401)-(-0.763)=1.164\,V$ As $E^{o}_{cell}$ is positive, the reaction is product-favored.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.