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: 31c


$E^{o}_{cell}=0.65\,V$ The reaction is product-favored.

Work Step by Step

Anode: $Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)+2e^{-}$ Cathode: $2Ag^{+}(aq)+2e^{-}\rightarrow 2Ag(s)$ $E^{o}_{cell}=E^{o}_{cathode}-E^{o}_{anode}$ $= (+0.7991)-(+0.15)=0.65\,V$ As $E^{o}_{cell}$ is positive, the reaction is product-favored.
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