## Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning

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

#### Answer

$E^{o}_{cell}=2.89\,V$ The reaction is product-favored.

#### Work Step by Step

Anode: $Mg(s)\rightarrow Mg^{2+}(aq)+2e^{-}$ Cathode: $I_{2}(s)+2e^{-}\rightarrow 2I^{-}$ $E^{o}_{cell}=E^{o}_{cathode}-E^{o}_{anode}$ $= (+0.535)-(-2.356)=2.89\,V$ As $E^{o}_{cell}$ is positive, the reaction is product-favored.

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