Chemistry: The Molecular Science (5th Edition)

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

Chapter 3 - Chemical Reactions - Questions for Review and Thought - General Questions - Page 148g: 107c


This is a redox reaction. Oxidizing agent: $Ti$ Reducing agent: $Mg$

Work Step by Step

Before the reaction, $Ti$ was in $TiCl_4$. Each chlorine has "-1" as an oxidation number, so, $Ti$ has +4. After the reaction $Ti$ is represented as $Ti(s)$ So, it has 0 as its oxidation number. $Ti$ reduced. (+4 --> 0), making it the oxidizing agent. Before the reaction $Mg$ was in $Mg(l)$ Therefore, it had an oxidation number of 0. After the reaction, $Mg$ is in $MgCl_2$ Each chlorine has "-1" as its oxidation number, so, $Mg$ has $+2$. $Mg$ is oxidized (0 --> +2), making it the reducing agent.
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