Introductory Chemistry (5th Edition)

Published by Pearson
ISBN 10: 032191029X
ISBN 13: 978-0-32191-029-5

Chapter 16 - Oxidation and Reduction - Exercises - Problems - Page 602: 33

Answer

a) $H_{2}$ (g) is oxidized. b) $Al$ (s) is oxidized. c) $Al$ (s) is oxidized.

Work Step by Step

$a)$ In this reaction, first find oxidation state of each atom in each side. In the reactant side, H and O atom both will have oxidation state '0' in the substance $H_{2}$ and $O_{2}$ respectively. In the product side, we have $H_{2}O$ molecule. In molecule, H have an oxidation state of +1 and O have an oxidation state of -2. So, here, oxidation state increases for H (0 to +1) and decreases for O (0 to -2). So, $H_{2}$ is oxidized (oxidation means increase in oxidation state) and $O_{2}$ is reduced (Reduction means decrease in oxidation state). $b)$ In this reaction, first find oxidation state of each atom in each side. In the reactant side, Al and O atom both will have oxidation state '0' in the substance $Al$ and $O_{2}$ respectively. In the product side, we have $Al_{2}O_{3}$ molecule. In molecule, O have an oxidation state of -2. Now, we have to find oxidation state of Al. Let's say oxidation state of Al = x. So, the equation is 2x + 3.(-2) = 0. By solving we get, x = +3. Hence, oxidation state of Al here is +3. So, here, oxidation state increases for Al (0 to +3) and decreases for O (0 to -2). So, $Al$ is oxidized (oxidation means increase in oxidation state) and $O_{2}$ is reduced (Reduction means decrease in oxidation state). c) In this reaction, first find oxidation state of each atom in each side. In the reactant side, Al and Cl atom both will have oxidation state '0' in the substance $Al$ and $Cl_{2}$ respectively. In the product side, we have $AlCl_{3}$ molecule. In molecule, Cl have an oxidation state of -1. Now, we have to find oxidation state of Al. Let's say oxidation state of Al = x. So, the equation is x + 3.(-1) = 0. By solving we get, x = +3. Hence, oxidation state of Al here is +3. So, here, oxidation state increases for Al (0 to +3) and decreases for Cl (0 to -1). So, $Al$ is oxidized (oxidation means increase in oxidation state) and $Cl_{2}$ is reduced (Reduction means decrease in oxidation state).
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