Chemistry: The Molecular Nature of Matter and Change 7th Edition

Published by McGraw-Hill Education
ISBN 10: 007351117X
ISBN 13: 978-0-07351-117-7

Chapter 21 - Problems: 21.44

Answer

See explanation below.

Work Step by Step

Oxidation reaction Cd(s) --> Cd2+(aq) + 2 e- -E° = +0.40 V Reduction: Cr2O7 2-(aq) + 14 H+ (aq) + 6 e- --> 2 Cr3+(aq) + 7 H2O(l) E° = +1.33 V Complete reaction is: Cr2O7 2-(aq) + 3 Cd(s) + 14 H+ (aq) --> 2 Cr3+(aq) + 3 Cd2+(aq) + 7 H2O(l) Ecell=1.73V Reaction is spontaneous as E cell is positive Second part. Oxidation reaction is Pb(s) --> Pb2+(aq) + 2 e- -E° = +0.13 V Reduction: Ni2+(aq) + 2 e- --> Ni(s) E° = -0.25 V Complete reaction is Pb(s) + Ni2+(aq) --> Pb2+(aq) + Ni(s) o E cell = -0.12 V Reaction is not spontaneous, as E cell is negative.
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