Answer
Equation is as under:
$Pb^{2+} (aq) + Cl_{2} (g) → Pb (s) + 2Cl^{+}(aq)$
Electrode potential of Cl = +1.36
Electrode potential of Pb = -0.126
$E^{o}(Cell) = 1.36+0.126 = 1.486$
Work Step by Step
Given above
Chemistry: A Molecular Approach (3rd Edition)
by Tro, Nivaldo J.
Published by
Prentice Hall
ISBN 10:
0321809246
ISBN 13:
978-0-32180-924-7
Chapter 18 - Sections 18.1-18.9 - Exercises - Problems by Topic - Page 905: 48b
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