Chemistry: A Molecular Approach (3rd Edition)

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

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
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