Pb(s) + Ni²+(ag) (1.0 M) = Pb²+ (aq) (1.0 M) + Ni(s). E°, Pb2+/Pb = -0.13V E°, %3D Ni²+/Ni = -0.25V At 25°C, calculate the free energy change and the equilibrium constant of the reaction above. AG° = |kJ/mol K =

Pb(s) + Ni²+(ag) (1.0 M) = Pb²+ (aq) (1.0 M) + Ni(s). E°, Pb2+/Pb = -0.13V E°, %3D Ni²+/Ni = -0.25V At 25°C, calculate the free energy change and the equilibrium constant of the reaction above. AG° = |kJ/mol K =

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

Pb(s) + Ni**(ag) (1.0 M) = Pb²* (aq) (1.0 M) + Ni(s)
E°.
Pb2*/Pb
= -0.13V
E°
Ni²+/Ni = -0.25V
At 25°C, calculate the free energy change and the equilibrium constant of the
reaction above.
AG° :
kJ/mol
K =

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