Cu2* (aq) + 2e → Cu (s) E°red = +0.34 V ½ O2 (g)+ H2O + 2e¨→ 2 OH¨(aq) E°red = +0.40 V Use the reduction potential above to solve for the cell potential E° cell in volts for the corrosion of copper: Cu (s) + ½ O2 (2) + H20 → Cu2* (aq) + 20H(aq)

Cu2* (aq) + 2e → Cu (s) E°red = +0.34 V ½ O2 (g)+ H2O + 2e¨→ 2 OH¨(aq) E°red = +0.40 V Use the reduction potential above to solve for the cell potential E° cell in volts for the corrosion of copper: Cu (s) + ½ O2 (2) + H20 → Cu2* (aq) + 20H(aq)

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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Concept explainers

Electrolysis

Electrolysis is defined as the process of breaking down ionic compounds into their constituent elements by passing a direct electric current through the compound in a fluid state. The cathode reduces cations, while the anode oxidizes anions. An electrolyte, electrodes, and some form of external power source are the main components required for conducting electrolysis. A partition, such as an ion-exchange membrane or a salt bridge, may also be used, but this is optional. These are primarily used to prevent the products from diffusing near the opposing electrode.

Question

Cu2+
(aq) + 2e" → Cu (s) E'red = +0.34 V
½ O2 (g)+ H2O + 2e¯→ 2 OH(aq) E°red = +0.40 V
Use the reduction potential above to solve for the cell potential E°cell in volts for the
corrosion of copper:
Cu (s) + ½ O2 (2) + H20 → Cu²* (aq) + 20H (ag)

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