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)
Chapter18: Introduction To Electrochemistry
Section: Chapter Questions
Problem 18.18QAP
Related questions
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)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa85b7a47-876a-498c-9458-2c63d2ad3b09%2F329cc726-f622-4dd5-9880-065e2c51ab4b%2F1qa4mte_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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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