Cu2* (aq) + 2e → Cu (s) E°red = +0.34 V ½ O2 (g)+ H2O + 2e* → 2 OH(aqg) 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 (g) + H2O → Cu²+ 20H´(aq) (aq) +
Cu2* (aq) + 2e → Cu (s) E°red = +0.34 V ½ O2 (g)+ H2O + 2e* → 2 OH(aqg) 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 (g) + H2O → Cu²+ 20H´(aq) (aq) +
Chapter18: Introduction To Electrochemistry
Section: Chapter Questions
Problem 18.19QAP
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Question
Please state the explanation and answers clearly.
Transcribed Image Text:Cu2* (aq) + 2e → Cu (s) E°red = +0.34 V
%3D
E°red
½ O2 (g)+ H2O + 2e¯ → 2 OH(aq)
+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 (g) + H2O → Cu²+
20H´(aq)
(aq) *
+
Transcribed Image Text:From the result above, the cell potential
for the corrosion of copper is very low
compared to iron. The
oxidation/corrosion of copper is a LESS
spontaneous reaction compared to the
oxidation/corrosion of iron. Therefore,
iron is more prone to oxidation
compared to copper. We can say that
iron is a more active metal than copper.
Using the result for Nail C:
Does wrapping Copper wire prevented
the Iron nail from undergoing
rusting/oxidation/corrosion?
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