Write the cell notation for an electrochemical cell consisting of an anode where Pb (s) is oxidized to Pb²+ (aq) and a cathode where Cu²+ (aq) is reduced to Cu (s). Assume all aqueous solutions have a concentration of 1 mol/L.
Write the cell notation for an electrochemical cell consisting of an anode where Pb (s) is oxidized to Pb²+ (aq) and a cathode where Cu²+ (aq) is reduced to Cu (s). Assume all aqueous solutions have a concentration of 1 mol/L.
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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![Write the cell notation for an electrochemical cell consisting of an anode where Pb (s) is oxidized to Pb²+ (aq) and a cathode where Cu²+ (aq)
is reduced to Cu (s). Assume all aqueous solutions have a concentration of 1 mol/L.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F216d8975-823a-443e-8abc-06f056adfa09%2Fe6170964-f1bb-4290-b116-71829f862127%2Fqmgjnus_processed.png&w=3840&q=75)
Transcribed Image Text:Write the cell notation for an electrochemical cell consisting of an anode where Pb (s) is oxidized to Pb²+ (aq) and a cathode where Cu²+ (aq)
is reduced to Cu (s). Assume all aqueous solutions have a concentration of 1 mol/L.
![Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction:
(aq) + 2Cu+ (aq) → Pb(s) + 2Cu²+ (aq)
AG°
Pb²+
=
Pb²+ (aq) + 2e¯ → Pb(s) Ee = -0.126 V
red
2+
Cu²+ (aq) + e
→ Cu+ (aq) Fº
red
kJ
K for this reaction would be
=
= 0.153 V
+ than one.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F216d8975-823a-443e-8abc-06f056adfa09%2Fe6170964-f1bb-4290-b116-71829f862127%2Fvt1zcg_processed.png&w=3840&q=75)
Transcribed Image Text:Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction:
(aq) + 2Cu+ (aq) → Pb(s) + 2Cu²+ (aq)
AG°
Pb²+
=
Pb²+ (aq) + 2e¯ → Pb(s) Ee = -0.126 V
red
2+
Cu²+ (aq) + e
→ Cu+ (aq) Fº
red
kJ
K for this reaction would be
=
= 0.153 V
+ than one.
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