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Although we frequently consider Cu²+ and Cu° as the most stable forms of copper, the intermediate oxidation state Cu* plays a key role in many metalloenzymes of biological importance in biology. For example in the enzyme superoxide dismutase, Cu²* helps regulate the redox potential of the cell by reacting with the superoxide radical 02 (a potent reducing agent) to form Cu* as an intermediate. More broadly, intermediate oxidation states of transitions metals frequently play key roles in catalysis on both organic/biological and inorganic systems. The two-electron reduction of Cu2+ to Cu° via Cu²+ + 2e→ Cu° has Eº= +0.337 V vs. SHE. The one-electron reduction of Cu* to Cu° via Cu* + e* -→ Cu° has Eº= +0.52 V vs. SHE . a) Based on the above, calculate E° for the one-electron reduction Cu2+ + 1e → Cu* . (Hint: While there are multiple ways to work this question, it might be easiest to write out the reactions, calculate free energies for the reactions as written, add or subtract equations (and free energies) as necessary, and then convert free energies back to potentials).

Although we frequently consider Cu²+ and Cu° as the most stable forms of copper, the intermediate oxidation state Cu* plays a key role in many metalloenzymes of biological importance in biology. For example in the enzyme superoxide dismutase, Cu²* helps regulate the redox potential of the cell by reacting with the superoxide radical 02 (a potent reducing agent) to form Cu* as an intermediate. More broadly, intermediate oxidation states of transitions metals frequently play key roles in catalysis on both organic/biological and inorganic systems. The two-electron reduction of Cu2+ to Cu° via Cu²+ + 2e→ Cu° has Eº= +0.337 V vs. SHE. The one-electron reduction of Cu* to Cu° via Cu* + e* -→ Cu° has Eº= +0.52 V vs. SHE . a) Based on the above, calculate E° for the one-electron reduction Cu2+ + 1e → Cu* . (Hint: While there are multiple ways to work this question, it might be easiest to write out the reactions, calculate free energies for the reactions as written, add or subtract equations (and free energies) as necessary, and then convert free energies back to potentials).

Chemistry
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
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Although we frequently consider Cu²+ and Cu° as the most stable forms of copper, the intermediate oxidation state Cu* plays a key role in many metalloenzymes of biological importance in biology. For example in the enzyme superoxide dismutase, Cu²* helps regulate the redox potential of the cell by reacting with the superoxide radical O2 (a potent reducing agent) to form Cu* as an intermediate. More broadly, intermediate oxidation states of transitions metals frequently play key roles in catalysis on both organic/biological and inorganic systems. The two-electron reduction of Cu2+ to Cu° via Cu²+ + 2e¯→ Cu° has Eº= +0.337 V vs. SHE. The one-electron reduction of Cu* to Cuº via Cu* + e* -→ Cu° has Eº= +0.52 V vs. SHE. a) Based on the above, calculate E° for the one-electron reduction Cu2+ + 1e → Cu* . (Hint: While there are multiple ways to work this question, it might be easiest to write out the reactions, calculate free energies for the reactions as written, add or subtract equations (and free energies) as necessary, and then convert free energies back to potentials). 4267 24 étv PDF
Transcribed Image Text:Although we frequently consider Cu²+ and Cu° as the most stable forms of copper, the intermediate oxidation state Cu* plays a key role in many metalloenzymes of biological importance in biology. For example in the enzyme superoxide dismutase, Cu²* helps regulate the redox potential of the cell by reacting with the superoxide radical O2 (a potent reducing agent) to form Cu* as an intermediate. More broadly, intermediate oxidation states of transitions metals frequently play key roles in catalysis on both organic/biological and inorganic systems. The two-electron reduction of Cu2+ to Cu° via Cu²+ + 2e¯→ Cu° has Eº= +0.337 V vs. SHE. The one-electron reduction of Cu* to Cuº via Cu* + e* -→ Cu° has Eº= +0.52 V vs. SHE. a) Based on the above, calculate E° for the one-electron reduction Cu2+ + 1e → Cu* . (Hint: While there are multiple ways to work this question, it might be easiest to write out the reactions, calculate free energies for the reactions as written, add or subtract equations (and free energies) as necessary, and then convert free energies back to potentials). 4267 24 étv PDF
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