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6. Suppose you wish to separate Cd²+ from Zn²+ by electrodeposition of metallic cadmium onto an inert cathode from an aqueous solution containing 0.015 M Cd(NH3)4²+, 0.010 M Zn(NH3)4²+, and 1.00 M uncomplexed NH3. You want to deposit 99.99% of the Cd while not depositing any Zn. What is the allowable range of potentials that you can use? Cd(NH3)4²+ + 2 e Cd(s) + 4 NH3 Zn(NH3)4²+ + 2e = Zn(s) + 4 NH3 E = -0.61 V vs. SHE E = -1.04 V vs. SHE

6. Suppose you wish to separate Cd²+ from Zn²+ by electrodeposition of metallic cadmium onto an inert cathode from an aqueous solution containing 0.015 M Cd(NH3)4²+, 0.010 M Zn(NH3)4²+, and 1.00 M uncomplexed NH3. You want to deposit 99.99% of the Cd while not depositing any Zn. What is the allowable range of potentials that you can use? Cd(NH3)4²+ + 2 e Cd(s) + 4 NH3 Zn(NH3)4²+ + 2e = Zn(s) + 4 NH3 E = -0.61 V vs. SHE E = -1.04 V vs. SHE

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
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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6. Suppose you wish to separate Cd²+ from Zn²+ by electrodeposition of metallic cadmium onto an inert cathode from an aqueous solution containing 0.015 M Cd(NH3)4²+, 0.010 M Zn(NH3)4²+, and 1.00 M uncomplexed NH3. You want to deposit 99.99% of the Cd while not depositing any Zn. What is the allowable range of potentials that you can use? Cd(NH3)4²+ + 2e = Cd(s) + 4 NH3 Zn(NH3)4²+ + 2e → Zn(s) + 4 NH3 E = -0.61 V vs. SHE E° -1.04 V vs. SHE
Transcribed Image Text:6. Suppose you wish to separate Cd²+ from Zn²+ by electrodeposition of metallic cadmium onto an inert cathode from an aqueous solution containing 0.015 M Cd(NH3)4²+, 0.010 M Zn(NH3)4²+, and 1.00 M uncomplexed NH3. You want to deposit 99.99% of the Cd while not depositing any Zn. What is the allowable range of potentials that you can use? Cd(NH3)4²+ + 2e = Cd(s) + 4 NH3 Zn(NH3)4²+ + 2e → Zn(s) + 4 NH3 E = -0.61 V vs. SHE E° -1.04 V vs. SHE
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