[TP] If you have two half-cells: Ag+ (1M) | Ag(s) with Eºred = 0.80 V and Cu²+(1M) | Cu(s) with Eºred = 0.34 V, which will be oxidized? =( Write the cell diagram that is spontaneous. E=E° 个 61% 1. Cu is oxidized: Cu(s) | Cu²+(1 M)|| Ag+(1 M) | Ag(s) 12% 2. Cu(s) | Cu²+(1 M)||Ag+ (1 M)|Ag(s) Ag is oxidized: Cu is oxidized: 15% 4. Ag is oxidized: 12% 3. Agts)|Ag (1 M) | | Cu² (1 M) | Cu(s) Ag(s) | Ag+ (1 M)||Cu²+(1 M) | Cu(s) E cell = Eox Ulu | (₂²4) + Ead (Ag+ / Ag) 27/₁1) ²0ut (0₂)
[TP] If you have two half-cells: Ag+ (1M) | Ag(s) with Eºred = 0.80 V and Cu²+(1M) | Cu(s) with Eºred = 0.34 V, which will be oxidized? =( Write the cell diagram that is spontaneous. E=E° 个 61% 1. Cu is oxidized: Cu(s) | Cu²+(1 M)|| Ag+(1 M) | Ag(s) 12% 2. Cu(s) | Cu²+(1 M)||Ag+ (1 M)|Ag(s) Ag is oxidized: Cu is oxidized: 15% 4. Ag is oxidized: 12% 3. Agts)|Ag (1 M) | | Cu² (1 M) | Cu(s) Ag(s) | Ag+ (1 M)||Cu²+(1 M) | Cu(s) E cell = Eox Ulu | (₂²4) + Ead (Ag+ / Ag) 27/₁1) ²0ut (0₂)
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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This is an image from my lecture. I am confused why Cu is being oxidized, in the formula why did Cu be converted to Eooxidation? When both values given were Eoreduction?
![red
[TP] If you have two half-cells: Ag+(1M) | Ag(s) with Eº = 0.80 V
Q=1
and Cu²+(1M) | Cu(s) with Eºred = 0.34 V, which will be oxidized? -
Write the cell diagram that is spontaneous.
E=E°
61% 1. Cu is oxidized: Cu(s) | Cu²+(1 M)|| Ag+(1 M) | Ag(s)
12% 2.
Ag is oxidized:
Cu(s) | Cu²+(1 M)||Ag+(1 M)|Ag(s)
12% 3.
Cu is oxidized:
Agts)|Ag (1 M)||Cu² (1 M)| Cu(s)
Ag(s) | Ag+(1 M)||Cu²+(1 M) | Cu(s)
15% 4. Ag is oxidized:
É cell = Eox
= Eox (lul lu²y) + Fad (Ag+lay)
Eind ((²³7) (u) + End (Ay+ /Ag)
=-
= -0.34V + 0.80V =0.46 U
165 of 702](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc61ac136-beb0-491a-99b0-c53254daee6a%2F0d3d6581-98da-49d4-b56f-c5eddda43755%2F9hp80z4_processed.png&w=3840&q=75)
Transcribed Image Text:red
[TP] If you have two half-cells: Ag+(1M) | Ag(s) with Eº = 0.80 V
Q=1
and Cu²+(1M) | Cu(s) with Eºred = 0.34 V, which will be oxidized? -
Write the cell diagram that is spontaneous.
E=E°
61% 1. Cu is oxidized: Cu(s) | Cu²+(1 M)|| Ag+(1 M) | Ag(s)
12% 2.
Ag is oxidized:
Cu(s) | Cu²+(1 M)||Ag+(1 M)|Ag(s)
12% 3.
Cu is oxidized:
Agts)|Ag (1 M)||Cu² (1 M)| Cu(s)
Ag(s) | Ag+(1 M)||Cu²+(1 M) | Cu(s)
15% 4. Ag is oxidized:
É cell = Eox
= Eox (lul lu²y) + Fad (Ag+lay)
Eind ((²³7) (u) + End (Ay+ /Ag)
=-
= -0.34V + 0.80V =0.46 U
165 of 702
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