Determine the cell potential for an electrochemical cell based on the following two half-reactions (assume T = 298K): Oxidation: Ni() → Ni²+ (aq, 0.050 M) Reduction: VO2(09, 1.909 M) +2 e E=0.23 V (aq, 1 M) + 2 H(aq, 1.818 M) + e- → VO²+ + H₂O(1) Ed= 1.00 V
Determine the cell potential for an electrochemical cell based on the following two half-reactions (assume T = 298K): Oxidation: Ni() → Ni²+ (aq, 0.050 M) Reduction: VO2(09, 1.909 M) +2 e E=0.23 V (aq, 1 M) + 2 H(aq, 1.818 M) + e- → VO²+ + H₂O(1) Ed= 1.00 V
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter17: Electrochemistry And Its Applications
Section: Chapter Questions
Problem 3QRT
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![Determine the cell potential for an electrochemical cell based on the following two half-reactions (assume
T = 298K):
Oxidation: Ni(s) → Ni
Reduction:
VO
(aq. 0.050 M) + 2e E=0.23 V
2(ag, 1.909 M)
+ 2 H+
(aq, 1 M)
(aq, 1.818 M) + e → VO²+
+ H₂O(1) Ed = 1.00 V](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F26b0b5a6-fa48-416c-ac4a-b5028772bd0b%2Fb2ec00aa-f137-4141-a1a9-7c091bfa198a%2Fua51thd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Determine the cell potential for an electrochemical cell based on the following two half-reactions (assume
T = 298K):
Oxidation: Ni(s) → Ni
Reduction:
VO
(aq. 0.050 M) + 2e E=0.23 V
2(ag, 1.909 M)
+ 2 H+
(aq, 1 M)
(aq, 1.818 M) + e → VO²+
+ H₂O(1) Ed = 1.00 V
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