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

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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Question

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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