Given the following standard reduction potentials: MnO4 (aq) + 8H*(aq) + 5 e - - Mn2*(aq) + 4 H20() E° = 1.51 V Fe3+ (aq) + e Fe<* (aq) E° = 0.77 V What is the cell potential E for the following reaction under the non-standard condition with 0.16 M of MnO4", 0.92 M of Mn2*, 0.96 M of Fe+, and 0.19 M of Fe2* in a buffered solution with 1.0 M of H* ? Please keep your answer to two decimal places. 5FE2* (aq) + MnO4 (aq) + 8H*(aq) (aq) + Mn2+ (aq) + 4H20 () --> 5FE3+
Given the following standard reduction potentials: MnO4 (aq) + 8H*(aq) + 5 e - - Mn2*(aq) + 4 H20() E° = 1.51 V Fe3+ (aq) + e Fe<* (aq) E° = 0.77 V What is the cell potential E for the following reaction under the non-standard condition with 0.16 M of MnO4", 0.92 M of Mn2*, 0.96 M of Fe+, and 0.19 M of Fe2* in a buffered solution with 1.0 M of H* ? Please keep your answer to two decimal places. 5FE2* (aq) + MnO4 (aq) + 8H*(aq) (aq) + Mn2+ (aq) + 4H20 () --> 5FE3+
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
![Given the following standard reduction potentials:
Mn04 (aq) +
8H*(aq) + 5 e
Mn2*(aq) +
4 H20()
E° = 1,51 V
Fe3+
(aq) + e –
Fe2* (aq)
E° = 0.77 V
What is the cell potential E for the following reaction under the non-standard condition with 0.16 M of MnO4", 0.92 M of Mn2+, 0.96 M of
Fe+, and 0.19 M of Fe2+ in a buffered solution with 1.0 M of H* ? Please keep your answer to two decimal places.
5FE2* (aq) + MnO4 (aq) + 8H*(aq)
5Fe3+ (ag) + Mn2* (aq) + 4H20 ()
-->](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3e9c522d-2490-4e14-b1ec-18afc9d188f8%2F77fb8d25-99f1-4de0-97de-4614e2a50114%2Fq74fdmj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Given the following standard reduction potentials:
Mn04 (aq) +
8H*(aq) + 5 e
Mn2*(aq) +
4 H20()
E° = 1,51 V
Fe3+
(aq) + e –
Fe2* (aq)
E° = 0.77 V
What is the cell potential E for the following reaction under the non-standard condition with 0.16 M of MnO4", 0.92 M of Mn2+, 0.96 M of
Fe+, and 0.19 M of Fe2+ in a buffered solution with 1.0 M of H* ? Please keep your answer to two decimal places.
5FE2* (aq) + MnO4 (aq) + 8H*(aq)
5Fe3+ (ag) + Mn2* (aq) + 4H20 ()
-->
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