Write the overall chemical reaction and calculate E° for the following reduction and oxidation half-reactions. 2H (aq) + 2e→ H2(g) E° = 0.00 V red Fe(s) → Fe2"(aq) + 2e E° +0.44 V %3D O Fe2 (aq) + H2(g) →Fe(s) + 2H*(aq) E° =+0.44 V O Fe(s) +2H"(ag) →F22*(aq) + H2(g) E° = -0.44 V O Fe2"(aq) + H2(g) →Fe(s) + 2H*(aq) E° = -0.44 V O Fe(s) +2H (aq) →FE2*(aq) + H2(g) E° =+0.44 V %3D
Write the overall chemical reaction and calculate E° for the following reduction and oxidation half-reactions. 2H (aq) + 2e→ H2(g) E° = 0.00 V red Fe(s) → Fe2"(aq) + 2e E° +0.44 V %3D O Fe2 (aq) + H2(g) →Fe(s) + 2H*(aq) E° =+0.44 V O Fe(s) +2H"(ag) →F22*(aq) + H2(g) E° = -0.44 V O Fe2"(aq) + H2(g) →Fe(s) + 2H*(aq) E° = -0.44 V O Fe(s) +2H (aq) →FE2*(aq) + H2(g) E° =+0.44 V %3D
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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![Write the overall chemical reaction and calculate E° for the following reduction and oxidation half-reactions.
2H (aq) + 2e H2(g)
E°
red
= 0.00 V
Fe(s) → Fe2"(aq) + 2e
E° =+0.44 V
OFF2 (aq) + H2(g) →Fe(s) + 2H"(aq) E° = +0.44 V
O Fe(s) + 2H"(aq)→FE2*(aq) + H2(g) E° = -0.44 V
OFE2"(ag) + H2(g)→Fe(s) + 2H*(aq) E° = -0.44 V
O Fe(s) +2H"(aq)→FE2"(aq) + H2(g) E° =+0.44 V](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F88d7f0e8-b5ca-4c28-98fd-8997c29f2013%2Fa3a36dfc-1e3a-4952-8cad-009021842daf%2Fxcupgtb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Write the overall chemical reaction and calculate E° for the following reduction and oxidation half-reactions.
2H (aq) + 2e H2(g)
E°
red
= 0.00 V
Fe(s) → Fe2"(aq) + 2e
E° =+0.44 V
OFF2 (aq) + H2(g) →Fe(s) + 2H"(aq) E° = +0.44 V
O Fe(s) + 2H"(aq)→FE2*(aq) + H2(g) E° = -0.44 V
OFE2"(ag) + H2(g)→Fe(s) + 2H*(aq) E° = -0.44 V
O Fe(s) +2H"(aq)→FE2"(aq) + H2(g) E° =+0.44 V
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