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8. The following redox reaction occurs at 25.0°C: 5F22*(aq) + MnO4 (aq) + 8H*(aq) → Mn2 (aq) + 4H20(1) + 5FE³*(aq) a. Calculate the E°cell for the redox reaction. Half Reaction E(V) MnOa (aq) + 8H*(aq) + 5e → Mn²*(aq) + 4H20(I) Fe3 (aq) + e → Fe2*(aq) +1.51 +0.77 b. Calculate the AG° in kJ/mol at 25.0°C.

8. The following redox reaction occurs at 25.0°C: 5F22*(aq) + MnO4 (aq) + 8H*(aq) → Mn2 (aq) + 4H20(1) + 5FE³*(aq) a. Calculate the E°cell for the redox reaction. Half Reaction E(V) MnOa (aq) + 8H*(aq) + 5e → Mn²*(aq) + 4H20(I) Fe3 (aq) + e → Fe2*(aq) +1.51 +0.77 b. Calculate the AG° in kJ/mol at 25.0°C.

General Chemistry - Standalone book (MindTap Course List)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter19: Electrochemistry
Section: Chapter Questions
Problem 19.140QP
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8. The following redox reaction occurs at 25.0°C: 5F22*(aq) + MnO4 (aq) + 8H*(aq) → Mn²*(aq) + 4H2O(1) + 5FE³*(aq) a. Calculate the E°cell for the redox reaction. Half Reaction E(V) MnO4 (aq) + 8H*(aq) + 5e¨ → Mn²*(aq) + 4H2O(1I) Fe3 (aq) + e → Fe2*(aq) +1.51 +0.77 b. Calculate the AG° in kJ/mol at 25.0°C. Calculate the AG when the following concentrations are used at 25.0°C. [Fe?*] = 0.200 M [MnO4] = 0.440M [H*] = 0.200 M [Mn2*] = 0.750 M [Fe3*] = 0.850 M с. %3D %3D %3D
Transcribed Image Text:8. The following redox reaction occurs at 25.0°C: 5F22*(aq) + MnO4 (aq) + 8H*(aq) → Mn²*(aq) + 4H2O(1) + 5FE³*(aq) a. Calculate the E°cell for the redox reaction. Half Reaction E(V) MnO4 (aq) + 8H*(aq) + 5e¨ → Mn²*(aq) + 4H2O(1I) Fe3 (aq) + e → Fe2*(aq) +1.51 +0.77 b. Calculate the AG° in kJ/mol at 25.0°C. Calculate the AG when the following concentrations are used at 25.0°C. [Fe?*] = 0.200 M [MnO4] = 0.440M [H*] = 0.200 M [Mn2*] = 0.750 M [Fe3*] = 0.850 M с. %3D %3D %3D
Expert Solution
Introduction

Electrochemistry is the branch of chemistry which studies the change of chemical energy into electrical energy during a chemical energy. Standard electrode potential is the potential of the individual electrode which can be determined experimentally. It is represented by. Its unit is Volt (V).

Given data

EMnO42-/Mn2+∘=+1.51 VEFe3+/Fe2+∘=+0.77 V

Faraday’s constant, F = 96500 C mol-1

No. of electrons, n = 5

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