(a) Take [Fe2+] = [Fe3+] = 10-2 mol/litre. Calculate logPO2 at pH = 1 for the following reactions:(i)(i)2Fe + 4H+ + O2 2FE2+ + 2H204FE2+ +4H* + O2 → 4FE3+ + 2H2OData: R = 8.314Jmol-1K-1, T = 298KGibbs energy of formation data.AGr (KJ mol1)SpeciesFe (s)Fe2+ (aq)Fe3- (aq)H2O (1)H*(aq)Ог (9)-85.0-10.5- 237

GIVEN: Fe2+ = Fe3+ = 10-2 mol/L pH= 1 R=8.314 J/mol K Temperature = 25oC = 298K To Calculate: log…

(a) Take [Fe2-] = [Fe3+] = 102 mol/litre. Calculate logPO2 at pH = 1 for the following reactions: (i) (i) %3D 2Fe + 4H+ + O2 2FE2+ + 2H2O 4FE2+ +4H* + O2 4FE3+ + 2H20 Data: R = 8.314Jmol-1K-1, T = 298K

(a) Take [Fe2-] = [Fe3+] = 102 mol/litre. Calculate logPO2 at pH = 1 for the following reactions: (i) (i) %3D 2Fe + 4H+ + O2 2FE2+ + 2H2O 4FE2+ +4H* + O2 4FE3+ + 2H20 Data: R = 8.314Jmol-1K-1, T = 298K

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

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Question

(a) Take [Fe2+] = [Fe3+] = 10-2 mol/litre. Calculate logPO2 at pH = 1 for the following reactions:
(i)
(i)
2Fe + 4H+ + O2 2FE2+ + 2H20
4FE2+ +4H* + O2 → 4FE3+ + 2H2O
Data: R = 8.314Jmol-1K-1, T = 298K
Gibbs energy of formation data.
AGr (KJ mol1)
Species
Fe (s)
Fe2+ (aq)
Fe3- (aq)
H2O (1)
H*(aq)
Ог (9)
-85.0
-10.5
- 237

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