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8 H* + 5 Fe²+ + MnO4¯ →5 Fe3+ + Mn²+ + 4 H₂O Answer the following questions: Question 4: The cathode half reaction is 5e¯ + 8H+ + MnO4 --> Mn2+ + 4H₂O 5Fe2+ --> 5Fe3+ + 5e Mn2+ + 4H2O --> 5e¯ + 8H+ + MnO4 5Fe3+ + 5e --> 5Fe2+ Check Reuse <> Embed Question 5: The anode half reaction is 5Fe3+ + 5e --> 5Fe²+ 5e¯ + 8H+ + MnO4 --> Mn2+ + 4H₂O Mn2+ + 4H2O --> 5e* + 8H+ + MnO4 5Fe2+ --> 5Fe3+ + 50* Check Reuse <> Embed H-P H-P

8 H* + 5 Fe²+ + MnO4¯ →5 Fe3+ + Mn²+ + 4 H₂O Answer the following questions: Question 4: The cathode half reaction is 5e¯ + 8H+ + MnO4 --> Mn2+ + 4H₂O 5Fe2+ --> 5Fe3+ + 5e Mn2+ + 4H2O --> 5e¯ + 8H+ + MnO4 5Fe3+ + 5e --> 5Fe2+ Check Reuse <> Embed Question 5: The anode half reaction is 5Fe3+ + 5e --> 5Fe²+ 5e¯ + 8H+ + MnO4 --> Mn2+ + 4H₂O Mn2+ + 4H2O --> 5e* + 8H+ + MnO4 5Fe2+ --> 5Fe3+ + 50* Check Reuse <> Embed H-P H-P

Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter17: Principles Of Chemical Reactivity: Other Aspects Of Aqueous Equilibria
Section17.6: Equilibria Involving Complex Ions
Problem 2.2ACP: A typical total phosphate concentration in a cell, [HPO42] + [H2PO4], is 2.0 102 M. What are the...
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8 H* + 5 Fe²+ + MnO4¯ →5 Fe3+ + Mn²+ + 4 H₂O Answer the following questions: Question 4: The cathode half reaction is 5e¯ + 8H+ + MnO4 --> Mn2+ + 4H₂O 5Fe2+ --> 5Fe3+ + 5e Mn2+ + 4H2O --> 5e¯ + 8H+ + MnO4 5Fe3+ + 5e --> 5Fe2+ Check Reuse <> Embed Question 5: The anode half reaction is 5Fe3+ + 5e --> 5Fe²+ 5e¯ + 8H+ + MnO4 --> Mn2+ + 4H₂O Mn2+ + 4H2O --> 5e* + 8H+ + MnO4 5Fe2+ --> 5Fe3+ + 50* Check Reuse <> Embed H-P H-P
Transcribed Image Text:8 H* + 5 Fe²+ + MnO4¯ →5 Fe3+ + Mn²+ + 4 H₂O Answer the following questions: Question 4: The cathode half reaction is 5e¯ + 8H+ + MnO4 --> Mn2+ + 4H₂O 5Fe2+ --> 5Fe3+ + 5e Mn2+ + 4H2O --> 5e¯ + 8H+ + MnO4 5Fe3+ + 5e --> 5Fe2+ Check Reuse <> Embed Question 5: The anode half reaction is 5Fe3+ + 5e --> 5Fe²+ 5e¯ + 8H+ + MnO4 --> Mn2+ + 4H₂O Mn2+ + 4H2O --> 5e* + 8H+ + MnO4 5Fe2+ --> 5Fe3+ + 50* Check Reuse <> Embed H-P H-P
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