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uestion 28 The half-reaction occurring at the anode in the balanced reaction shown below is 3MNO 4 (aq) + 24H (aq) + 5Fe (5) → 3Mn 2+ (aq) + 5Fe 3+ (aq) + 12H 20 () O Fe (5) → Fe2+ (aq) + 2e O 2MN04 (aq) + 12H (aq) + 6e 2MN2+ (aq) + 3H20 (I) Fe2+ (aq) Fe3+ (aq) + e Fe (s) Fe3+ (aq) + 3e" O Mn04 (aq) + 8H* (aq) + 5e Mn* (aq) + 4H2O (I)

uestion 28 The half-reaction occurring at the anode in the balanced reaction shown below is 3MNO 4 (aq) + 24H (aq) + 5Fe (5) → 3Mn 2+ (aq) + 5Fe 3+ (aq) + 12H 20 () O Fe (5) → Fe2+ (aq) + 2e O 2MN04 (aq) + 12H (aq) + 6e 2MN2+ (aq) + 3H20 (I) Fe2+ (aq) Fe3+ (aq) + e Fe (s) Fe3+ (aq) + 3e" O Mn04 (aq) + 8H* (aq) + 5e Mn* (aq) + 4H2O (I)

Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter18: Electrochemistry
Section: Chapter Questions
Problem 18.101QE: At 298 K, the solubility product constant for PbC2O4 is 8.5 1010, and the standard reduction...
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> A Moving to the next question prevents changes to this answer. uestion 28 The half-reaction occurring at the anode in the balanced reaction shown below is 3MNO 4 (aq) + 24H * (aq) + 5Fe (s) → 3Mn 2+ 3+ (aq) + 5Fe (aq) + 12H 20 (O O Fe (s) → Fe2+ (aq) + 2e O 2MN04 (aq) + 12H (aq) + 6e 2MN2+ (aq) + 3H20 (1) Fe2+ (aq) – Fe3+ (aq) + e Fe (5) → Fe3+ (aq) + 3e" O Mn04 (aq) + 8H* (aq) + 5e → Mn2+ (aq) + 4H20 (1) A Moving to the next question prevents changes to this answer. 93 OType here to search
Transcribed Image Text:> A Moving to the next question prevents changes to this answer. uestion 28 The half-reaction occurring at the anode in the balanced reaction shown below is 3MNO 4 (aq) + 24H * (aq) + 5Fe (s) → 3Mn 2+ 3+ (aq) + 5Fe (aq) + 12H 20 (O O Fe (s) → Fe2+ (aq) + 2e O 2MN04 (aq) + 12H (aq) + 6e 2MN2+ (aq) + 3H20 (1) Fe2+ (aq) – Fe3+ (aq) + e Fe (5) → Fe3+ (aq) + 3e" O Mn04 (aq) + 8H* (aq) + 5e → Mn2+ (aq) + 4H20 (1) A Moving to the next question prevents changes to this answer. 93 OType here to search
Expert Solution
Step 1

We know that in the half cell reaction 

The reduction occurs at cathode and oxidation occurs at anode. 

Based on this we can say that which is going to oxides in the half cell reaction in the above chemical equation.we can see that iron is oxidizing to Iron(III) 

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