Half-Reactions: Find Strongest Oxidizing Agent

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SUMMARY

The strongest oxidizing agent among the half-reactions discussed is Cd2+, as confirmed by the standard reduction potentials. The half-reaction for fluorine, 2F- → F2 + 2e-, has a potential of -2.87 V and represents a reducing agent, not an oxidizing agent. The confusion arises from the need to interpret half-reactions correctly, where the oxidizing agent is identified by its ability to gain electrons, not by the reduction potential of its corresponding reducing agent. Therefore, the assertion that F- is the strongest oxidizing agent is incorrect.

PREREQUISITES
  • Understanding of electrochemistry concepts, specifically oxidation and reduction.
  • Familiarity with standard reduction potentials and how to interpret them.
  • Knowledge of half-reaction notation and its implications in redox reactions.
  • Ability to differentiate between oxidizing and reducing agents based on half-reactions.
NEXT STEPS
  • Study standard reduction potentials for various half-reactions in electrochemistry.
  • Learn how to construct and interpret half-reaction equations accurately.
  • Explore the concept of Gibbs free energy in relation to redox reactions.
  • Investigate common misconceptions in identifying oxidizing and reducing agents.
USEFUL FOR

Chemistry students, educators, and anyone studying electrochemistry or redox reactions will benefit from this discussion, particularly those seeking to clarify the concepts of oxidizing and reducing agents.

brake4country
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Homework Statement


Based only on the half-reactions in the table, which of the following is the strongest oxidizing agent?
(See attachment)

Homework Equations

The Attempt at a Solution


I noticed in this table that not all elements are in the reduction form. I flipped them around and changed the sign, however, the correct answer states that Cd2+ is the correct answer. Shouldn't it be F-? Thanks in advance.
 

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brake4country said:
F-?
... an oxidizing agent?
 
Bystander said:
... an oxidizing agent?
Ok, I typed that wrong. According to the chart, we know that the the half reaction: 2F-→F2 + 2e- has potential of -2.87 V. But this is being oxidized. Don't we have to flip it and change it to the reduction half reaction to compare it to the rest of the agents listed?
 
No. F/F2 is an oxidizing agent; F- is a reducing agent.
 
Right, so wouldn't so wouldn't 2F-→F2+ 2e- be the strongest oxidizing agent and not Cd2++2e-→Cd? I think the book is wrong in this one.
 
NO. The half reactions are written with both species; you are not being asked to pick a half reaction, just a single species.
 

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