Final answer:
Fluorine (F2) is the strongest oxidizing agent with an E° value of 2.87 V, indicating its exceptional capacity to accept electrons and oxidize other substances. The correct answer is A. Fluorine; F2 (2e^– → 2F^–), E° = 2.87 V.
Step-by-step explanation:
The strongest oxidizing agent among the options provided is Fluorine; F2 (2e– → 2F–), E° = 2.87 V. This is evidenced by its very high standard electrode potential (E°) value compared to the other options given.
Lithium has a negative standard electrode potential, signifying that it is a reducing agent rather than an oxidizing one, while Hydrogen, with an E° = 0.00 V, is the standard reference but not as strong an oxidizing agent as fluorine.
Fluorine is not only the most powerful oxidizing agent of the known elements, it reacts directly and vigorously with most other substances, including elements to its left in the periodic table and even with some noble gases under certain conditions. A practical example of its strength is how it can oxidize bromide ions to molecular bromine, with an E° = 1.09 V.
It's essential to note that the principle guiding the strength of an oxidizing agent is its ability to gain electrons (be reduced) - the more positive the standard electrode potential, the stronger the oxidizing agent.