Question

The following equations are half reactions and reduction potentials.

Li+ (aq) + e– mc009-1.jpg Li(s) has a reduction potential of –3.04 V
F2(g) + 2e– mc009-2.jpg 2F–(aq) has a reduction potential of +2.87 V

Now consider lithium (LI+) and fluoride (F2) as oxidizing agents. How do these compare as oxidizing agents?
Lithium is a stronger oxidizing agent than fluoride.
Fluoride is a stronger oxidizing agent than lithium.
Fluoride and lithium have the same oxidizing strength.
Reduction potential and oxidizing/reducing strength are unrelated.

Answer 1

Fluorine (F2) is the stronger oxidizing agent. A more positive reduction potential means a greater tendency to be reduced (i.e., to oxidize something else): +2.87 V for F2 vs −3.04 V for Li+, so F2 is the much stronger oxidizer.