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

B. Fluoride is a stronger oxidizing agent than lithium.

Answer 2

The higher the reduction potential of a substance, the greater is its tendency to become reduced.
Moreover, an oxidizing agent is a chemical substance that oxidizes another chemical substance while the oxidizing agent itself is reduced. Therefore, a good good oxidizing agent has a high tendency to be reduced, meaning it has a high reduction potential. We can now conclude that
Fluoride is a stronger oxidizing agent than lithium.