Why is the conjugate of a strong oxidizing agent a weak reducing agent?

I am a grade 12 student. While I was reviewing some electrochemistry, I had the following question. Take this reduction half-reaction:

$$\ce{F2 (g) + 2e- <=> 2F- (aq) + \pu{2.87V}}$$

I know that $\ce{F2 (g)}$ is a strong oxidizing agent, but why does this make $\ce{F- {(aq)}}$ a weak reducing agent? I thought that since the electrical potential of the reduction half-reaction:

$$\ce{2 F- (aq) <=> F2 (g) + 2 e-}$$

was $\pu{-2.87 V}$. Therefore, should $\ce{F- (aq)}$ not be a strong reducing agent?

• You're always looking for a positive potential as an indicator of spontaneity. $-2.87\ \mathrm{V}$ is not strong... – Zhe Jun 19 '18 at 17:57

It is not the magnitude of the standard cell potential that indicates whether or not the cell reaction will be spontaneous, but the sign. In the same way a reaction is only spontaneous if $$\Delta G < 0$$, a substance only oxidizes/reduces if the corresponding cell potential is positive.