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:

$$\ce{F2 (g) + 2e- &lt;=&gt; 2F- (aq) + $\pu{2.87V}$}$$

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:

$$\ce{2 F- (aq) &lt;=&gt; F2 (g) + 2 e-}$$

was $\pu{-2.87 V}$. Therefore, should $\ce{F- (aq)}$ not be a strong reducing agent?
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 $\Delta G &lt; 0$, a substance only oxidizes/reduces if the corresponding cell potential is positive.