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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?

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    $\begingroup$ You're always looking for a positive potential as an indicator of spontaneity. $-2.87\ \mathrm{V}$ is not strong... $\endgroup$
    – Zhe
    Jun 19, 2018 at 17:57

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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.

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