Why is I- a stronger reducing agent than F-?
2F^- --> F2 + 2e -2.87
2I^- ==> I2 + 2e -0.535
But somehow I don't think this is the answer your are looking for. As to WHY the potentials are that way-- I suppose that's the nature of the beast. F^- being such a small ion isn't likely to give up electrons. Those electrons on I^- are farther out because of the size and more easily given up.
Interesting. Thank you.
In order to understand why I- (iodide ion) is a stronger reducing agent than F- (fluoride ion), we need to consider their reduction potentials. Reduction potential is a measure of the tendency of a species to gain electrons and undergo reduction.
To determine the relative strength of reducing agents, we compare their reduction potentials. A more negative reduction potential indicates a stronger reducing agent.
To find the reduction potentials, we can refer to standard reference tables such as the Standard Reduction Potential table. This table lists reduction potentials for various half-cell reactions, including the reduction of different species.
For the reduction of F- and I-, we look for their respective half-cell reactions:
F2(g) + 2e- → 2F-(aq) (1)
I2(g) + 2e- → 2I-(aq) (2)
From these reactions, we can observe that F- is formed from F2, while I- is formed from I2, with the gain of two electrons in both cases.
By looking up the standard reduction potentials for these reactions, we find that the reduction potential for reaction (1) is higher than that of reaction (2). In other words, F- has a less negative reduction potential compared to I-.
This means that F- is a weaker reducing agent compared to I-. F- is less likely to donate electrons because it has a higher tendency to be reduced itself. On the other hand, I- has a stronger tendency to donate electrons and undergo reduction, thus making it a stronger reducing agent.
So, in summary, I- is a stronger reducing agent than F- because it has a more negative reduction potential, indicating a higher tendency to donate electrons and undergo reduction.