Fluorine has very high charge/mass ratio as it is very small in size, its electron gain enthalpy is very high. Oxidizing agent pulls the electron cloud of the substance being oxidized towards itself, for example:

$$\ce{F2 + 2 X- → 2F^- + X2}\qquad (\ce{X} = \ce{Cl}, \ce{Br}, \ce{I})$$

Even in this post it is mentioned that electron cloud is repelled.

Shouldn’t the electron cloud of fluorine atom repel the electron cloud? It seems paradoxical the fluorine pulls electron cloud towards itself so easily.

Also, many sources state that fluorine

1.Has low bond dissociation enthalpy and high hydration enthalpy due to which fluorine is a good oxidizing agent.

2.standard electrode potential for fluorine is the highest +2.87 E-/V, this indicates maximum tendency of F2 gas to 2F- (fluoride ions)

My questions

1.What has hydration enthalpy and bond dissociation enthalpy got to do with oxidizing power?

As lithium is the best reducing agent , and also has high hydration enthalpy , we can’t say it is a oxidising agent

2.What am I missing here? What is the reason for such high standard electrode potential for fluorine ?

  • $\begingroup$ look up Born-Haber cycle, hydration energy effects will make sense. $\endgroup$ – William R. Ebenezer Mar 10 at 17:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.