[CentralFC]

Hi everyone! I'm new to this forum and I came with hopes of having a random chemistry-related question answered.

I'm in Chemistry-IB and after working on an anode/cathode assignment, I started wondering why Lithium has a higher ease of oxidation than Francium. It would make sense for Francium to give up electrons easier than Lithium because of its greater atomic radius, resulting in a weaker grasp of it's valence electrons; however, Lithium has a higher ease of oxidation according to my textbook and "activity series" (sheet with random information).

Does anyone have a guess to why this is? If so, thanks in advance!

[DevaDevil]

Standard reduction potentials:
Li⁺ + e^- <--> Li;  E⁰ = -3.02V
Na⁺ + e^- <--> Na;  E⁰ = -2.71V
K⁺ + e^- <--> K;  E⁰ = -2.92V
Rb⁺ + e^- <--> Rb;  E⁰ = -2.92V
Cs⁺ + e^- <--> Cs;  E⁰ = -2.92V

Fr⁺ + e^- <--> Fr;  E⁰ =estimated -2.92V
Francium is very radioactive, and I do not know of anyone who isolated the stable element yet.

The difference is small, but indeed present. Lithium oxidises slightly easier than any of the larger alkali metals.
One of the reasons (likely) in watery solutions is the higher hydration energy of Lithium (hydration energy increases with smaller atomic radius).


now if you are talking in ionization energies (releasing an electron in vacuum), then lithium has a far higher ionization energy due to its smaller radius. I do not know francium's E_i (radioactive), but assume it would be much lower than Li (see for example Li E_i = 520 kJ/mol; Cs E_i = 376 kJ/mol), thus in vacuum I would expect your reasoning to be true and Litium to lose its valence electron less easily.