Why does electrode potential depend upon enthalpy of hydration and sublimation? It's given in text that "Electrode potential among other things depends upon: enthalpy of sublimation,ionization enthalpy and hydration enthalpy." I could understand why it would depend upon ionization enthalpy as less is the IE easily we could remove the electron thereby reducing the other reactant. But I couldn't figure out why the other two enthalpy comes. But after some thinking I thought maybe we are adding the enthalpy of the three with sign and substituting in ∆G°=∆H°-T∆S° and obtaining ∆G° to substitute in ∆G°=-nFE°cell to find E°cell value. I couldn't think further clearly after this. Could you help me? ( I'm trying to understand why lithium because of its high hydration enthalpy overweighs its high ionization enthalpy and becomes the strongest reducing agent in aqueous solution)
In a thought experiment, you can substitute the electrochemical half reaction by three steps:


Removing atoms from the electrode into the gas phase (sublimation)
Removing electrons from the atoms (ionization)
Solvating the resulting ions


If you have quantitative descriptions of these processes, you get a quantitative description of what is actually going on. This is because many thermodynamic functions (such a enthalpy) are functions of state, i.e. they don't depend on the path from A to B, just on the state of A and B.

I would have said these quantities are related rather than one depends on the other. There is really no causality here.