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 $$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$ and obtaining $$\Delta G^\circ$$ to substitute in $$\Delta G^\circ = -n\mathcal{F}E^\circ$$ 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)

• You might find this helpful: K. Schmidt-Rohr, "How Batteries Store and Release Energy: Explaining Basic Electrochemistry", Journal of Chemical Education, 95 (2018) 1801-1810. It addresses some of the issues you raise, but is not for total novices.
– Ed V
Mar 19, 2020 at 17:07

In a thought experiment, you can substitute the electrochemical half reaction by three steps:

1. Removing atoms from the electrode into the gas phase (sublimation)
2. Removing electrons from the atoms (ionization)
3. 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.

Copper has the positive standard electrode potential because when we use copper as an electrode. copper would not convert into copper + 2 ion because the energy which is needed to convert copper to the copper + 2 iron is not full filled by the hydration energy, which is ivolve when we emerged copper into the electrolytic solution. so that's why copper will not converted into copper + 2 because of this less hydration energy it does not came into the copper + 2 ion it means copper will not give electron and converted into positive + 2 ion, as it not give electron so it will not have negative electrode potential because only such elements which can donate electrons for example Lithium or sodium or potassium can have (-)electorde pottential.thus the only option which is left for the copper is for having the positive standard electrode potential as it could not give electrons and converted into positive ion, (only due to low hydration energy)

the root behind this is due to the low hydration energy. Only due to this low hydration energy electron does not convert into copper + 2 and due to which it does not give electrons and due to which it does not have negative electrode potential and due to which the copper has the positive standard electrode potential.

First, I present an example, which provides a quantitative description of how the hydration enthalpy could impact the electrode potential. In particular, select salts can undergo hydrolysis resulting in the liberation of H+, for example with copper or iron:

$$\ce{Fe[(H2O)6](3+) (aq) + H2O (l) = [Fe(H2O)5(OH)](2+) (aq) + H3O+ (aq)}$$

Reference: one of many possible links .

This produces a pH effect, which per a source on the question: 'What is the equation that connects pH and its effect on electric potential of an electrochemical cell?', the answer employs the Nernst equation to derive an explanation. I will quote but a small summary part of the results relating to acidic conditions:

As the pH decreases, the solution is more acidic, so 10−pH=[H+] increases and 10pH−14=[OH−] decreases.

• If H+ is a product, Q therefore increases, and the nonstandard cell potential decreases.
• If H+ is a reactant, Q therefore decreases, and the nonstandard cell potential increases.

Next, with respect to the impact of sublimation (and relatedly, ionization), I note that this can be energy (and especially temperature) related. Quoting a reference:

As you supply heat (or any form of energy) to the system containing the atom, it's kinetic energy will increase and it will move faster through space. ... Continue to supply even more energy and more electrons will be removed from the atom.

On the connection of temperature to the electrode potential, I cite an answer from The Student Room:

Increasing the temperature AFFECTS electrode potentials, not increases them.

The electrode potential is a measure of the extent of a redox process - negative and the equilibrium lies relatively to the LHS (compared to the standard hydrogen electrode)

example: $$\ce{Zn(2+)(aq) + 2e- -> Zn(s)}$$

The process is not quite as simple as it appears. To move from right to left the equilibrium has to have the zinc atomised and then ionised and then the ions solvated. Each of these processes is either exo or endothermic and is affected by the temperature. The equilibrium between the ions and the pure metal is also affected by the temperature. If the equilibrium is exothermic overall from left to right then an increase in the temperature will make the equilibrium move more to the left hand side making the electrode potential more negative. The reverse is also true - if the equilibrium is endothermic overall from left to right then increasing the temperature makes the electrode potential more positive.

Note, both provided answers, are in effect, very situational, with, for example, is the removal of atoms via sublimation an exothermic or endothermic process.