Question

Relation between electrode potential hydration enthaply and enthaply of atomisation

Answer 1

Well, reduction potentials also depend on enthalpy of hydration, sublimation and stuff.

Think of an enthalpy cycle (Hess's Law assuming standard conditions).

The first step will be enthalpy of atomisation (∆H_atom)

Solid metal → gaseous metal atoms

The second step would be to find the ionisation energy (IE),

Gaseous metal atoms → gaseous metal ions

Then you find enthalpy of hydration (∆H_hydration).

Gaseous metal ions → gaseous metal ions in solution

The electrode potential is the net sum of the energy of these changes. It can be measured in a hydrogen cell.

2H+(aq) + M(s) reversible M2+(aq) + H2 (g)

By calculating ∆H of atomisation we can actually see how stable the oxidation state of that particular transition element is in aqueous solution. The smaller the value of energy change for that particular oxidation state, he greater its stability.

Transition elements generally have large enthalpies of atomisation and ionisation energies. Therefore they are very stable.

Answer 2

The standard electrode potential of a reaction (Eⁿ) is related to the Gibbs free energy change (ΔGⁿ) and the enthalpy change (ΔHⁿ) by:  

1. Gⁿ = -nFEⁿ = ΔH-TΔSⁿ

For example, The reaction of metallic lithium with water is more exothermic than any other alkali metal (many times more exothermic for a given mass or volume), but it is less vigorous for kinetic reasons. The enthalpy change of atomisation is twice as high for lithium than for potassium, indicating stronger bonding in the metal.

{you can write ⁿ = powerθ (note that it will be in power)  dont written bcz there is no power theta}

HOPE IT WAS HELPFUL