These are physical properties of an element
(A) Sublimation enthalpy
(B) Ionisation enthalpy
(C) Hydration enthalpy
(D) Electron gain enthalpy
The total number of above properties that affect the reduction potential is ____________ (Integer answer)

Solution

Chemistry ma’am · 3 steps

Break the electrode reaction into gas-phase steps and see which enthalpies actually appear.

1. Build the cycle. For a metal electrode, think of the oxidation M(s) → M⁺(aq) + e⁻ as three steps. The reduction potential is set by the total energy of this path, run in reverse.

   M(s) →(Δ_(sub)H) M(g) →(ΔᵢH) M⁺(g) →(Δ_(hyd)H) M⁺(aq)

   Property	Step it controls	Affects reduction potential?
   (A) Sublimation enthalpy	solid metal to gaseous atoms	Yes
   (B) Ionisation enthalpy	gaseous atom to gaseous cation	Yes
   (C) Hydration enthalpy	gaseous cation to aqueous cation	Yes
   (D) Electron gain enthalpy	forming an anion, which never happens in this path	No

2. Why not D. A metal electrode forms a cation, never an anion. Electron gain enthalpy describes adding an electron to a gaseous atom, and that step does not occur here.
3. So. Sublimation, ionisation and hydration enthalpies decide the reduction potential, giving 3.

The common slip: 4. Counting electron gain enthalpy as well gives 4. That property applies only to anion formation, which is not part of the metal electrode reaction.

The idea to keep. For metal electrode potentials, remember the trio: sublimation, ionisation, hydration. This is exactly why Li has the most negative E°: its huge hydration enthalpy outweighs its ionisation cost.

Answer from the NTA final answer key. Solution by PrepWiser.