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)

We wish to know which of the listed thermodynamic quantities actually enter the calculation of the standard reduction potential of an element in aqueous solution. For a simple monovalent metal, the relevant redox half-reaction is usually written as

$$\text{M}^{+}(aq)+e^{-}\;\longrightarrow\;\text{M}(s)$$

The standard reduction potential $$E^{\circ}$$ for this process is derived from the overall Gibbs free-energy change $$\Delta G^{\circ}$$ using the familiar relation

$$\Delta G^{\circ}\;=\;-\,nF\,E^{\circ},$$

where $$n$$ is the number of electrons transferred (here $$n=1$$) and $$F$$ is the Faraday constant. To express $$\Delta G^{\circ}$$ we construct a Born-Haber-type thermodynamic cycle that converts solid metal to aqueous ion in several clearly defined steps. The successive steps, with their corresponding enthalpy changes, are written out explicitly below so that every algebraic manipulation is visible.

1. Sublimation of the solid metal to the gaseous atom: $$\text{M}(s)\;\xrightarrow{\;\Delta_{\text{sub}}H\;}\;\text{M}(g)$$ Here $$\Delta_{\text{sub}}H$$ is called the sublimation enthalpy.

2. Ionisation of the gaseous atom to the gaseous cation: $$\text{M}(g)\;\xrightarrow{\;I_{1}\;}\;\text{M}^{+}(g)+e^{-}$$ $$I_{1}$$ (or more generally $$I_{n}$$) is the ionisation enthalpy.

3. Hydration of the gaseous cation to the aqueous cation: $$\text{M}^{+}(g)\;\xrightarrow{\;\Delta_{\text{hyd}}H\;}\;\text{M}^{+}(aq)$$ $$\Delta_{\text{hyd}}H$$ is the hydration enthalpy (negative in sign because the process is exothermic).

Adding the three steps, we move from $$\text{M}(s)$$ on the right-hand side of the desired redox equation back to $$\text{M}^{+}(aq)$$ on the left-hand side. The algebraic sum of their enthalpy changes therefore equals the enthalpy change for the reverse of the redox reaction, and from it we obtain $$\Delta G^{\circ}$$ and thence $$E^{\circ}$$. Hence:

$$\boxed{\Delta G^{\circ}= \Delta_{\text{sub}}H + I_{1} + \Delta_{\text{hyd}}H}$$

We now inspect the list of properties given in the question:

(A) Sublimation enthalpy — appears explicitly in the equation above, so it affects $$E^{\circ}$$.

(B) Ionisation enthalpy — also appears in the equation, so it affects $$E^{\circ}$$.

(C) Hydration enthalpy — appears in the equation, so it affects $$E^{\circ}$$.

(D) Electron gain enthalpy — this quantity is used when a neutral atom gains an electron to form an anion (e.g., $$\text{Cl}(g)+e^{-}\to\text{Cl}^{-}(g)$$). In the metal-ion reduction process under discussion, no such step is involved, so electron gain enthalpy does not contribute.

Therefore, exactly three of the four listed properties influence the reduction potential.

So, the answer is $$3$$.