What determines reduction potential?

Reduction potential of an electrode depends on the energy change involved in converting the species from its standard state into the reduced form.

For a metal ion half-cell, $M^{n+}(aq) + ne^- \rightarrow M(s)$ the energetic factors connected with this process can be understood using a Born–Haber type cycle.

Relevant energetic terms

To relate metal and ion, the important steps are:

• converting metal atom in solid to gaseous atom: sublimation enthalpy
• removing electron(s) from gaseous atom: ionisation enthalpy
• converting gaseous ion to aqueous ion: hydration enthalpy

These three directly affect the tendency of $M^{n+}(aq) + ne^- \rightarrow M(s)$ and hence affect the standard reduction potential.

Check each given property

(A) Sublimation enthalpy — Yes, affects reduction potential.

(B) Ionisation enthalpy — Yes, affects reduction potential.

(C) Hydration enthalpy — Yes, affects reduction potential.

(D) Electron gain enthalpy — No, this is relevant to the tendency of an isolated gaseous atom to gain an electron, $X(g) + e^- \rightarrow X^-(g)$ It is not one of the standard terms used to determine the reduction potential of a metal ion/metal electrode in this context.

Total number of properties affecting reduction potential

$3$

Comparison with stored answer

Stored correct answer = $3$

This matches the derived answer.