What does standard electrode potential E^ of $\mathrm{M^+ / M}$ represent?

It corresponds to the reaction $\mathrm{M^+(aq) + e^- \rightarrow M(s)}$ or equivalently its reverse oxidation process: $\mathrm{M(s) \rightarrow M^+(aq) + e^-}$

The energy changes involved in converting $\mathrm{M(s)}$ to $\mathrm{M^+(aq)}$ can be understood through a BornHaber-type cycle.

Steps involved in forming $\mathrm{M^+(aq)}$ from $\mathrm{M(s)}$

To convert solid metal into aqueous ion, the following processes are involved:

1. Sublimation of the solid metal $\mathrm{M(s) \rightarrow M(g)}$ This contributes to the electrode potential.

2. Ionisation of the gaseous metal atom $\mathrm{M(g) \rightarrow M^+(g) + e^-}$ This also contributes.

3. Hydration of the gaseous metal ion $\mathrm{M^+(g) \rightarrow M^+(aq)}$ This also contributes.

Which process is not relevant?

The option says: $\text{Ionisation of a solid metal atom}$

This is not a standard thermodynamic step used in the cycle. A solid metal atom is not ionised directly in the BornHaber description; first the metal is sublimed to gaseous atoms, then those gaseous atoms are ionised.

So, standard electrode potential does not depend on ionisation of a solid metal atom.

Evaluate the options

• A: Ionisation of a gaseous metal atom  Depends on this.
• B: Sublimation of a solid metal  Depends on this.
• C: Ionisation of a solid metal atom  Does not depend on this.
• D: Hydration of a gaseous metal ion  Depends on this.

Final answer

The correct option is: $\boxed{C}$