Reduction potential of `M^(2+)//M` will depend on

Options

A. `IE_(1)+IE_(2)`
B. `DeltaH_("atomisation")`
C. `Hydration energy
D. All of these

Answer

To determine the factors that affect the reduction potential of the half-cell reaction \( M^{2+} + 2e^- \rightarrow M \), we can analyze the contributions of three key thermodynamic parameters: enthalpy of atomization, ionization enthalpy, and hydration enthalpy.

### Step-by-Step Solution:

1. **Understanding Reduction Potential**:
The reduction potential (\( E^\circ \)) of the half-reaction \( M^{2+} + 2e^- \rightarrow M \) indicates how readily the metal ion \( M^{2+} \) can gain electrons to form the neutral metal \( M \).

2. **Enthalpy of Atomization**:
- This is the energy required to dissociate a mole of a metal into its gaseous atoms.
- A higher enthalpy of atomization indicates that more energy is needed to break the bonds in the metal, which affects the stability of the metal in its elemental form.
- Thus, a higher enthalpy of atomization contributes positively to the reduction potential.

3. **Ionization Enthalpy**:
- This is the energy required to remove an electron from a gaseous atom of the metal.
- A higher ionization enthalpy means that it is more difficult to remove electrons from the metal, which can also affect the reduction potential.
- Like the enthalpy of atomization, a higher ionization enthalpy contributes positively to the reduction potential.

4. **Hydration Enthalpy**:
- This is the energy released when gaseous ions are solvated by water molecules.
- A higher (more negative) hydration enthalpy implies that more energy is released when the ion is hydrated, which stabilizes the ion in solution.
- Since this is a release of energy, it contributes negatively to the reduction potential.

5. **Combining the Factors**:
- The overall standard electrode potential can be expressed as:
\[ E^\circ = \text{Enthalpy of Atomization} + \text{Ionization Enthalpy} - \text{Hydration Enthalpy} \]
- Here, the enthalpy of atomization and ionization enthalpy contribute positively, while the hydration enthalpy contributes negatively.

6. **Conclusion**:
- The reduction potential of \( M^{2+} // M \) depends on all three factors: enthalpy of atomization, ionization enthalpy, and hydration enthalpy.
- Therefore, the correct answer is that the reduction potential depends on all of these factors.

### Final Answer:
The reduction potential of \( M^{2+} // M \) will depend on **all of these factors** (option 4).