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

Options

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

Answer

To determine the factors that influence the reduction potential of `Mn^(2+) // M`, we can break down the process into several steps based on the enthalpy cycle. Here’s a step-by-step solution:

### Step 1: Understand Reduction Potential
The reduction potential of a species is a measure of its tendency to gain electrons and be reduced. For `Mn^(2+)`, this potential will depend on several thermodynamic factors.

### Step 2: Consider the Enthalpy Cycle
We can analyze the reduction potential using an enthalpy cycle, which involves several steps:

1. **Enthalpy of Atomization (ΔH_atomization)**: This is the energy required to convert solid metal (M) into gaseous metal atoms. This step is crucial as it determines how much energy is needed to break the metallic bonds.

2. **Ionization Energies (IE)**: The ionization energy is the energy required to remove electrons from gaseous metal atoms to form gaseous metal ions. For `M`, we need to consider:
- **IE1**: Energy to remove the first electron to form `M^+`.
- **IE2**: Energy to remove the second electron to form `M^(2+)`.
The total energy required to form `M^(2+)` from `M` will be the sum of these two ionization energies: `IE1 + IE2`.

3. **Enthalpy of Hydration (ΔH_hydration)**: This is the energy change when gaseous metal ions are solvated in water to form aqueous ions. The hydration energy affects how stable the ions are in solution.

### Step 3: Combine the Factors
The reduction potential of `Mn^(2+)` will depend on:
- The enthalpy of atomization (ΔH_atomization)
- The sum of the ionization energies (IE1 + IE2)
- The enthalpy of hydration (ΔH_hydration)

### Conclusion
The reduction potential of `Mn^(2+) // M` will depend on:
- ΔH_atomization
- IE1 + IE2
- ΔH_hydration

Thus, all these factors are important in determining the reduction potential.