Step 1: List the energy terms involved.
The value of \(E^\circ_{M^{2+}/M}\) depends on three energies acting together: enthalpy of atomisation, ionisation enthalpy (for the two electrons), and hydration enthalpy of the \(M^{2+}\) ion.
Step 2: Explain part (I) the irregular trend.
These three energy terms do not change smoothly across the series. Their changes do not cancel evenly, so the overall \(E^\circ\) value goes up and down rather than following a steady pattern.
Step 3: Explain part (II) the high positive value for copper.
For copper, the sum of the atomisation enthalpy and the ionisation enthalpy is very high (much energy is needed to make \(Cu^{2+}\)), and this is not balanced by its hydration enthalpy. So the process is not favourable, and \(E^\circ_{Cu^{2+}/Cu}\) comes out exceptionally positive.
Step 4: Explain part (III) the very negative value for manganese.
Manganese forms \(Mn^{2+}\), which has a stable half-filled \(3d^5\) configuration. This extra stability makes the formation of \(Mn^{2+}\) favourable, so its electrode potential is highly negative.
Step 5: Tie it together.
In each case it is the balance of these enthalpy terms (and the special stability of \(d^5\)) that decides the value.
Answer: (I) The irregular trend arises because the atomisation, ionisation and hydration enthalpies do not vary smoothly and do not cancel out. (II) \(Cu^{2+}/Cu\) is exceptionally positive because the high atomisation plus ionisation enthalpy of copper is not offset by its hydration enthalpy. (III) \(Mn^{2+}/Mn\) is highly negative because the half-filled \(d^5\) stability of \(Mn^{2+}\) favours its formation.