Step 1: What \(E^\circ_{M^{2+}/M}\) depends on.
This value is fixed by three energy terms taken together: the enthalpy of atomisation (sublimation) of the metal, the sum of the first two ionisation enthalpies, and the hydration enthalpy of the \(M^{2+}\) ion.
Step 2: Why the trend is irregular (I).
Across the series these three terms do not change smoothly. Small ups and downs in ionisation enthalpy and hydration enthalpy, linked to extra stability of half-filled and filled d sub-shells, make the overall \(E^\circ\) zig-zag instead of following one smooth line.
Step 3: Why Cu is exceptionally positive (II).
Copper has a high enthalpy of atomisation and a high sum of the first two ionisation enthalpies, while its hydration enthalpy is not enough to balance these large energy inputs. So converting Cu to \(Cu^{2+}(aq)\) is unfavourable, which makes \(E^\circ_{Cu^{2+}/Cu}\) positive.
Step 4: Why Mn is highly negative (III).
Forming \(Mn^{2+}\) gives a stable half-filled \(3d^5\) configuration. This extra stability makes \(Mn^{2+}\) form easily, so the reduction \(Mn^{2+} + 2e^- \rightarrow Mn\) is hard, and \(E^\circ_{Mn^{2+}/Mn}\) becomes highly negative.
Step 5: Summary.
All three answers come from comparing atomisation, ionisation and hydration energies plus the stability of \(d^5\) and \(d^{10}\).
Answer: (I) The irregular trend arises from uneven changes in atomisation, ionisation and hydration enthalpies across the series. (II) \(E^\circ_{Cu^{2+}/Cu}\) is positive because Cu's high atomisation and ionisation energies are not offset by hydration energy. (III) \(E^\circ_{Mn^{2+}/Mn}\) is highly negative due to the stable half-filled \(3d^5\) configuration of \(Mn^{2+}\).