The $E^{\Theta} (M^{2+} / M)$ value for copper is positive $(+0.34 \ V)$. What is the possible reason for this? (Hint: consider its high $\Delta_{a} H^{\Theta}$ and low $\Delta_{hyd} H^{\Theta}$)

(N/A) The $E^{\Theta} (M^{2+} / M)$ value of a metal depends on the energy changes involved in the following steps:
$1.$ Sublimation: The energy required for converting one mole of an atom from the solid state to the gaseous state.
$M_{(s)} \longrightarrow M_{(g)}$ $\Delta_{a} H^{\Theta}$ (Atomization energy)
$2.$ Ionization: The energy required to remove electrons from one mole of atoms in the gaseous state.
$M_{(g)} \longrightarrow M^{2+}_{(g)} + 2e^-$ $\Delta_{i} H^{\Theta}$ (Ionization energy)
$3.$ Hydration: The energy released when one mole of ions are hydrated.
$M^{2+}_{(g)} + aq \longrightarrow M^{2+}_{(aq)}$ $\Delta_{hyd} H^{\Theta}$ (Hydration energy)
The net energy change is the sum of these energies. Copper has a very high enthalpy of atomization $(\Delta_{a} H^{\Theta})$ and a relatively low enthalpy of hydration $(\Delta_{hyd} H^{\Theta})$. The sum of these energy terms is positive,which makes the overall $E^{\Theta} (M^{2+} / M)$ value for copper positive.