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

The $E^{\theta}\left( M ^{2+} / M \right)$ value of a metal depends on the energy changes involved in the following:

$1.$ Sublimation: The energy required for converting one mole of an atom from the solid state to the qaseous state.

$M _{(x)} \longrightarrow M _{( g )}$    $\Delta_{s} H$ ( Sublimation energy )

$2.$ Ionization: The energy required to take out electrons from one mole of atoms in the gaseous state.

$M _{(g)} \longrightarrow M ^{2+}_{(g)}$     $\Delta_{i} H$ (Ionization energy)

$3.$ Hydration: The energy released when one mole of ions are hydrated.

$M _{(g)}^{2+} \longrightarrow M ^{2+}_{(a q)}$   $\Delta_{ hyd } H$ (Hydration energy)

Now, copper has a high energy of atomization and low hydration energy. Hence, the $E^{\theta\left(M^{2+} / M\right)}$ value for copper is positive.