Question

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

Answer 1

$E^{\ominus }\left(M^{2+}/M\right)$ $\text{for copper}$

$E^{\ominus }\left(M^{2+}/M\right)$ for any metal is related to the sum of enthalpy changes taking place in following steps:

$M\left(s\right)+{\Delta }_{a}H\to M\left(g\right)$

$M\left(g\right)+{\Delta }_{i}H\to M^{2+}\left(g\right)+2e^{-}$

$M^{2+}\left(g\right)+aq\to M^{2+}\left(aq\right)+{\Delta }_{hyd}H$


The $E^{\ominus }\left(M^{2+}/M\right)$ value for metal depends on atomization enthalpy, ionization enthalpy and hydration enthalpy.

Copper has a high value of atomization enthalpy and low hydration enthalpy. Thus, as a result, the overall effect $E^{\ominus }\left(M^{2+}/M\right)$ for copper is positive.