$\mathrm{E}^{\circ}$ of $\mathrm{Cu}$ is $+0.34 \mathrm{~V}$ while that of $\mathrm{Zn}$ is $-0.76 \mathrm{~V}$. Explain.

Vedclass pdf generator app on play store
Vedclass iOS app on app store

In case of copper, the hydration enthalpy of $\mathrm{Cu}^{2+}$ does not balance the summation of sublimation enthalpy and ionization enthalpies and so it has positive $\mathrm{E}^{\circ}$ values.

However, in case of $\mathrm{Zn}$ the second ionization enthalpy is low because electrons is to be removed from $4 s$-orbital to attain stable $3 d^{10}$ configuration. Hence $\mathrm{Zn}$ has negative $\mathrm{E}^{\circ}$ value.

Similar Questions

How many unpaired electrons are there in $N{i^{2 + }}$

When ammonia is added to a cupric salt solution, the deep blue colour is observed it is due to the formation of 

The solubility of silver bromide in hypo solution is due to the formation of :

The correct formula of permanganic acid is

$\mu = \sqrt{15}$ is true for the pair :-