$E^{\circ}$ of $Cu$ is $+0.34 \ V$ while that of $Zn$ is $-0.76 \ V$. Explain.
(N/A) The standard electrode potential $(E^{\circ})$ of a metal depends on the sum of its enthalpy of sublimation,ionization enthalpy,and hydration enthalpy.
For $Cu$,the sum of sublimation and ionization enthalpies is very high,which is not compensated by the hydration enthalpy of $Cu^{2+}$. Thus,$Cu$ has a positive $E^{\circ}$ value $(+0.34 \ V)$.
For $Zn$,the second ionization enthalpy is relatively low because the removal of electrons from the $4s$-orbital leads to a stable $3d^{10}$ configuration. This low energy requirement makes the overall process favorable,resulting in a negative $E^{\circ}$ value $(-0.76 \ V)$.
For $Cu$,the sum of sublimation and ionization enthalpies is very high,which is not compensated by the hydration enthalpy of $Cu^{2+}$. Thus,$Cu$ has a positive $E^{\circ}$ value $(+0.34 \ V)$.
For $Zn$,the second ionization enthalpy is relatively low because the removal of electrons from the $4s$-orbital leads to a stable $3d^{10}$ configuration. This low energy requirement makes the overall process favorable,resulting in a negative $E^{\circ}$ value $(-0.76 \ V)$.
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