Question

The \[\ce{E^{\circ}_{(M^{2+}/M)}}\] value for copper is positive (+0.34 V). What is possible reason for this? (Hint: consider its high Δ_aH° and low Δ_hydH°)

Answer 1

For a metal, \[\ce{E^{\circ}_{(M^{2+}/M)}}\] is related to the sum of the enthalpy changes in the following terms:

\[\ce{M_{(s)} + \Delta_{{a}}H -> M_{(g)}}\] (Δ_aH = atomic enthalpy = positive)

\[\ce{M_{(g)} + \Delta_{{i}}H -> M^{2+}_{ (g)}}\] (Δ_iH = ionization enthalpy = positive)

\[\ce{M^{2+}_{ (g)} + \Delta_{hyd}H -> M^{2+}_{ (aq)}}\] (Δ_hydH = hydration enthalpy = negative)

Copper has a high atomic enthalpy and a low hydration enthalpy. Hence the value of \[\ce{E^{\circ}_{(Cu^{2+}/Cu)}}\] is positive. Hence, the high energy of transformation of Cu_(s) into \[\ce{Cu^{2+}_{ (aq)}}\] is not balanced by its hydration enthalpy.