Question #44138

The standard reduction potentials $E^\circ(\ce{M^{2+}/M})$ for \ce{Mn} and \ce{Zn} at $298\text{ K}$ are $-1.18\text{ V}$ and $-0.76\text{ V}$, respectively. The thermodynamic parameters for these metals are given below: Enthalpy of atomization ($\Delta_{a}H^\circ$): \ce{Mn} = $281\text{ kJ mol}^{-1}$, \ce{Zn} = $130\text{ kJ mol}^{-1}$. Sum of first and second ionization enthalpies ($IE_1 + IE_2$): \ce{Mn} = $2226\text{ kJ mol}^{-1}$, \ce{Zn} = $2639\text{ kJ mol}^{-1}$. Assuming that the standard entropy change ($\Delta S^\circ$) for the reduction half-reaction $\ce{M^{2+}(aq) + 2e^- -> M(s)}$ is identical for both metals, calculate the difference in their standard hydration enthalpies, $\Delta_{hyd}H^\circ(\ce{Zn^{2+}}) - \Delta_{hyd}H^\circ(\ce{Mn^{2+}})$. (Use $F = 96500\text{ C mol}^{-1}$)