How to find enthalpy of solvation for gas phase cation using Hess's law? Find enthalpy of solvation $\Delta_\mathrm{solv}H$ using Hess's law, and show the expression that allows to calculate the enthalpy of solvation.
$$\ce{Ca^2+(g) -&gt; Ca2+(aq)}$$

I am not told which enthalpies of reaction to use. I don't need the exact value, but I do need to show the sum of each enthalpy. I managed to find the enthalpy of the following reactions:
$$
\begin{align}
\ce{Ca(g) &amp;-&gt; Ca+(g) + e-} &amp;\quad \Delta H &amp;= \pu{590 kJ mol-1} \\
\ce{Ca+(g) &amp;-&gt; Ca^2+(g) + e-} &amp;\quad \Delta H &amp;= \pu{1145 kJ mol-1} \\
\ce{Ca(s) + 2 H+(aq) &amp;-&gt; Ca^2+(aq) + H2(g)} &amp;\quad \Delta H &amp;= \pu{1925.9 kJ mol-1} \\
\ce{Ca(s) &amp;-&gt; Ca(g)} &amp;\quad \Delta H &amp;= \pu{178 kJ mol-1}
\end{align}
$$
I nearly have the reactions I need, but I am still needing
$$\ce{H2 -&gt; 2 H+ + 2 e^-}$$
I haven't been able to find the answer.
P.S. I am aware of the enthalpy of formation of $\ce{H2}$, but don't know how/if I can use it here.
One approach to compute the enthalpy change for the reaction
$$\ce{H2(g)-&gt;2H+(aq) + 2e-}$$
is to combine the bond dissociation enthalpy and hydrogen ionization energy (both refer to gas phase species) and the hydration enthalpy of the hydrogen ion:
$$\ce{H2(g)-&gt;2H(g)}~~~~~~~~{\Delta H = \pu{439 kJ/mol}  }\\
\ce{2H(g)-&gt;2H+(g) +2e-(g)}~~~~~~~~{\Delta H = \pu{262 kJ/mol}  }\\
\ce{2H+(g)-&gt;2H+(aq)}~~~~~~~~{\Delta H = \pu{-2260 kJ/mol}  }$$
If you add up the enthalpies you obtain as the solvation enthalpy of $\ce{Ca^{2+}}$ the value $\Delta_{\text{solv}} H = -1546 \pu{kJ/mol}$ which is quite reasonable (see eg a table here)
The result is clearly very different from my earlier answer which I am still pondering:
The catch is that $\Delta H^\circ=0$ for this reaction. The reason is because $\Delta G^\circ=0$ for the reaction (by definition, see standard reduction potential) at all temperatures, and then, since
$$\Delta H = -T^2\left( \frac{\partial (\Delta G/T)}{\partial T}\right)_p$$
it follows that  $\Delta H^\circ=0$. I suspect the problem with this approach is due to a missing term involving enthalpy of solvation of the electrons (or a similar term involving the electrons) but I haven't put my finger on it.