= Solution
Use the balanced aqueous-acid reaction
$$
\mathrm{Ln(s)+3H^+(aq)\longrightarrow Ln^{3+}(aq)+\tfrac32H_2(g)}.
$$
A Hess cycle first atomizes the metal, then ionizes its gas-phase atoms, then hydrates the triply charged gas ions:
$$
\begin{aligned}
\mathrm{Ln(s)}&\longrightarrow\mathrm{Ln(g)}&&\Delta_{\rm at}H^\circ>0,\\
\mathrm{Ln(g)}&\longrightarrow\mathrm{Ln^{3+}(g)+3e^-(g)}&&I_1+I_2+I_3>0,\\
\mathrm{Ln^{3+}(g)}&\longrightarrow\mathrm{Ln^{3+}(aq)}&&\Delta_{\rm hyd}H^\circ<0.
\end{aligned}
$$
Close the reference leg by combining the three electrons with $3\mathrm{H^+(aq)}$ to make $\tfrac32\mathrm{H_2(g)}$. Calling that element-independent contribution $C_H$ gives
$$
\boxed{\Delta H^\circ_{\rm acid}=\Delta_{\rm at}H^\circ+I_1+I_2+I_3+\Delta_{\rm hyd}H^\circ+C_H.}
$$
A single-ion <standard enthalpy of formation> uses the conventional hydrogen-ion reference; it must not be treated as a balanced creation of net charge from neutral metal without an <electron>/reference term. The constant <hydrogen> leg does not affect comparisons along the <lanthanide> series.
\Image[/past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1-lanthanide-cycle.png]
{title=Original Hess cycle separating atomization, three ionizations, hydration and the common aqueous hydrogen reference leg}
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