The first ionization enthalpy ($IE_1$) corresponds to the energy required to remove an electron from a neutral gaseous atom: $Na(g) \rightarrow Na^+(g) + e^-; \quad \Delta H = +5.1 \text{ eV}$ The electron gain enthalpy of the cation ($Na^+$) corresponds to the energy change when an electron is added to the ion to form the neutral atom. This is the exact reverse of the ionization process: $Na^+(g) + e^- \rightarrow Na(g)$ According to thermodynamic principles, reversing a chemical reaction reverses the sign of the enthalpy change ($\Delta H$) while the magnitude remains the same. Therefore, the electron gain enthalpy of $Na^+$ is $-5.1 \text{ eV}$.