Question about calculating enthalpy change $$\Delta H = \left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds broken}\end{array}\right)-\left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds made}\end{array}\right)$$

$$\Delta H = \sum \Delta H_\mathrm{F} \left( \text{products}\right)-\Delta H_\mathrm{F} \left( \text{reactants}\right)$$

Both $\Delta H$'s given by the above formulas should be equivalent.
However, I found one special case that the above 2 formulas give 2 different results.
$$\ce{2 H+ + 2 e- → H2}$$
$$\Delta H = \sum \Delta H_\mathrm{F} \left( \text{products}\right)-\Delta H_\mathrm{F} \left( \text{reactants}\right)= 0-0=0$$
$$\Delta H = \left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds broken}\end{array}\right)-\left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds made}\end{array}\right)&lt;0$$ because there is only bond formation, no bond break.
How to explain this difference? Can I say the heat is released during this process, because $$ \Delta H $$ is smaller than 0?
The approach to estimating the enthalpy change in a reaction by using bond dissociation energies assumes homolytic cleavage of the bonds: each atom gets one electron from the covalent bond. 

The bond dissociation energy of the $\ce{H-H}$ bond is reported as 436 kJ/mol.  The relevant reaction is:

$$\ce{H2(g) -&gt; 2H.(g)}$$

Note that this reaction is not the same as $\ce{H2 -&gt; 2H+ + 2e-}$. The homolytic cleavage   reaction produces hydrogen atoms, not hydrogen ions. If we look up the standard enthalpy of formation for hydrogen atoms, it is +218 kJ/mol. Since the standard enthalpy of formation for $\ce{H2}$ is defined as zero, using the enthalpies of formation we also get $\Delta H = +436$ kJ/mol for this reaction. 

It is generally hard to use bond dissociation energies for heterolytic reactions (reactions that produce ions from molecules).