On the subject of methane pyrolysis, it turns out if you look at the Gibbs free energy calculation, about half of the energy of methane combustion is released from the formation of water, and the other half from the formation of carbon dioxide.

About 70% of the energy is in hydrogen, 30% is in carbon. 1 GJ of methane weighs about 20 kg, 5 kg of which comprise hydrogen. At 142 MJ/kgH2 (higher heating value, which implies condensation of the produced water), 710 MJ out of that 1 GJ is due to hydrogen.

With a 60%-70% efficient hydrogen fuel cell, about 50% of the electricity generated from hydrogen from pyrolysis of methane would drive the process, and 50% could go into the grid.

You have to account for the energy required to break the bonds of the CH4, though. This means if you burn methane the usual way you get (CH4 + 2O2 --> CO2 + 2H2O + 803 kJ/mol); if you burn it with an ideal zero-emissions reaction, you get (CH4 + O2 --> C + 2H2O + 409 kJ/mol), or just a little more than half the energy from the same gas.

Your accounting works if someone else does the pyrolysis for you and you're left with just the H2 and C at the end, but mine includes the energy consumed by the pyrolysis step that breaks the methane molecule (albeit neglecting any thermodynamic losses, which there will be several -- for example you need to recapture the heat carried away by the hot carbon atoms). On the other hand, you can hardly wish for a better feedstock for CVD diamond production...