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H2+½O2→H2O(liquid) <> ΔHf° = -285.8kJ/mol
H2+½O2→H2O(gas) <> ΔHf° = -241.8kJ/mol
...multiplied by 50% (at best) times efficiency of alternator (35%) means something like 1382kJ/mol are needed to dissociate water at the engine, that's less the 18% efficiency:
% = 100%*(P.out/P.in)
% = 100*(241.8kJ/mol / 1382kJ/mol) = 100%*(0.175) < 18%
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