I was looking at the enthalpy change for water-splitting reaction:

$$ \Delta H^o_R = [\Delta H^0_{H_2(g)} +\frac{1}{2}\Delta H^0_{O_2(g)}]-\Delta H^0_{H_2O(l)} = 285.83 kJ/mol$$

According to the book "Thermal physics"by Schroeder; at constant T and P; if there are no other forms of work on the system besides compression/expansion, then $\Delta H^o_R = Q$. However, when there are other forms of work being done we then have $\Delta Gº_R \leq W_{other}$ and $\Delta H^o_R = Q + W_{other}$. The value for the Gibbs free energy in this case is $ \Delta Gº_R = 237 kJ/mol $. We can relate $\Delta G$ and $\Delta H$ by $\Delta G =\Delta H -T\Delta S$ .

My confusion arises first from reading that the gibbs free energy is the work we need to drive the reaction, say electrical work. However, the enthalpy change shows that the energy required could be done via heat and/or another form of work is higher than that of the Gibbs free energy? My guess so far is that we can take some energy from the environment for "free", but what happens when we drive this reaction only via heat, such that $\Delta H = Q$, would $\Delta G = 0 $ ?