Why does hydrogen burn spontaneously in fuel cell at room temperature?

It's well known that if you mix $$\ce{H2}$$ and $$\ce{O2}$$ in ambient conditions, the $$\ce{H2}$$ will be stable indefinitely, and will only react with $$\ce{O2}$$ to form water when ignited by a flame. The reason being that for the reaction

$$\ce{2H2 + O2 -> 2H2O} \label{rxn:1}\tag{1}$$

to occur, first the bonds must be broken in $$\ce{H2}$$ and in $$\ce{O2},$$ which is an insurmountable activation energy at room temperature.

So I am confused why a hydrogen fuel cell works at room temperature without ignition, as it is powered by the same chemical reaction \eqref{rxn:1}.

In particular, why can the strong double bond in $$\ce{O2}$$ suddenly be broken in this context without a flame?

My best guess is that somehow the electron sea in the copper wire acts as a catalyst. If this is true, what are the details of this mechanism that allows the $$\ce{O2}$$ bond to break at room temperature?

• My guess would be that the reaction you've stated isn't the actual reactions occurring but the overall reaction of the two half equations that take place in an alkaline hydrogen-oxygen fuel cell. $$\ce{H2 + 2OH- ->2H2O + 2e- }$$ $$\ce{O2 + 2H2O + 4e- ->4OH-}$$ – H.Linkhorn Jun 9 '19 at 16:58