$\ce{H2O}$ molecule is dissociated into $\ce{OH-}$ and $\ce{H+}$ ions on reduction at the cathode in electrolysis. But what happens at the atomic level to $\ce{H2O}$ that proton and hydroxide ions get separated after gaining electrons from cathode? What really happens when water molecule gains electrons from cathode?

$$\ce{2H2O(l) + 2 e- -> H2(g) + 2 OH-(aq)}$$


1 Answer 1


The dissociation of water to $\ce{OH-}$ and $\ce{H+}$ happens first before any interaction with the cathode. $\ce{H+}$ ions then gain an electron at the cathode to produce hydrogen. At the anode, water is oxidized (loses electrons) to produce oxygen and more $\ce{H+}$ ions.

In pure water there is a very small concentration of $\ce{H+}$ and $\ce{OH-}$ which is why the reaction is very slow and not thermodynamically favorable. It is also the reason why pure water is such a good insulator. But when you add an electrolyte the process becomes much more favourable kinetically and thermodynamically. Further proof that $\ce{H+}$ and $\ce{OH-}$ are generated before any interaction with the cathode.

  • $\begingroup$ A simple way of understanding what Answer 3 is saying is that one of the electrons from a hydrogen atom in a water molecule is so strongly attracted to the Anode, that it binds to the anode rather than staying with its proton. The proton (aka H+) is then repelled away from the anode (both are positively charged) and at the same time attracted to the Cathode. $\endgroup$ Commented Jun 10, 2019 at 6:56
  • $\begingroup$ "$\ce{H+}$ ions then gain an electron at the cathode to produce hydrogen".I think that the hydrogen produced at the cathode is the reduction of water molecules and not the reduction of $\ce{H+}$ ions because the amount of $\ce{H+}$ ions is very low $\endgroup$ Commented Jun 10, 2019 at 7:50

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