If electrolysis splits water, then that means that $\ce{H2O}$ is split into $\ce{H}$ and $\ce{OH}$ or $\ce{O}$.

How come that if a water molecule is split at e.g. the negative pole (anode), only the $\ce{O}$ bubbles up. What happens to the $\ce{H}$ and/or $\ce{OH}$? Do they react with the pole and stay there?


Water, as you may know, has a dualist nature between covalent and ionic bonding; oxygen is the second most electronegative element in the periodic table, while hydrogen's simplistic construction makes it very zen about how it forms bonds (it defines the center point of most electronegativity scales). While the bond between the hydrogens and oxygen of a water molecule are termed covalent, the electrophilic nature of the oxygen makes the bonds border on the ionic, and so water is just as easily thought of as $\ce{2H^+ + O^2-}$, or in more stable terms as $\ce{H^+ + OH^-}$.

In electrolysis, this ionic character of water is exploited in a series of "half-reactions"; at each node, a gas is formed by adding or removing electrons from water to liberate one of its two components, and when that gas precipitates out, what's left remains dissolved in the water as charged ions, which are repelled by the node that created it and attracted to the other node.

When electrons are added to water at the cathode, they are attracted to the partial positive charge of the "hydrogen side" of water, and can then form an orbit around a hydrogen. That liberates the hydrogen; it no longer needs the electron it's sharing with the oxygen to balance its charge, but it still needs a second electron to stabilize its 1s shell, and the best bet is to pair with another hydrogen to form hydrogen gas, which precipitates out. That leaves soluble hydroxide ions in the water, which are repelled by the cathode (and each other, so these ions wouldn't form any compounds) and move toward the anode. It's highly unlikely that both hydrogen atoms would be liberated from a single oxygen, and even if it did, the oxygen, with extra electrons, couldn't bond with another negative-charge oxygen to form gas; it would be much more likely to deprotonate another water molecule forming two hydroxides.

$$\ce{4H2O + 4e^- -> 2H2 (g) (^) + 4OH^- (aq)}$$

At the anode, the positive potential causes water to give up electrons, liberating the hydrogens again, this time as naked protons, which are attracted to water's partial negative pole to form hydronium ($\ce{H_3O^+}$ or simply $\ce{H^+ (aq)}$). The now chargeless but bond-seeking oxygen is left to pair up as oxygen gas, which precipitates out. Hydrogen peroxide is a possible intermediate, but very short-lived, as this peroxide will either find another peroxide and pair up the extra oxygens, or it will give up its hydrogens as protons under the positive voltage potential the same way water would, and become oxygen gas.

$$\ce{2H_2O - 4e^- \to O_2(g)\uparrow + 4H^+ (aq)}$$

... Then, somewhere in the middle, the hydronium and hydroxide meet, and as these proton donors and proton acceptors will do, they reform water:

$$\ce{4H^+(aq) + 4OH^- (aq) \to 4H_2O }$$

You'll notice that only one gas precipitates out at each node; the other atoms are trapped as soluble ions, either protons/hydronium or hydroxide, which cannot form other compounds because their like charges repel them from each other. Instead, they remain dissolved as local acidic and basic solutions. If the potential charge in the nodes is high enough, dipping a litmus strip at each end of the water bath (or dropping in a few mL of pH indicator dye) will yield an acidic reading at the anode and a basic reading at the cathode.


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