Electrolysis Without Ions

Electrolysis...

$M+H_2O\rightarrow MH+OH$

$2MH\rightarrow 2M+H_2$

$MH+H_2O\rightarrow M+H_2+OH$

where $M$ is a metal with an unpaired orbit and,

$M+OH\rightarrow MOH$

If copper, $Cu$ is used as electrodes, the formation of $CuH$ will turn both electrodes red.  However, since the equations above do not involve charged ions explicitly, they can occur at both the cathode and the anode during electrolysis.

But...only at the anode (+),

$2OH\rightarrow O_2+2H^{+}$

the $H$ in $OH$ looses an electron, turns positive and is ejected.  Two $O$ atoms forms $O_2$.  The positively charged $H^{+}$,

$H^{+}+OH\rightarrow H_2O^{+}$

forms water with a positive charge, $H_2O^{+}$ and moves to the cathode.

Only at the cathode (-), because of the positively charged $H_2O^{+}$

$M+H_2O^{+}+e^{-}\rightarrow MH+OH$

then,

$2MH\rightarrow 2M+H_2$

and more importantly, the main $H_2$ producer,

$MH+H_2O^{+}+e^{-}\rightarrow M+H_2+OH$

In this scenario, $O_2$ gas starts to form at the anode first, after the positively charged water, $H_2O^{+}$ has migrated to the cathode does $H_2$ starts to evolve there.

Because of $OH$ forming $H_2O^{+}$ with $H^{+}$, a membrane to allow just $H^{+}$ is futile as it is $H_2O^{+}$ that transports the charges in the electrolyte.  Metals without an unpaired orbit and so do not form metal hydride, $MH$, cannot be used as electrode because the the formation of metal hydride $MH$ is part of the $H_2$ production process.

It is possible to block just the formation of $H_2$ by blocking the passage of $H_2O^{+}$, as the needed $OH$ are produce at both electrodes.

The metal hydride $MH$, must remain structurally on the electrode to act as a catalyst and not contribute to electrode corrosion.  Since $OH$ is produced at the cathode and $MH$ is reverted to $M$ with the evolution of $H_2$ there, and the production of $O_2$ does not involve $MH$, the anode can be made of any conductive and oxidation resistant metal without the concern for metal hydride formation.  In this case $OH$ will have to first migrate to the anode for $O_2$ production to start.  (The metal hydride formed at the anode does not revert back to metal, and will eventually insulate the electrode completely.  In practice, the electrodes are swapped periodically.)  To facilitate $O_2$ production then, $OH$ can be introduced into the electrolyte as $NaOH$.  An alkaline is added to water instead of acid or a alkaline metal salt.

It is likely that the production of $OH$ is the reason behind the increase in efficiency when acid or alkaline salt is added.  For an acid,

$Hsalt\_part+H_2O\rightarrow OH+H_2+salt\_part$

and for an alkaline metal salt,

$Msalt\_part+H_2O\rightarrow OH+MH+salt\_part$

It is not the conductivity of the electrolyte that matters but the availability of $OH$.

Next stop, fuel cells...