# How does an electrolyte react to voltages below its electrolysis potential?

An electrolyte can be electrolysed by a voltage higher than the reverse reaction’s cell potential, while conducting a current in the process. But what happens at the electrodes when the applied voltage is below this cell potential, such as below 1.23 V for water or 4.07 V for NaCl? Does the reverse reaction occur simultaneously, or is the electrolyte unable to conduct a current with such a low voltage at all?

Electrolysing water is NOT required for it to conduct electricity. Water ionizes into $\ce{H+}$ and $\ce{OH-}$. Pure water will contain $10^{-7}\ \mathrm{M}$ of either ion at room temperature at equilibrium. This will result in $18\ \mathrm{M\Omega\ cm}$ of resistivity. If and when potentials above the electrolysis potential is applied, the local pH around the electrodes will change and more ions will be created.
$\ce{NaCl}$ dissociates as soon as it is dissolved in water. The $\ce{Na+}$ and $\ce{Cl-}$ ions created in the solution carry current and increase the conductivity.
• I think the OP also wants to know what will happen to the water or NaCl outside the window. In short, in water, you'll evolve $\ce{H2}$ or $\ce{O2}$ at the electrodes, or $\ce{Na(s)}$ or $\ce{Cl2}$ with $\ce{NaCl}$ depending on oxidizing or reducing conditions. Apr 30, 2015 at 15:56