How do electrolytes allow electrolysis to occur?

Question

From what I know, when two electrodes are places in a electrolyte (say $\ce{H2SO4}$, which decomposes into $\ce{H}^{+}$ and $\ce{SO4}^{2-}$ in aqueous solution), the negative ion is attracted to the positive electrode and is oxidized while the cation is attracted to the negative electrode and is reduced.

This transfer of electrons from the negative electrode to the cation and from the anion to the positive electrode "completes" the circuit and electricity is conducted through the wire.

What I don't understand is how the electrolyte allows for the electrolysis of water, since the electrons are used in the redox reaction of the ions in the electrolyte in which water isn't even involved.

Answer 1

In the electrolysis of $\ce{H2O}$, a small amount of sulphuric acid is added. Why? Because pure $\ce{H2O}$ doesn't ionize much on its own ($K_w=10^{-14}$ approximately). So traces of an electrolyte ($\ce{H2SO4}$ in this case) is added so that it can help break one $\ce{O-H}$ bond and $\ce{H}^{+}$ released forms a coordinate bond with one $\ce{H2O}$ to form $\ce{H3O}^{+}$ ion. So now the ions present in the solution are: $\ce{H3O}^{+}$, $\ce{OH}^{-}$ and $\ce{SO4}^{2-}$. Of these, the oxidation potential of $\ce{OH}^{-}$ is higher than that of $\ce{SO4}^{2-}$ so at the anode $\ce{OH}^{-}$ gets oxidized to $\ce{O2}$. In this case only one cation goes to the cathode so it gets reduced to $\ce{H2}$. In case you use an electrolyte that gives a cation $\ce{M}^{n+}$ other than$\ce{H}^{+}$ then $\ce{H2}$ would be formed only if the reduction potential of $\ce{M}^{n+}$ is lower than that of $\ce{H3O}^{+}$. So, if the electrolyte used is $\ce{ZnSO4}$ then $\ce{H2}$ will be formed but if $\ce{CuSO4}$ is used then $\ce{H2}$ is not formed, but $\ce{Cu}$ is deposited at the cathode.

So the ionization of water doesn't have anything to do with the electrons used in the redox process. The ionization depends simply on the electrolyte used.