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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.

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  • $\begingroup$ Hydrolysis and electrolysis of water are quite different, nearly unrelated things. Which one are you interested in? $\endgroup$ – Ivan Neretin Oct 12 '16 at 2:29
  • $\begingroup$ @IvanNeretin I meant electrolysis (sorry) $\endgroup$ – Aniekan Umoren Oct 12 '16 at 3:12
  • $\begingroup$ You mean to ask that how does an aqueous electrolytic solution also contain H+ and OH- ions? $\endgroup$ – EdmDroid Oct 12 '16 at 3:16
  • $\begingroup$ OK, then what would be the ultimate result of electrolytic oxidation of $\ce{SO4^2-}$? $\endgroup$ – Ivan Neretin Oct 12 '16 at 4:11
  • $\begingroup$ @EdmDroid no I mean to ask how adding an electrolyte to allows electrolysis to occur since the e- are used to redox the ions in the electrolyte instead of breaking the bonds between the atoms of h2o $\endgroup$ – Aniekan Umoren Oct 12 '16 at 4:13
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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.

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