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For water electrolysis in acidic media ($\ce{TiO2}$ anode and $\ce{Pt}$ cathode) the half reactions for water splitting and $\ce{O2/H2}$ evolution are as follows:

  1. Anode: $\ce{H2O + 4 H+ -> 4 H+ + O2}~~(E = \pu{-1.23 V}\text{ vs NHE})$

  2. Cathode: $\ce{4H+ + 4 e- -> 2 H2} \quad\quad~~(E = \pu{0.00 V}\text{ vs NHE)}$

If I use $\pu{0.1 M}$ $\ce{HCl}$ as my acidic electrolyte, what reaction will occur at the anode (the standard oxidation potential for $\ce{2Cl-}$ to $\ce{Cl2}$ is $\pu{-1.36 V}$ vs NHE). In $\pu{0.1 M}$ $\ce{HCl}$ electrolyte, there should be $\ce{H_2}$ evolution at the cathode. Will the anode maintain $\ce{O2}$ or $\ce{Cl2}$ evolution (or maybe both?)

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Possible reactions :

At Cathode:

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

$\ce{2H+{(aq)} + 2e- -> H2{(g)}}$

At anode:

$\ce{2H2O_{(l)} -> O2_{(g)} + 4H+_{(aq)} + 4e-}$

$\ce{2Cl-{(aq)} -> Cl2{(g)} + 2e-}$

General rules of electrolysis of aqueous solution using inert electrodes :-

  1. Reduction potential of $\ce{H2O}$ > reduction potential of cations of 1, 2 and 13 groups of periodic table. So reduction of $\ce{H2O}$ takes place at cathode.

  2. Reduction potential of $\ce{H2O}$ < reduction potential of other cations (than cations of 1, 2 and 13 group). So reduction of $\ce{H2O}$ does not occur at cathode.

  3. Oxidation potential of $\ce{H2O}$ > oxidation potential of fluoride, sulphite, sulphate, nitrate, etc. So oxidation of $\ce{H2O}$ occurs at anode.

  4. Oxidation potential of $\ce{H2O}$ < oxidation potential of chloride, bromide and iodide, etc. So oxidation of these ions takes place.

Keeping these rules in mind, at cathode reduction of hydrogen ions will take place and hence $\ce{H2}$ gas will be evolved. At anode, oxidation of chloride ions will take place and chlorine gas will be evolved and $\ce{O2}$ won't be evolved.

Although the oxidation potential of $\ce{H2O}$ > oxidation potential of chloride ion, chloride ion gets oxidized because of over potential of $\ce{O2}$ i.e. extra potential is required to initiate a reaction at a required rate. And formation of $\ce{O2}$ from $\ce{H2O}$ is kinetically very slow.

Therefore oxidation of chloride ion takes place.

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  • $\begingroup$ Thanks for your comment. The oxidation potential of H2O is -1.23 V. The oxidation potential of Cl- is -1.36 V. The oxidation potential of SO42- is -2.01 V. Surely then H2O oxidation potential of H2O is less than Cl- and SO42- and so H2O oxidation should be favoured? $\endgroup$ – Guest Feb 12 '17 at 15:43
  • $\begingroup$ Modified my answer. $\endgroup$ – Mitchell Feb 12 '17 at 15:50

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