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Can anyone explain why in the electrolysis of aqueous solutions such as zinc nitrate, the oxidation half equation shows hydroxide ions forming oxygen and water - can this half equation be balanced producing oxygen and hydrogen ions - if not why not?

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  • $\begingroup$ Indeed it can. It might even be more appropriate that way. Also, welcome to Chem.SE. $\endgroup$ – Ivan Neretin Mar 7 '16 at 6:27
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You mean, why we write $\ce{4OH- -> O2 + 2H2O + 4e-}$ and not $\ce{4H2O -> O2 + 4H+ + 4e-}$?

Indeed, you may find both these reactions in the table of standard electrode potentials (at $E^0=+0.401$ and $+1.229$, correspondingly; also, note the reversed direction). In some sense, the second equation may even be considered more appropriate, since $\ce{H+}$ is a more abundant species in the solution than $\ce{OH-}$ (zinc salts would produce slightly acidic solutions due to hydrolysis).

But the first reaction was chosen not without reason. See, when we electrolyze something, it is anions that rush to the anode to get oxidized. Water molecules are neutral; electric field would not drag them in either direction. So this might be the reason why the equation is written this way, even if the amount of $\ce{OH-}$ is minuscule and the overall reaction looks more like the second one.

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  • $\begingroup$ Thanks for that, but what about 2OH−⟶O2+2H+ 2e− $\endgroup$ – B West Mar 8 '16 at 20:06
  • $\begingroup$ That's out of question. There is no such thing as H, to begin with. If you mean $\ce{H2}$, then it is still impossible. You just can't have both oxidation and reduction in one half-reaction. $\endgroup$ – Ivan Neretin Mar 8 '16 at 20:32
  • $\begingroup$ I really appreciate your help. The question that is bothering me is a “selective discharge” question . The example has used aqueous copper sulfate. They have given the reduction of Cu2+ as one half equation and the oxidation of OH- as the other. However the text has written the oxidation half equation as 4OH- →O2 + 2H2O + 4e My query is can the oxidation equation be written as 2OH- →O2 + 2H+ + 4e $\endgroup$ – B West Mar 10 '16 at 23:15
  • $\begingroup$ Now you propose a mix between the two. Well, no, nobody writes it this way; it would look kinda unnatural. $\endgroup$ – Ivan Neretin Mar 11 '16 at 5:53

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