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The reduction of water at a cathode is represented by the following equation:

Cathode (reduction): $\ce{2 H2O_{(l)} + 2e^{−} → H2_{(g)} + 2 OH^{-}_{(aq)}}$

If this is done in a divided cell, what would happen to the $\ce{OH-}$ anions? The hydrogen gas would accumulate and bubble would form. Would the $\ce{OH-}$ anions just build up in the solution? Do they interact with each other naturally forming $\ce{H2}$ and $\ce{O2}$?

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  • $\begingroup$ Will you observe evolution of hydrogen at all if your half cells are separated by a impermeable wall? $\endgroup$ – Klaus-Dieter Warzecha Jan 29 '14 at 23:52
  • $\begingroup$ Would you please post species involved in both the half cells. $\endgroup$ – Immortal Player Jan 30 '14 at 13:06
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You only show one a half-reaction, reduction at the cathode. Write the other half-reaction, oxidation at the anode to give oxygen. A real world water electrolysis cell requires a salt electrolyte for conductivity. $\ce{NaCl}$ could be electrolyzed to $\ce{Cl2}$ and $\ce{Na}$, the latter reacting with water to give $\ce{H2}$ and $\ce{NaOH}$.

http://www.ineris.fr/ippc/sites/default/interactive/brefca/image16.gif

If you want $\ce{H2}$ and $\ce{O2}$, make the electrolyte $\ce{NaOH}$. $\ce{Na2CO3}$ can be interesting.

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