There is a list of standard electrode potentials at 298 K from the p. 23 of IB Data Booklet 2016. Which of the following equations (forward/backward reactions), from the two possible ones involving the discharge of hydrogen gas and the other two with oxygen gas discharge, should I use for the oxidation and reduction of water in electrolytic cells?

$$ \begin{array}{cc} \hline \ce{\text{Oxidized species} <=> \text{Reduced species}} & E^⦵(\pu{V}) \\ \hline \begin{align} \ce{H2O(l) + e- &<=> 0.5 H2(g) + OH-(aq)} \\ \ce{H+(aq) + e- &<=> 0.5 H2(g)} \\ \ce{0.5 O2(g) + H2O(l) + 2 e- &<=> 2 OH-(aq)} \\ \ce{0.5 O2(g) + 2 H+(aq) + 2 e- &<=> H2O(l)} \end{align} & \begin{array}{r} -0.83 \\ 0.00 \\ +0.40 \\ +1.23 \end{array} \\ \hline \end{array} $$

(Unless the use of any of these equations cannot be generalized — for a concise explanation of why this is so and what to do then I would be equally grateful.)


1 Answer 1


For the acidic electrolysis, use the reactions where $\ce{H+}$ occurs.

As $\ce{OH-}$ is not available in considerable amount there as a reagent, neither it is created as a product.

Generally, for a reaction choice, apply the principle of availability and stability, allowing for a reagent to exist in (relative) abundance.
$\ce{OH-}$ or anions of weak acids like $\ce{ClO-}$ do not survive in acids. Acids do not survive in hydroxides.

But note that using reactions with half of a molecule is not necessery.

$$\begin{align} \ce{O2(g) + 4H+(aq) + 4e- &<=> 2 H2O(l)}\\ \ce{2H+(aq) + 2e- &<=> H2(g)} \end{align}$$

For the alkaline electrolysis, similarly, use the reactions where $\ce{OH}$- occurs.

$$\begin{align} \ce{2 H2O(l) + 2e- &<=> H2(g) + 2 OH^-(aq)}\\ \ce{O2(g) + 2 H2O(l) + 4e- &<=> 4 OH^-(aq)} \end{align}$$

  • $\begingroup$ Thank you, @Poutnik. Could you also explain to me how I could apply this knowledge to, for example, the electrolysis of aqueous sodium chloride? A textbook example uses equations from both categories you mentioned. $\endgroup$
    – w_w
    Apr 19, 2019 at 16:06
  • 1
    $\begingroup$ Apply the same principle of availability and stability. $\endgroup$
    – Poutnik
    Apr 19, 2019 at 16:27
  • $\begingroup$ @Poutnik, I suppose that by "availability" you mean concentration of the salt, but I am not sure how to tackle "stability"... Does "stability" have to do with electrode potential values? I am too verdant to figure this out by myself. $\endgroup$
    – w_w
    Apr 19, 2019 at 17:07
  • $\begingroup$ @w_w I mean ability to survive. $\ce{OH-}$ or anions of weak acids like $\ce{ClO-}$ do not survive in acids. Acids do not survive in hydroxides. Generally, choice duch a reaction form, where reactants on both sides may exist in abundance. $\endgroup$
    – Poutnik
    Apr 19, 2019 at 17:10
  • $\begingroup$ I forgot to ask: If we have a neutral solution can we act and choose the equation as if we want to keep it neutral or is this not necessarily the case? $\endgroup$
    – w_w
    Apr 19, 2019 at 22:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.