The electrode potential for Bromine is $\pu{+1.07V}$ whereas for Oxygen to be produced it is $\pu{+0.40V}$.

$$\ce{O2 +2H2O + 4e- -> 4OH-}$$

As the electrode potential for bromine is higher, it means bromine should be reduced rather than $\ce{OH-}$ ions.

I'm aware that the concentration affects the electrode potentials, but the question simply states 'an aqueous solution of NaBr' with no reference to concentration whatsoever, so I can only assume standard solutions.

Also, why do we not look at the electrode potentials of other possible reactions at the anode, like;

$$\ce{O2 + 4H+ + 4e– -> 2H2O}~~~~~~~(E=\pu{+1.23V})$$ or

$$\ce{HO2- + H2O + 2e– ->3OH–}~~~~~~(E=+\pu{0.88V})$$ or

$$\ce{H2O2 + 2H+ + 2e– -> 2H2O}~~~~~~~ (E=\pu{+1.77V})$$.

Why are these not viable?

  • $\begingroup$ You don't have a standard solution of $\ce{OH-}$ in there, otherwise things might have turned differently. $\endgroup$ Commented Apr 27, 2017 at 6:56
  • $\begingroup$ But all I'm told is that it is aqueous. How do I judge if it is a standard or non-standard solution? $\endgroup$
    – John
    Commented Apr 27, 2017 at 7:18
  • $\begingroup$ What is a standard solution, come to think of it? $\endgroup$ Commented Apr 27, 2017 at 7:35
  • $\begingroup$ A solution where the concentrations of any ions is 1M $\endgroup$
    – John
    Commented Apr 27, 2017 at 7:47
  • 2
    $\begingroup$ Sodium won't be reduced; you can't make the concentration of H+ low enough for that. As for the other reactions, some of them might occur to some minor extent, depending on the conditions. $\endgroup$ Commented Apr 27, 2017 at 11:11

1 Answer 1


This is how I see it:

In the electrolysis of NaBr, water is reduced at the cathode. This occurs because water is more easily reduced than are sodium ions. This is reflected in their standard reduction potential.

At cathode reduction of water occurs:

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

And hydrogen gas is produced

At the anode, where oxidation occurs, the standard oxidation potential of water is –1.23 volts, while that for bromide ions is -1.07 volts.

The production of bromine itself has a negative electrode potential. One of the half-reactions must be reversed to yield an oxidation.

$$\ce{Br^2- -> Br2 + 2e-}~~~~~~~ Eº -1.07$$

Remember that when one reverses a reaction, the sign of Eº (+ or –) for that reaction is also reversed.

This means that bromide ions are more easily oxidized than water.

It is important to note: When current begins to flow, the distribution of ions around the electrodes changes, and the equilibrium electrode potentials no longer accurately apply.

I think other factors also apply;

  • Concentration
  • Temperature
  • Nature of ions
  • $\begingroup$ Why is water reduced at the cathode? Shouldn't it be Hydrogen ions since they have an electrode potential of 0 V? Also, could you write the reaction in which water is reduced? The 1.23V reaction I know has Oxygen being reduced, not water. $\endgroup$
    – John
    Commented Apr 27, 2017 at 10:00
  • $\begingroup$ I added the equation showing reduction of water at cathode $\endgroup$ Commented Apr 27, 2017 at 11:10
  • $\begingroup$ The electrode potential for the reduction of water is -0.83V whereas for H+ ions, it is 0.00V. As such, shouldn't Hydrogen ions be reduced? Also, why are you considering the +1.23V reaction of water rather than the +0.40V one? $\endgroup$
    – John
    Commented Apr 27, 2017 at 11:45
  • $\begingroup$ And where are you generating the hydrogen ions from in the first place? Please show the reaction that you are referring to. Also read Ivan Neretin's comments they essentially answer your additional questions $\endgroup$ Commented Apr 27, 2017 at 11:57
  • $\begingroup$ The hydrogen ions come from the water present H2O <---> H+ +OH- $\endgroup$
    – John
    Commented Apr 27, 2017 at 12:21

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