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Hydrogen and hydroxide both exceed sodium and chlorine in terms of reduction and oxidation potential respectively.

While electrolyzing a concentrated solution of aqueous NaCl, it is known that chlorine is discharged at the anode (contaminated with traces of oxygen). However, at the cathode, hydrogen is charged in place of sodium. Why is this so? I have read many explanations, but they never provide the reasoning.

I suspect these particular properties of factoring in preferential discharge:

  • Atomic mass (and consequently weight).
  • Magnitude of oxidation & reduction potentials.

I'm looking for a thorough, quantitative answer. An example experiment would be lovely, but reasoning is a must.

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Let's take a look at a table of standard electrode potentials and find the relevant half-reactions:

Reductions (there is also a half-reaction for water reduction that doesn't involve a proton at -0.83 V, but for the sake of simplicity, we can ignore it):

$$\begin{align*} &\ce{Na+ + e- <=> Na} & &E°=-2.71\ \mathrm{V}\\ &\ce{2H+ + 2e- <=> H2} & &E°=0\ \mathrm{V} \end{align*}$$

Oxidations (reduction reactions reversed):

$$\begin{align*} &\ce{2H2O <=> O2 + 4H+ + 4e-} & &E°=-1.229\ \mathrm{V}\\ &\ce{2Cl- <=> Cl2 + 2e-} & &E°=-1.36\ \mathrm{V} \end{align*}$$

Now, if you know the concentrations of the species involved, you can use the Nernst equation to figure out the actual potentials, but we can reason around it without knowing the exact numbers.

The big thing to notice is how close the potentials of each of the reductions and oxidations are to one another—the reduction reactions are separated by almost 3 V, but the oxidation reactions are very similar. This means that for reduction, it is far less energetically costly to produce hydrogen gas than to reduce sodium, and in fact, if any sodium were produced, it would instantly react with the water to produce hydrogen gas, so if there is any water around, that half-reaction will be favoured. In the real world, the exact potential hydrogen is produced depends a great deal on the electrode used (see overpotential), but it's more or less impossible to electrodeposit metals with much lower reduction potentials than zinc in aqueous solutions. To actually produce metallic sodium, electrolysis of molten NaCl without any water present is the most common method.

Looking at the oxidations, the story's quite different. It doesn't take much more energy to oxidize chloride than it does water (and if the chloride concentration is high, the oxidation potential will go up). This means that (subject to the effects of the electrodes on overpotential), some amount of both half-reactions will occur and because chlorine is a gas and therefore leaves the solution and the oxidations potentials are so close (leaving little energy to overcome kinetic barriers), the chlorine produced will not significantly oxidize the water.

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  • $\begingroup$ In your answer, you mentioned another possible reduction half reaction for water with a having a standard potential of -0.83V. But that can be ignored anyway as it is less than 0V. Similarly, there is an oxidation reaction directly from Hydroxide ions, i.e. 4OH->O2 + 2H2O + 4e. Why don't we consider THIS reaction at the anode? Comparing the standard electrode potentials, this one is much more likely to occur than the oxidation of chloride that you mentioned. So why does Chlorine still get liberated? $\endgroup$ – Newton Feb 23 '17 at 14:38
  • $\begingroup$ Should I explain my doubt better with a separate question or could you understand what I was saying? Thanks $\endgroup$ – Newton Feb 23 '17 at 14:44
  • $\begingroup$ @newton Remember that these are standard electrode potentials. OH- is going to be at much lower concentration than either H2O or Cl-. This is where you must use the Nernst equation with the actual activities of all the species to get the right potentials. $\endgroup$ – Michael DM Dryden Feb 24 '17 at 18:46
  • $\begingroup$ I get your point, but you used the H+ reaction at the cathode right? And not the one involving H2O. We just got lucky that using either results in the same conclusion at the cathode. $\endgroup$ – Newton Feb 25 '17 at 13:50
  • $\begingroup$ @Newton Yeah, I just picked that one because its potential is 0 V so it's easy to see the huge difference versus sodium reduction. It will really be a mix of several reactions, including the H+ one and in the real world, things like the electrode material you use (which determines the overpotential for each reaction) can make a big difference. $\endgroup$ – Michael DM Dryden Feb 26 '17 at 2:15

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