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When you electrolyse an aqueous solution of $\ce{NaCl},$ the product formed at cathode is $\ce{H2}$ gas (preferential discharging), but the product formed at anode is $\ce{Cl2}$ gas. According to standard electrode potential, reduction of $\ce{Cl2}$ is 1.36 V and $\ce{O2}$ is 1.23 V.

Keeping the values in mind, $\ce{O2}$ gas should be formed at anode because it has a smaller electrode potential. But it is given that chloride ions is preferred by anode because of a phenomena called overvoltage.

I don't understand how overvoltage can effect the products of electrolysis.

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    $\begingroup$ Basically, there is a kinetic hindrance to the production of oxygen gas at the anode. Overcoming that requires increasing the voltage over what would be required if there was no hindrance. The extra voltage is the over-voltage. The over-voltage depends on factors such as anode composition and surface characteristics. $\endgroup$ – Ed V Jun 3 at 18:04
  • $\begingroup$ @EdV Where does this kinetic hindrance come from? $\endgroup$ – Leah Jun 3 at 18:07
  • $\begingroup$ Good question! I have no idea about the exact mechanistic details that are ultimately responsible for the hindrance. It is likely very complicated and hard to study. But the oxidation of chloride is evidently relatively easy: take electrons from a chloride ions and then chlorine atoms form chlorine molecules. $\endgroup$ – Ed V Jun 3 at 18:11
  • $\begingroup$ @EdV But I just feel like I can't grasp the entirety of preference of O2 gas over Cl2 gas. I have so many questions! My textbook says the slowness of the electrode reactions create a resistance on the electrode surface which the over voltage needs to overcome? How is this resistance being set up? How do I predict products of electrolysis of other salts like PbBr2 without knowing if it is either preferential discharge or over voltage the priority that the ion is following? Its all very confusing for me. $\endgroup$ – Leah Jun 3 at 18:20
  • $\begingroup$ The resistance is just another way of looking at it: there is a reaction rate “bottleneck”, i.e., the rate of oxidation is hindered. So the current flow, involved in oxygen production, is reduced by this impedance. Over-voltage effects are basically empirical and you just have to deal with not having certainty. Sorry, but that is how it plays out. $\endgroup$ – Ed V Jun 3 at 18:26
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When you electrolyse an aqueous solution of NaCl, the product formed at cathode is H2 gas (preferential discharging), but the product formed at anode is Cl2 gas.

There is another complication. The book is missing the fact whether chlorine or oxygen is evolved depends on the concentration of the salt. At low salt concentrations, oxygen will be formed and at high NaCl concentrations $\ce{Cl2}$ will be produced at the anode.

Keep in mind the thermodynamic "competition" is between the oxidation of water and oxidation of chloride ions. Electrochemists have spent their lives and careers in understanding the oxidation reduction of water for hundreds of years. So if you are slightly confused and you have a lot of questions that is a sign of a healthy scientific mind. I am glad that you did not digest what the textbook was saying right away. Wikipedia hints as how complex is the process of hydrogen and oxygen production at a electrode.

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The numbers which you are quoting are like comparing apples and oranges. This half cell $$\ce{2H2O(l)} → \ce{O2(g) + 4 H+(aq) + 4e−}$$ $E^o$= +1.23 V is valid for pH of 0, i.e., 1 M H$^+$ ion. Your aqueous NaCl solution is not present in 1 M H$^+$.

You might need to use the Nernst equation to see what is the true potential at a neutral pH.

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  • $\begingroup$ So the given electrode potentials aren't correct because here NaCl is more in concentration? Then why would my textbook give the reason of over-voltage? I think the experiment not taking place under standard conditions is a better explanation than over-voltage. $\endgroup$ – Leah Jun 4 at 5:53
  • $\begingroup$ No my point is that we cannot ignore concentrations. Read about Nernst equation first. Overpotential is another story but first think about what is the concentration of NaCl. $\endgroup$ – M. Farooq Jun 4 at 5:57
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First one comment on "overvoltage": It doesn´t mean that there is zero reaction below. Even below the theoretical electrochemical potential, electrolysis runs, very slowly. Around the potential plus overvoltage, the slope of the U-I plot has a bend and goes from close to zero to a constant, finite value.

Very simplistic explanations (or rationalisations, rather) for the overvoltage:

To create a diatomic gas, you need to oxidise or reduce two ions in very close proximity. That requires a little extra energy/potential, which turns into heat once the two single atoms recombine. Depends on the mobility of the reduced atom, the adsorbance of which also is an equillibrium process.

There are more possible contributions. The extra "dipolar" charge on the ion by its deformed hydrate hull requires a small extra potential to overcome. In the moment the ion discharges, the ordered water molecules suddenly depolarise, again creating heat.

To adsorb the ion on the surface, you need to push aside other species, which might have a weaker or stronger interaction with the electrode.

Etc.

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    $\begingroup$ I am slightly unsure about it "To create a diatomic gas, you need to oxidise or reduce two ions in very close proximity. That requires a little extra energy/potential". The occurrence of two Cl radicals nearby should be a completely random process and it would to be dependent on the concentration of the ions next to the electrode. Why would appearance of two oxidized Cl radicals next to each require an additional energy? $\endgroup$ – M. Farooq Jun 4 at 2:43
  • $\begingroup$ @M.Farooq Good point. Perhaps if you put it this way: The discharge of the ions is an equillibrium process. Either you push it in one direction with a larger potential, or the probability of two radicals occuring next to each other becomes very small. If the radicals are very mobile on the surface, the overvoltage would be smaller, and we know it varies a lot with different electrode materials. $\endgroup$ – Karl Jun 4 at 6:59

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