I've created an electrolysis setup by connecting a $6~\mathrm{V}$ battery to a cup filled with saline water via pencils; I am confused as to why only the negative pencil bubbles though. After running the setup for longer than five minutes, the water turns brown but so far that information has not been useful. Why is the negative pole the only one bubbling and if possible, why is the water turning brown?


Your setup, using table salt ($\ce{NaCl}$ plus additives) will initially electrolyse according to the following two reactions, separated by cathodic reduction and aniodic oxidation:

$$\ce{2 H+ + 2 e- -> H2 ^}\tag{CatRed1}$$

$$\ce{2 Cl- -> Cl2 + 2 e-}\tag{AnOx1}$$

These reactions assume no counterions. However, you don’t have dissolved protons in your solution because you aren’t electrolysing hydrochloric acid. Instead, the proton is taken from water molecules and if we include them (and sodium counterions) we get the following improved cathodic reduction equation:

$$\ce{2 Na+ + 2 H2O + 2 e- -> H2 ^ + 2 NaOH}\tag{CatRed2}$$

This still displays the hydrogen gas bubbling away at the cathode but chlorine gas at the anode should also bubble after some time (or at least evolve noticeably). However, with the increase of $\mathrm{pH}$ due to the generated hydroxide, the chlorine produced at the anode will react further:

$$\ce{Cl2 + 2 OH- -> Cl- + OCl- + H2O}\tag{AnOx2}$$

And both of these are soluble ions. The overall reaction of your electrolysis setup is:

$$\ce{NaCl + H2O -> H2 ^ + NaOCl}\tag{RedOx}$$

There will be hydrogen evolution at the cathode but no gas evolution at the anode. The browning will likely be due to a number of side reactions induced by the oxidant hypochlorite anion.

In case you’re wondering why water’s oxygen is not oxidised, this is due to its higher standard electrode potential.

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    $\begingroup$ are there any books you would recommend reading to learn about this sort of stuff? $\endgroup$ – Rob Dawson Nov 14 '16 at 5:19
  • $\begingroup$ @JakeWickham I generally don’t recommend books because I never use them, but there is a question here on chem.SE about good books for certain topics. $\endgroup$ – Jan Nov 15 '16 at 13:42
  • $\begingroup$ I've done this experiment using steel clips (from a car battery charger) and got a brown colour due to the steel being reduced to Fe3+ salts. After the experiment you could clearly see where the clip had been corroded. $\endgroup$ – John Rennie Nov 21 '16 at 12:00

Depends a bit on your voltage, but you might just be making chlorine at the anode. It's pretty soluble in water. I've always used sodium sulfate as the electrolyte.


Try collecting the gas at the negative electrode and testing it. Hold a match to the gas and my guess is that it will go pop. (Positive test for hydrogen) If there is a group 1 metal salt in your solution like $\ce{NaCl}$ the ions would be Na+ and $\ce{Cl^-}$. When the Na+ is reduced to Na by accepting the electron from the cathode (negative terminal) it becomes the metal atom. $$\ce{Na+ + 1e^- \rightarrow Na}$$ This then reacts with the water in the solution to release hydrogen bubbles at the negative electrode. $$\ce{2Na + 2H2O \rightarrow 2NaOH + H2}$$

This is probably why you have bubbles at the negative electrode. If the cup vessel you are doing the experiment in has any imperfections in the ceramics or has any metals in the surface then there is probably a displacement reaction from the Na ions or the Halide ions floating around. There could also be displacement of other ions from inside the saline (saline is a mix of different salts)

That's my best argument for your observations

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    $\begingroup$ This is incorrect. Sodium shall not even touch the cathode as long as species exist that can be reduced more easily (such as hydrogen in water)! $\endgroup$ – Jan Nov 13 '16 at 17:29
  • $\begingroup$ It's saline. Salts (ionic substances)would be reduced far more easily than water (simple covalent structure). $\endgroup$ – ZeroEntropy Nov 13 '16 at 17:35
  • $\begingroup$ Add indicator the negative electrode would show the presence of an alkali. $\endgroup$ – ZeroEntropy Nov 13 '16 at 17:37
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    $\begingroup$ Re first comment: Wrong. Please go back to your books! $\endgroup$ – Jan Nov 13 '16 at 17:40
  • $\begingroup$ I’ll agree with you on the presence of a base near the cathode; see my answer as to how this is formed. $\endgroup$ – Jan Nov 13 '16 at 17:40

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