2
$\begingroup$

In the electrolysis of water, $\ce{H2O}$ molecules are being reduced at the cathode (forming $\ce{H2}$ and $\ce{OH-}$) and oxidized at the anode (forming $\ce{O2}$ and $\ce{H_3O+}$).

It is said that:

The electrolysis of water usually involves dilute, or moderately concentrated, salt solutions in order to reduce the power loss driving the current through the solution, but the presence of salt is not a requirement for electrolysis.

What is the reason for "power loss"? How does the presence of foreign ions change that?

The electrolysis seems possible to happen in pure water alone (as the quote also says).

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ Steeven, notice that I changed the mathrm formatting to ce (chemical equation). That is the more proper format, it's simpler to use for chemical equations and has a slightly different appearance upon rendering. $\endgroup$
    – airhuff
    Feb 11, 2017 at 19:41

2 Answers 2

Reset to default
4
$\begingroup$

The terrible conductivity of water can explain this.

Pure water is a really bad electric medium, since it has a low tendency to ionize by itself (we know the low concentrations of ions involved by looking at the Kw).

This means that the voltage needed to get a current flow from anode to cathode would be really high, with consequences both in term of safety and energetic waste.

A salt dissolved provides ions, which make the diluted solution a good electric medium, and allow a current to easily flow between anode and cathode.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Water molecules reach the cathode > They split into $H_3O^+$ to receive the $e^-$ > The result is $H_2$. I understand that water is very reluctant to split into it's ionic components $H_3O^+$ and $OH^-$, but why does the salt ions do any difference to that? $\endgroup$
    – Steeven
    Feb 11, 2017 at 17:26
  • 1
    $\begingroup$ Why would water molecules reach the cathode in the first place? $\endgroup$
    – Ivan Neretin
    Feb 11, 2017 at 17:44
  • $\begingroup$ @IvanNeretin Is the purpose of other ions to "carry" the water ions away from the electrodes after a reaction? $\endgroup$
    – Steeven
    Feb 11, 2017 at 18:49
4
$\begingroup$

Pure water is a very good insulator. Due to it's very low self ionization, there is not a sufficient amount of ions to carry current unless a significant overpotential is applied in order to overcome this kinetic barrier.

Sea water, which has a salinity of about 3.5%, is about one million times as conductive as pure water. According to this Wikipedia article:

Electrolysis of pure water requires excess energy in the form of overpotential to overcome various activation barriers. Without the excess energy the electrolysis of pure water occurs very slowly or not at all. This is in part due to the limited self-ionization of water.

The article sited by the OP states that the resistivity of pure water is 18 MΩ cm while that of sea water is 20 Ω cm. This implies that a relatively huge overpotential is required to electrolyze pure water. At these high potentials more high-energy side products like ozone are likely to be produced, as well as excess heat. This results in power lost to the production of both chemical and heat energy.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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