Re: Why do water molecules separate when electric current is passed thru

Date: Thu Mar 15 10:29:15 2001
Posted By: Ewen McLaughlin, Lecturer, Chemistry, Swansea College
Area of science: Chemistry
ID: 984347620.Ch

Message:

What is discharged when you electrolyse water depends on the pH of the solution. In acid solution, H+ and H2O are prevalent, but the hydroxide ion (OH-) is not. In alkaline solution H+ is essentially absent and H2O and OH- prevail.

Acid cathode: 4H+ + 4e- = 2H2
Acid anode: 2H2O = O2 + 4H+ + 4e-

Basic cathode: 4H2O + 4e- = 2H2 + 4OH-
Basic anode: 4OH- = O2 + 2H2O + 4e-

If you add either anode reaction to the relevant cathode reaction you get: 2H2O = 2H2 + O2

How do the ions and water molecules ‘know’ which electrode to arrive at? The ions are attracted to the electrodes by electrostatic forces (i.e. positive ions are attracted to negative electrodes and vice-versa). The water is found throughout the solution so there will be lots at either electrode.

The reactions above are highly unlikely, as they involve large numbers of electrons, ions and molecules reacting simultaneously. The ions and molecules seem to break apart in a series of simpler steps. The simplest are the hydrogen ions, H+. The cathode can transfer an electron to the ion to make a hydrogen radical (another phrase for hydrogen atom, but emphasising the single electron it has):

H+ + e- = H*

Hydrogen radicals are very reactive: their electron shells are neither full or empty. Two hydrogen radicals will combine to make a molecule, H2. In this way, the two electrons required to make the hydrogen molecule are transferred one by one to hydrogen ions which then pair up.

2 H* = H2

Water will accept an electron to form a hydroxide ion and a hydrogen radical:

H2O + e- = H* + OH-

Water can also lose electrons in a series of reactions that produce oxygen eventually and hydrogen peroxide (H2O2) as an intermediate:

H2O = OH* + H+ + e-
OH* + H2O = H2O2 + H+ + e-
H2O2 = HO2* + H+ + e-
HO2* = O2 + H+ + e-

The above is based on the electrode potentials found in most advanced inorganic or physical chemistry texts (e.g. Huheey’s “Inorganic Chemistry” or Webelements – see:
http://www.webelements.com/webelements/elements/text/O/redn.html and
http://www.webelements.com/webelements/elements/text/H/redn.html for a fine summary and reference list).

I couldn’t find a similar series for loss of electrons from hydroxide ions, but my guess is this:

OH- = OH* + e-
OH* + 2OH- = HO2- + H2O + e-
HO2- = HO2* + e-
HO2* + OH- = O2 + H2O + e-

Basically, the potential differences at the electrodes force the ions and molecules to accept or release electrons. This forms unstable intermediates (radicals) that react to form the familiar products. Might this be the answer to your question? Let’s hope so!

Other references:
What is electrolysis? Simple summary
Rigorous discussion for school students in India. Some knowledge of cricket required to understand analogies (e.g. Sachin Tendulkar is a good batsman, Debashish Mohanty is not). Don’t let that put you off!

Where will the products of the electrolysis of water form?
Details of hydro/electro-lysis
No-oxygen problem
Intramolecular bonding in water.

Ewen McLaughlin

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