On The Production of
Hydrogen Gas by The
Electrolysis of Water
Johnston
Page 2
In the diagram (Figure 1) is a vessel filled with water and an
electrolyte. I chose sulfuric acid (H2SO4) as the electrolyte because
it seemed to be Faraday's favorite. When the electrolyte (which could
also be represented as SO3, before hydration) is placed into the
water it links up with water molecules to become H2SO4 (sulfuric
acid). It apparently does this by attaching /attracting to itself the
oxygen atom from a water molecule.
Figure 2:
As you can see, even with three oxygen atoms attached to the
sulfur atom (SO3), there are two "holes" remaining in the valence
shell of the sulfur. All three oxygen atoms have their shells
entirely filled though. This being accomplished by each oxygen atom
sharing two of the sulfur's electrons. So then, next the oxygen atom
from an H2O molecule is attracted into the SO3 atom to fill those
last two holes in the shell of the sulfur atom.
This happens because the oxygen is more strongly attracted to the
sulfur atom than it is to the hydrogen atoms. Interestingly enough
the ionization potentials of both hydrogen and oxygen are an
identical 13.6 electron volts. "The oxidation potential of an atom is
an expression of the tightness with which an atom holds on to an
electron and is defined as the energy required to pull an electron
off an atom. One electron volt (3.8 x 10 to the negative 20th
power,cal) is equivalent to 23 kcal per Avogadro number of
atoms."(1)(Section 3.14, Page 66)
The result of this bonding creates a problem for the hydrogen
atoms, which are still attached to the oxygen atom, which has now
linked into the SO3 (now SO4) molecule. They (the H2 atoms) would now
have to "share" their single electrons with the oxygen, which is also
sharing two of it's electrons with the sulfur atom (which has an
ionization potential of 10.4e.v.). This double sharing is apparently
the force which weakens the H2 bond to the oxygen enough to
ultimately break it. "In other words, the formation of a chemical
bond indicates that the resulting molecule represents a state of
lower energy than the isolated atoms." (1)(Section 4.1, Page 76) And
also; "Since oxygen is the more electronegative (in a water
molecule), the shared electron pair is not shared equally but belongs
more to the oxygen than to the hydrogen. In the course of combining
H2(g) with O2(g) there is a change in electron sharing." (1)(Section
5.4, Page 104)
Part of the reason for this is that the oxygen, coupled to the
sulfur atom now has a complete octet (8) of electrons in it's outer
shell, as does the sulfur. "A shell of eight electrons, the so-called
octet is especially difficult to break up. It is especially hard to
pull an electron off an atom having eight electrons in it's outermost
shell." (1)(Section 3.14, Page 68) But, conversely, the oxygen atom
while in the water molecule also had a full octet shell by sharing
one of it's electrons with each hydrogen atom, which in turn each
shared it's one electron with the oxygen atom; "By sharing electrons,
the oxygen atom gets a complete octet of electrons in it's outer
shell, and each hydrogen atom gets two electrons, all that it's
valence shell can accomodate. Since the molecule as a whole is
electrically neutral and since all valence shells are filled with
shared electrons, no other atoms can bind to the molecule. The
valence is saturated." (1)(Section 4.7, Page 90) So it is aparrently
easier to for the oxygen atom to pull the hydrogen's electrons away
from them, when it joins the SO3 molecule than it would be for the
hydrogen to take them back from the oxygen.
The electrons that originally belonged to the hydrogen atoms now
seem to be more strongly attached to the oxygen atoms due to their
being shared than they are to the hydrogen atoms to whom they
initially belonged. This, by necessity, due to the octet rule, forces
the two H atoms (minus their electrons) away from the complete SO4
molecule and, in so doing, creates two H+ ions, which are the two
hydrogen atoms without their single electrons. Am I trying to say
that these H+ ions (which amount to bare protons) now go racing off
through the solution on their own? No. Chemistry invokes something
called "Water of Hydration" to take care of that problem. "The
ionization of a solute by water can be considered a chemical reaction
and can be described by a chemical equation such as:
HCl + H2O ------> H3O+ and Cl-"
"Since the hydronium ion, H3O+, can be considered as a hydrated
proton, and since water of hydration is often omitted from chemical
equations, the above reaction can be written more simply as:
HCl-----> H+ and Cl-"
"with the tacit understanding that all species are hydrated."
"The practice of omitting water of hydration from chemical
equations is dangerous unless we constantly bear in mind that, in
aqueous solution, water is always associated with any dissolved
species and may affect it's properties. The danger is greatest in the
case of the hydrogen ion, because the hydrogen ion is nothing but a
bare proton (nucleus of H atom). Whereas H+ is essentially of zero
size, H3O+ has a volume which is about 10 to the fifteenth power as
great and is comparable in size to other ions." (1)(Sections 10.6
& 10.7, page 204)
Note: The above example assumes that the HCl which is added to the
water is already formed. To form HCl, Cl gas can be dissolved in
water where it breaks apart water molecules to form HCl and an OH-
ion. So technically the above example ignores the presence of the OH-
ion. In practice, the gasses evolved in a cell containing water and
HCl will be H2 at the cathode and either O2 or Cl2 at the anode with
the evolution of Chlorine gas being the most common result,
indicating that Cl is more easily oxidized than the OH- ion.mj
Very simply then, H+ ions, once separated from their original
water molecule by the action of the electrolyte (in this case the
oxidation of the sulfur atom in the SO3 molecule by the oxygen atom
in the water molecule), are then picked up and carried by sympathetic
H2O molecules. So then; "The only way by which electrons can be
shifted away from an atom is for them to be pulled toward another
atom. In this process (oxidation) the oxidation number of the first
atom increases, and the oxidation number of the second atom
decreases. Oxidation and reduction must always occur together and
must just compensate each other. It should be evident that when a
substance acts as a reducing agent, it itself must be oxidized in the
process" (1)(Section 5.6, Page 108)
Ionic Bonds then can be described as bonds in which electrons are
completely transferred from one atom to another. So the formation of
an ionic bond can be thought of in three steps (using NaCl as an
example):
Na -----> Na+ and e-
Cl(with 7 electrons) + e- ------> CL(with 8 electrons)- (minus
sign after "e" meaning one "extra")
Na+ plus Cl- ---------> Na+[Cl]-
Step (1) Requires energy equal to the ionization potential of
sodium (5.1e.v.)
Step (2) Releases energy equal to the electron affinity of
chlorine (3.75e.v.)
Step (3) Releases energy because of the attraction between
positive and negative ions. The ionic bond is formed only because the
energy released in steps (2) and (3) is greater than that required in
step (1).
So when the oxygen links with the sulfur in the SO3 molecule it
pulls the electrons completely away from the hydrogen atoms, creating
the H+ ions. Please note that all of this happens when SO3 is mixed
into water and that no addition of energy from an outside source is
required to split the water molecules. Whether in an electrolysis
cell or a primary (voltaic) cell, the splitting of water into it's
component parts is accomplished by chemical reactions. Once all of
the SO3 has reacted with available water molecules a state of
equilibrium is reached and no more reaction occurs.
The fact that energy is used up in the process of adding a solute
to a solvent can also be seen, on a more macroscopic level, by noting
that; "When sugar is placed in a beaker of water, the sugar
disappears. in the solution process, the sugar lattice is ripped
apart, and the individual sugar molecules are distributed throughout
the solution. The process can be envisioned as follows: First, a
sugar molecule must be pulled from it's neighbors. Since molecules
attract each other in the crystal lattice, work must be done against
the attractions of the molecules remaining in the lattice; i.e.,
solute-solute attraction must be overcome. Second, the water
molecules must be pushed aside to make a hole to accomodate the sugar
molecule. This process also requires the expenditure of energy, since
the molecules of water also attract each other: i.e., solvent-solvent
attraction must also be overcome. Since both of these processes
require energy, they of themselves cannot account for the fact that
solution occurs.
There must be a third step which provides the energy that is
needed. This third step is called solvation and arises from an
attraction between the sugar molecules and the water molecules.
Sugar-water attraction supplies almost enough energy to overcome the
sugar-sugar attraction and water-water attraction. The last bit of
energy required is supplied by a reduction in the average kinetic
energy of all of the molecules. In this case solution is accompanied
by a slight drop in temperature."
"The very fact that water dissolves sugar implies that H2O
molecules exist preferentially in the solution state rather than in
the pure water state."
"There are many cases in which the solution process is accompanied
by the disassociation or breaking apart of molecules. The
disassociated fragments are usually electrically charged, so that
electrical measurements can show whether disassociation has occured.
Charged particles, or ions, in solution carry electrical current,
whereas water conducts electricity very poorly. Electrical
conductivity requires charged particles. The greater the number of
charges available for carrying electricity, the greater the
conductivity observed. Strong electrolytes are essentially 100 per
cent disassociated into ions, whereas weak electrolytes may be
disassociated only a few percent."
For ionic substances, it is not surprising that there are charged
particles in solution, because the solid itself is made of charged
particles. The solvent rips the lattics apart into it's constitiuent
pieces. Figure 10.10 (not reproduced) shows what is thought to happen
when the ionic solid NaCl dissolves in water. Since the chloride ion
is negative (like the SO4 -2 ion in sulfuric acid), the positive ends
of water molecules cluster about the chloride (or SO4 -2) ion. The
chloride ion surrounded by it's cluster of water molecules now moves
off into the solution. It is now a hydrated chloride ion. The species
is negatively charged because the chloride ion itself is negatively
charged. At the sane time the sodium ion undergoes similar hydration
(like the H+ ion in sulfuric acid), with the difference being that
the negative or oxygen end of the water molecule faces the positive
ion. Since the solution as a whole must remain electrically neutral,
an equal number of hydrated sodium ions and hydrated chloride ions is
formed. When positive and negative electrodes are inserted into this
solution, the positively charged hydrated sodium ions (or H+ in
sulfuric acid) are attracted to the negative electrode, the
negatively charged hydrated chloride ions are attracted to the
positive electrode. There is a net transport of electrical charge as
the positive charge moves in one direction and the negative charge
moves in the opposite direction." (1)(Section 10.6, Pages
200-203)
Or, as another example, consider electrically neutral HCL. The HCL
molecules interact with the solvent to form ions;
HCL + H2O --------> H30Cl -------->H30+ and Cl-
HCL molecules collide with water molecules to give an unstable
intermediate.This immediately forms H30+ and Cl- as shown.