Electrolysis of dilute sulfuric
acid - the products of electrolysing water acidified with sulfuric acid are hydrogen
gas and oxygen gas
 Two experimental setups are described, the
Hofmann voltammeter demonstration (left diagram) and a simple cell (right
diagram) for use in schools and colleges for pupils to use. Dilute sulfuric acid
is used as the electrolyte in this investigation. The Hofmann voltammeter is
filled with the electrolyte (dilute sulfuric acid) by opening the taps at the top of
the outer tubes to allow any gas to escape. The gases formed on the electrolysis
of the dilute sulfuric acid can be collected via the same taps. The platinum or
carbon electrodes are inert.
You need inert (nonreactive)
electrodes like platinum (left) and much cheaper carbon (graphite electrodes, right). In
the simple electrolysis cell, the graphite (carbon) electrodes are inserted through a
large rubber bung, 'upwardly' dipped into an solution of acidified water. In
this cheap and simple apparatus the gaseous products (hydrogen and oxygen) are
collected in small test tubes inverted over the carbon electrodes. You have to
fill the little test tubes with the electrolyte (dil. sulfuric acid), hold the
liquid in with your finger and carefully invert them over the nearly full
electrolysis cell. The gases can
be collected and tested.
On
the right is another simple apparatus for doing experiments like the
electrolysis of water. The electrodes must be made of an inert wire.
Students should note that the
electrolysis will only take place when electricity is passed through the dilute
sulfuric acid solution.
The electrolyte is dilute
sulfuric acid (= acidified water) which, during electrolysis is
split into hydrogen and oxygen gases. Theoretically into a 2 : 1 ratio by gas volume
- which you can measure with the Hofmann Voltammeter.
This is one experimental method of showing water is a compound composed
of the elements hydrogen and oxygen atoms i.e. by splitting liquid water into
two gaseous element molecules of hydrogen (H2) and oxygen (O2).
Water only ionises to a tiny extent giving minute
concentrations of hydrogen ions and hydroxide ions, so the presence of high
concentrations of hydrogen ions (H+ or H3O+) and
sulfate ions (SO42) from the sulfuric acid, makes water a much better
electrical conductor (a much better electrolyte solution) than pure water.
These small extra ion concentrations of hydrogen ions (H+) and hydroxide ions
(OH)
are from the self-ionisation of water itself.
The majority of liquid water
consists of covalent H2O molecules, but there
are trace quantities of H+ and OH ions from the
reversible selfionisation of water:
H2O(l)
H+(aq) + OH(aq)

Diagram
explaining the electrolysis of water showing the ion movement to the electrodes..
The electrode reactions and products of the
electrolysis of acidified water are illustrated by the theory diagram above
The half-equations for the electrolysis of water
electrode equations (electrolyte of acidified with dilute sulfuric acid) are
described and explained below.
(a) The negative cathode
electrode reaction for the electrolysis of water
The negative cathode electrode
reaction is a reduction
(electron gain).
The
hydrogen ions (H+) are attracted to the negative cathode and are discharged as
hydrogen gas. The hydrogen
ion or water molecules are reduced to hydrogen gas molecules by
electron gain at the negative electrode
2H+(aq)
+ 2e ==> H2(g)
(cathode half reaction) Hydrogen gas
bubbles seen on the negative cathode electrode)
The positive hydrogen ion is reduced by
electron gain to form hydrogen molecules.
You can
also, at a higher level, write the reduction half reaction as ...
2H3O+(aq)
+ 2e ==> H2(g) + 2H2O(l)
(cathode half reaction) The two equations amount to the
same overall change i.e. the formation of hydrogen gas molecules
and as far as I know any is acceptable in an exam? The electrolysis
of many salts (of reactive metals) or acid solutions produce hydrogen at
the negative cathode electrode.
All acids give hydrogen at the cathode.
Hydrogen ions are reduced to hydrogen in preference
to the ions of reactive metals.
(b) The positive anode electrode reaction for the electrolysis of water
The positive anode reaction is an oxidation
electrode reaction (electron loss).
The negative sulfate ions (SO42-)
or the traces of hydroxide ions (OH) are attracted to the
positive electrode, but the sulfate ion is too stable and nothing happens (its
not discharged).
Instead either hydroxide ions or water molecules are discharged and oxidised to form
oxygen. The hydroxide ions or water molecules are oxidised to oxygen gas molecules
by electron loss at the positive electrode
(i) (+)
2H2O(l)
4e ==> 4H+(aq)
+ O2(g)
(anode half reaction)
Oxygen gas bubbles are seen at the positive anode
electrode.
or
2H2O(l) ==> 4H+(aq)
+ O2(g) + 4e
The water molecule is oxidised by electron
loss, resulting in the formation of oxygen.
(ii)
You can also write the half reaction at the anode as the oxidation of
the hydroxide leading to the formation of oxygen molecules.
4OH(aq)
4e ==> 2H2O(l)
+ O2(g)
(anode half reaction)
negative ion oxidation by
electron loss
or
4OH(aq) ==> 2H2O(l)
+ O2(g) + 4e
(anode half reaction)
There are two
possible electrode equations that describe the formation of oxygen in the electrolysis of
water, but in strongly acid solution equation (i) is probably more
appropriate, whereas in alkali, (ii) is the more likely electrode
equation.
Both equations amount to the same
overall change i.e. the formation of oxygen gas molecules and as
far as I know either is acceptable in an exam?
The overall equation for the electrolysis of
water is:
2H2O(l) ==> 2H2(g) + O2(g)
Note the 2:1 molar
ratio of H2:O2 gases.
Extra COMMENTS on the electrolysis of
water
1. The electrolysis
of many salts (e.g. sulfates of reactive metals) or sulfuric acid
produces hydrogen at the negative cathode electrode and oxygen at the positive anode electrode.
Whereas hydrochloric acid
gives chlorine at the anode (as will all chloride salts), the sulfate ion does nothing and instead oxygen is
formed.
See Part 7.
Electrolysis of hydrochloric acid
So for most sulfate salts of reactive metals e.g.
sodium sulfate, magnesium sulfate and potassium sulfate, the electrolysis products of the aqueous salt
solution are hydrogen
at the negative () cathode electrode and
oxygen at the positive (+) anode
with inert electrodes such as carbon or platinum.
So the electrode
equations are the same as above e.g.
(i) 2H+(aq)
+ 2e ==> H2(g) (at the -ve
cathode)
(ii)
2H2O(l)
4e ==> 4H+(aq)
+ O2(g) (at the +ve anode)
or
4OH(aq)
4e ==> 2H2O(l)
+ O2(g)
The electrolysis of sodium hydroxide gives
exactly the same products, hydrogen and oxygen in exactly the same proportions
(2:1 by volume) so, as with the salt examples quoted above, the electrode equations are the
same as above.
Aqueous solutions containing the cations of reactive
metals like potassium, sodium or magnesium are NOT discharged at the
negative cathode electrode, but hydrogen ions are, so you get hydrogen
instead of a metal formed on the electrode surface.
Generally speaking, the
less reactive a metal, the more easily its ion is reduced to
the metal on the electrode surface.
A general rule with reference to the
reactivity series of metals:
If the metal in the salt solution
is more reactive than hydrogen, then the
hydrogen ion is most likely to be discharged at the negative cathode giving
hydrogen.
If the metal in the salt solution
is less reactive than hydrogen it is the metal ion that
is likely to be discharged forming a deposit of the metal
on the electrode surface.
These rules apply if a mixture of ions is present in the electrolyte.
2.
Theoretically the gas volume ratio for H2:O2
is 2:1 which you see with the Hofmann Voltammeter.
This isn't just because the ratio of hydrogen atoms
to oxygen atoms is 2 : 1 in the molecule
It is equally important that you
realise for every four electrons that flow through the circuit four hydrogen
ions are reduced to two molecules of hydrogen AND two molecules of water (or
four hydroxide ions) are oxidised to give one molecule of oxygen.
Therefore, passage of the same quantity of electrical current (electron flow), you
would expect two molecules of hydrogen to one of oxygen in the electrolysis of
water.
Since both gases are only very slightly soluble in water, you should
observe an approximate 2:1 ratio for H2:O2 gas
volumes,
3. Chemical Tests
for the gases formed from electrolysis of water
experiment
You can collect samples of gases
through the taps on the Hofmann voltammeter or from the little test tubes in the
simple school electrolysis cell.
At the negative cathode electrode a colourless
gas that gives a squeaky pop when ignited with a lit splint -
hydrogen.
At the positive anode a
colourless
gas that relights a glowing splint - oxygen.
4.
What is left in the electrolyte
solution after prolonged electrolysis?
As the electrolysis proceeds the dilute
sulfuric acid gets slightly more concentrated as only water is removed
from the electrolyte, albeit as hydrogen and oxygen in a 2:1 ratio.
SUMMARY OF PRODUCTS FROM THE ELECTROLYSIS
OF acidified WATER
with inert carbon (graphite) electrodes
or inert platinum electrodes |
Electrolyte |
negative cathode
product |
(-) negative electrode
cathode half-equation reduction - electron gain |
positive anode
product |
(+) positive electrode
anode
half-equation oxidation - electron loss |
dilute solution of
sulfuric acid
sulfuric acid
H2SO4(aq)
but can be sulfate salts of reactive
metals whose ions are not discharged, and alkaline hydroxides of
reactive metals like sodium hydroxide |
hydrogen gas |
2H+(aq)
+ 2e ==> H2(g)
or
2H3O+(aq)
+ 2e ==>
H2(g) + 2H2O(l) |
oxygen gas |
(i) 4OH(aq)
4e ==> 2H2O(l)
+ O2(g)
or
4OH(aq) ==> 2H2O(l)
+ O2(g) + 4e
(ii)
2H2O(l)
4e ==>
4H+(aq)
+ O2(g)
or
2H2O(l) ==>
4H+(aq)
+ O2(g) + 4e |
Learning
objectives for
the
electrolysis of acidified water and other solutions
Know that electrolysis requires a conducting solution of ions (electrolyte
of acidified water)
and two inert solid conducting electrodes e.g. graphite (carbon) or platinum
(expensive!).
Know that the electrolyte here contains free moving metal ions and
non-metal ions from dilute sulfuric acid added to water.
Know that electrolysis will only happen if a d.c. electrical current is
passed through the acidified water and reduction and oxidation reactions
occur on passage of the electric current and the ions discharged to give the
products hydrogen and oxygen
Where practicable in a school or college laboratory, be able to describe
the apparatus required to electrolyse acidified water
knowing that the positive ions are reduced by electron gain and discharged at
the negative cathode as hydrogen gas molecules
knowing that the negative hydroxide ions or neutral water molecules ions are
oxidised by electron loss and discharged at the positive anode as oxygen
molecules.
and be able to write out the electrode equations (half equations) for the formation of
hydrogen by electron gain of hydrogen ions - reduction and hydroxide ions or
water molecules oxidised by electron loss to give oxygen.
From the electrode equations, be able to explain why the mole ratio e.g.
of hydrogen to oxygen molecules is
theoretically 2 : 1 in the electrolysis of acidified water.
Know how to test for the gases formed and be able to recognise from
observations of testing the gases to confirm the products are hydrogen and
oxygen.
Know the electrolysis of water is an important process to produce the green
fuel hydrogen from surplus electricity from renewable energy sources like
hydroelectric power of wind turbines.
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keywords and phrases:
revision study notes for AQA Edexcel OCR IGCSE/GCSE chemistry
topics modules on electrolysis of water acidified dilute sulfuric acid,
the electrolyte needed, apparatus needed for investigating the
electrolysis of water, the formation of products from the electrolysis
of dilute sulphuric acid, the anode and cathode electrode equations for
the electrolysis of water, a labelled and explained diagram for the
electrolysis of water, the products of the electrolysis of sulfates,
magnesium sulfate, sodium sulfate and sodium hydroxide
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