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ElectrolysisElectrolysis is using an electric current to break up an ionic compound to form elements. Covalent compounds can’t be split up by electrolysis. Terms used in electrolysis:Electrolyte
- the compound which is being broken down. Must contain ions, and the ions
must be free to move (i.e. the substance must be a liquid/solution. Test: the solution ormolten substance will conduct electricity if it is an electrolyte. Anode
- the positive electrode, to which negative ions, referred to as
anions
will be
attracted. Cathode
- the negative electrode, to which positive ions, known as
cations
will be attracted
Electrodes are normally madeout of
inert
(unreactive)
materials. Graphite
and
platinum
are
common electrode materials.
O
xidation
I
s^
L
oss
R
eduction
I
s^
G
ain
of electrons
Because both reduction and oxidation take place at the same time, electrolysis is a
REDOX
During electrolysis, the positive ions travel to the negative electrode. Here they receiveelectrons which turns them from positive ions back into atoms. The negative ions travel tothe positive electrode. Here they give up electrons to become atoms.^ reaction.
Practice
:^
(answers at the end of the topic)
What would the products be when aluminium chloride (which contains Al
3+
and Cl
-^ ions) is
melted and electrolysed? Write half equations to show what goes on at each electrode.Explain the following observations: When lead(II) bromide is heated until it melts and anelectric current passed through, a silvery coloured liquid is found under the negativeelectrode (cathode) and a brown gas appears at the positive electrode (anode). Use halfequations to support your answer. Electrolysis of solutions of ionic compounds Ionic compounds will often dissolve in water, so the ions are free to move around in thesolution. Water
also contains
ions
H
+^
and
OH
- , as well as molecules of
H
O 2
. These ions are
also attracted to the electrodes.e.g. When we make a solution of sodium chloride (NaCl), we get:
•^
Na
+^ ions and
Cl
-^ ions in the solution (from the NaCl)
•^
H
+^ and
OH
-^ ions in the solution (from the water)
The
Na
+^ and
H
+^ cations are both attracted to the cathode
The
Cl
-^ and
OH
-^ anions are both attracted to the anode
What happens at the electrodes during electrolysis of a solution: At the cathodeThere is competition between the two positively charged ions. RULE:The ion of the more reactive element stays in the solution.The ion of the less reactive element is given electrons andreduced to atoms of that element.^ e.g.• In a sodium chloride solution,
Na
is more reactive than
H
Na
+^ cations stay in the solution
H
+^ cations each gain an electron to become
H
atoms
H
atoms bond in pairs to form molecules of
H
2(g)
2H
+(aq)
+ 2e
-^
H
2(g)
At the anodeIn dilute solutions, oxygen is formed from the hydroxide ions. In more concentratedsolutions of halides, the halogen can be produced in preference to oxygen: RULE:If Cl
- , Br -^ or I -^ anions are present in sufficient concentration, they give up an electron and
become Cl
, Br 2
or I 2
Otherwise, the
OH
-^ anions from water give up an electron, and oxygen gas is formed at the
anode:
4 OH
O
2(g)
+ 2 H
O 2
(l)
e.g.
2 Cl
Cl
2(g)
+ 2e
Example: What would be produced during electrolysis of a sulphuric acid
H
SO 2
4(aq)
solution?
sulphuric acid solution (electrolyte)
H
+^ cations (from acid and water) attracted^ H
given off 2 2H
+(aq)
H
2(g)
OH
-^ and
SO
2- 4
anions attracted SO
2- 4 remains in solution,
O
given off 2 4OH
2H
O 2
(l)
+ O
2(g)
In this reaction, water is being split up into hydrogen andoxygen.
2H
O 2
(l)^
2H
2(g)
+ O
2(g)
We would observe that
twice the volume of hydrogen
is
collected
compared to oxygen
. This proves that the chemical
formula for water is H
O. 2
Practice: Similar experiments can be used to determine the formula ofother simple compounds that split up into gaseous elementsduring electrolysis.
What would you observe during electrolysis of (i) silver nitrate solution; (ii)
magnesium iodide solution? Give relevant half equations.
(answers at end of topic)
Moles of electrons
One Faraday
is the amount of electricity that corresponds to
one
mole of electrons
flowing around an electrical circuit (such as an
electrolysis cell)
In electrolysis we use half-equations to show what happens at each electrode. For examplein electrolysis of brine: ANODE:
2Cl
-^
Cl
“Two moles of chloride ions are oxidised to form one mole of chlorine molecules, and twoFaradays of electrons flows from the anode around the circuit to the power supply” CATHODE:
2H
+^ + 2e
-^
H
2
“Two Faradays of electrons flows into the cathode from the power supply, allowing twomoles of hydrogen ions to be reduced to form one mole of hydrogen molecules” Calculations involving Faradays e.g. 0.5 Faradays of electrons flow into the cathode during electrolysis of a copper sulphatesolution. What mass of copper could be deposited on the cathode?At the cathode:
Cu
2+
-^
Cu
Ratio
:^
:^
Moles of electrons = 0.
(1 Faraday = 1 mole of electrons)
Mole ratio
electrons : Cu
So moles of Cu
Mass of Cu = moles of Cu x A
of Cur
= 0.25 x 64
= 16g
Electrolysis of brine is carried out in a
diaphragm cell
(or sometimes in a membrane cell).
The diaphragm cell is designed so that all the products are kept separate:If chlorine mixes with hydrogen, it produces a mixture which will
explode violently
on
exposure to sunlight or heat, so the
hydrogen and chlorine gases need to be kept apart
Chlorine also reacts with sodium hydroxide solution. Therefore, if we are trying tomanufacture
chlorine and sodium hydroxide these must be kept apart
as well.
Answers 1)
What would the products be when aluminium chloride (which contains Al
3+
and Cl
ions) is melted and electrolysed? Write half equations to show what goes on at eachelectrode.Aluminium would be produced at the cathode:
Al
3+
-^
Al
Chlorine would be produced at the anode:
2Cl
-^
Cl
Explain the following observations: When lead(II) bromide is heated until it melts andan electric current passed through, a silvery coloured liquid is found under thenegative electrode (cathode) and a brown gas appears at the positive electrode(anode). Use half equations to support your answer.Molten lead forms under the cathode:
Pb
2+
-^
Pb
Bromine is given off at the anode:
2Br
-^
Br
What would you observe during electrolysis of (i) silver nitrate solution; (ii)magnesium iodide solution? Give relevant half equations.Silver would be deposited on the cathode:
Ag
+^ + e
-^
Ag
Oxygen would be given off at the anode:
4OH
-^
O
+ 2H 2
O + 2e 2
Hydrogen would be produced at the cathode:
2H
+^ + 2e
-^
H
2
Iodine would be formed at the anode:
2I
-^
I
