ELECTROLYSIS AND
ELECTROLYTIC CELLS
ELECTROLYTIC CELLS VS. GALVANIC CELLS
• Electrolytic Cells
• Non-spontaneous cells that
need a battery, electron
pump, or any DC source to
run
• Work is done ON the system
• Reactions occur in a single
container
• AN OX and RED CAT still
apply BUT, polarity of the
electrodes is reversed
• Cathode is negative and
anode is positive
• Electrons still flow FAT CAT
• Usually use inert electrodes
• Galvanic Cells
• Spontaneous cells
• Used as a battery
• Reactions are separated
into two half-cells to
generate electricity
• AN OX and RED CAT apply
• Cathode is positive and
anode is negative
SO, WHAT EXACTLY IS ELECTROLYSIS?
• Defined as the use of electricity to bring about a
chemical change
• Works by forcing a current through a cell to produce a
chemical change for which the cell potential is negative
• Literal interpretation
• “Split with electricity”
• Great practical importance
• Charging a battery
• Producing aluminum metal
• Chrome plating
• Two main types of electrolysis:
• Electrolysis of molten ionic compounds
• Electrolysis with active electrodes
PREDICTING THE PRODUCTS OF
ELECTROLYSIS
• If there is no water present and you have a pure
molten ionic compound, then:
• The cation will be reduced (gain electrons/go down in charge)
• The anion will be oxidized (lose electrons/go up in charge)
ELECTROLYSIS OF MOLTEN COMPOUNDS
• Example – decomposition of molten NaCl
• The half-reactions that take place are as
follows:
• Cathode - 2Na+(l) + 2e-  2Na(l)
• Anode - 2Cl-(l)  Cl2(g) + 2e-
• Overall electrolysis reaction:
2 NaCl l → 2 Na l + Cl2 (g)
• Industrially, this type of electrolysis is used
to produce metals like Al!
ELECTROLYSIS WITH ACTIVE
ELECTRODES
• This type of electrolysis is where active electrodes
take part in electrolysis
• It is highly used in electrolytic plating, like chroming!
PREDICTING THE PRODUCTS OF AN
ACTIVE ELECTRODE ELECTROLYSIS
• If water is present and you have an aqueous solution of
an ionic compound, then:
• You’ll need to figure out if the ions are reacting or the water is
reacting
• You can always look at the standard reduction potential table to
figure it out
• OR use the following as a rule of thumb:
NO GROUP I OR II METAL WILL BE REDUCED IN AN AQUEOUS
SOLUTION
(WATER WILL BE REDUCED INSTEAD)
NO POLYATOMIC ION WILL BE OXIDIZED IN AN AQUEOUS
SOLUTION
(WATER WILL BE OXIDIZED INSTEAD)
HALF-REACTIONS FOR THE
ELECTROLYSIS OF WATER
• If oxidized:
2 H2 O l → O2 g + 4 H+ aq + 4 e−
• If reduced:
2 H2 O l + 2 e− → H2 g + 2 OH− aq
ELECTROLYSIS OF AQUEOUS SODIUM
CHLORIDE
Half reaction
E o, V
Cl2 (g) + 2e-
2Cl- (aq)
1.36
Na+ + e-
Na (s)
-2.71
2H2O (l) + 2e-
H2 (g) + 2OH-
-0.83
2H+ + 2e-
H2 (g)
0.000
O2 (g) + 4H+ + 4e-
2H2O (l)
1.23
• Which substances are actually present during the
electrolysis reactions?
• Cl-(aq), Na+(aq), and H2O (l) are the only
substances present in significant concentrations
• Trace amounts of O2 (g) and H+ (aq) are present
ELECTROLYSIS OF AQUEOUS SODIUM
CHLORIDE
Half reaction
E o, V
Na+ + e-
Na (s)
-2.71
2H2O (l) + 2e-
H2 (g) + 2OH-
-0.83
• Which reduction reaction is most likely to occur?
H2O (l) has the most positive E0red value
• So, the only product formed at the cathode is
hydrogen gas
• Half-reaction of the cathode (-):
2 H2 O l + 2 e− → H2 g + 2 OH− aq
ELECTROLYSIS OF AQUEOUS SODIUM
CHLORIDE
Half reaction
E o, V
2Cl- (aq)
Cl2 (g) + 2e-
-1.36
2H2O (l)
O2 (g) + 4H+ + 4e-
-1.23
• Which oxidation reaction is most likely to occur?
• H2O has the most positive E0ox value as an active
electrode, although Cl- is very close
• Therefore, you would think that both would
oxidize
• In the lab, only chlorine gas is produced at
the anode
• Half-reactions occurring at the anode (+):
2Cl- (aq)
↔
Cl2 (g) + 2e-
OVERALL ELECTROLYTIC REACTION
2H2O (l) + 2Cl-
↔
Cl2(aq)+ H2(g)+ 2 OH-
Ecell = -0.83 v + -1.36 v = -2.19 V
IN SUMMARY…
• Electrolysis of aqueous solutions of sodium chloride
doesn’t give the same products as electroysis of molten
sodium chloride
• Molten NaCl Electrolysis:
2 NaCl l → 2 Na l + Cl2 (g)
• Aqueous NaCl Electrolysis:
2 NaCl aq + 2 H2 O l
→ 2 Na+ aq + 2 OH− aq + H2 g + Cl2 (g)
CALCULATING THE ELECTRICAL
ENERGY OF ELECTROLYSIS
• Questions take the form of:
• How much metal could be plated out?
• How long would it take to plate out?
• Electrolysis follows Faraday’s Law
• The amount of a substance being oxidized or reduced at
each electrode during electrolysis is directly proportional to
the amount of electricity that passes through the cell
• In order to use Faraday’s law, we need to recognize
the relationship between current, time, and the
amount of electric charge that flows through a circuit
• In other words…
Use dimensional analysis for these calculations!
THE STOICHIOMETRY OF ELECTROLYSIS
• Solutions to these problems follow the general steps:
Current and time → quantity of charge → mole of electrons
→ moles of analyte → grams of analyte
• Helpful tips:
• # coulombs = I (amps) x t (sec)
• 1 Amp = 1 coulomb/second
• 1 volt = 1 Joule/coulomb
• Faraday = 96, 500 coulombs/mole of electrons
• Balanced redox equation gives # moles of electrons per mole of
substance
• Molar mass gives grams/mole
USING FARADAY’S LAW TO MEASURE
PRODUCT AMOUNT OF ELECTROLYSIS
PRACTICE!
• Determine the mass of copper that is plated out when a current
of 10.0 amps is passed for 30.0 minutes through a solution
containing Cu2+
• Since an amp is a coulomb of charge per second, we multiply the current
by the time in seconds to obtain the total coulombs of charge passed into
the Cu2+ solution at the cathode:
C
60.0 sec
10.0
× 30.0 minutes ×
= 1.80 × 104 C (coulombs)
s
1 minute
• Since 1 mole of electrons carries a charge of 1 faraday (96,485 coulombs),
we can calculate the number of moles of electrons required to carry 1.80 x
104 coulombs of charge:
1 mol e−
4
1.80 × 10 C ×
= 1.87 × 10−1 mol e−
96,485 C
• Each Cu2+ ion requires two electrons to become a copper atom. Thus,
each mole of electrons produces ½ mole of copper metal:
1 mol Cu
−1
−
1.87 × 10 mol e ×
= 9.35 × 10−2 mol Cu
−
2 mol e
• We mow know the moles of copper metal plated onto the cathode, and
we can calculate the mass of copper formed by multiplying by its molar
mass:
63.546 g
9.35 × 10−2 mol Cu ×
= 5.94 g Cu
1 mol Cu
ALL STEPS COMBINED
C
60.0 sec
1 mol e−
10.0
× 30.0 minutes ×
×
s
1 minute
96,485 C
1 mol Cu
63.546 g
×
×
−
2 mol e
1 mol Cu
= 5.94 g Cu
ELECTROLYSIS OF KI (AQ) LAB
Problem: You are to separate a KI solutions by electrolysis,
and determine what products are formed. Then you must
write the oxidation and reduction half-reactions to
describe the changes, and the net ionic reaction for the
entire process.
1. Place 20 drops of 0.5 M KI into two separate wells of a
24-well plate. Place two small pieces of Cu wire into one
well as active electrodes, and two pieces of pencil
lead(carbon) as inactive electrodes in the other well. Use
the jumper cables with alligator clips to connect the
active electrodes to a lab power supply. Observe and
describe any action in the well. Repeat with the inert
electrodes.
ELECTROLYSIS OF KI(AQ) LAB
2. A few drops of phenolphthalein can be added to any solutions to
detect the presence of excess OH1- ions. Add a drop of starch to
check for the blue/black color formed when molecular iodine is
present. If Cu2+ ions form in a high enough concentration, the
electrolyzed solution will have a blue color before any starch is
added.
Electrolysis of Potassium Iodide Solution
LAB HALF-REACTIONS: ( AN SRP TABLE)
Half reaction
E o, V
I2 (g) + 2e-
2I- (aq)
0.54
K+ + e -
K (s)
-2.92
2H2O (l) + 2e-
H2 (g) + 2OH-
-0.83
Cu2+ + 2e-
Cu0
0.34
2H+ + 2e-
H2 (g)
0.000
O2 (g) + 4H+ + 4e-
2H2O (l)
1.23
Which substances are actually present during the electrolysis
reactions?
I-(aq), K+(aq), H2O (l) and Cu0 are the only
substances present in significant concentrations.
Trace amounts of O2 (g) and H+ (aq) are present.
ELECTROLYSIS OF KI(AQ) LAB
Half reaction
E o, V
K+ + e -
K (s)
-2.92
2H2O (l) + 2e-
H2 (g) + 2OH-
-0.83
• Which reduction reaction is most likely to occur?
H2O (l) has the most positive E0red value
• So, the only product formed at the cathode is
hydrogen gas
• Half-reaction of the cathode (-):
2 H2 O l + 2 e− → H2 g + 2 OH− aq
ELECTROLYSIS OF KI(AQ) LAB
Half reaction
E o, V
2I- (aq)
I2 (g) + 2e-
-0.54
2H2O (l)
O2 (g) + 4H+ + 4e-
-1.23
Cu0
Cu2+ + 2e-
-0.34
• Which oxidation reaction is most likely to occur?
• Cu0 has the most positive E0ox value as an active
electrode, although I- is very close
• Therefore, both will oxidize
• Half-reactions occurring at the anode (+):
2I- (aq)
↔
l2 (g) + 2eCu0 (s)
↔
Cu2+(aq) + 2e-
LAB OVERALL CELL POTENTIALS
• Reactions and voltage for a water-iodide cell:
2H2O (l) + 2I-
I2 (aq) + H2 (g) + 2OH-
Ecell = -0.83 v + -0.54 v = -1.37 V
• Reactions and voltage for a water-copper cell:
2H2O (l) + Cu0
Cu2+ + H2 (g) + 2OH-
Ecell = -0.83 v + -0.34 v = -1.17 V
An external voltage of at least 1.17 volts must be
added, and usually a little more to initiate the
reaction and overcome resistance
Electrolysis, the Faraday & Avogadro’s #
Set up the apparatus shown below:
Electrolysis, the Faraday & Avogadro’s #
1. Connect the negative side of the battery to the negative side of the ammeter and
then connect the positive side of the ammeter to the copper wire. Record the
starting time for the experiment. Hydrogen gas bubbles will be generated above
the copper wire and collect in the eudiometer.
2. Record the initial amp reading at the start of the electrolysis, and then every
minute thereafter. Calculate the average flow of electricity in amps per minute.
3. Collect the gas until ~50 mL have been produced. At this point, stop the
electrolysis by disconnecting the copper electrode from the power source.
Record the elapsed time for the experiment.
4. Use the Ideal Gas Law to calculate the moles of hydrogen gas produced. Use
this quantity to calculate the value for 1 faraday and the value for Avogadro’s
number.
5. As a comparison, calculate the moles of zinc metal that was oxidized, then use
this value to also determine the value for 1 faraday and for Avogadro’s number.
Determine the percent error for both methods and discuss sources of error and
the accuracy of each method.