Electrolytic cells: contains nonnonspontaneous reactions.
Electrochemistry 1.2
Topic: Electrolytic cells
Electrolysis
Electrical energy is provided to the cell from an external
source and this energy causes the chemical reactions
to occur.
Electrolytic Cell
requires V-source
Could be a
1-compartment or
2-compartment
system
Negative cathode
Positive anode
Electrolytic Cell: Typical Setup,
one--compartment
one
Electrolytic Cells
A container for the
reaction mixture.
Two electrodes
immersed in the
reaction mixture**.
A voltage source of
direct current.
Electrical energy is used to force nonspontaneous (reactant-favored) chemical
reactions to occur.
The process is called electrolysis.
Two examples of commercial electrolytic
reactions are:
The motion of the ions through the solution = electric current.
Electrolytic conduction
1. Positive ions migrate toward the negative electrode (cathode).
2. Negative ions migrate toward positive electrode (anode).
Electroplating of jewelry and auto parts.
Electrolysis of chemical compounds.
Electrolytic Cell:
Molten NaCl (Down Cell)
Solid NaCl does NOT conduct electricity. The ions vibrate
about fixed positions and they are not free to move
throughout the crystal.
Electrolytic Cell
Electrolytic Cells/Electrolysis
Anode
Liquid Na is produced at the
cathode (-).
2{Na+ + e Na(l)*}
Gaseous Cl2 is produced at
the anode (+).
2Cl- Cl2 + 2e--------------------------------2Na++2Cl-2Na(l)+Cl2(g)
e-
*Na(l) Na(s)
e-
Molten (melted) NaCl = Electrical Conductor.
Consists of Na+ and Cl- ions.
If direct current is applied (by way of two inert electrodes) through
the cell containing the molten NaCl, we observe the following:
Pale green gas (Cl2) is formed in one electrode.
Molten, silvery Na forms at the other electrode. The Na
particles floats on top of the molten NaCl.
eCathode
This is non-spontaneous except at
very high T (>801°C)
Direct current (dc) source supplies
energy to force the reaction
forward.
1
Electrolytic Cell: Molten NaCl
Downs Cell: Industrial electrolysis
of NaCl
Expensive due to costs in
the construction, electricity
and heating for melting of
the NaCl.
Commercial Production of
Na (metal).
The liquid Na is drained off,
cooled and cast into blocks,
then stored in inert mineral
oil to prevent reaction with
atmosphere (O2) or H2O.
The Cl2 product can also be
collected (but there are
cheaper alternatives to
producing Cl2 gas).
Electrons are used in the cathode half-reaction
(reduction) and produced in the anode half-reaction
(oxidation).
Travel of e- : from ANODE(+) to CATHODE(-).
The dc source forces the e- to flow non-spontaneously
from the positive electrode to the negative electrode.
Na and Cl must NOT be allowed to come into contact
with each other because they react spontaneously,
rapidly and explosively to form NaCl.
Case: Electrolysis of molten KI
Consider standard reduction potentials:
K+ (aq) + e- K(s)
I2(s) + 2e- 2I- (aq)
-2.924 V
+0.535 V
Spontaneous reaction is:
2K(s) + I2(s) 2I- (aq) + 2K+
Eocell = 3.459 V
Opposite reaction: non-spontaneous
2I- (aq) + 2K 2K(s) + I2(s) Eocell = –3.459 V
Thus, to produce K(s) from KI(aq),
Applied voltage > 3.459 V is required.
Electrolysis products of MX salts
may be different from molten
solution vs. aqueous solutions.
CAUTION: In aqueous solutions, the more stable species
(the ones which are easily reduced or oxidized species
spontaneously) will be undergoing the electrolytic
reactions last.
Example: The more easily reduced species in aqueous
solution will be reduced first before K+ (because K wants
to exist as K +).
Thus, H2O might be a reactive redox specie which can
either produce O2 or H2.
Case: Electrolysis of KI(l)
Non-spontaneous since negative.
2I- (aq) + 2K 2K(s) + I2(s) Ecell = –3.459 V
Thus, to produce K(s) from KI(aq),
Applied voltage > 3.459 V is required.
LESSON: To find out the required applied potential for
the electrolysis to start, use the equation
E= Ecathode- Eanode (do not change the signs of the
standard reduction potential!) The result will be
negative.
E= (EK+/K) - (EI2/2I-) = (-2.924 V) – (+0.535 V)= –3.459 V
Total applied potential should be 3.459 V + overpotential.
Electrolysis of Aqueous IAIA-salts
(e.g. KI (aq)
Electrolysis of KI
H2 bubbles
phenolpthalein
turns to pink.
Brown I2 forms in
the other electrode
So, what are the electrolytic reactions which
occurred? Which is anode/cathode?
2
Electrolysis of Aqueous KI soln.
Cathode: H20 is
reduced to
H2(g)+OH-,
turning the
phenolpthalein
to pink.
Brown soluble I2 (s)
forms in the anode.
Cathode: 2H2O + 2e- 2H2 (g) + OHAnode: 2I- I2 (s) + 2e-
Electrolysis of aqueous KI:
H2O might also be reduced or oxidized.
K+ (aq)
+ e- K(s)
2H2O + 2e- 2H2 (g) + OHI2(s) + 2e- 2I- (aq)
O2 (g) + 4H+ +4e- 2H2O
-2.924 V
-0.828 V
+0.535 V
+1.229 V
Calculations of different redox combinations and the
minimum applied voltages required for electrolysis:
E1 = E(K+/K) – E(I2/I) = (-2.924 V) – (+0.535 V) = -3.459V
E2 = E(H2O/H2,OH) – E(I2/I) = (-0.828V) – (0.535 V) = -1.363V
E3 = E(H2O/H2,OH) – E(H2O/O2, H+)=(-0.828V) – (1.229) = -2.057
Thus, we can see from the absolute magnitudes of the required
applied potential (theoretical), that the 2nd is the smallest and
hence, this combination will happen first.
Electrolysis of aqueous NaCl
NaCl:: Again,
H2O might also be reduced or oxidized first.
Na+ (aq) + e- Na(s)
2H2O + 2e- 2H2 (g) + OHCl2(s) + 2e- Cl- (aq)
O2 (g) + 4H+ +4e- 2H2O
-2.714 V
-0.828 V
+1.360 V
+1.229 V
Calculations of different redox combinations and the
minimum applied voltages required for electrolysis:
E1 = E(Na+/Na) – E(Cl-/Cl2) = (-2.714 V) – (+1.360 V) = -4.074 V
E2 = E(H2O/H2,OH) – E(Cl2/Cl-) = (-0.828V) – (1.360 V) = -2.188 V
E3 = E(H2O/H2,OH) – E(H2O/O2, H+)=(-0.828V) – (1.229) = -2.057
WARNING: Smallest potential does not mean first reaction to occur!
The third is the smallest so we expect that O2(g) would be generated
first before Cl2(g). However, the overpotential is higher for the
former (--O2) which explains the experimental results.
Electrolysis of KI(aq)
Consider standard reduction potentials:
K+ (aq) + e- K(s)
-2.924 V
I2(s) + 2e- 2I- (aq)
+0.535 V
2H2O + 2e- 2H2 (g) + OH-0.828 V
Applied voltages required:
2I- (aq) + 2K 2K(s) + I2(s);Eoapplied = –3.459 V
2H2O + 2I- 2H2 (g) + OH- + I2(s) Eoapplied = -1.363
Water is reduced first before K+, since a lower voltage is
required. There are other factors which affect this order
of reduction/oxidation (i.e. the overpotential).
Electrolysis of Aqueous IAIA-salts
(e.g. KI, NaCl)
Observation for the Electrolysis of NaCl:
H2 gas at one electrode. Basic solution near the vicinity
green Cl2 gas produced at the other electrode.
Conclusions:
Oxidation at the Anode:
2Cl- Cl2 + 2eReduction at the Cathode: Na+ ions NOT reduced to
Na(metal)**, rather:
2H2O + 2e- 2H2 (g) + OH**active metals Li, K, Ca, Na can displace H2 from aqueous solutions,
hence they are not produced.
Electrolysis of Aqueous NaCl
Cathode: 2H2O(l) + 2e- 2H2 (g) + OHAnode: 2Cl-(aq) Cl2 (g) + 2e-------------------------------------------------Overall reaction: 2H2O(l) + 2Cl-(aq) 2H2
(g) + OH-(aq)
Required WORK (applied potential) should
be greater than 2.188 V
Solid NaOH can be obtained by
evaporation of the residual
solution. This is an important
commercial preparation for NaOH
and Cl2 gas.
Less expensive that electrolysis of molten
NaCl since heating is not required.
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Electrolysis of Aqueous Na2SO4
Electrolysis of Aqueous Na2SO4
Observation during electrolysis of Na2SO4 (aq):
H2 gas produced at one electrode. Solution becomes basic around that electrode.
O2 gas produced at the other electrode. Solution becomes acidic around that
electrode.
How come the SO4-2 is neither reduced nor oxidized?
One possible reaction is:
-2
+
(1) SO4 + 4H + 2e- ==> SO2(g) + 2H2O(l) : 0.20 V
Observation during electrolysis of Na2SO4 (aq):
H2 gas produced at one electrode. Solution becomes basic around that electrode.
O2 gas produced at the other electrode. Solution becomes acidic around that
electrode.
Electrolysis of Aqueous Na2SO4
How come the SO4-2 is neither reduced nor oxidized?
Na2SO4 (aq) is a slightly basic solution,hence, electrolysis of Sulfate
will not happen here!
E2 = E(SO4-2/SO2,H2O) – E(H2O/O2, H+)=(0.20V) – (1.229 V) = -1.429 V
E3 = E(H2O/H2,OH) – E(H2O/O2, H+)=(-0.828V) – (1.229) = -2.057
Electrolysis of Aqueous Na2SO4
How come the SO4-2 is neither reduced nor oxidized?
Another possible reaction is:
(2) SO4-2(aq) +H2O(l) +2e- SO3-2 (aq)+2OH (aq) : -0.93 V
Another possible reaction is:
(2) S2O8-2(aq) +2e- 2SO4-2 (aq) : -2.01 V
E = E(SO4-2/SO2,H2O) – E(H2O/O2, H+)=(-0.93V) – (1.229 V) = -2.159 V
E = E(SO4-2/SO2,H2O) – E(H2O/O2, H+)=(-2.01V) – (1.229 V) = -3.239 V
E = E(H2O/H2,OH) – E(H2O/O2, H+)=(-0.828V) – (1.229) = -2.057
E = E(H2O/H2,OH) – E(H2O/O2, H+)=(-0.828V) – (1.229) = -2.057
The reduction of Sulfate to sulfite requires a higher potential,
compared to the reduction of water to O2, hence, water reduction
occurs first.
The reduction of Sulfate requires a higher potential, compared to the
reduction of water to O2, hence, water reduction occurs first.
Electrolysis of Aqueous Na2SO4
is plain electrolysis of water
Electrolysis of Water
Anode: 2H2O O2 (g) + 4H+ + 4eCathode: 2{2H2O + 2e- H2 + 2OH-}
-------------------------------------------------Overall reaction: 6H2O 2H2 + 4H+ + 4OHNet Reaction: 2H2O 2H2 + O2
Water is more readily reduced than Na+ and SO42.
Na+ and SO4-2 ions conduct the current through
the solution, but they do NOT take part directly on
the reaction (they are just supporting electrolytes).
OBSERVATIONS:
Cathode: H2 bubbles (popping sound),
vicinity of electrode becomes basic
Anode: O2 bubbles (ember glows much brighter);
vicinity of electrode becomes acidic
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Electrolysis of Water
Cathode: 2{2H2O + 2e- H2 + 2OH-}
Anode: 2H2O O2 (g) + 4H+ + 4e-------------------------------------------------Overall reaction: 6H2O 2H2 + 4H+ + 4OHNet Reaction: 2H2O 2H2 + O2
Electrolytic Cell: Summary
Products can be predicted based on comparing
reduction potentials associated with possible
redox processes.
The species which is most favored to be oxidized
would be oxidized. The species most favored to be
reduced would be reduced.
The electrode itself could be an “active”
participant in the electrolysis (e.g. Cu used for
electroplating).
For problems involving Ecell
which requires overpotential
Add the overpotential to the required absolute
magnitude of the Eapplied.
The one which requires the lowest total potential will
be the first to undergo the electrolysis.
It is possible that the other electrolytic reactions
to occur even before the “first” electrolytic
reactions to finish. Therefore, there would be
cross-contamination of products (e.g. Of gases
or solids).
Electrolytic Cell: Summary
Employs an external source of electricity (dc) to
drive the electrochemical reaction forward.
(-) terminal of source is connected to Cathode,
(+) terminal of the source is connected to the
Anode.
Current carrying medium: molten salt or
electrolyte solution.
The electrolysis reactions occuring in these two
media might be different.
WARNING !!!
It is not enough to calculate the required applied
potential (I.e. least amount of Eapplied will be
the first reaction to occur).
You also have to consider the
“OVERPOTENTIAL”: additional work/voltage
required for the reaction to proceed.
Example: generation of gas, adsorption of solids unto
the electrode, migration of ions from solution towards
the electrode, etc….
Counting Electrons:
Coulometry and Faraday’s Law of Electrolysis
Michael Faraday’s
Conclusion for his
Electrolysis Studies
(Faraday’s Law of
Electrolysis):
“The amount of substance that
undergoes oxidation or
reduction at each electrode
during electrolysis is directly
proportional to the amount of
electricity that passes through
the cell.”
Michael Faraday(1781-1867)
“greatest scientist of the 19th century”
Self-educated by extensive reading
Apprentice of Sir Humphry Davy (Royal
Institution/London).
5
Counting Electrons:
Coulometry/Electrolysis
Faraday’s Law of Electrolysis
A faraday is the amount of electricity that reduces one equivalent of
a species at the cathode and oxidizes one equivalent of a species at
the anode.
1 faraday of electricity = 6.02 x1023 electrons = 1mol eCoulomb (C) = amount of charge that passes a given point when 1
ampere (A) of electric current flows for 1 second.
1 ampere = 1 coulomb/second 1C = 1A*s
1 Faraday = 6.022 x 10 23 electrons = 96,485 C (~96500 C)
Coulometry: the use of electrochemical cells to relate the amount of reactant or product to
the amount of the current passed.
Coulometry: Practice Exercise
Example 21-1: Calculate the mass of Pd produced by the
reduction of Pd(II) ions during the passage of 3.20 A of
current through a solution of Pd(II)sulfate for 30.0 min.
Pd+2 + 2e1 mol
2mol e-
Pd
1 mol
(a) The Electrical charge passed through the cell:
#Coulombs = (A) x (s) = (3.20 A) (30min) (60s/min) = 5760 C
(b)The moles of electrons
# mol e- (faradays) = 5760 C x (1mol e-/96485 C) = 5.97x10-2 mol e-
Practice Exercises
Calculate the mass of Al produced in 30.0 min by
the electrolysis of molten Al if the electrical
current is 10.0 A.
Al+3 + 3e- Al
(a)no. of Coulombs
= 10.0 A x 30.0min x 60.0s/min
= 1.80x104+ C
(b)no. of mole e= 1.80x104+ C x (1 mol e-/96485)
= 0.187 mol e(c)no. of grams Al= 0.187 mol e- x (1mol Al/3mol e-)x (27.0g
Al/mol Al)
= 1.68 g Al
Faraday’s Law states that during electrolysis, one faraday of
electricity (96,485 coulombs) reduces and oxidizes, respectively,
one equivalent of the oxidizing agent and the reducing agent.
This corresponds to the passage of one mole of electrons through
the electrolytic cell.
1 equivalent of oxidizing agent = gain of 6.022x1023 electrons
1 equivalent of reducing agent = loss of 6.022x1023 electrons
Coulometry: Practice Exercise
Example 21-1: Calculate the mass of Pd produced by the
reduction of Pd(II) ions during the passage of 3.20 A of
current through a solution of Pd(II)sulfate for 30.0 min.
(c) The moles of Pd metal formed.
# mol Pd = (5.97x10-2 mol e-) * (1mol Pd/2 mol e-)
= 2.98x10-2 mol Pd
(d) The no. of grams Pd = 2.98x10-2 *106.4 g/mol
= 3.17 g Pd
Practice Exercises
Elemental Ca is produced by the electrolysis of
molten Calcium Chloride. What mass of Ca can
be produced by the electrolysis if a current of
5500 A is applied for 48 hours?
Ca+2 + 2e- Ca
a)no. of Coulombs
= 5500 A x 48 hours x 3600 s/hour
= 9.504x108 C
(b)no. of mol e- = 9.504x108C x (1mol e-/96500)=9848.7 mol e(c)no. of grams Ca= 9848.7 x (1mol Ca/2mol e-)x (40.1g Ca/mol)
= 197467 g or 197.5kg
6
Practice Exercises:
The mass of Ag deposited on a spoon during electrolysis
was 0.904mg. How much electric charge is passed
through cell?
Atomic Mass Ag = 107.9 g/mol
An electrolytic cell contains 50.0 ml of a 0.152M FeCl3
solution. A current of 0.620 is passed through the cell,
causing deposition of Fe(s) at the cathode.
What is the concentration of the Fe+3 in the cell after this
current has run for 20.0 min?
Coulometry:: Practice Exercises
Coulometry
Example W21-2: Calculate the volume of
oxygen (measured at STP) produced by the
oxidation of water in previous example (21-1).
2H2O O2 + 4H+ +
4e2mol
1mol 4mol
4mol eVol. Of O2,STP = (3.2A*30.0min*60s/min)* (4mol e/96485C)*(1mol O2/mol e-) *22.4L* {1/4(96485)
= 0.334 L or 334ml O2 at STP
STP: 22.41 L at 0°Celsius and 1 atm, 1 mol gas = 22.4 liters
Electrolysis: Still More Practice
Exercises
Ex.25/834. What mass of Pt that could be plated onto a ring from
the electrolysis of a Pt(II) salt with a current of 0.415A current for
142s.
Pt+2 + 2e- Pt(metal)
Mass Pt=0.415 A * 142s * (1mol e-/96500C) * (1mol Pt/2 mol e-)
* 195.1g Pt/mol
Mass of plated Pt = 0.119g Pt
Ex.26/834. What mass of Ag could be plated onto a spoon from
the electrolysis of AgNO3 with a current of 3.86A current for 45.0
min?
Ag+ + 1e- Ag(metal)
Mass of Ag = 3.86A*45.0min*60s/min*(1 mol e/96500)* (1 mol/1 mol e-)*
107.9 g Ag/mol
Mass of Ag = 11.7 g Ag
Electrolysis: Still More Practice
Exercises
Ex. 28/834. A current is passed through a 500.0ml solution of CaI2. The
following electrode reactions occur:
Anode: 2I- I2 + 2e- , Cathode: 2H2O +2e- H2 + 2OH(c) What volume of dry H2 at STP has been formed?
(d) What is the pH of the solution?
Vol-H2@STP = 7.24x103C * (1F/96500C)* (1mol H2/1mol e)*(22.4LH2/molH2)
= 1.68 Liters of H2
pH = ?
mMoles (OH-) = 37.5 mol I2 x (2 mol OH-/mol I2) = 75.0 mmol
[OH-]= 75.0 mmol/500.0ml = 0.150 M
pOH=0.823, pH=13.2
Electrolysis: Still More Practice
Exercises
Ex. 28/834. A current is passed through a 500.0ml solution of CaI2. The following
electrode reactions occur:
Anode: 2I- I2 + 2e-; Cathode: 2H2O +2e- H2 + 2OHAfter some time, analysis of the solution shows that 37.5 mmol of I2 has been
formed.
(a) How many faradays of charge have passed through the solution?
(b) How many coulombs?
(c) What volume of dry H2 at STP has been formed?
d) What is the pH of the solution?
Answer:
(a)
37.5mmol I2 * (2 mmol e-/mmol I2) = 75.0 mmol e- or 0.075Faradays
(b)
No. of coulombs = 0.075 Faradays * (96500C/F) =7.24x103C
Work in Electrolytic Cells
WORK = total applied potential required for the
electrolysis to happen; thus, this is the theoretical
| Ecell | (absolute value) for the redox
combination and the overpotential.
∆G is naturally not asked because electrolytic
reactions are not spontaneous reactions. ☺
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Commercial Application of Electrolytic
Cells : Som examples
Commercial Application of
Electrolytic Cells
Commercial Application of
Electrolytic Cells
Many elements are produced commercially by
electrolysis of solutions. (e.g. Na, Cl2, H2, and O2).
Electrolysis of molten compounds for production of Al,
Group 1A, 2A metals (ex. Ba).
Refining of impure metals
Metal platings (jewelries, electrical contacts).
Copper plating on many objects (art objects and the
billion $$$ semicon <chips> industry)
Automobiles: example, steel bumpers (Cr)
Tin Cans (steel cans plated with some Sn or Cr).
Steel/Metal Industry
Electroplating with Copper
Electrolytic Refining and Electroplating of Metals
Impure metallic Cu can be purified “electrolytically” to
100% pure Cu.
**The impurities (e.g. metals such as: Ag, Au, and Pt.)
The cathode is a thin sheet of copper metal connected to
the negative terminal of a direct current source.
The anode is large impure bars of copper
Anode=pure Cu dissolves during the electroplating process.
This replenishes the Cu+2 ion which are removed from the
solution as Cu plates out to the cathode.
Commercial Application of
Electrolytic Cells: Cu refining
Commercial Application of
Electrolytic Cells: Cu refining
The electrolytic solution is CuSO4 and H2SO4
The impure Cu dissolves to form Cu2+.
The Cu2+ ions are reduced to Cu at the cathode.
Any active metal impurities are oxidized to cations
that are more difficult to reduce than Cu2+.
This effectively removes them from the Cu metal.
Anode
impure
Cu(0s) Cu (2aq+ ) + 2e −
Cathode very pure Cu(2+aq ) + 2e − Cu(0s)
Net rxn.
No net rxn.
Z n
0
Z n
2 +
+
2 e
−
F e 0 F e 2+ + 2 e −
A n d s o fo rth fo r o th e r
a c tiv e m e ta ls
The less active metals are not oxidized and precipitate
to the bottom of the cell. These metal impurities can
be isolated and separated after the cell is
disconnected. Metals that precipitate include:
**Ag, Au, Pt, Pd, Se, Te
8