ELECTROCHEMISTRY - Wits Structural Chemistry

4/1/2010
Electrochemistry
Relating electricity and chemical reactions
ELECTROCHEMISTRY
Transfer of electrons
CHAPTER 20
Chemical reaction
occurs to produce
electricity
Electricity applied
to produce
chemical reaction
Manufacturing
of chemicals
Electroplating
and refining
of metals
Study of redox
reactions
Applications of
Electrochemistry
Bioelectrochemistry:
study of electron
transfer in biological
regulation of
organisms
Batteries
Fuel cells
Study and
control for
corrosion
Recall
Oxidation – LOSS of electrons
Reduction – GAIN of electrons
Oxidation number –
For a monatomic ion the oxidation no. = the
actual charge of the atom
or it is the hypothetical charge assigned to the
atom using a set of rules.
Make sure YOU know how to assign oxidation
numbers!!!!
Oxidation occurs at the ANODE
Reduction occurs at the CATHODE
Oxidising agent/oxidant – The substance that
causes oxidation of another substance and
hence it is reduced.
Reducing agent/reductant – The substance that
cause reduction of another substance and
hence it is oxidised.
ELECTRIC CURRENT = transfer of charge
METALLIC CONDUCTION = flow of electrons
with no movement of the atoms of the metal
ELECTROLYTIC (IONIC) CONDUCTION =
electric current by movement of ions through a
solution or pure liquid
How do we know if a chemical
reaction is a REDOX reaction
Look to see if there is a change in oxidation
numbers of elements.
Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g)
0 +1 +2 0
Both oxidation and reduction must occur.
Electrons lost by one component is gained
by another.
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Oxidation Number Guidelines
1. Atoms in elemental form, oxidation number is zero.
(Cl 2 , H 2 , P 4 , Ne are all zero)
2. Monoatomic ion, the oxidation number is the charge on
the ion. (Na + : +1; Al 3+ : +3; Cl - : -1)
3. O is usually -2. But in peroxides (like H 2 O 2 and Na 2 O 2 ) it
has an oxidation number of -1.
4. H is +1 when bonded to nonmetals and -1 when bonded
to metals.
(+1 in H 2 O, NH 3 and CH 4 ; -1 in NaH, CaH 2 and AlH 3 )
5. The oxidation number of F is -1.
6. The sum of the oxidation numbers for the molecule is the
charge on the molecule (zero for a neutral molecule).
Example:
A nickel–cadmium battery (NiCad) is a
rechargeable “dry cell” that uses the following
redox reaction to generate electricity:
Cd(s) + NiO 2 (s) +2H 2 O(l) Cd(OH) 2 (s) + Ni(OH) 2 (s)
Identify the substances oxidised & reduced and
the oxidising & reducing agents.
Balancing Redox Reactions
In acidic solutions:
1. Write the 2 half-reactions
2. Balance elements other than O and H
3. Balance O by adding H 2 O
4. Balance H by adding H +
5. Balance electrons
6. Multiply half-reactions by integer so that no.
of electrons in both are equal
7. Add the 2 half-reactions
EXAMPLE:
Mn 2+ (aq) + NaBiO 3 (s) Bi 3+ (aq) + MnO 4- (aq)
1. Write the 2 half-reactions
2. Balance elements other than O and H
3. Balance O by adding H 2 O
4. Balance H by adding H +
5. Balance electrons
6. Multiply half-reactions by integer so that no. of
electrons in both are equal
7. Add the 2 half-reactions
In basic solutions:
1. Write the 2 half-reactions
2. Balance elements other than O and H
3. Balance O by adding H 2 O
4. Balance H by adding H +
5. Add an equal number of OH - to both sides of
the reaction so that all H + is “neutralised”
Recall: H + + OH - H 2 O
6. Balance electrons
7. Multiply half-reactions by integer so that no.
of electrons in both are equal
8. Add the 2 half-reactions
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EXAMPLE:
Mn 2+ (aq) + NaBiO 3 (s) Bi 3+ + MnO 4- (aq)
1-4. Same as before
5. Add an equal number of OH - to both sides of the
reaction so that all H + is “neutralised”
For you to do!
Balance the following:
MnO 4- (aq) + C 2 O
2-
4 (aq) Mn 2+ (aq) + CO 2 (g)
Answers:
6-8. Same as before
Voltaic/Galvanic Cells
A spontaneous redox reaction produces energy
which can do electrical work.
Example:
Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s)
Spontaneous reaction!
Half-reactions:
Zn(s) Zn 2+ (aq) + 2e -
Cu 2+ (aq) + 2e - Cu(s)
Electrons lost by Zn are gained by Cu 2+ .
Zn strip inserted into
CuSO 4 solution
Zn(s) Zn 2+ (aq) + 2e -
Cu 2+ (aq) + 2e - Cu(s)
Molecular view:
In a galvanic/voltaic cell – the transfer of
electrons takes place through an external
circuit rather than directly between reactants.
Cu 2+ ion must collide with
the Zn strip for electron
transfer to occur.
The Zn 2+ ion formed
enters the solution and
the Cu o is deposited on
the Zn strip.
Reaction at cathode =
Reaction at anode =
Zn(s) Zn 2+ (aq) + 2e - Cu 2+ (aq) + 2e - Cu(s)
Oxidation:
Reduction:
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Reaction at cathode =
Reaction at anode =
Zn(s) Zn 2+ (aq) + 2e - Cu2+(aq) + 2e- Cu(s)
Oxidation:
Reduction:
Salt bridge:
‣ Completes the circuit.
‣ Maintains electrical neutrality in the two
compartments by migration of ions
through the porous material.
‣ No further redox reaction will take place if
electrical neutrality is not maintained.
‣Prevents mixing of the electrode solutions.
Molecular view:
2
The 2e - ’s move through the circuit to the Cu strip.
Cell Notation:
Oxidation at anode
Reduction at cathode
Separate
phases
electrode in soln
Salt
bridge
Separate
phases
in soln electrode
Zn(s) Zn 2+ (1 M) Cu 2+ (1 M) Cu(s)
1
On the Zn strip, a Zn atom
“looses” 2e - ’s to form Zn 2+
which moves into solution.
When the Cu 2+ ions collide with
the Cu strip, 2e - ’s are
transferred to form Cu o
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Cell EMF
E cell depends on:
Why do electrons flow in the direction they do
Due to the potential difference between the 2
electrodes (or difference in potential energy)
Electrons at the anode have a higher potential
energy than the electrons at the cathode
electrons spontaneously flow from
anode to cathode
The potential difference is the driving force that
pushes electrons through the circuit
= electromotive force = EMF
= cell voltage
= cell potential = E cell
• the reactions that occur at the electrodes
• the concentrations of reactants and products
• temperature (assume 25 o C unless otherwise stated)
E cell > 0 for a SPONTANEOUS cell reaction!
By convention
E o cell = standard cell potential
Standard conditions:
1 M concentration for reactants & products
1 atm pressure for gases
Zn(s) + Cu 2+ (aq,1M) Zn 2+ (aq,1M) + Cu(s)
E o cell = +1.10 V
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Standard Reduction Potentials
Standard hydrogen electrode (SHE)
Recall: the cell potential is the difference
between 2 electrode potentials (the anode and
the cathode)
E o cell = E o red(cathode) - E o red(anode)
Table of Standard Reduction Potentials
Potentials all measured relative to the
standard hydrogen electrode (SHE)
2H + (aq, 1M) + 2e - H 2 (g, 1atm)
E o = 0 V
2H + (aq, 1M) + 2e - H 2 (g, 1atm)
H 2 (g, 1atm) 2H + (aq, 1M) + 2e-
E o (Cath) = 0 V
E o (An) = 0 V
Standard Reduction Potentials in H 2 O at 25 o C
The more positive the value for E o red, the
stronger the driving force for reduction.
Note:
Since electrical potential measures potential
energy per electrical charge, the standard
reduction potential is a intensive property.
1 V = 1 J 1 C
changing a stoichiometric coefficient in a
half-reaction does NOT affect the value of the
standard reduction potential.
Zn 2+ (aq, 1M) + 2e - Zn(s)
2Zn 2+ (aq, 1M) + 4e - 2Zn(s)
E o = -0.76 V
E o = -0.76 V
Example:
List according to increasing strength of
reducing agent:
Ag(s), Zn(s), Cu(s)
Which is the weakest oxidising agent:
Ag + , Zn 2+ , Cu 2+
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Spontaneity of Redox Reactions
So far we know:
E o cell = E o red(cathode) - E o red(anode)
AND
From the standard reduction potentials table:
E o red(cathode) > E o red(anode)
for a spontaneous reaction
THEREFORE
E o cell > 0 for a SPONTANEOUS cell reaction!
We also know that:
G < 0 spontaneous reaction
Relationship between EMF and free energy
change:
G = -nFE
where n = no. of e - ’s transferred
F = Faraday’s constant
= 96 500 C mol -1 96 500 J V -1 mol -1
For reactants and products in their standard
state:
G o = -nFE o
Example:
Use standard reduction potentials to calculate standard
free energy change for the following reaction:
4Ag(s) + O 2 (g) + 4H + (aq) 4Ag + (aq) + 2H 2 O(l)
From the table:
Reduction:
O 2 (g) + 4H + (aq) + 4e - 2H 2 O(l)
Oxidation:
Ag + (aq) + e - Ag(s)
E o = 1.23 V
E o = 0.80 V
Also: 4Ag + (aq) + 4e - 4Ag(s) E o = 0.80 V
Repeat for:
2Ag(s) + 1 / 2 O 2 (g) + 2H + (aq) 2Ag + (aq) + H 2 O(l)
Ans:
Recall:
G
G
Effect of concentration
o
Since G o = -nFE o
E E
E E
E E
o
o
o
RTlnQ
RT
lnQ
nF
and G = -nFE
2.303RT
logQ
nF
0.05916
logQ
n
Nernst equation
at 25 o C
R=8.315 J K -1 mol -1
F=96500 J V -1 mol -1
T = 298 K
E E
o
Use the Nernst equation to:
0.05916
logQ
n
- find the EMF produced by a cell under nonstandard
conditions.
- determine the concentration of reactant or
product by measuring the EMF of the cell.
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Example:
Consider the reaction:
Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s)
Calculate the cell emf when:
[Cu 2+ ] = 5.0 M and [Zn 2+ ] = 0.5 M
Since emf depends on concentration, a voltaic cell with
a non-zero emf can exist using the same species in both
the anode and cathode compartments.
CONCENTRATION CELL
Anode: Ni(s) Ni 2+ (aq) + 2e - E o red = -0.28V
Cathode: Ni 2+ (aq) + 2e - Ni(s) E o red = -0.28V
E o cell = E o red(cathode) - E o red(anode)
= (-0.28 V) – (-0.28 V) = 0V
But the cell is operating under non-standard conditions
since concentrations are 1 M.
Driving force of cell due to the difference in
concentration tries to equalise
concentrations in both compartments.
Anode: Ni(s) Ni 2+ (aq, dil) + 2e -
Cathode:
Ni 2+ (aq, conc) + 2e - Ni(s)
Ni 2+ (aq, conc) Ni 2+ (aq, dil)
Anode: Ni(s) Ni 2+ (aq, dil) + 2e -
Cathode:
o 0.05916 [Ni
E E log
n [Ni
]
2
dil
2
]
conc
0.05916 (1.00
10
E (0V) log
2 (1.00)
E 0.0887 V
NOTE:
Ni 2+ (aq, conc) + 2e - Ni(s)
Ni 2+ (aq, conc) Ni 2+ (aq, dil)
When the concentrations in the 2 compartments
become equal,
Q = 1 and E = 0 V.
3
)
EMF and Equilibrium
Why does the emf drop as a voltaic cell
discharges
0.05916
Look at Nernst equation: E E
o logQ
n
As reactants are converted to products, Q
increases.
Eventually E = 0 V.
Since G = -nFE, G = 0 kJ/mol
equilibrium!
i.e. when E = 0 V, equilibrium
no net reaction
The equilibrium constant can be calculated for
a redox reaction as follows:
At equilibrium:
E = 0 V and Q = K
E E
0 E
o
o
0.05916
logQ
n
0.05916
logK
n
o
nE
logK
0.05916
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Nernst equation
E E
o
RT
lnQ
nF
BATTERIES
A battery is a portable , self contained
electrochemical power source that consist of
one or more voltaic cells.
The Nobel Prize in Chemistry 1920
Nobel Lecture, December 12, 1921
Studies in Chemical Thermodynamics
Greater voltages can be achieved if cells are
placed in series e.g. in a torch.
Cathode = “+” Anode = “-”
Walther Hermann Nernst
(1864 - 1941)
German physicist and chemist
Separate anode and cathode compartments
with a porous barrier
Cell emf depends on the substances that are
oxidised and reduced.
Cell life depends on the quantities of the
substances.
Primary cell – cannot be recharged
Secondary cell – can be recharged
Lead-acid battery
Used in cars
12 V = 6 x 2 V cells
Anode = Pb
Cathode = PbO 2 packed on metal grid
Electrolyte = H 2 SO 4
Spacers prevent cathode and anode from touching.
Cathode: PbO 2 + 4H + + SO
2-
4 + 2e - PbSO 4 + 2H 2 O
Anode: Pb + SO
2-
4 PbSO 4 + 2e-
Overall: PbO 2 + Pb + 4H + + 2SO
2-
4 2PbSO 4 + 2H 2 O
E o cell = (+1.685)-(-0.356) = +2.041 V
Solid products and reactants
not appear in expression for Q
maintain relatively constant emf while discharging
variation due to [H 2 SO 4 ] decreases during discharge
Can be recharged since PbSO 4 adheres to the electrode
surface.
External source of power = alternator (generator) driven
by the engine.
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Alkaline battery
Most common primary battery
Anode = Zn powder immobilised in gel surrounded by
electrolyte
Cathode = MnO 2 + graphite
Electrolyte = conc. solution of KOH
Separator = porous fabric
Cathode: 2MnO 2 +2H 2 O + 2e - 2MnO(OH) + 2OH -
Anode: Zn + 2OH - Zn(OH) 2 + 2e-
Overall: 2MnO 2 + Zn + 2H 2 O 2MnO(OH) + Zn(OH) 2
E o cell = +1.55 V
Dry cell or Zinc carbon battery
Cathode: 2MnO 2 + NH 4+ + 2e - Mn 2 O 3 + NH 3 +
Anode: Zn Zn 2+ + 2e-
Overall: 2MnO 2 + Zn + NH 4+ Mn 2 O 3 + Zn 2+ +NH 3 + H 2 O
E o cell = +1.55 V
Nickel-cadmium, Nickel-metalhydride
and Lithium-ion batteries
Used in high-power-demand portable devices
e.g. cellphones, PC’s etc
Lightweight and readily rechargeable
Alkaline batteries better performance than the old dry
cells which are also based on MnO 2 and Zn .
NiCad batteries:
Cathode: 2NiO(OH) + 2H 2 O + 2e - 2Ni(OH) 2 + 2OH -
Anode: Cd + 2OH - Cd(OH) 2 + 2e -
Overall: 2NiO(OH) + Cd + 2H 2 O 2Ni(OH) 2 + Cd(OH) 2
NiMH batteries:
Same cathode reaction as in NiCad batteries.
Anode reaction different.
E o cell = = 1.30 V
use cells in series
Anode = metal alloy (eg ZrNi 2 ) that has the ability to
absorb hydrogen atoms
-Solid adheres to electrodes rechargeable
-Disadvantage:
Cd = toxic heavy metal
increases mass of battery
+ environmental hazard
During oxidation process H-atoms are released
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Li-ion batteries:
Advantages:
- Li has the most negative standard reduction potential,
thus makes a good anode.
- Li is a lightweight metal.
- Cell emf can be as high a 3 V.
Solid electrolyte used.
Based on ability of Li + ions to insert themselves into
certain layers solids.
Disadvantages:
- Batteries do not yet have high reliability and long
lifetime
Fuel cells
Thermal energy released during combustion of
fuels is converted to electrical energy.
This process is only about 40% efficient.
Because combustion reactions are redox
reactions, the process can be carried out
electrochemically
found efficiency is almost doubled
Unlike batteries, fuel cells do not store chemical
energy. Reactants must be constantly supplied
and products removed from the fuel cell.
Example: The hydrogen fuel cell
Cathode: O 2 + 2H 2 O + 4e - 4OH -
Anode: 2H 2 + 4OH - 4H 2 O + 4e-
Overall:
2H 2 + O 2 2H 2 O
2H 2 + O 2 2H 2 O
CORROSION
The deterioration of metals by an
electrochemical process.
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Examples of corrosion:
- rust on iron
- tarnish on silver
- green patina on copper and brass
For almost all metals oxidation is a
thermodynamically favourable process at room
temperature.
Can be extremely destructive
or can form a protective oxide layer.
Oxidation of Aluminium
Expect Al to be readily oxidised
Al 3+ + 3e - Al
E o = -1.66 V
However, we find it is exceptionally stable in air.
WHY
A thin protective layer of a hydrated form of
Al 2 O 3 forms on the surface of the metal.
This oxide layer is then impermeable to O 2 and
H 2 O, thus protecting the underlying metal from
further corrosion.
Corrosion of Iron
Rusting
Need both O 2 and H 2 O.
Cathode: O 2 (g) + 4H + (aq) + 4e - 2H 2 O(l) E o red = 1.23 V
Anode: Fe(s) Fe 2+ (aq) + 4e- E o red = -0.44 V
Fe + O 2 + 4H + Fe 2+ + 2H 2 O
Need H + for corrosion
above pH 9 Fe does not corrode
Cathode:
O 2 (g) + 4H + (aq) + 4e - 2H 2 O(l)
Anode: Fe(s) Fe 2+ (aq) + 4e-
Fe + O 2 + 4H + Fe 2+ + 2H 2 O
Further oxidation of Fe 2+ to Fe 3+ :
4Fe 2+ + O 2 + (4+2x)H 2 O 2Fe 2 O 3 .xH 2 O + 8H +
Rust
Part of the Fe surface is the anode at which
oxidation occurs.
e - ’s migrate through the Fe metal to another
part of the Fe surface, the cathode, where O 2 is
reduced.
Other factors can accelerate rusting:
- pH of the soln
(lower pH faster corrosion)
- presence of salts
(improves conductivity of electrolyte)
- contact with metals more difficult to oxidise
than Fe
- stress on Fe
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Preventing the corrosion of iron
Covering of iron with e.g. paint or tin
- Prevents O 2 and H 2 O from reaching surface.
Galvanised iron coated with layer of zinc
- Protects Fe by electrochemical means even
after coating is damaged
Easier to oxidise Zn (anode) than Fe (cathode)
Zn rather corrodes
- Coating damaged corrosion
Cathodic protection
Zn = sacrificial anode
E o red/V
Reduction Half-reaction
+0.77 Fe 3+ (aq) + e - Fe 2+ (aq)
Cathodic protection is used for e.g. the
protection of iron pipelines or storage tanks
-0.44 Fe 2+ (aq) + 2e - Fe(s)
-0.76 Zn 2+ (aq) + 2e - Zn(s)
-2.36 Mg 2+ (aq) + 2e - Mg(s)
Mg is the sacrificial anode for the Fe tank.
ELECTROLYSIS
Electrolysis of molten NaCl
Use electrical energy to
cause non-spontaneous
redox reaction to occur
- Takes place in electrolytic
cells
- Need an external DC source
e.g. a battery
= electron pump
Reduction occurs at the
cathode (negative) and
oxidation occurs at the
anode (positive)
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Electrolysis of aqueous solutions
Example: Electrolysis of aqueous NaF solution
Need high temp’s to produce molten salts.
Can also produce ions by dissolving salts in
water at room temp.
2 possible cathode reactions:
Na + (aq) + e - Na(s)
2H 2 O(l) + 2e - H 2 (g) + 2OH - (aq)
E o red = -2.71 V
E o red = -0.83 V
Problem:
Electrolysis of aqueous salts complicated
by the presence of water.
Favoured reduction reaction
more +ve or less –ve E o red
Cathode:
2H 2 O(l) + 2e - H 2 (g) + 2OH - (aq)
2 possible anode reactions:
2F - (aq) F 2 (g) + 2e -
2H 2 O(l) O 2 (g) + 4H + (aq) + 4e -
Favoured oxidation reaction
more -ve or less +ve E o red
2H 2 O(l) O 2 (g) + 4H + (aq) + 4e -
E o red = +2.78 V
E o red = +1.23 V
In fact it is easier still to oxidise OH - (aq) that is
produced at the cathode:
Anode: 4OH - (aq) O 2 (g) + 2H 2 O(l) + 4e -
E o red = +0.40 V
Example: Electrolysis of aqueous NaCl solution
Cathode reactions as in previous example.
2 possible anode reactions:
2Cl - (aq) Cl 2 (g) + 2e -
2H 2 O(l) O 2 (g) + 4H + (aq) + 4e -
Expect favoured reaction to be:
2H 2 O(l) O 2 (g) + 4H + (aq) + 4e -
E o red = +1.36 V
E o red = +1.23 V
Thermodynamics
HOWEVER, in experiments Cl - is oxidised.
Due to kinetics of electrode process!
Activation energy is lower for Cl - oxidation, so it
is kinetically favoured.
Electrolysis with active electrodes
Example: Ni electroplating
Inert electrodes do not undergo reaction but
only act as a surface where oxidation and
reduction occurs.
Active electrodes participate in the
electrolysis process
E.g. electroplating
Find the favoured reaction to be:
Cathode (steel strip):
Ni 2+ (aq) + 2e - Ni(s)
Anode (nickel strip):
Ni(s) Ni 2+ (aq) + 2e -
Pic of
cell
Overall reaction: “transferring Ni atoms” from
the anode to form a thin layer on the steel
cathode.
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Quantitative aspects of electrolysis
The stoichiometry of a half-reaction shows us
how many electrons needed in electrolysis.
E.g.
Na + + e - Na
1 mol e - ’s will plate out 1 mol Na metal
Or 2 mol e-’s will plate out 2 mol Na metal
E.g.
Ni 2+ + 2e - Ni
2 mol e-’s will plate out 1 mol Ni metal
Recall:
FARADAY`S LAW
The amount of substance that undergoes
oxidation or reduction at each electrode is
directly proportional to the amount of electricity
that passes through the cell.
Unit of charge (Q) = coulomb (C)
The charge in 1 mole of electrons = 96 500 C
Faraday’s constant, F = 96 500 C/mol
Q = It (Current in A, time in s)
Example:
Calculate the mass of aluminium produced in 1.0 hour
by the electrolysis of molten AlCl 3 if the electrical
current is 20 A.
Where does the Al come from
Al 3+ + 3e - Al
1 Calculate the amount of charge transferred:
Q = It = (20 A)(3600s) = 7.2 x 10 4 C
2 Calculate the no. of moles of e - ’s transferred: Q=nF
7.2 x 10 4 C
Moles of e - ’s =
= 0.75 mol
96 500 C/mol
3 Relate no. of moles of e - ’s to no. of moles of Al formed:
3 mol e - 1 mol Al
0.75 mol e - 0.25 mol Al
4 Convert moles to grams:
m = nM = (0.25 mol)(27.0 g/mol) = 6.7 g Al
Recall:
Electrical work
G < 0 and E cell > 0 spontaneous process
G = -nFE
G = w max
Thus the maximum useful work obtainable from
a voltaic cell is:
w max = -nFE
For an electrolytic cell, work has to be done on
the system for the reaction to occur,
w = nFE ext
where E ext = external potential applied.
Example:
Calculate the number of kWh of electricity required to
produce 1.0x10 3 kg of aluminium by electrolysis of Al 3+
if the applied emf is 4.50 V.
Electrical power is the rate of energy
expenditure.
Unit: Watt (W) = J/s
1 kWh = 3.6x10 6 J
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