• For the SHE, we assign
2H+(aq, 1M) + 2e-  H2(g, 1 atm)
• Ered = 0.
E cell  E red cathode   E red anode 
Prentice Hall © 2003
Chapter 20
• For Zn:
Ecell = Ered(cathode) - Ered(anode)
0.76 V = 0 V - Ered(anode).
• Therefore, Ered(anode) = -0.76 V.
• Standard reduction potentials must be written as
reduction reactions:
Zn2+(aq) + 2e-  Zn(s), Ered = -0.76 V.
Prentice Hall © 2003
Chapter 20
• Changing the stoichiometric coefficient does not affect
Ered.
• Therefore,
2Zn2+(aq) + 4e-  2Zn(s), Ered = -0.76 V.
Prentice Hall © 2003
Chapter 20
• Reactions with Ered < 0 are spontaneous oxidations
relative to the SHE.
• The larger the difference between Ered values, the larger
Ecell.
Prentice Hall © 2003
Chapter 20
Oxidizing and Reducing Agents
• The more positive Ered the stronger the oxidizing agent
on the left.
• The more negative Ered the stronger the reducing agent
on the right.
Prentice Hall © 2003
Chapter 20
Prentice Hall © 2003
Chapter 20
• More generally, for any electrochemical process
Ecell  Ered reduction process   Ered oxidation process 
Prentice Hall © 2003
Chapter 20
Example: For the following cell, what is the cell
reaction and Eocell?
Al(s)|Al3+(aq)||Fe2+(aq)|Fe(s)
Al3+(aq) + 3e- → Al(s); EoAl = -1.66 V
Fe2+(aq) + 2e- → Fe(s); EoFe = -0.41 V
Prentice Hall © 2003
Chapter 20
Spontaneity of Redox
Reactions
EMF and Free-Energy Change
• We can show that
G  nFE
• G is the change in free-energy, n is the number of moles
of electrons transferred, F is Faraday’s constant, and E is
the emf of the cell.
• We define
1F  96,500 C/mol  96,500 J/V·mol
• Since n and F are positive, if G > 0 then E < 0.
Prentice Hall © 2003
Chapter 20
The Nernst Equation
G  G  RT ln Q
 nFE  nFE   RT ln Q
Prentice Hall © 2003
Chapter 20
This gives:
RT
E  E 
ln Q
nF
• At a temperature of 298 K:
0.0592
E  E 
log Q
n
Prentice Hall © 2003
Chapter 20
Cell EMF and Chemical Equilibrium
• A system is at equilibrium when G = 0.
• E = 0 V and Q = Keq:
0.0592
0  E 
log K eq
n
nE 
log K eq 
0.0592
Prentice Hall © 2003
Chapter 20
In the problem presented before, calculate ΔGo
and Keq
Al(s)|Al3+(aq)||Fe2+(aq)|Fe(s)
Al3+(aq) + 3e- → Al(s); EoAl = -1.66 V
Fe2+(aq) + 2e- → Fe(s); EoFe = -0.41 V
Prentice Hall © 2003
Chapter 20
Example: For the reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Calculate the Ecell given that Eocell = +1.10 V, [Cu2+] = 5.0 M
and [Zn2+] = 0.050 M
Prentice Hall © 2003
Chapter 20
Exercise: Calculate the emf generated by the cell that employs the
following cell reaction:
2Al(s) + 3I2(s) → 2Al3+(aq) + 6I-(aq)
Take Eocell to be +2.20 V; [Al3+] = 4.0 x 10-3M and [I-] = 0.010M
Ans: +2.36 V
Prentice Hall © 2003
Chapter 20
Batteries
• Battery- self-contained
electrochemical power source
with one or more voltaic cell.
• When the cells are connected in
series, greater emfs can be
achieved.
• Primary cell – nonrechargeable
• Secondary cell-rechargeable
Prentice Hall © 2003
Chapter 20
• Lead-acid battery: Pb and PbO2 are electrodes and are
immersed in sulphuric acid. Electrodes are separated by glass
fibres or wood.
• Alkaline battery: anode is Zn in contact with a concentrated
solution of KOH. The cathode is a mixture of MnO2(s) and
graphite separated from the anode by a porous fabric
Prentice Hall © 2003
Chapter 20
• Nickel-cadmium, Ni-metal hydride and Li-ion batteries
Prentice Hall © 2003
Chapter 20
Fuel cells
Fuel cells differ from batteries in that they are not self-contained
systems.
They use conventional fuels e.g. H2 and CH4 to produce
electricity.
Prentice Hall © 2003
Chapter 20
The most common fuel cell is the one involving the reaction of
H2(g) and O2(g) to produce H2O(l).
This is based on that if H2O can be split by electricity, then
combining H2 and O2 should produce water and electricity.
Prentice Hall © 2003
Chapter 20
Prentice Hall © 2003
Chapter 20
Corrosion
•
•
•
•
•
Corrosion of Iron
Since Ered(Fe2+) < Ered(O2) iron can be oxidized by
oxygen.
Cathode: O2(g) + 4H+(aq) + 4e-  2H2O(l).
Anode: Fe(s)  Fe2+(aq) + 2e-.
Dissolved oxygen in water usually causes the oxidation
of iron.
Fe2+ initially formed can be further oxidized to Fe3+
which forms rust, Fe2O3.xH2O(s).
Prentice Hall © 2003
Chapter 20
Galvanised iron, i.e. Fe coated with Zn uses the principle of
electrochemistry to protect the iron from corrosion even after
the surface coat is broken.
The standard red. Pot. for Fe and Zn are:
Fe2+(aq) + 2e- → Fe(s) Eored = -0.44 V
Zn2+(aq) + 2e- → Zn(s) Eored = -0.76 V
Prentice Hall © 2003
Chapter 20
Protecting a metal from corrosion by making it the cathode in
an electrochemical cell is called cathodic protection.
The metal that is oxidised while protecting the cathode is called
the sacrificial anode.
Prentice Hall © 2003
Chapter 20
Electrolysis of Aqueous Solutions
• Nonspontaneous reactions require an external current in
order to force the reaction to proceed.
• Electrolysis reactions are nonspontaneous.
• In voltaic and electrolytic cells:
–
–
–
reduction occurs at the cathode, and
oxidation occurs at the anode.
However, in electrolytic cells, electrons are forced to flow
from the anode to cathode.
Prentice Hall © 2003
Chapter 20
–
In electrolytic cells the anode is positive and the cathode is
negative. (In galvanic cells the anode is negative and the
cathode is positive.)
Prentice Hall © 2003
Chapter 20
•
•
•
•
Example, decomposition of molten NaCl.
Cathode: 2Na+(l) + 2e-  2Na(l)
Anode: 2Cl-(l)  Cl2(g) + 2e-.
Industrially, electrolysis is used to produce metals like Al.
Prentice Hall © 2003
Chapter 20
Electroplating
• Active electrodes: electrodes that take part in electrolysis.
• Example: electrolytic plating.
Prentice Hall © 2003
Chapter 20
Prentice Hall © 2003
Chapter 20
•
•
•
•
•
Electroplating
Consider an active Ni electrode and another metallic
electrode placed in an aqueous solution of NiSO4:
Anode: Ni(s)  Ni2+(aq) + 2eCathode: Ni2+(aq) + 2e-  Ni(s).
Ni plates on the inert electrode.
Electroplating is important in protecting objects from
corrosion.
Prentice Hall © 2003
Chapter 20
Quantitative Aspects of Electrolysis
• We want to know how much material we obtain with
electrolysis.
• Consider the reduction of Cu2+ to Cu.
–
–
–
–
Cu2+(aq) + 2e-  Cu(s).
2 mol of electrons will plate 1 mol of Cu.
The charge of 1 mol of electrons is 96,500 C (1 F).
Since Q = It, the amount of Cu can be calculated from the
current (I) and time (t) taken to plate.
Prentice Hall © 2003
Chapter 20
Example: When an aqueous solution of CuSO4 is
electrolysed, Cu metal is deposited:
Cu2+(aq) + 2e- → Cu(s)
If a constant current was passed for 5.00 h and 404 mg
of Cu metal was deposited, what was the current?
Ans: 6.81 x 10-2 A
Prentice Hall © 2003
Chapter 20
Example: The half reaction for formation of Mg metal upon
electrolysis of molten MgCl2 is:
Mg2+(aq) + 2e- → Mg(s)
a) Calculate the mass of Mg formed upon passage of a
current of 60.0A for a period of 4.00 x 103s
b) How many seconds would be required to produce 50.0g of
Mg from MgCl2 if the current is 100.0A?
Ans: a) 30.2g, b) 3.97 x 103s
Prentice Hall © 2003
Chapter 20
Electrical Work
∆G is a measure of maximum useful work, wmax, that can be
extracted from the process: ∆ G = wmax = -nFE
For an electrolytic cell, an external source of energy is used to
bring about a non-spontaneous electrochemical process. In this
case,
w = nFEext
Prentice Hall © 2003
Chapter 20
Electrical work can be expressed in energy units of watts x time:
1 W = 1 J/s
Usually, the kWh is used
Example: Calculate the no. of kWh of electricity required to
produce 1.00 kg of Mg from electrolysis of molten MgCl2 if the
applied emf is 5.00 V
Ans: 11.0 kWh
Prentice Hall © 2003
Chapter 20