Download Electrochemistry

Electrochemistry
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Electrochemical cells
 A chemical system in which oxidation
and reduction can occur – often a
single displacement reaction
Zn + Cu+2  Zn+2 + Cu
 Oxidation reaction and reduction
reaction are physically separated so
that useable work can be obtained
from the reaction
Electrochemical cells
 Voltaic cells (galvanic cells) – redox
occurs spontaneously
 Anode – where oxidation takes place
(solid metal becomes aqueous
positive ions)
 Cathode – where reduction takes
place (metal ions deposit as solid
metal)
 AN OX RED CAT
Electrochemical cells
Electrons flow from anode
to cathode through wire
cathode
anode
Salt bridge
Electrochemical cells
 Salt bridge – usually KNO3 or a
semipermeable membrane–
completes circuit by providing mobile
ions
 Conditions for spontaneity – one
metal must lose electrons more easily
than another (metals can be identical
if one is warmer) p. 288 (activity
series)
Electric current
 Rate of flow – amperes
1 amp = 1 coulomb/sec
1 coulomb = 1/96490 of a mole of
electrons (6.24x1018e-)
 Faraday’s number = 96490coul/mole
 Potential – volts (joules/coulomb)
Cell potential
 Half cell potential (E) – the voltage
contribution of a half reaction to the
cell
 Standard half cell potential (Eº) voltage when solutions are 1M, gases
are 1atm and temp. is 25ºC.
 Half cell potentials are given as
reductions relative to the reduction of
H+ to H2, which is assigned 0 volts.
Cell potential
 Half cell potential is a measure of the
tendency of a particle to gain electrons.
 The cell potential is the sum of the two
half cell potentials.
 For oxidation, the sign of the reduction
potential is reversed.
 A positive cell potential means a
spontaneous reaction, i.e. a galvanic
cell.
Cell potential
 Example. Find the cell potential for a
cell with a mercury/mercury (II)
cathode and a lead/lead (II) anode.
Standard reduction potentials:
Hg+2 +2e-  Hg Eº = 0.851V
Pb+2 + 2e-  Pb Eº = -0.1262V
 Lead is the anode, so it is oxidized
(reduction equation is reversed)
Cell potential
Hg+2 +2e-  Hg Eº = 0.851V
Pb  Pb+2 + 2e- Eº = 0.1262V
 Cell potential is the sum of the half
cell potentials
Hg+2 + Pb  Pb+2 + Hg
Eº = 0.851V + 0.1262V = 0.977V
 Potential is intensive, so it’s not
affected by coefficients
Cell potential
 A positive cell potential means a
spontaneous reaction, i.e. a galvanic
cell.
 Galvanic cell calculations
 Cell notation
Zn|Zn+2║Cu+2|Cu
anode
salt bridge cathode
Galvanic cell calculations
 Vertical lines represent the barrier
between two different states of
matter.
 Two different materials in the same
part of the cell are separated by
commas.
H2,Pt|H+║Ag+|Ag+
Galvanic cell calculations
 Calculate the voltage of this cell:
Mg|Mg+2║Au+3|Au
Mg  Mg+2 + 2eE º = +2.37V
cathode: Au+3 + 3e-  Au
E º = +1.50V
 Total cell voltage = 3.87V
Current calculations
 Extent of oxidation/reduction and
amount of material oxidized or
reduced is directly related to the
number of electrons transferred
 Moles e- = current x time / Faraday’s #
= It/F F = 96485coul/mole
Current calculations
 A zinc anode with mass 2.30g is used
in a copper/zinc cell. The cell has a
current of 0.00140 amps. How long
will the electrode last?
 Solution: 2.30 g Zn is 0.0352 mole
zinc.
 Each mole zinc requires 2e-, so
0.0704 mole e- are required to
completely oxidize the anode.
Current calculations
 There are 96,485 coulombs/mole, so
6790 coulombs are needed.
0.0704mol e- x 96485 coul/mol = 6790c
 At 0.00140 coulombs/sec, the
electrode will last 4.85x106 sec, or
1350 hours.
Types of cells
 Concentration
cell – uses
identical
electrodes –
potential
difference due
to
concentration
differences in
the cell
Nonstandard cells
 Ecell = Eºcell – (RT/nF )lnQ
 Q = reaction quotient
 For the reaction aA + bB  cC + dD,
Q = [C]c[D]d/[A]a[B]b
 Find the voltage for a zinc/copper cell
at 55ºC where the concentrations of
zinc and copper are 0.10 and 0.75M
respectively.
Types of cells
 Leclanché cell
(dry cell)
MnO2, water,
NH4Cl in a
paste around
a graphite
cathode, with
a zinc anode
Types of cells
 Overall Leclanché cell equation:
Zn+MnO2+NH4ClZnCl2+Mn2O3+NH3+H2O
Balance it!
Zn  Zn+2 + 2e2H+ + 2MnO2 + 2e-  Mn2O3 + H2O
2H++2MnO2+ZnZn+2+Mn2O3+H2O
Add spectators (NH3 and Cl-):
2NH4Cl+2MnO2+ZnZnCl2+Mn2O3+H2O+2NH3
Types of cells
 Alkaline
manganese cell
– used in
alkaline
batteries – uses
KOH as
electrolyte
 Zn+MnO2+H2O
Zn(OH)2+
Mn2O3
Types of cells
 Electrolytic cells –
nonspontaneous redox
reaction is forced to
proceed by application
of an electric current
 Electrolysis of water –
Hoffman apparatus
2H2O  2H2+ O2
Electrolysis of aluminum oxide
 Alumina (Al2O3) from bauxite is
dissolved in molten cryolite (Na3AlF6).
Electrolysis of aluminum oxide
 The steel container is coated with
carbon (graphite) and serves as the
negative electrode (anode).
 Electrolysis of the alumina/cryolite
solution gives aluminum at the
cathode and oxygen at the anode.
 Aluminum is more dense than the
alumina/cryolite solution, and so falls
to the bottom of the cell.
Electrolysis of aluminum oxide
 Aluminum can be tapped off the bottom as
pure liquid metal.
 The overall reaction is:
2Al2O3(l)  4Al(l) + 3O2(g)
 Oxygen is discharged at the positive carbon
(graphite) anode.
 Oxygen reacts with the carbon anode to
form carbon dioxide gas. The carbon anode
slowly disappears as carbon dioxide and
needs to be replaced regularly.
Anodic protection