Download Electrochemistry

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
Electrochemistry
Guaranteed to
give you a jolt
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