Transcript AP Notes Chapter 20

AP Chemistry
Chapter 20 Notes
Electrochemistry
Applications of Redox
Review
 Oxidation
reduction reactions involve a
transfer of electrons.
 OIL- RIG
 Oxidation Involves Loss
 Reduction Involves Gain
 LEO-GER
 Lose Electrons Oxidation
 Gain Electrons Reduction
Applications
 Moving
electrons is electric current.
 8H++MnO4-+
 Helps
5Fe+2 +5e-  Mn+2 + 5Fe+3 +4H2O
to break the reactions into half rxns.
 8H++MnO4-+5e-  Mn+2 +4H2O
 5Fe+2  5Fe+3 + 5e- )
 In
the same mixture it happens without doing
useful work, but if separate
 Connected
this way the reaction starts
 Stops immediately because charge builds
up.
H+
MnO4-
Fe+2
Galvanic Cell
Salt
Bridge
allows
current
to flow
H+
MnO4-
Fe+2
 Electricity
travels in a complete circuit
 Instead of a salt bridge
H+
MnO4-
Fe+2
Porous
Disk
H+
MnO4-
Fe+2
e-
e-
e-
e-
Anode
e-
Reducing
Agent
Cathode
e-
Oxidizing
Agent
Cell Potential
 Oxidizing
agent pushes the electron.
 Reducing agent pulls the electron.
 The push or pull (“driving force”) is called
the cell potential Ecell
 Also
called the electromotive force (emf)
 Unit is the volt(V)
 = 1 joule of work/coulomb of charge
 Measured with a voltmeter
0.76
H2 in
Anode
Zn+2
SO4-2
1 M ZnSO4
Cathode
H+
Cl1 M HCl
Standard Hydrogen
Electrode
H2 in
 This
is the reference
all other oxidations
are compared to
 Eº = 0
 º indicates standard
states of 25ºC, 1 atm,
1 M solutions.
H+
Cl1 M HCl
Cell Potential
+ Cu+2 (aq)  Zn+2(aq) + Cu(s)
 The total cell potential is the sum of the
potential at each electrode.
 Zn(s)

Eº cell = EºZn Zn+2 + Eº Cu+2 Cu
 We
can look up reduction potentials in a
table.
 One of the reactions must be reversed, so
change it sign.
Cell Potential
 Determine
the cell potential for a galvanic
cell based on the redox reaction.
 Cu(s) + Fe+3(aq)  Cu+2(aq) + Fe+2(aq)
e- Fe+2(aq)
Eº = 0.77 V
 Cu+2(aq)+2e- Cu(s)
Eº = 0.34 V
 Cu(s) Cu+2(aq)+2eEº = -0.34 V
 2Fe+3(aq) + 2e- 2Fe+2(aq) Eº = 0.77 V
 Fe+3(aq) +
Line Notation
solidAqueousAqueoussolid
 Anode on the leftCathode on the right
 Single line different phases.
 Double line porous disk or salt bridge.
 If all the substances on one side are
aqueous, a platinum electrode is indicated.
 For the last reaction
 Cu(s)Cu+2(aq)Fe+2(aq),Fe+3(aq)Pt(s)

Galvanic Cell


1)
2)
3)
4)
The reaction always runs spontaneously
in the direction that produced a positive
cell potential.
Four things for a complete description.
Cell Potential
Direction of flow
Designation of anode and cathode
Nature of all the components- electrodes
and ions
Practice
 Completely
describe the galvanic cell
based on the following half-reactions under
standard conditions.
 MnO4- + 8 H+ +5e-  Mn+2 + 4H2O
Eº=1.51
 Fe+3 +3e-  Fe(s)
Eº=0.036V
Potential, Work and DG
 emf
= potential (V) = work (J) / Charge(C)
 E = work done by system / charge
 E = -w/q
 Charge
 -w
is measured in coulombs.
= qE
 Faraday
= 96,485 C/mol e-
= nF = moles of e- x charge/mole e w = -qE = -nFE = DG
q
Potential, Work and DG
DGº = -nFE º
FIXED
 if E º > 0, then DGº < 0 spontaneous
 if E º < 0, then DGº > 0 nonspontaneous

 In
fact, reverse is spontaneous.
 Calculate DGº for the following reaction:
 Cu+2(aq)+ Fe(s) Cu(s)+ Fe+2(aq)
 Fe+2(aq) +
e-Fe(s)
 Cu+2(aq)+2e- Cu(s)
Eº = 0.44 V
Eº = 0.34 V
Cell Potential and
Concentration
 Qualitatively
- Can predict direction of
change in E from LeChâtelier.
 2Al(s)
+ 3Mn+2(aq)  2Al+3(aq) + 3Mn(s)
if Ecell will be greater or less than
Eºcell if [Al+3] = 1.5 M and [Mn+2] = 1.0 M
 Predict
[Al+3] = 1.0 M and [Mn+2] = 1.5M
 if [Al+3] = 1.5 M and [Mn+2] = 1.5 M
 if
The Nernst Equation
 DG
= DGº +RTln(Q)
 -nFE = -nFEº + RTln(Q)
E = Eº - RTln(Q)
nF
 2Al(s) + 3Mn+2(aq)  2Al+3(aq) + 3Mn(s)
Eº = 0.48 V
 Always
have to figure out n by balancing.
 If concentration can gives voltage, then
from voltage we can tell concentration.
The Nernst Equation
 As
reactions proceed concentrations of
products increase and reactants decrease.
 Reach
equilibrium where Q = K and
=0
0 = Eº - RTln(K)
nF
Eº = RTln(K)
nF
 nFEº = ln(K)
RT
Ecell
Batteries are Galvanic Cells
 Car
batteries are lead storage batteries.
 Pb +PbO2 +H2SO4 PbSO4(s) +H2O
 Dry Cell
Zn + NH4+ +MnO2  Zn+2 + NH3 + H2O
 Alkaline
Zn +MnO2  ZnO+ Mn2O3 (in base)
 NiCad
 NiO2 + Cd + 2H2O  Cd(OH)2 +Ni(OH)2
Corrosion
 Rusting
- spontaneous oxidation.
 Most structural metals have reduction
potentials that are less positive than O2 .
 Fe  Fe+2 +2eEº= 0.44 V
 O2 + 2H2O + 4e- 4OHEº= 0.40 V
+ O2 + H2O Fe2 O3 + H+
 Reaction happens in two places.
 Fe+2
Salt speeds up process by increasing
conductivity
Water
Rust
e-
Iron Dissolves- Fe  Fe+2
Preventing Corrosion
 Coating
to keep out air and water.
 Galvanizing - Putting on a zinc coat
 Has a lower reduction potential, so it is
more. easily oxidized.
 Alloying with metals that form oxide coats.
 Cathodic Protection - Attaching large
pieces of an active metal like magnesium
that get oxidized instead.
Electrolysis
 Running
a galvanic cell backwards.
 Put a voltage bigger than the potential and
reverse the direction of the redox reaction.
 Used for electroplating.
1.10
e-
e-
Zn
Cu
1.0 M
Zn+2
Anode
1.0 M
Cu+2
Cathode
e-
A battery
>1.10V
Zn
e-
Cu
1.0 M
Zn+2
Cathode
1.0 M
Cu+2
Anode
Calculating plating
 Have
to count charge.
 Measure current I (in amperes)
 1 amp = 1 coulomb of charge per second
q = I x t
 q/nF = moles of metal
 Mass of plated metal
 How long must 5.00 amp current be
applied to produce 15.5 g of Ag from Ag+
Other uses
 Electroysis
of water.
 Seperating mixtures of ions.
 More positive reduction potential means
the reaction proceeds forward.
 We want the reverse.
 Most negative reduction potential is easiest
to plate out of solution.
Balancing
Redox
Equations
2. in base
Am3+(aq) + S2O82-(aq) ---->
AmO2+(aq) + SO42-(aq)
3. MnO4
-(aq)
+ H2C2O4(aq) 
Mn2+(aq) + CO2(g)
4.
2Bi(OH)3 + SnO2
2Bi(s) + SnO3

ELECTROLYTIC
CELLS
Electrolytic Cell
a cell that uses
electrical energy to
produce a chemical
change that would
otherwise NOT occur
spontaneously
Process
referred to
as
electrolysis
(+)
(-)
M+(aq)
M
X-(aq)
M
e-
(+)
(-)
e
M+(aq)
M
Anode
M M+ + eoxidation
X-(aq)
M
Cathode
M+ + e M
reduction
Ampere
a unit of electrical
current equal to one
coulomb of charge
per second
coul
1 amp = 1
sec
Coulomb
a unit of electric
charge equal to the
quantity of charge in
19
about 6 x 10
electrons
Faraday
a constant
representing the
charge on one mole
of electrons
1 F = 96,485 C
96,500 C
3: It is necessary to
replate a silver teapot
with 15.0 g of silver. If
the electrolytic cell
runs at 2.00 amps, how
long will it take to plate
the teapot?
4: Sodium metal and chlorine
gas are prepared industrially
in a Down’s Cell from the
electrolysis of molten NaCl.
What mass of metal and
volume of gas can be made
per day if the cell operates at
7.0 volts and 4.0 x 104 amps
if the cell is 75% efficient?
5: At what current must
a cell be run in order to
produce 5.0 kg of
aluminum in 8.0 hours if
the cell produces solid
aluminum from molten
aluminum chloride?
ELECTROCHEMISTRY,
FREE ENERGY,
& EQUILIBRIUM
w ork(J)
em f( V ) 
ch arge(C)
w
E
q
.
thus: wmax = - q Emax
but:
wmax = DG
and
q = nF
.
thus if: wmax = - q Emax
then
DG = - nFE
DG =
0
DG
+ RT ln Q
DG = - nFE
- nFE = -
0
nFE
+ RT ln Q
RT
EE 
ln Q
nF
0
RT
EE 
ln Q
nF
0
NERNST EQUATION
RT
EE 
ln Q
nF
0
if: aA + bB  cC + dD

C D
Q
a
b
A B
c
d
RT
EE 
ln Q
nF
0
0
25 C
IF T =
= 298.15 K
ln Q = 2.303 log Q
.
R = 8.314 J/mol K
F = 96,485 C/mol
0.0592
EE 
log Q
n
0
RT
EE 
ln Q
nF
0
what if : Q = Keq ?
then: E = 0.0 V
RT
E 
ln K
nF
0
0.0592
E 
log K
n
0
6: Calculate the
equilibrium constant
0
at 40 C for the cell:
2+
Cd(s) Cd
(1M)
2+
Pb
(1M) Pb(s)
7a: Calculate the
standard free energy
for the cell:
Cr(s) Cr3+ (1M) Fe2+ (1M) Fe(s)
7a: Calculate the
standard free energy for
the cell:
Cr(s) Cr3+ (1M) Fe2+ (1M) Fe(s)
7b: What will be the
2+
voltage if [Fe ] = 0.50M
and [Cr3+] = 0.30M at 200C?
8: Through
electrochemical
calculations, determine
the Ksp for silver bromide.
AgBr + e  Ag + Br
0
E = 0.10 V
Review of Redox &
Electrochemical Cells
Review
e
Oxidation: loss of
[increase in ox #]
[reducing agent]
Reduction: gain of e[decrease in ox #]
[oxidizing agent]
Reduction Potential
The ease with which a
chemical species can
be reduced
Standard Reduction
Potential
Appendix M
Table 20.1 in text
1. Which of the following
elements listed is the
best reducing agent?
Cu
Zn
Fe
Ag
Cr
2a. Choosing from
among the reactants in
the given half reactions,
identify the strongest
and weakest oxidizing
agents.
Anode and Cathode
 OXIDATION
occurs at the ANODE.
 REDuction occurs at the CAThode.
Electrochemical Cell
device in which
chemical energy is
spontaneously
changed to electrical
energy
battery
voltaic cell
galvanic cell
An electrochemical
cell consists of ???
M1+(aq)
M1
X-(aq)
M2+(aq)
X-(aq)
M2
M1+(aq)
M1
X-(aq)
Anode
M1  M1+ + e-
M2+(aq)
X-(aq)
M2
Cathode
M2+ + e-  M2
K+(aq) NO3-(aq)
M1+(aq)
M1
X-(aq)
Anode
M1  M1+ + e-
M2+(aq)
X-(aq)
M2
Cathode
M2+ + e-  M2
e
flow is from
source of high
“concentration” to
source of low
“concentration”
ee-
K+(aq) NO3-(aq)
M1+(aq)
M1
X-(aq)
Anode
M1  M1+ + e-
M2+(aq)
X-(aq)
M2
Cathode
M2+ + e-  M2
shorthand notation
oxidation reduction
+
+
M1 | M1 || M2 | M2
anode  cathode

e
flow 
e
this
flow can
accomplish work
w ork(J)
em f( V ) 
ch arge(C)
Electrochemical
Standard State
Conditions
[ions] = 1 M
0
T = 25 C
Pgas = 1 atm
An electrochemical cell
is spontaneous if:
Oxidation-reduction occurs
Ered + Eox > 0
2b. Which of the
oxidizing agents listed
is (are) capable of
oxidizing Br to BrO3 ?
Line Notation:
ANODE
CATHODE
Ni(s)|Ni2+ (aq, 1 M)||Au3+(aq, 1 M)|Au(s)
oxidation
reduction
Line Notation:
ANODE
CATHODE
Al(s) | Al3+(aq, 1 M) || Ni2+(aq, 1 M) | Ni (s)
oxidation
reduction