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Electrochemistry
Applications of Redox
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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
 Moving
Applications
electrons is electric current.
+
+2 +5e 8H +MnO4 + 5Fe
 Mn+2 + 5Fe+3 +4H2O
 Helps to break the reactions into half
reactions.
 8H++MnO4-+5e-  Mn+2 +4H2O
 5(Fe+2  Fe+3 + e- )
 In the same mixture it happens without
doing useful work. If separate, current
can flow.
 Connected
this way the reaction starts
 Stops immediately because charge builds
up.
Fe2+
MnO4H+
Galvanic Cell
Salt
Bridge
allows
current
to flow
Fe+2
H+
MnO4-
 Electricity
travels in a complete circuit
 Instead of a salt bridge
Fe+2
H+
MnO4-
Porous
Disk
Fe+2
H+
MnO4-
e-
e-
e-
e-
Anode
e-
Oxidation supplies electrons
Cathode
e-
Reduction uses electrons
17.2 Cell Potential
 Oxidizing
agent pulls the electron.
 Reducing agent pushes 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 (BDVD)half-reactions
0.76
H2 in
Anode
Zn+2
SO4-2
1 M ZnSO4
Cathode
H+
Cl1 M HCl
Pt
Standard Hydrogen Electrode
 This
is the reference
all other oxidations
are compared to
 Eº = 0
 º indicates standard
states of 25ºC,
1 H+
atm, 1 M solutions. Cl1 M HCl
H2 in
Cell Potential
Zn(s) + Cu+2 (aq)  Zn+2(aq) + Cu(s)
 The total cell potential is the sum of the
potential at each electrode.


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 its sign. Overall cell must be +
 Potentials are not multiplies by the coefficients
in the balanced equation. (BDVD) Galvanic cells

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)
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) + e- 
Fe+2(aq)
Cell Potential
A
galvanic cell based on the reaction:
 Al3+(aq) + Mg(s)
Al(s) + Mg2+(aq)
 Give the balanced cell reaction and
calculate E° for the cell. Ex.17.1
 MnO4-(aq) + H+(aq) + ClO3-(aq)
ClO42+
+
Mn
(aq)
(aq) + H2O (l)

(ClO4- + 2H+ + 2e- --> ClO3- + H2O = 1.19V)
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 Description


The reaction always runs
spontaneously in the direction that
produced a positive cell potential.
Four things for a complete description.
1. Cell Potential.
2. Direction of flow.
3. Designation of anode and cathode.
4. Nature of all the componentselectrodes and ions.
QuickTime™ and a
Sorenson Video decompressor
are needed to see this picture.
Practice
 Completely
describe (sketch) the galvanic
cell based on the following half-reactions
under standard conditions. Ex.17.2
 Ag+(aq) + e-  Ag(s)
+3
 Fe (aq) + e  Fe2+(aq)
 Describe the cell with this reaction:
 IO3-(aq) + Fe+2(aq)
Fe3+(aq) + I2(aq) #26a
Practice
 Completely
describe (sketch) the galvanic
cell based on the following half-reactions
under standard conditions. #30
 H2O2 + 2H+ + 2e- 2H2O
1.78 V
 O2 + 2H+ + 2e- H2O2
0.68 V
 Give the standard line notation for the
previous problems.
17.3 Potential, Work and G
 emf
= potential (V) = work (J)/Charge(C)
 E = work done by system / charge
 E = -w/q
 Charge is measured in coulombs.
 -w = qE (- sign means work is leaving the
system)
 Faraday
(F) = 96,485 C/mol e-
= nF = moles of e- x charge/mole e w = -qE = -nFE = G
q
Potential, Work and G
Gº = -nFE º
 if E º < 0, then Gº > 0 nonspontaneous
 if E º > 0, then Gº < 0 spontaneous
 Calculate Gº for the following reaction:
 Cu+2(aq)+ Fe(s)  Cu(s) + Fe+2(aq)

Eº = -0.44 V
 Cu+2(aq)+2e-  Cu(s) Eº = 0.34 V
 Fe+2(aq) +
2e- Fe
Ex.17.3
 The
Practice
amount of manganese in steel is
determined by changing it to
permanganate ion. The steel is first
dissolved in nitric acid, producing Mn2+
ions. These ions are then oxidized to the
deeply colored MnO4- ions by periodate
ion (IO4-) in acid solution. #38
 Complete and balance an equation
describing each of the above reactions.
 Calculate E° and G° at 25°C for each
reaction.
Practice
G° formula can also be used for
half-reactions. Use standard reduction
potentials to estimate G°f for Fe2+(aq)
and Fe3+(aq). (G°f for e- = 0.) #42
 Using data from Table 17.1 place the
following in order of increasing strength
as reducing agents. #44
 Cr3+, H2, Zn, Li, F-, Fe2+
 The
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
17.4 Cell Potential and
Concentration
Qualitatively - We can predict direction of
change in E from LeChâtelier. Shift to the left
reduces potential. To the right increases
potential.
+2
+3
 2Al(s) + 3Mn (aq)  2Al (aq) + 3Mn(s)

Predict if Ecell will be greater or less than Eºcell
if [Al+3] = 2.0 M and [Mn+2] = 1.0 M Ex. 17.5
+3
+2
 if [Al ] = 1.0 M and [Mn ] = 3.0M
+3
+2
 if [Al ] = 1.5 M and [Mn ] = 1.5 M

A
Concentration Cells
potential can be generated with the same
components but in different concentrations.
 The reaction proceeds in the direction that
will equalize the ion concentration in the
components.
 These voltages are typically small.
 By using the Nernst Equation, we can
calculate the potential of a cell in which
some or all of the components are not in
their standard states.
Concentration Cells
This cell has a driving
force that transfers
electrons from left to
right. Silver will be
deposited on the right
electrode, thus lowering
the concentration of Ag+
in the right compartment.
In the left, the silver
electrode dissolves
(producing Ag+ ions) to
raise the concentration of
Ag+ in solution.
Practice
Consider this cell. Predict the direction of electron flow,
and designate the anode and cathode. If the Fe2+
concentration in the right compartment is changed from
0.1 M to 1 x 10-7 M Fe2+, what would change. #52
 G
The Nernst Equation
= Gº + RTln(Q)
 -nFE = -nFEº + RTln(Q)
 E = Eº - RTln(Q)
nF E  E  0.0592 logQ
n
 At 25°C, We use
 Always have to figure out n by balancing.
 If concentration can give voltage, then
from the voltage we can tell the
concentration.
Example
2Al(s) + 3Mn+2(aq)  2Al+3(aq) + 3Mn(s)
Eº = 0.48 V under standard conditions.
 Consider a cell [Mn2+] = 0.50 M and [Al3+] =
1.50 M
 Q= [Al3+]2 = (1.50)2 = 18
[Mn2+]3 (0.50)3
 n = 6 from the balanced equation = 6e Using Nernst Ecell =0.48 - 0.0591 log(18)
6
= 0.48 - 0.01 = 0.47V (p.844)
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
The Nernst Equation

As reactions proceed concentrations of products
increase and reactants decrease.
Reach equilibrium where Q = K and Ecell = 0,
this is a “dead” cell.
 At equilibrium, the components in the two cell
compartments have the same free energy and
G = 0
 0 = Eº - 0.0591 log(K)
n
 log(K) = nEº
0.0591

Practice
For the redox reaction:
S4O62-(aq) + Cr2+(aq)  Cr3+(aq) + S2O32-(aq)
Half reactions are:
S4O62- + 2e- S2O32- E°= 0.17 V
Cr3+ + e-  Cr2+
E°= -0.50 V
Balance the redox reaction, and calculate E° and
K at 25°C Ex. 17.8
log(K) = nEº
0.0591
Practice
The overall reaction in a lead storage battery is
 Pb(s) + PbO2(s) + 2H+(aq) + 2HSO4-(aq) 
2PbSO4(s) + 2H2O(l)
 Calculate E at 25°C for this battery when
[H2SO4] = 4.5 M. At 25°C, E° = 2.04V for the
lead storage battery. #56
 Al|Al3+(1.00 M)||Pb2+(1.00M)|Pb
 In the above cell, calculate the cell potential
after the reaction has operated long enough for
the [Al3+] to have changed by 0.60 mol/L. #58

17.5 Batteries are Galvanic Cells
A
battery is a galvanic cell or, more
commonly, a group of galvanic cells
connected in series.
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

17.6 Corrosion
 Corrosion
(rusting) - spontaneous
oxidation.
 Many metals develop a metal oxide
coating that prevents further 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
 Fe+2 + O2 + H2O  Fe2 O3 + H+
 Reaction happens in two places.
Salt speeds up process by increasing
conductivity
Water
Rust
Anodic area
Cathodic area
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.
17.7 Electrolysis
 Forcing
a current through a cell to
produce a chemical change for which the
cell potential is negative.
 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
Stoichiometry of Electrolysis
How much chemical change occurs with the
flow of a given current for a specified time?
 Measure current I (in amperes)
 1 amp = 1 coulomb of charge per second

1 Faraday (F) = 96,485 C/mol e Problems are essentially a unit conversion.
 current and time  quantity of charge moles
of electrons  moles of substance grams of
substance
 How long must 5.00 amp current be applied to
produce 10.5 g of Ag from Ag+? Ex.17.9

 How
Practice
long would it take to produce 10.0 g of
Bi by the electrolysis of a BiO+ solution
using a current of 25.0A? #74
 Aluminum is produced commercially by the
electrolysis of Al2O3 in the presence of a
molten salt. If a plant has a continuous
capacity of 1.00 million amp, what mass of
aluminum can be produced in 2.00 hours? #76
 Electrolysis of an alkaline earth metal
chloride using a current of 5.00A for 748
seconds deposit 0.471 g of metal at the
cathode. What is the identity of the alkaline
earth metal? #78
Other uses
 Electrolysis
of water.
 Separating mixtures of ions.
 More positive reduction potential means
the reaction proceeds forward.
 A solution of Ag+, Cu2+, and Zn2+ were
introduced to a low voltage that was
slowly turned up. The most positive
reduction potential is easiest to plate out
of solution. Therefore Ag+> Cu2+ >Zn2+