Chemistry: A Molecular Approach, 1st Ed.
Nivaldo Tro
Chapter 18
Electrochemistry
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2007, Prentice Hall
Redox Reaction
• one or more elements change oxidation number
•
•
•
•
all single displacement, and combustion,
some synthesis and decomposition
always have both oxidation and reduction
split reaction into oxidation half-reaction and a
reduction half-reaction
aka electron transfer reactions
half-reactions include electrons
oxidizing agent is reactant molecule that causes oxidation
contains element reduced
reducing agent is reactant molecule that causes reduction
contains the element oxidized
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Oxidation & Reduction
• oxidation is the process that occurs when
oxidation number of an element increases
element loses electrons
compound adds oxygen
compound loses hydrogen
half-reaction has electrons as products
• reduction is the process that occurs when
oxidation number of an element decreases
element gains electrons
compound loses oxygen
compound gains hydrogen
half-reactions have electrons as reactants
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Rules for Assigning Oxidation States
• rules are in order of priority
1. free elements have an oxidation state = 0
 Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g)
2. monatomic ions have an oxidation state equal
to their charge
 Na = +1 and Cl = -1 in NaCl
3. (a) the sum of the oxidation states of all the
atoms in a compound is 0
 Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0
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Rules for Assigning Oxidation States
3. (b) the sum of the oxidation states of all the atoms in
a polyatomic ion equals the charge on the ion
 N = +5 and O = -2 in NO3–, (+5) + 3(-2) = -1
4. (a) Group I metals have an oxidation state of +1 in all
their compounds
 Na = +1 in NaCl
4. (b) Group II metals have an oxidation state of +2 in
all their compounds
 Mg = +2 in MgCl2
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Rules for Assigning Oxidation States
5. in their compounds, nonmetals have oxidation
states according to the table below
 nonmetals higher on the table take priority
Nonmetal
Oxidation State
Example
F
-1
CF4
H
+1
CH4
O
-2
CO2
Group 7A
-1
CCl4
Group 6A
-2
CS2
Group 5A
-3
NH3
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Oxidation and Reduction
• oxidation occurs when an atom’s oxidation state
increases during a reaction
• reduction occurs when an atom’s oxidation state
decreases during a reaction
CH4 + 2 O2 → CO2 + 2 H2O
-4 +1
0
+4 –2
+1 -2
oxidation
reduction
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Oxidation–Reduction
• oxidation and reduction must occur simultaneously
 if an atom loses electrons another atom must take them
• the reactant that reduces an element in another reactant
is called the reducing agent
 the reducing agent contains the element that is oxidized
• the reactant that oxidizes an element in another reactant
is called the oxidizing agent
 the oxidizing agent contains the element that is reduced
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
Na is oxidized, Cl is reduced
Na is the reducing agent, Cl2 is the oxidizing agent
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Identify the Oxidizing and Reducing Agents
in Each of the Following
3 H2S + 2 NO3– + 2 H+ 3 S + 2 NO + 4 H2O
MnO2 + 4 HBr MnBr2 + Br2 + 2 H2O
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Identify the Oxidizing and Reducing Agents
in Each of the Following
red ag
ox ag
+1 -2
+5 -2
3 H2S + 2 NO3– + 2 H+ 3 S + 2 NO + 4 H2O
+1
0
+2 -2
+1 -2
oxidation
reduction
ox ag
red ag
+4 -2
+1 -1
MnO2 + 4 HBr MnBr2 + Br2 + 2 H2O
+2 -1
0
+1 -2
oxidation
reduction
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Common Oxidizing Agents
Oxidizing Agent
Product when Reduced
-2
O2
O
H2O2
H2O
-1
F2, Cl2, Br2, I2
-1
-1
-1
-1
F , Cl , Br , I
-1
-1
-1
-1
-1
ClO3 (BrO3 , IO3 )
Cl , (Br , I )
H2SO4 (conc)
SO2 or S or H2S
SO3
-2
-2
S2O3 , or S or H2S
HNO3 (conc) or NO3
-1
MnO4 (base)
-1
MnO4 (acid)
-2
CrO4 (base)
-2
Cr2O7 (acid)
-1
NO2, or NO, or N2O, or N2, or NH3
MnO2
Mn
+2
Cr(OH)3
Cr
+3
11
Common Reducing Agents
Reducing Agent
Product when Oxidized
H2
H
H2O2
O2
I
+1
-1
I2
NH3, N2H4
-2
S , H2S
SO3
-2
NO2
-1
C (as coke or charcoal)
+2
Fe (acid)
Cr
+2
Sn
+2
metals
N2
S
SO4
-2
NO3
-1
CO or CO2
Fe
Cr
+3
+3
Sn
+4
metal ions
12
Balancing Redox Reactions
1) assign oxidation numbers
a) determine element oxidized and element reduced
2) write ox. & red. half-reactions, including electrons
a)
ox. electrons on right, red. electrons on left of arrow
3) balance half-reactions by mass
a)
b)
c)
d)
first balance elements other than H and O
add H2O where need O
add H+1 where need H
neutralize H+ with OH- in base
4) balance half-reactions by charge
a)
balance charge by adjusting electrons
5) balance electrons between half-reactions
6) add half-reactions
7) check
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Ex 18.3 – Balance the equation:
I(aq) + MnO4(aq)  I2(aq) + MnO2(s) in basic solution
I(aq) + MnO4(aq)  I2(aq) + MnO2(s)
Assign
Oxidation
States
Separate
into halfreactions
ox: I(aq)  I2(aq)
red: MnO4(aq)  MnO2(s)
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Ex 18.3 – Balance the equation:
I(aq) + MnO4(aq)  I2(aq) + MnO2(s) in basic solution
Balance
Balancehalf-ox: ox:
2 I(aq)
I(aq)
2
I

I2(aq)
I2(aq)
I2(aq)
(aq)

 MnO
reactions
half- by red: red:
4 H+MnO
+
MnO

MnO
+ 22(s)H2+O2(l)H2O(l)
(aq) 4 (aq) 4 (aq)  2(s)
mass
reactions
+



4
H
+
4
OH
+
MnO

MnO
+
2
H
O
+
4
OH
by
mass
(aq)
(aq)
4
(aq)
2(s)
2
(l)
(aq)
then O by
adding
then
in
base,
HHby
2O 4 H2O(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l) + 4 OH(aq)
adding H+
neutralize
MnO4(aq) + 2 H2O(l)  MnO2(s) + 4 OH(aq)
+
the H
with OH-
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Ex 18.3 – Balance the equation:
I(aq) + MnO4(aq)  I2(aq) + MnO2(s) in basic solution
Balance
Halfreactions
by
charge
Balance
electrons
between
halfreactions
ox: 2 I(aq)  I2(aq) + 2 e
red: MnO4(aq) + 2 H2O(l) + 3 e  MnO2(s) + 4 OH(aq)
ox: 2 I(aq)  I2(aq) + 2 e } x3
red: MnO4(aq) + 2 H2O(l) + 3 e  MnO2(s) + 4 OH(aq) }x2
ox: 6 I(aq)  3 I2(aq) + 6 e
red: 2 MnO4(aq) + 4 H2O(l) + 6 e  2 MnO2(s) + 8 OH(aq)
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Ex 18.3 – Balance the equation:
I(aq) + MnO4(aq)  I2(aq) + MnO2(s) in basic solution
Add the ox: 6 I(aq)  3 I2(aq) + 6 e
Halfred: 2 MnO4(aq) + 4 H2O(l) + 6 e  2 MnO2(s) + 8 OH(aq)
reactions tot: 6 I(aq)+ 2 MnO4(aq) + 4 H2O(l)  3 I2(aq)+ 2 MnO2(s) + 8 OH(aq)
Check
Reactant
Count
Element
Product
Count
6
I
6
2
Mn
2
12
O
12
8
H
8
2
charge
2
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Practice - Balance the Equation
H2O2 + KI + H2SO4  K2SO4 + I2 + H2O
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Practice - Balance the Equation
H2O2 + KI + H2SO4  K2SO4 + I2 + H2O
+1 -1
+1 -1 +1 +6 -2
+1 +6 -2
0
+1 -2
oxidation
reduction
ox:
red:
tot
2 I-1 I2 + 2e-1
H2O2 + 2e-1 + 2 H+  2 H2O
2 I-1 + H2O2 + 2 H+  I2 + 2 H2O
1 H2O2 + 2 KI + H2SO4  K2SO4 + 1 I2 + 2 H2O
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Practice - Balance the Equation
ClO3-1 + Cl-1  Cl2 (in acid)
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Practice - Balance the Equation
ClO3-1 + Cl-1  Cl2 (in acid)
+5 -2
-1
0
oxidation
reduction
ox:
red:
tot
2 Cl-1 Cl2 + 2 e-1 } x5
2 ClO3-1 + 10 e-1 + 12 H+  Cl2 + 6 H2O} x1
10 Cl-1 + 2 ClO3-1 + 12 H+  6 Cl2 + 6 H2O
1 ClO3-1 + 5 Cl-1 + 6 H+1  3 Cl2 + 3 H2O
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Electrical Current
• when we talk about the current
•
of a liquid in a stream, we are
discussing the amount of water
that passes by in a given period
of time
when we discuss electric
current, we are discussing the
amount of electric charge that
passes a point in a given period
of time
 whether as electrons flowing
through a wire or ions flowing
through a solution
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Redox Reactions & Current
• redox reactions involve the transfer of electrons
from one substance to another
• therefore, redox reactions have the potential to
generate an electric current
• in order to use that current, we need to separate
the place where oxidation is occurring from the
place that reduction is occurring
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Electric Current Flowing
Directly Between Atoms
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Electric Current Flowing
Indirectly Between Atoms
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Electrochemical Cells
• electrochemistry is the study of redox reactions
that produce or require an electric current
• the conversion between chemical energy and
electrical energy is carried out in an
electrochemical cell
• spontaneous redox reactions take place in a
voltaic cell
aka galvanic cells
• nonspontaneous redox reactions can be made to
occur in an electrolytic cell by the addition of
electrical energy
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Electrochemical Cells
• oxidation and reduction reactions kept separate
 half-cells
• electron flow through a wire along with ion flow
•
through a solution constitutes an electric circuit
requires a conductive solid (metal or graphite)
electrode to allow the transfer of electrons
 through external circuit
• ion exchange between the two halves of the system
 electrolyte
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Electrodes
• Anode
•
electrode where oxidation occurs
anions attracted to it
connected to positive end of battery in electrolytic
cell
loses weight in electrolytic cell
Cathode
electrode where reduction occurs
cations attracted to it
connected to negative end of battery in electrolytic
cell
gains weight in electrolytic cell
electrode where plating takes place in electroplating
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Voltaic Cell
the salt bridge is
required to complete
the circuit and
maintain charge
balance
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Current and Voltage
• the number of electrons that flow through the system per
second is the current
 unit = Ampere
 1 A of current = 1 Coulomb of charge flowing by each second
 1 A = 6.242 x 1018 electrons/second
 Electrode surface area dictates the number of electrons that can
flow
• the difference in potential energy between the reactants
and products is the potential difference
 unit = Volt
 1 V of force = 1 J of energy/Coulomb of charge
 the voltage needed to drive electrons through the external circuit
 amount of force pushing the electrons through the wire is called
the electromotive force, emf
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Cell Potential
• the difference in potential energy between the
anode the cathode in a voltaic cell is called the
cell potential
• the cell potential depends on the relative ease
with which the oxidizing agent is reduced at the
cathode and the reducing agent is oxidized at the
anode
• the cell potential under standard conditions is
called the standard emf, E°cell
25°C, 1 atm for gases, 1 M concentration of solution
sum of the cell potentials for the half-reactions
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Cell Notation
• shorthand description of Voltaic cell
• electrode | electrolyte || electrolyte | electrode
• oxidation half-cell on left, reduction half-cell on
the right
• single | = phase barrier
if multiple electrolytes in same phase, a comma is
used rather than |
often use an inert electrode
• double line || = salt bridge
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Fe(s) | Fe2+(aq) || MnO4(aq), Mn2+(aq), H+(aq) | Pt(s)
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Standard Reduction Potential
• a half-reaction with a strong tendency to
•
•
•
occur has a large + half-cell potential
when two half-cells are connected, the
electrons will flow so that the half-reaction
with the stronger tendency will occur
we cannot measure the absolute tendency
of a half-reaction, we can only measure it
relative to another half-reaction
we select as a standard half-reaction the
reduction of H+ to H2 under standard
conditions, which we assign a potential
difference = 0 v
 standard hydrogen electrode, SHE
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Half-Cell Potentials
• SHE reduction potential is defined to be exactly 0 v
• half-reactions with a stronger tendency toward
•
•
reduction than the SHE have a + value for E°red
half-reactions with a stronger tendency toward
oxidation than the SHE have a  value for E°red
E°cell = E°oxidation + E°reduction
 E°oxidation = E°reduction
 when adding E° values for the half-cells, do not multiply the
half-cell E° values, even if you need to multiply the halfreactions to balance the equation
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Ex 18.4 – Calculate Ecell for the reaction at 25C
Al(s) + NO3−(aq) + 4 H+(aq)  Al3+(aq) + NO(g) + 2 H2O(l)
Separate the
reaction into
the oxidation
and reduction
half-reactions
ox:
Al(s)  Al3+(aq) + 3 e−
red:
NO3−(aq) + 4 H+(aq) + 3 e−  NO(g) + 2 H2O(l)
find the E for
each halfreaction and
sum to get
Ecell
Eox = −Ered = +1.66 v
Ered = +0.96 v
Ecell = (+1.66 v) + (+0.96 v) = +2.62 v
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Ex 18.4a – Predict if the following reaction is
spontaneous under standard conditions
Fe(s) + Mg2+(aq)  Fe2+(aq) + Mg(s)
ox:
Fe(s)  Fe2+(aq) + 2 e−
Separate the
reaction into
red:
Mg2+(aq) + 2 e−  Mg(s)
the oxidation
and reduction
half-reactions
look up the
relative
positions of the
reduction halfreactions
red:
red:
Mg2+(aq) + 2 e−  Mg(s)
Fe2+(aq) + 2 e−  Fe(s)
since Mg2+ reduction is below Fe2+
reduction, the reaction is NOT spontaneous
as written
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the reaction is
spontaneous in the
reverse direction
Mg(s) + Fe2+(aq)  Mg2+(aq) + Fe(s)
ox:
red:
Mg(s)  Mg2+(aq) + 2 e−
Fe2+(aq) + 2 e−  Fe(s)
sketch the cell and
label the parts –
oxidation occurs at
the anode; electrons
flow from anode to
cathode
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Practice - Sketch and Label the Voltaic Cell
Fe(s)  Fe2+(aq) Pb2+(aq)  Pb(s) , Write the
Half-Reactions and Overall Reaction, and Determine
the Cell Potential under Standard Conditions.
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ox: Fe(s)  Fe2+(aq) + 2 e−
E = +0.45 V
red: Pb2+(aq) + 2 e−  Pb(s)
E = −0.13 V
tot: Pb2+(aq) + Fe(s)  Fe2+(aq) + Pb(s)
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E = +0.32 V
43
Predicting Whether a Metal Will
Dissolve in an Acid
• acids dissolve in metals if the
reduction of the metal ion is
easier than the reduction of
H+(aq)
• metals whose ion reduction
reaction lies below H+ reduction
on the table will dissolve in acid
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E°cell, DG° and K
• for a spontaneous reaction
one the proceeds in the forward direction with the
chemicals in their standard states
DG° < 1 (negative)
E° > 1 (positive)
K > 1
• DG° = −RTlnK = −nFE°cell
n is the number of electrons
F = Faraday’s Constant = 96,485 C/mol e−
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Example 18.6- Calculate DG° for the reaction
I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq)
Given: I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq)
Find: DG, (J)
Concept Plan:
E°ox, E°red
E°cell
E cell  E ox  E red
DG°
DG   nFEcell
Relationships:
− E° = −1.09 v
2 Br−
→ Br
+
2
e
Solve: ox:
(aq)
2(l)
DG  nFE


cell

e− →
2 I−(aq)
DG  2 mol e 96,485
red: I2(l) + 2
 0.55 
C = +0.54 v
E°
mol e 
J
C
tot:  I2(l) + 2Br−(aq)5 → 2I−(aq) + Br2(l) E° = −0.55 v
DG  1.110 J
Answer: since DG° is +, the reaction is not spontaneous in
the forward direction under standard conditions
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Example 18.7- Calculate Kat 25°C for the reaction
Cu(s) + 2 H+(aq) → H2(g) + Cu2+(aq)
Given: Cu(s) + 2 H+(aq) → H2(g) + Cu2+(aq)
Find: K
Concept Plan:
E°ox, E°red
E°cell
E

cell
E E

ox

red
E

cell
K
0.0592 V

log K
n
Relationships:
0.0592
V 2+ + 2 e−

Cu
→
Cu
E° = −0.34 v
Solve: Eox:
log(aq)K
cell  (s)
n

+
red: 2 H (aq) + 2 e−2→
E° = +0.00 v
molHe2(aq)
log K   0.34 V 
 11.5
0.0592 2+
V
+
tot: Cu11
Cu (aq) + H2(g) E° = −0.34 v
(s).5+ 2H (aq) →
12
K  10
 3.2 10
Answer: since K < 1, the position of equilibrium lies far to
the left under standard conditions

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
47
Nonstandard Conditions the Nernst Equation
•
•
•
•
DG = DG° + RT ln Q
E = E° - (0.0592/n) log Q at 25°C
when Q = K, E = 0
use to calculate E when concentrations not 1 M
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E at Nonstandard Conditions
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Example 18.8- Calculate Ecell at 25°C for the reaction
3 Cu(s) + 2 MnO4−(aq) + 8 H+(aq) → 2 MnO2(s) + Cu2+(aq) + 4 H2O(l)
3 Cu(s) + 2 MnO4−(aq) + 8 H+(aq) → 2 MnO2(s) + Cu2+(aq) + 4 H2O(l)
Given:
[Cu2+] = 0.010 M, [MnO4−] = 2.0 M, [H+] = 1.0 M
Ecell
Find:
Concept Plan:
E°ox, E°red
E cell  E ox  E red
Relationships:
Solve:
ox:
Cu(s)V→ Cu2+E(aq)
cell + 2
0.0592
E cell  E cell 
Ecell
0.0592 V
log Q
n
2 3
0
.
0592
V
[
Cu
]
e− }x3E° =log−0.34 v 3  8
n
[MnO 4 ] [H ]
E cell
E cell  E cell
− 
red: MnO4
E°cell
log Q
+
−
2 H2VO(l) }x2
(aq) + 4 H
n (aq) + 3 e → MnO2(s) +0.0592
[0.010E°
]3 = +1.68 v
E
 1.34 V 
log
tot: 3 Cu(s) + 2 MnO4−(aq) + 8 H+(aq) →cell2 MnO2(s) + Cu6 2+(aq) + [42.0]
H23O
[1(l))
.0]8E° = +1.34 v
E cell  1.41 V
Check:
units are correct, Ecell > E°cell as expected because
[MnO4−] > 1 M and [Cu2+] < 1 M
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Concentration Cells
• it is possible to get a spontaneous reaction when the oxidation
•
and reduction reactions are the same, as long as the electrolyte
concentrations are different
the difference in energy is due to the entropic difference in the
solutions
 the more concentrated solution has lower entropy than the less
concentrated
• electrons will flow from the electrode in the less concentrated
solution to the electrode in the more concentrated solution
 oxidation of the electrode in the less concentrated solution will increase
the ion concentration in the solution – the less concentrated solution has
the anode
 reduction of the solution ions at the electrode in the more concentrated
solution reduces the ion concentration – the more concentrated solution
has the cathode
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Concentration Cell
when
cell
when the
cellthe
concentrations
concentrations
are different,
electrons flow
areside
equal
there
from the
with
theisless
no difference
in
concentrated
solution
energy
(anode)
to thebetween
side with the
the half-cellssolution
and
more concentrated
no electrons flow
(cathode)
Cu(s) Cu2+(aq) (0.010 M)  Cu2+(aq) (2.0 M) Cu(s)
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LeClanche’ Acidic Dry Cell
• electrolyte in paste form
 ZnCl2 + NH4Cl
 or MgBr2
• anode = Zn (or Mg)
Zn(s)  Zn2+(aq) + 2 e-
• cathode = graphite rod
• MnO2 is reduced
2 MnO2(s) + 2 NH4+(aq) + 2 H2O(l) + 2 e 2 NH4OH(aq) + 2 Mn(O)OH(s)
• cell voltage = 1.5 v
• expensive, nonrechargeable, heavy,
easily corroded
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Alkaline Dry Cell
• same basic cell as acidic dry cell, except
•
electrolyte is alkaline KOH paste
anode = Zn (or Mg)
Zn(s)  Zn2+(aq) + 2 e-
• cathode = brass rod
• MnO2 is reduced
2 MnO2(s) + 2 NH4+(aq) + 2 H2O(l) + 2 e 2 NH4OH(aq) + 2 Mn(O)OH(s)
• cell voltage = 1.54 v
• longer shelf life than acidic dry cells and
rechargeable, little corrosion of zinc
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Lead Storage Battery
• 6 cells in series
• electrolyte = 30% H2SO4
• anode = Pb
Pb(s) + SO42-(aq)  PbSO4(s) + 2 e-
• cathode = Pb coated with PbO2
• PbO2 is reduced
PbO2(s) + 4 H+(aq) + SO42-(aq) + 2 e PbSO4(s) + 2 H2O(l)
• cell voltage = 2.09 v
• rechargeable, heavy
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NiCad Battery
• electrolyte is concentrated KOH solution
• anode = Cd
Cd(s) + 2 OH-1(aq)  Cd(OH)2(s) + 2 e-1
• cathode = Ni coated with NiO2
• NiO2 is reduced
NiO2(s) + 2 H2O(l) + 2 e-1  Ni(OH)2(s) + 2OH-1
E0 = 0.81 v
E0 = 0.49 v
• cell voltage = 1.30 v
• rechargeable, long life, light – however
recharging incorrectly can lead to battery
breakdown
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Ni-MH Battery
• electrolyte is concentrated KOH solution
• anode = metal alloy with dissolved hydrogen
 oxidation of H from H0 to H+1
M∙H(s) + OH-1(aq)  M(s) + H2O(l) + e-1
E° = 0.89 v
• cathode = Ni coated with NiO2
• NiO2 is reduced
NiO2(s) + 2 H2O(l) + 2 e-1  Ni(OH)2(s) + 2OH-1
E0 = 0.49 v
• cell voltage = 1.30 v
• rechargeable, long life, light, more environmentally
friendly than NiCad, greater energy density than
NiCad
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Lithium Ion Battery
• electrolyte is concentrated KOH
•
•
solution
anode = graphite impregnated with Li
ions
cathode = Li - transition metal oxide
 reduction of transition metal
• work on Li ion migration from anode
•
to cathode causing a corresponding
migration of electrons from anode to
cathode
rechargeable, long life, very light,
more environmentally friendly,
greater energy density
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Fuel Cells
• like batteries in which
reactants are constantly
being added
 so it never runs down!
• Anode and Cathode
both Pt coated metal
• Electrolyte is OH–
solution
• Anode Reaction:
2 H2 + 4 OH–
→ 4 H2O(l) + 4 e• Cathode Reaction:
O2 + 4 H2O + 4 e→ 4 OH–
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Electrolytic Cell
• uses electrical energy to overcome the energy barrier
and cause a non-spontaneous reaction
 must be DC source
•
•
•
•
•
the + terminal of the battery = anode
the - terminal of the battery = cathode
cations attracted to the cathode, anions to the anode
cations pick up electrons from the cathode and are
reduced, anions release electrons to the anode and are
oxidized
some electrolysis reactions require more voltage than
Etot, called the overvoltage
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electroplating
In electroplating, the work
piece is the cathode.
Cations are reduced at
cathode and plate to
the surface of the work
piece.
The anode is made of
the plate metal. The
anode oxidizes and
replaces the metal
cations in the solution
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Electrochemical Cells
• in all electrochemical cells, oxidation occurs at the
•
anode, reduction occurs at the cathode
in voltaic cells,
 anode is the source of electrons and has a (−) charge
 cathode draws electrons and has a (+) charge
• in electrolytic cells
 electrons are drawn off the anode, so it must have a place to
release the electrons, the + terminal of the battery
 electrons are forced toward the anode, so it must have a
source of electrons, the − terminal of the battery
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Electrolysis
• electrolysis is the process of using
•
•
electricity to break a compound
apart
electrolysis is done in an
electrolytic cell
electrolytic cells can be used to
separate elements from their
compounds
 generate H2 from water for fuel cells
 recover metals from their ores
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Electrolysis of Water
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Electrolysis of Pure Compounds
• must be in molten (liquid) state
• electrodes normally graphite
• cations are reduced at the cathode to metal
element
• anions oxidized at anode to nonmetal element
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Electrolysis of NaCl(l)
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Mixtures of Ions
• when more than one cation is present, the cation
that is easiest to reduce will be reduced first at
the cathode
least negative or most positive E°red
• when more than one anion is present, the anion
that is easiest to oxidize will be oxidized first at
the anode
least negative or most positive E°ox
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Electrolysis of Aqueous Solutions
• Complicated by more than one possible oxidation and reduction
• possible cathode reactions
 reduction of cation to metal
 reduction of water to H2
2 H2O + 2 e-1  H2 + 2 OH-1
• possible anode reactions
E° = -0.83 v @ stand. cond.
E° = -0.41 v @ pH 7
 oxidation of anion to element
 oxidation of H2O to O2
2 H2O  O2 + 4e-1 + 4H+1
 oxidation of electrode
E° = -1.23 v @ stand. cond.
E° = -0.82 v @ pH 7
particularly Cu
graphite doesn’t oxidize
• half-reactions that lead to least negative Etot will occur
 unless overvoltage changes the conditions
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Electrolysis of NaI(aq)
with Inert Electrodes
possible oxidations
2 I-1  I2 + 2 e-1
2 H2O  O2 + 4e-1 + 4H+1
E° = −0.54 v
E° = −0.82 v
possible reductions
Na+1 + 1e-1  Na0
E° = −2.71 v
2 H2O + 2 e-1  H2 + 2 OH-1 E° = −0.41 v
overall reaction
2 I−(aq) + 2 H2O(l)  I2(aq) + H2(g) + 2 OH-1(aq)
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Faraday’s Law
• the amount of metal deposited during
electrolysis is directly proportional to the charge
on the cation, the current, and the length of time
the cell runs
charge that flows through the cell = current x time
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Example 18.10- Calculate the mass of Au that can be plated in
25 min using 5.5 A for the half-reaction
Au3+(aq) + 3 e− → Au(s)
Given:
Find:
Concept Plan:
3 mol e− : 1 mol Au, current = 5.5 amps, time = 25 min
mass Au, g
t(s), amp
charge (C)
5 .5 C
1s
Relationships:
mol e−
1 mol e 
96,485 C
mol Au
1 mol Au
3 mol e 
g Au
196.97 g
1 mol Au
Solve:
60 s 5.5 C 1 mol e  1 mol Au 196.97 g
25 min 





1 min
1s
96,485 C 3 mol e 1 mol Au
 5.6 g Au
Check:
units are correct, answer is reasonable since 10 A
running for 1 hr ~ 1/3 mol e−
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Corrosion
• corrosion is the spontaneous oxidation of a
metal by chemicals in the environment
• since many materials we use are active
metals, corrosion can be a very big problem
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Rusting
• rust is hydrated iron(III) oxide
• moisture must be present
water is a reactant
required for flow between cathode and anode
• electrolytes promote rusting
enhances current flow
• acids promote rusting
lower pH = lower E°red
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Preventing Corrosion
• one way to reduce or slow corrosion is to coat
the metal surface to keep it from contacting
corrosive chemicals in the environment
paint
some metals, like Al, form an oxide that strongly
attaches to the metal surface, preventing the rest
from corroding
• another method to protect one metal is to attach
it to a more reactive metal that is cheap
sacrificial electrode
galvanized nails
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Sacrificial Anode
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