E_-_102_F11_Notes

E_102_F11_Notes - E)
Electrochemistry
 Petrucci
et
al,
10th
Edition:
3.4,
5.1‐5.2,
5.4‐5.6,
20.1‐20.7
 


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Unformatted text preview: E)
Electrochemistry
 Petrucci
et
al.,
10th
Edition:
3.4,
5.1‐5.2,
5.4‐5.6,
20.1‐20.7
 
 Oxidation
States
 • Oxidation
state
(or
number)
is
related
to
the
number
of
electrons
that
an
atom
gains,
 loses,
or
otherwise
uses
when
combining
with
other
atoms
in
a
compound
 • The
concept
of
oxidation
state
is
arbitrary
 • it
does
not
necessarily
have
a
fundamental
meaning
 • it
is
useful
when
understanding
oxidation‐reduction
reactions
 
 NaCl:

 Na
loses
1
electron
→
Na+
(Na
has
an
oxidation
state
+1)
 
 Cl
gains
1
electron
→
Cl‐
(Cl
has
an
oxidation
state
‐1)
 
 MgBr2:
 Mg
loses
2
electrons
→
Mg2+
 
 Each
Br
gains
1
electron
→
2
Br
‐
 
 [see
Petrucci
et
al.,
front
cover
for
periodic
table]
 
 Rules
for
Assigning
Oxidation
States
(O.S.)
 • The
O.S.
of
an
individual
atom
in
a
free
element
is
0
(zero)
 • Cl,
Cl2,
N2,
O2,
O,
etc.
 • The
total
of
the
O.S.
of
all
atoms
in
a
neutral
species
is
0
(zero)
 • H2O
(H:
+1,
O:
‐2)
 • The
total
of
the
O.S.
of
all
atoms
in
an
ion
equals
the
charge
of
the
ion
 • Fe3+
(Fe:
+3),
MnO4‐
(Mn:
+7,
O:
‐2)
 • Group
1
metals
(Li,
Na,
K,
Rb,
Cs,
Fr)
have
O.S.
=
+1
 • Group
2
metals
(Be,
Mg,
Ca,
Sr,
Ba,
Ra)
have
O.S.
=
+2
 • F
has
O.S.
=
‐1

 • Cl,
Br,
I,
and
At
have
O.S.
=
‐1
except
when
combined
with
oxygen
and
other
halogens
 • H
has
O.S.
=
+1,
except
when
bonded
to
metals
when
its
O.S.
=
‐1
(LiH,
for
instance)
 • O
has
O.S.
=
­2
except
for
peroxides
(H2O2
and
N2O2
where
O.S.
=
‐1
for
O)
or
when
 combine
with
F
(OF2
where
O.S.
=
+2
for
O)
 • • • • • In
binary
compounds
with
metals:
 group
17
elements
(F,
Cl,
Br,
I,
At)
have
O.S.
=
‐1
 group
16
elements
(O,
S,
Se,
Te,
Po)
have
O.S.
=
‐2
 group
15
elements
(N,
P,
As,
Sb,
Bi)
have
O.S.
=
‐3
 All
other
oxidation
numbers
are
selected
to
make
the
algebraic
sum
of
the
oxidation
 numbers
equal
to
the
net
charge
of
the
molecule
or
ion

 ChE102
Fall
2011
Class
Notes
‐
Electrochemistry
 
 
 
 
 
 
 
 E ‐1 
 Reduction­Oxidation
(REDOX)
Reactions
 • In
a
reduction‐oxidation
reaction,
one
of
the
reagents
is
oxidized
while
the
other
is
 reduced
 • A
chemical
substance
is
oxidized
when
it
loses
electrons:

 












Zn
→
Zn2+
+
2e‐

(O.S.:
0
→
+2)
 • A
chemical
substance
is
reduced
when
it
gains
electrons:

 












Cu2+
+
2e‐
→
Cu

(O.S.:
+2
→
0)
 • Substance
that
causes
the
oxidation
of
another
is
called
the
oxidizing
agent
or
 ______________________.
 • Substance
that
causes
the
reduction
of
another
is
called
the
reducing
agent
or
 ______________________.
 
 
 Examples
of
REDOX
Reactions
 
 Zn
(s)
+
2
H+
(aq)
→
Zn2+
(aq)
+
H2
(g)
 Zn
(s)
→
Zn2+
(aq)
+
___
e‐
 
 (reducing
agent)
 ___
H+
(aq)
+
___
e‐
→
H2
(g)

 
 (oxidizing
agent)
 
 Zn
(s)
+
Cu2+
(aq)
→
Zn2+
(aq)
+
Cu
(s)
 Zn
(s)
→
Zn2+
(aq)
+
___
e‐
 
 (reducing
agent)
 Cu2+
(aq)
+
___
e‐
→
Cu
(s)
 
 (oxidizing
agent)
 
 Cu
(s)
+
2
Ag+
(aq)
→
Cu2+
(aq)
+
2
Ag
(s)
 Cu
(s)
→
Cu2+
(aq)
+
___
e‐
 
 (reducing
agent)
 __
Ag+
(aq)
+
___
e‐
→
___
Ag
(s)
 (oxidizing
agent)
 
 Zn
(s)
+
SnCl2
(aq)
→
Zn2+
(aq)
+
Sn
(s)
 Zn
(s)
→
Zn2+
(aq)
+
___
e‐
 
 (reducing
agent)
 Sn2+
(aq)
+
___
e‐
→
Sn
(s)

 
 (oxidizing
agent)
 ___
Zn
(s)
→
___
Zn2+
(s)
+
___
e‐
 (reducing
agent)
 O2
(g)
+
____
e‐
→
2
O2‐
(s)

 (oxidizing
agent)
 
 2
Zn
(s)
+
O2
(g)
→
2
ZnO
(s)
 
 ChE102
Fall
2011
Class
Notes
‐
Electrochemistry
 
 
 
 
 
 
 
 E ‐2 
 Hydronium
Ion
(H3O+)
 Zn
(s)
+
2
H+
(aq)
→
Zn2+
(aq)
+
H2
(g)
 Zn
(s)
→
Zn2+
(aq)
+
2
e‐
 2
H+
(aq)
+
2
e‐
→
H2
(g)
 
 Zn
(s)
+
2
H3O+
(aq)
→
Zn2+
(aq)
+
H2
(g)
+
2H2O
(l)
 Zn
(s)
→
Zn2+
(aq)
+
2
e‐
 2
H3O+
(aq)
+
2
e‐
→
H2
(g)
+
2
H2O
(l)
 
 [see
Petrucci
et
al.,
Figure
5.5]
 These
two
chemical
equations
are
equivalent
 
 
 Balancing
REDOX
Reactions
 1. Write
separate
half‐reactions
for
oxidation
and
reduction
 2. Balance
all
atoms,
except
H
and
O
 3. Balance
O
atoms
by
adding
H2O
 4. Balance
H
atoms:
 • Acidic
medium:
add
H+
to
the
hydrogen‐deficient
side
 • Basic
medium:
add
H2O
to
the
hydrogen‐deficient
side
and
OH‐
to
the
other
side
 5. Balance
charges
by
adding
e‐
 6. Add
the
half‐reactions
to
cancel
e‐
 
 Final
equation
should
be
balanced
for
both
number
of
atoms
and
electrical
charges
 
 
 Question
E1
 Balance
the
following
REDOX
reactions:
 a) I2
+
H2S

→
H+
+
I‐
+
S

 
 (acidic
medium)
 
 b) CuS
+
NO3‐
→
Cu2+
+
SO42‐
+
NO
 (acidic
medium)
 c) N2H4
+
Cu(OH)2
→
N2
+
Cu

 
 (basic
medium)
 
 
 ChE102
Fall
2011
Class
Notes
‐
Electrochemistry
 
 
 
 
 
 
 
 E ‐3 
 
 Electrode
Potential
 Galvanic
=
produce
electricity
from
spontaneous
chemical
reactions
 [see Petrucci et al., Figure 20.3] 
 2+ 
 Cu(s) → Cu (aq) + 2 e + 
 Ag (aq) + e → Ag(s) 
 Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s) 
 
 A
schematic
representation
of
this
galvanic
cell
is:

 Cu|Cu2+
||
Ag+|Ag
 
 
 __________________________
occurs
at
the
anode.
 • Electrons
flow
from
the
anode
so
it
shows
a
negative
charge.
 Reduction
occurs
at
the
cathode.
 • Electrons
flow
to
the
cathode
so
it
shows
























charge.
 Salt
bridge
contains
a
salt
such
as
NaNO3
 • Permits
ions
to
diffuse
from
one
side
to
the
other
but
avoids
bulk
mixing
of
the
 solutions
(which
would
stop
the
flow
of
electrons
through
the
circuit).

 • Without
the
salt
bridge
there
would
be
an
accumulation
of
positive
charges
at
the
 anode
side
and
negative
charges
at
the
cathode
side,
interrupting
the
electrical
 current
of
the
cell.
 Ammeter
measures
the
magnitude
of
the
electrical
current.
 Voltmeter
measures
the
cell’s
voltage,
which
is
also
called
the
































.
 Electrons
flow
from
the
anode
to
the
cathode.;
ions
in
solution
are
free
to
flow
in
any
 direction.
 These
electrodes
are
called

























because
they
either
dissolve
(Cu)
or
form
(Ag)
 during
the
reaction.

 
 
 Batteries:
Electricity
from
Chemical
Reactions
 Leclanché
(Dry)
Cell

 [see
Petrucci
et
al.,
Figure
20.14]
 • Max
voltage
1.55V
 • Some
problems:

produces
ammonia
(NH3)
under
high
draw
conditions,
Zn
dissolves
 under
acidic
conditions
 • Lead
to
creation
of
alkaline
battery
 ChE102
Fall
2011
Class
Notes
‐
Electrochemistry
 
 
 
 
 
 
 
 E ‐4 
 Car
Battery
 [see
Petrucci
et
al.,
Figure
20.15]
 • Most
batteries
are
six
cell,
capable
of
12
V
 
 
 Ag/Zn
Button
Battery
 [see
Petrucci
et
al.,
Figure
20.16]
 
 
 Hydrogen
Fuel
Cell
 [see
Petrucci
et
al.,
Figure
20.17]
 
 
 
 
 Corrosion
Prevention
and
Sacrificial
Anodes
 


[see
Petrucci
et
al.,
Figure
20.21]
 
 
 Faraday’s
Law
 “The
mass
of
a
given
substance
that
is
produced
or
consumed
at
an
electrode
is
 proportional
to
the
quantity
of
electric
charge
passed
through
the
cell.”

 
 Charge
of
a
single
electron:
 
 e
=
1.6021773
x
10‐19
C
(coulombs)
 
 Charge
of
one
mole
of
electrons
(Faraday
constant):
 
 F
=
(6.022137
x
1023
mol‐1)
(1.6021773
x
10‐19
C)
 
 



=
96
485.31
C
/
mol
 
 Electric
current
(I)
is
the
amount
of
charge
(Q)
flowing
through
circuit
per
unit
of
time
(t):
 
 
 
 
 The
number
of
moles
of
electrons
that
flow
through
a
circuit
can
be
calculated
by
the
 equation:

 It total ch arg e pas sin g thru the circuit 
 n= = F ch arg e of a mole of electrons ChE102
Fall
2011
Class
Notes
‐
Electrochemistry
 € 
 
 
 
 
 
 
 E ‐5 
 Application
of
Faraday’s
Law
 • From
the
number
of
moles
passing
through
the
circuit
 • We
can
calculate
the
mass
of
a
given
substance
being
produced
or
consumed
at
an
 electrode.
 • For
the
previous
Cu|Cu2+||Ag+|Ag
cell,
for
each
mole
of
electrons
that
passes
through
 the
circuit:
 • ½
mole
Cu
(s)
is
oxidized
=
(63.5
g/mol)
/
(2
mol)
=
_________
g
of
Cu
(s)
are
 dissolved
at
the
anode.
 • 1
mole
of
Ag+
is
reduced
=
_______
g
of
Ag
(s)
are
deposited
at
the
cathode.

 
 
 Question
E2
 A
galvanic
cell
consists
of
a
cadmium
cathode
immersed
in
a
CdSO4
solution
and
a
zinc
 anode
immersed
in
a
ZnSO4
solution.

A
salt
bridge
connects
the
two
half‐cells.

A
current
 of
1.45
A
is
observed
to
flow
for
a
period
of
2.6
hours.

How
many
mols
of
electrons
have
 passed
through
the
circuit
during
this
time?


 
 Standard
Electrode
Potentials
 
 
 H2
(g)
→
2
H+
(aq)
+
2
e‐
 
 a=1
is
activity
of
1
if
[H+]
=
1M
 
 
 
 
 
 
 
 
 Petrucci
et
al.,
Figure
20.5
 
 
 Question
E3
 A
galvanic
cell
is
constructed
that
has
a
zinc
anode
immersed
in
a
Zn(NO3)2
solution
and
a
 platinum
cathode
immersed
in
a
NaCl
solution
equilibrated
with
Cl2
(g)
at
1
atm
and
25°C.

 A
salt
bridge
connects
the
two
cells.

A
current
of
0.80A
is
observed
to
flow
for
a
period
of
 25
minutes.

Calculate
the
volume
of
gaseous
chlorine
generated
or
consumed
as
a
result
 of
the
reaction.
 ChE102
Fall
2011
Class
Notes
‐
Electrochemistry
 
 
 
 
 
 
 
 E ‐6 
 Standard
Cell
Potentials
 Galvanic
cell
voltage
depends
on:
 • Reactants
in
the
two
half‐cells
 • Concentrations
in
the
two
half‐cells
 • Temperature
 
 We
need
to
define
a
_______________

__________
in
order
to
compare
the
voltages
of
different
 galvanic
cells:
 • 1.0
M
concentration
for
dissolved
species
 • 1
atm
pressure
for
gases
 • Most
stable
form
at
25˚C
for
solids
 
 Standard
cell
potential
(E˚cell)
is
defined
as
the
voltage
measured
at
the
standard
state
 
 
 Potentials
for
Half­Cell
Reactions
 Galvanic
cells
can
be
constructed
from
a
very
large
number
of
different
combinations
of
 two
half‐cells.

 Therefore,
it
isn’t
necessary
to
measure
the
standard
cell
potentials
for
all
these
possible
 combinations
 
 Define
half­cell
potentials:
 • Half‐cell
potentials
are
assigned
to
a
half‐cell
only.

 • When
two
half‐cells
are
combined
to
form
a
galvanic
cell,
the
standard
cell
potential
 of
the
resulting
galvanic
cell
can
be
calculated
from
the
two
half‐cell
potentials.
 
 Measuring
E˚Cu2+/Cu
and
E˚Zn2+/Zn

 [see
Petrucci
et
al.,
Figure
20.6]
 
 How
to
Measure
Half­Cell
Potentials
 Half‐cell
potential
for
the
SHE
cell
is
arbitrarily
set
to
zero:
 (SHE
=
Standard
_____________________
Electrode)
 
 H2
(1
atm)
→
2
H+
+
2
e‐
 
 E°H+/H2
=
0
V
 Half‐cell
potentials
for
other
half‐reactions
are
measured
relative
to
hydrogen
half‐cell:
 Zn(s)
+
2
H+
→
Zn2+
+
H2
 Zn
→
Zn2+
+
2
e‐
 
 (oxidation)
 Voltage
=
____
V
 Cu2++
H2
→
Cu(s)
+
2
H+
 Cu2+
+
2
e‐
→
Cu

 
 (reduction)
 Voltage
=
____
V

 ChE102
Fall
2011
Class
Notes
‐
Electrochemistry
 
 
 
 
 
 
 
 E ‐7 
 Sign
Convention
for
E°cell
 Hydrogen
is
the
anode
(on
the
left)
while
the
cell
of
interest
is
the
cathode
(on
the
right)
 
 
E°cell
=
E°
(cathode)
–
E°
(anode)

 
 Zn
(s)
+
Cu2+
→
Cu
(s)
+
Zn2+
 
 E°cell
=
+1.10
V
 
 

 spontaneous
direction
 

 Cu
(s)
+
Zn2+
→
Zn
(s)
+
Cu2+
 
 E°cell
=
‐1.10
V
 
 

 spontaneous
direction

 
 
 Measuring
E°Cu2+/Cu
and
E°Zn2+/Zn
 Positive
potentials
measure
the
tendency
of
the
reduction
reaction
to
occur
when
a
half‐ cell
is
combined
with
the
hydrogen
half‐cell:
 
 Zn2+
+
2e‐
→
Zn

 EºZn2+/Zn
=
‐0.76
V

 
 
 


 
 
 
 (Zn
does
not
abstract
e‐
from
H2)
 
 2
H+
+
2
e‐
→
H2
 EºH+/H2
=
0
V
 2+
+
2
e‐
→
Cu
 
 Cu EºCu+/Cu
=
0.34
V

 
 
 


 
 
 
 (Cu
abstracts
e‐
from
H2)
 
 Combine
the
half
cell
potentials
to
obtain
the
Eºcell:
 
 Zn
→
Zn2+
+
2
e‐
 
 
‐
(EºZn2+/Zn
)
=
‐
(‐0.76
V)
 
 Cu2+
+
2
e‐
→
Cu








.
 
 


EºCu+/Cu


=


0.34
V
 2+
→
Zn2+
+
Cu
 
 
 Zn
+
Cu 


Eºcell






=

1.10
V

 
 (see
Petrucci
et
al.,
Table
20.1
for
more
standard
reduction
potentials)
 
 
 Question
E4
 A
Cu(s)
strip
is
immersed
in
a
AgNO3
solution
and
in
a
Zn(NO3)2
solution.

Which
flask
 contains
the
AgNO3
solution?

(see
Petrucci
et
al.,
Figure
20.1)
 
 
 Question
E5
 In
a
galvanic
cell,
one
half‐cell
consists
of
a
zinc
strip
dipped
into
a
1.0M
solution
of
 Zn(NO3)2.

In
the
second
half‐cell,
solid
indium
adsorbed
on
graphite
is
in
contact
with
a
 1.0M
solution
of
In(NO3)3.

Indium
is
observed
to
plate
out
as
the
galvanic
cell
operates
 and
the
initial
cell
voltage
is
measured
to
be
0.425V
at
25˚C.

Calculate
the
standard
 reduction
potential
of
an
In3+|In
half‐cell.

 ChE102
Fall
2011
Class
Notes
‐
Electrochemistry
 
 
 
 
 
 
 
 E ‐8 
 Non­standard
Conditions
 [see
Petrucci
et
al.,
Figure
20.10]
 
 The
Nernst
equation
quantifies
the
dependency
of
cell
voltage
on
the
concentration
of
 chemical
species
and
temperature
in
a
galvanic
cell:
 
 
 
 
 R:
 gas
constant
=
8.3145
J
/
mol.
K
 
 T:
 temperature
in
K
 
 F:
 Faraday
constant
=
96485
C
/
mol
electrons
 
 n:
 number
of
electrons
transferred
in
the
reaction
as
written
 Q:

 reaction
quotient
 
 
 
 for
the
generic
reaction:


aA
+
bB
↔
cC
+
dD
 
 
 
Change
from
ln
to
log:
 
 
 
 
Nernst
equation
at
T
=
25oC
=
298
K:
 
 
 
 
 Question
E6
 A
cell
is
assembled
by
placing
a
silver
electrode
in
a
0.020
M
Ag+
solution
and
a
copper
 electrode
in
a
0.050
M
Cu2+
solution
at
25˚C.

What
is
the
cell’s
voltage?

 
 
 Ecell
as
a
Function
of
Concentration
 [see
Petrucci
et
al.,
Figure
20.11]
 
 Apply
the
Nernst
equation
to
a
cell
operating
with
the
same
half‐reactions
(E°cell
=
0)
at
 different
concentrations
(Q
≠
1):
 
 
 
 For
the
Pt|H2(g,
1
atm)|H+(x
mol)||H+(1
mol)|H2(g,
1
atm)|Pt
cell
at
25°C:
 
 
 
 ChE102
Fall
2011
Class
Notes
‐
Electrochemistry
 
 
 
 
 
 
 
 E ‐9 
 Question
E7
 Two
hydrogen‐hydrogen
ion
half‐cells
are
connected
to
make
a
single
galvanic
cell.

In
one
 of
the
half‐cells
the
pH
is
1.0,
but
the
pH
in
the
other
half‐cell
is
not
known.

The
measured
 voltage
delivered
by
the
combination
is
0.16V
and
the
electrode
in
the
half‐cell
of
known
 concentration
is
positive.

Assume
T
=
25°C.

What
is
the
unknown
concentration
of
H+?

 
 
 Cell
Potentials
&
Equilibrium
Constants
 What
happens
when
a
galvanic
cell
achieves
equilibrium?
 
 Nernst
equation
for
non‐equilibrium
conditions:
 
 
 
 Nernst
equation
when
chemical
equilibrium
is
reached
(Q=Kc):
 
 
 
 
 
 
 
 Question
E8
 Knowing
that

 
 Fe3+
+
e‐
→
Fe2+

 EoFe3+/Fe2+

=

0.771
V
 
 Fe2+
+
2
e‐
→
Fe

 Eo
Fe2+/Fe


=

‐0.440
V
 calculate
the
value
of
E°Fe3+/Fe
at
25°C
for:
 
 Fe3+
+
3
e‐
→
Fe
 
 
 Electrolysis:
Causing
Non­spontaneous
Reactions
to
Occur
 Electrolytic
cell
‐
when
an
opposing
external
potential
causes
the
reaction
in
a
galvanic
 cell
to
occur
in
the
direction
opposite
to
the
spontaneous
direction
 
 [see
Petrucci
et
al.,
Figure
20.22]
 Uses
electrical
energy
from
an
external
circuit
to
promote
a
chemical
reaction
that
 otherwise
would
not
occur.
 
 [see
Petrucci
et
al.,
Figure
20.23]
 Electrolytic
cells
can
be
used
to
recover
pure
metals.

In
this
electrolytic
cell,
the
anode
is
 impure
Cu.

Pure
Cu
is
deposited
in
the
cathode.

Notice
that
there
is
no
salt
bridge ChE102
Fall
2011
Class
Notes
‐
Electrochemistry
 
 
 
 
 
 
 
 E‐10
 Electrolysis
of
Water
 











 
 
 
 Faraday’s
Law
of
Electrolysis
 “The
number
of
moles
of
material
oxidized
or
reduced
at
an
electrode
is
related
by
the
 stoichiometry
of
the
electrode
reaction
to
the
amount
of
electricity
passed
through
the
 cell.”
 
 
 1
mole
e‐

 →
 1
mole
Ag+
 
 

 
 →
 ½
mole
Cu2+
 
 

 
 →
 1/3
mole
Al3+
 
 Remember:

charge
of
1
mole
e‐
=
1
F
=
96,485
C
 
 
 Question
E9
 Consider
three
electrolytic
cells
containing
ZnSO4,
AgNO3
and
CuSO4
as
electrolytes.

A
 steady
current
of
1.50A
is
passed
through
all
the
cells
until
1.45g
of
Ag
are
deposited
at
 the
cathode
of
the
second
cell.

What
mass
of
Cu
and
Zn
were
deposited
in
the
other
cells?
 ChE102
Fall
2011
Class
Notes
‐
Electrochemistry
 
 
 
 
 
 
 
 E‐11
 ...
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This note was uploaded on 12/28/2011 for the course CHE 102 taught by Professor Simon during the Winter '08 term at Waterloo.

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