e- e- Electrochemistry
Galvanic Cells
Chapter 17 Sections 1, 2 &4
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e- e- Electrochemistry- Standard Reduction Potentials
Activity Series
Metals
Halogens
Li
F
Rb
Cl
K
Br
I
Ba
Ca
Na
Mg
Al
Mn
Zn
Cr
Fe
Ni
Sn
Pb
H
Cu
Hg
Ag
Pt
Au
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e- e- Electrochemistry- Standard Reduction Potentials
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e- e- Electrochemistry- Standard Reduction Potentials
ep. 795
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e- e- Electrochemistry – Galvanic Cells
Electrons are transferred directly when reactants collide
No work is obtained – instead heat is released
How can we obtain work?
Separate the oxidizing and reducing agents
require the e- to go through a wire
MnO41- + 8H+ + 5e-  Mn2+ + 4H2O
Fe2+  Fe3+ + 1e- Oxidation - Fe
Reduction - Mn
Oxidizing agent - MnO41-
Reducing Agent – Fe2+
Current flows for a second then stops
Something more needs to added.
ep. 792
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e- e- Electrochemistry – Galvanic Cells
It needs a salt bridge
Solutions must be connected so that ions can flow and keep the net
charge in each container at zero.
A salt bridge is a U-tube with electrolyte (pastey stuff) or porous disk.
Choose a substance that would be noninteractive (something made of
spectator ions if it’s a paste)
It allows ions to flow without mixing the solutions
ep. 792
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e- e- Electrochemistry – Galvanic Cells
A galvanic cell changes chemical energy to electrical energy.
Components
1. Two separate solutions (oxidizing & reducing agent)
2. A wire
3. A salt bridge
e- 
REDUCING
AGENT
p. 793
ANODE
oxidation
X  Y + e-
OXIDIZING
AGENT
CATHODE
reduction
X + e-  Y
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e- e- Electrochemistry – Galvanic Cells
What is the
cathode?
Be able to diagram a cell, label the parts and the flow.
Which way
do e- flow?
e- 
Anode to Cathode
Reduction
occurs at the
cathode.
What is
happening at
the salt bridge?
What are the
agents?
ZnSO4
REDUCING
AGENT
Copper has a
greater
reduction
potential
cations 
Zn
SO4
2-
Zn2+
see
the
atoms
p. 793
ANODE
oxidation
Zn2+  Zn + e-
anions
CuSO4
Cu
Cu2+
OXIDIZING
AGENT
SO4 2CATHODE
reduction
Cu + 2e-  2Cu2+
standard
reduction
potentials
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e- e- Electrochemistry- Standard Reduction Potentials
Line Notation for the cell – so you don’t have to draw the cell
• Anode on the left; Cathode on the right
• Separate the half cells with a
||
• Separate the electrode from the solution with a
ANODE
Zn(s) | Zn
CATHODE
2+
(aq)
Oxidation chamber
electrode
|
| soln
|| Cu3+(aq) | Cu(s)
|| Reduction chamber
||
soln
| electrode
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e- e- Electrochemistry – Galvanic Cells
Cell Potential – the “pull” or driving force that makes electrons go from the
reducing agent to the oxidizing agent
•The potential to lose eThe push from the element mostly likely to
loose to the one that will gain
•E
cell
•Also called electromotive force, emf, of the cell
•Unit = volt = 1 joule of work per columb of charge transferred
– V=1J/1C
•Measured with a voltmeter
– Would measure less than cell potential
– Because it doesn’t measure the frictional heating of the wire
– New voltmeters use a negligible amount of current so they are used
•Measured with a potentiometer
– Variable voltage device (powered from a cell circuit)
– Adjusted so no current flows in the cell circuit
e– Then cell potential = voltage setting but opposite sign
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e- e- Electrochemistry – Galvanic Cells
Water only spontaneously flows one way in a waterfall.
Likewise, electrons only spontaneously flow one way in a redox
reaction—from higher to lower potential energy.
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e- e- Electrochemistry- Standard Reduction Potentials
If you don’t want the electrode to participate in the reaction pick a
“chemically inert” metal or element.
Like Au or Pt. But what would be cheaper?
Carbon. Really? Does carbon conduct electricity?
Graphite (gr) does!
p. 794
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e- e- Electrochemistry- Standard Reduction Potentials
How would I know how any metal
compares against any other?
We need a standard – one that can be
oxidized or reduced.
2H+ + 2e-  H20 reduction
H20  2H+ + 2e- oxidation
H+  H2+ possible but not common
The standard’s E 
cell
would be zero.
Standard Hydrogen Electrode or SHE
half cell
H2 can’t be solid so use a Pt electrode
p. 794
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e- e- Electrochemistry- Standard Reduction Potentials
Zn  Zn2+ + 2e-
2H+ + 2e-  H2
Can measure total potential of the cell E  cell = 0.76 V
Can’t measure the potential of the half reactions ( or half cells)
Setting the standard potential for the hydrogen half reaction to zero
Allows us to assign values to all other half reactions
p. 794
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e- e- Electrochemistry- Standard Reduction Potentials
 Denotes “standard state”
not the same as STP
Standard conditions p. 246
Compounds
gases = 1 atm
soln = 1M
Liquid or solid = pure
Element
1 atm
25 C
Zn  Zn2+ + 2e-
E cell = E  2H
+
 H2
2H+ + 2e-  H2
+ E  Zn  Zn
0.76 = 0.000V
E  Zn  Zn = 0.76 V
2+
p. 794
2+
+
x
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e- e- Oxidation Reduction Reactions
Oxidation States
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e- e- Electrochemistry- Standard Reduction Potentials
ep. 795
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e- e- Electrochemistry- Standard Reduction Potentials
Zn  Zn2+ + 2e-
Cu2+ + 2e-  Cu
E cell = E  Zn  Zn + E  Cu  Cu
1.10V = 0.76 V + E  Cu  Cu
E  Cu  Cu = 0.34 V
2+
2+
2+
2+
p. 795
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e- e- Electrochemistry- Standard Reduction Potentials
 The half reaction with the largest potential will run as a
reduction (as written)
The other will be oxidation; so it will run in reverse
So…
F2 + 2e-  2F1- E = 2.87 V (reduction)
but
2F1-  F2 + 2e- E = - 2.87 V (oxidation)
E cell = E cathode + E anode reverse the sign of the anode
E
cell = E cathode - E anode
 Multiplying the half reaction so that the e- are equal doesn’t
change the E 
E  is an intensive property
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e- e- Electrochemistry- Standard Reduction Potentials
About Standard Reduction Potentials
•A negative reduction potential means it will most likely oxidize
•If the E cell is greater than zero you will get current,
spontaneously.
•If the E cell is zero then the system has reached equilibrium.
The cell is “dead”.
•Values in the table are predicting perfect conditions. You
probably won’t get this in the lab. Why?
–
Voltmeter
•
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e- e- Electrochemistry- Standard Reduction Potentials
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e- e- Electrochemistry- Standard Reduction Potentials
This is the chart given with the AP
Exam.
Half reactions are written as
reductions
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e- e- Electrochemistry- Standard Reduction Potentials
The strongest oxidizers
(oxidizing agents) have the
most positive reduction
potentials.
The strongest reducers
(reducing agents) have the
most negative reduction
potentials.
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e- e- Electrochemistry- Standard Reduction Potentials
The greater the difference
between the two, the
greater the voltage of the
cell.
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e- e- Electrochemistry- Standard Reduction Potentials
Will Br2 oxidize H2O2?
yes
Will Cd reduce Ag+?
yes
Can Al oxidize Au?
What are common metals in
nature?
What are common metal ions in
nature?
What are common non metal
ions in nature?
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e- e- Electrochemistry- Sample Exercise 17.1 Page 797
a.
Consider a galvanic cell based on the reaction
Al3+ (aq) + Mg (s)  Al (s) + Mg2+(aq)
Find the half reactions, balance the cell reaction and calculate E 
for the cell.
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e- e- Electrochemistry- Sample Exercise 17.1 Page 797
b.
Consider a galvanic cell based on the reaction
MnO4 1- (aq) + H+ (aq) + ClO3 1- (aq)  ClO4 1- (aq) + Mn2+(aq) + H2O (l)
Find the half reactions, balance the cell reaction and calculate E 
for the cell.
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e- e- Electrochemistry- Standard Reduction Potentials
What if you have a substance that is not solid for the electrode?
Look at the example on page 799.
It is the same reaction as our Redox Titration Lab Fe2+ + MnO4 1In the example iron is solid but MnO4 1- / Mn2+ is a solution.
Fe2+  Fe
MnO4 1-  Mn2+
Include the
inert
electrode
What is the
line notation?
Don’t
include
the water
Fe ǀ Fe2+ ǁ
H+, MnO4 1-, Mn2+ ǀ Pt
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e- e- Electrochemistry- Try #29 Page 831
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e- e- Electrochemistry- Gold/Nickel Voltaic Cell
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© 2009, Prentice-Hall, Inc.
e- e- Electrochemistry- Standard Reduction Potentials
e-
?
K+(aq) NO3-(aq)
Au3+(aq)
Au
X-(aq)
Ni2+(aq)
Ni
X (aq)
Au3+ + 3e-  Au
E0 = +1.50 V
Ni2+ + 2e-  Ni
0 = -0.25 V
E
A cell will always run spontaneously in the direction that produces a positive cell
potential
Nickel is the anode E cell = E cathode - E anode flip the nickel
Current flows from anode to cathode
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e- e- Electrochemistry- Standard Reduction Potentials
e-
K+(aq)
Au3+(aq)
Au
X-(aq)
Au3+ + 3e-  Au
E0 = 1.50 V
reduction: cathode
NO3
-(aq)
e-
Ni2+(aq)
Ni
X (aq)
Ni  Ni2+ + 2eE0 = 0.25 V
oxidation: anode
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e- e- Electrochemistry- Standard Reduction Potentials
Au3+ + 3e-  Au
Ni  Ni2+ + 2e-
E0(V)
1.50
0.25
2 Au3+ + 3 Ni  2 Au + 3 Ni2+ 1.75
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e- e- Electrochemistry- Standard Reduction Potentials
Line Notation:
ANODE
CATHODE
Ni(s)|Ni2+ (aq, 1 M)||Au3+ (aq, 1 M)|Au(s)
oxidation
reduction
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e- e- Electrochemistry- Aluminum/Nickel Cell
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© 2009, Prentice-Hall, Inc.
e- e- Electrochemistry- Standard Reduction Potentials
e-
?
K+(aq) NO3-(aq)
Al3+(aq)
Al
X-(aq)
Al3+ + 3e-  Al
E0 = -1.66 V
Ni2+(aq)
X-(aq)
Ni
Ni2+ + 2e-  Ni
E0 = -0.25 V
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e- e- Electrochemistry- Standard Reduction Potentials
Al  Al3+ + 3eNi2+ + 2e-  Ni
E0(V)
1.66
-0.25
2Al + 3Ni 2+  3Ni + 2Al3+ 1.41
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e- e- Electrochemistry- Standard Reduction Potentials
Line Notation:
ANODE
CATHODE
Al(s) | Al3+(aq, 1 M) || Ni2+(aq, 1 M) | Ni (s)
oxidation
reduction
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e- e- Electrochemistry- Extra Problem
Sketch the Ag/Ag+ & Zn/Zn2+ galvanic cell. Show the
direction of e- flow and the direction of ion migration.
Calculate E0 for the cell. Give the line notation for the cell.
e-
?
K+(aq) NO3-(aq)
Zn2+(aq)
Zn X-(aq)
Zn2+ + 2eZn
E0 = - 0.76 V
Ag+(aq)
Ag
X (aq)
Ag+ + eAg
E0 = + 0.80 V
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e- e- Electrochemistry- Extra Problem
Sketch a system having a solid lead electrode and a lead(IV) oxide
electrode immersed in sulfuric acid. Show the direction of e- flow
and the direction of ion migration. Calculate E0 for the cell. Give
the line notation for the cell.
e-
?
PbSO4 + 2ePb + SO42E0 = - 0.35 V
H2SO4(aq)
PbO2
Pb
PbO2 + 4H+ + SO42- + 2e-  PbSO4 + 2H2O
E0 = + 1.69V
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e- e- Electrochemistry- Standard Reduction Potentials
The Lead-Acid Car Battery
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e- e- Electrochemistry- Concentration on E 
Cu(s) + 2Ce 4+ (aq)  2Cu 2+ (aq) + Ce 3+ (aq)
Remember E  is positive. Treat this like an equilibrium.
Increase [Ce
4+
], which direction is favored?
Forward. Increase the driving force on e-, increase the E 
Increase [Cu
2+
] or [Ce
3+],
which direction is favored?
Reverse. Decrease the driving force on e-, decrease the E 
Sample Exercise 17.5 page 803
2Al + 3 Mn2+  2Al3+ + 3 Mn
a. [Al3+] = 2.0M; [Mn2+] = 1.0M
 Al3+
 decrease E  < 0.48V
b. [Al3+] = 1.0M; [Mn2+] = 3.0M
 Mn2+  increase E  > 0.48V
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e- e- Electrochemistry- Concentration on E 
Cell Potentials depend on concentration
Can construct cells with same components but different concentrations
Ag ǀ 0.10M Ag2+ ǁ 1.0 M Ag2+ ǀ Ag
E cell =0.80V – 0.80 V = 0
However since the concentration is not equal,
the half cell potentials are not 0.80V.
This is a concentration cell. Voltages are typically small
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e- e- Electrochemistry- Concentration on E 
The Nernst Equation
Derived from the dependence of free energy on concentration
*** Will learn about this after thermodynamics***
E = E cell -
0.0591
log (Q)
n
moles of e-
Quotient = Keq
[P]x
[R]y
Like Keq but not necessarily at equilibrium
Remember (l) and (s) don’t go into the Keq
Why not? It’s all about concentration – solids and liquids have a definite volume
•The M for solids and liquids is 1
M= mol/L
•If you change moles for solids and liquids then the volume changes
proportionally
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e- e- Electrochemistry- Concentration on E 
The Nernst Equation
Derived from the dependence of free energy on concentration
*** Will learn about this after thermodynamics***
E = E cell -
0.0591
log (Q)
n
moles of e-
Quotient = Keq
[P]x
Trying to get the max voltage?
[R]y
•Choose the half reactions that give max volts
– be practical Li or F won’t work
•Make the reactant concentration greater than 1
& the product concentration less than 1
That way [P]x / [R]y is 0.XX and then log 0.XX is a negative
number and you’re adding to the E cell
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e- e- Electrochemistry- Sample Exercise 17.7 page 806
Handy ideas about logs
log x
2
= 2 log x
log 100 = 2
log x y = y log x
log
1
x
= log 1 – log x = -log x
log xy = log x + log y
log y√ x = log x1/y = (1/y ) log x
Sig. Figs for Logarithms
For any log, the number to the left of the decimal point is called the
characteristic, and the number to the right of the decimal point is
called the mantissa. The characteristic only locates the decimal point
of the number, so it is usually not included when determining the
number of significant figures. The mantissa has as many significant
figures as the number whose log was found.
Example 1: log 5.43 x 1010 = 10.735
The number has 3 significant figures, but its log ends up with 5
significant figures, since the mantissa has 3 and the characteristic
has 2.
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e- e- Electrochemistry- Sample Exercise 17.7 page 806
Describe the cell based on the following half reactions:
VO2+(aq) + 2H1+ (aq) + e-  VO 2+ (aq) + H2O
Zn2+ (aq) + 2e-  Zn (s)
T = 25 C
[VO2+] = 2.0 M
[H1+] = 0.50M
[VO 2+ ] = 1.0 x 10-2 M
[Zn2+ ] = 1.0 x 10-1 M
(l)
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e- e- Electrochemistry- Sample Exercise 17.7 page 806
Describe the cell based on the following half reactions:
VO2+(aq) + 2H1+ (aq) + e-  VO 2+ (aq) + H2O
Zn2+ (aq) + 2e-  Zn (s)
Where T= 25 C
[VO2+] = 2.0 M
[H1+] = 0.50M
[VO 2+ ] = 1.0 x 10-2 M
[Zn2+ ] = 1.0 x 10-1 M
(l)
Solution
1. Balance First
2. Calculate the standard cell potential
3. Use the Nernst Equation
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e- e- Electrochemistry- Try #59 Page 832
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e- e- Electrochemistry- Try #57 Page 832
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