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Determine the oxidation number for each
atom in the following molecules
1.H2S
2.P2O5
3.S8
4.SCl2
5.Na2SO3
6. SO4-2
7. NaH
8. Cr2O7-2
9. SnBr4
10. Ba(OH)2
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Intersection 14
12/05/06
Electrochemistry
19.9-19.13 p 941-955
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December in Studio
S
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12/11 Poster
session, paper
due
12/17
Review
session 79 pm
Tu
W
12/5
Exam 3
12/6
Studio
12/12 final 12/13 InIS
class
assignment
12/19
Final exam
8-10 am
Th
F
12/8
Polymers;
check out
S
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Watershed Poster Session
•
•
•
•
Monday, December 11 in USB 2165
Board (4 ft x 4ft), easel, pins
Set up by 1:10 and 3:10
One person stationed at poster; others
evaluate
• Rubric available
• Paper due same time
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Last In-Class Assignment
• Wednesday, December 13th in studio
• Available on-line
• Read papers before coming to class; bring
them with you.
• May make any notes you like on the papers
• Goal: to evaluate scientific method and
data
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Outline
• Ed’s demos
• Balancing Redox Reactions
• Electrochemistry
– Electrochemical cells and Standard Hydrogen
Electrodes
– Nernst
– Quantifying current
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Ed’s Demos
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Oxidation States of Vanadium:
Reduction of V5+ to V+2
• Reaction 1
– Zn (s) + 2 VO3- (aq) + 8 H3O+ (aq) ↔ 2 VO2+ (aq) + Zn+2 (aq) + 12 H2O (l)
• Reaction 2
– Zn (s) + 2 VO2+ (aq) + 8 H3O+ (aq) ↔ 2 V3+ (aq) + Zn+2 (aq) + 6 H2O (l)
• Reaction 3
– Zn (s) + 2 V3+ (aq) ↔ 2 V2+ + Zn+2 (aq)
V+5 (aq) → V+4 (aq) yellow to green
V+4 (aq) → V+3 (aq) green to blue
V+3 (aq) → V+2 (aq) blue to violet
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Oxidation States of Manganese:
Mn+7, Mn+6, Mn+4, and Mn+2
• +7 (purple) to +2 (colorless)
– 2 MnO4- (aq) + H+ (aq) + 5 HSO3- (aq) ↔ 2 Mn+2 (aq) + 5 SO4-2 (aq) + 3 H2O(l)
• + 7 (purple) to +4 (brown)
– OH- + 2 MnO4- (aq) + 3 HSO3- (aq) ↔ 2 MnO2 (s) + 3 SO4-2 (aq) + 2 H2O(l)
• + 7 (purple) to + 6 (green)
– 2 MnO4- (aq) + 3 OH- + HSO3- (aq) ↔ 2 MnO4-2(aq) + SO4-2 (aq) + 2 H2O(l)
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Thinking back….
• What happened when Na(s) was added to
water?
Na(s) + H2O(l)  Na+ (aq) + H2(g) + OH-(aq)
• Determine the oxidation state of each reactant and
product
• What was oxidized?
• What was reduced?
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Balancing Redox Reactions
When KMnO4 (potassium permanganate) is mixed with Na2C2O4
(sodium oxalate) under acidic conditions, Mn+2(aq) ions and
CO2(g) form.
The unbalanced chemical equation is:
KMnO4(aq) + Na2C2O4(aq)  Mn+2(aq) + CO2(g) + K+(aq) + Na+(aq)
K+ and Na+ are spectator ions, so we can ignore them at this point.
MnO4- (aq) + C2O4-2(aq)  Mn+2(aq) + CO2(g)
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Half-Reactions
MnO4- (aq) + C2O4-2(aq)  Mn+2(aq) + CO2(g)
• Reduction reaction
• Oxidation reaction
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Reduction reaction
MnO4-
 Mn+2
Step 1: Balance all elements other than oxygen and hydrogen.
Step 2: Balance the oxygens by adding water.
Step 3: Balance the hydrogens using H+
Step 4: Balance the electrons
Mn+7 on reactants side
Mn+2 on products side
Step 5: Check charge balance and elemental balance
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Oxidation reaction
C2O4-2  CO2
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Combine Half Reactions
5 e- + 8H+ + MnO4-  Mn+2 + 4 H2O
C2O4-2  2 CO2 + 2e-
We are assuming the reaction takes place under acidic conditions!
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Balancing in Base
Change H+ to water by adding OH- to each side
5 e- + 8H+ + MnO4-  Mn+2 + 4 H2O
C2O4-2  2 CO2 + 2e-
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Electrochemical Cells
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Definitions
• Electrochemical cell: A combination of anode,
cathode, and other materials arranged so that a
product-favored redox reaction can cause a current
to flow or an electric current can cause a reactantfavored redox reaction to occur
• Voltaic cell (battery): An electrochemical cell or
group of cells in which a product-favored redox
reaction is used to produce an electric current.
• Galvanic cell: A cell in which an irreversible
chemical reaction produces electrical current
• Electrolytic cell: electrochemical reactions are
produced by applying electrical energy
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A Copper-Zinc battery – What Matters?
Consider reduction potentials:
Cu+2 + 2e- → Cu(s)
Zn+2 + 2e- → Zn(s)
0.3419 V
-0.7618 V
Place Zn electrode in copper sulfate solution – What happens?
Copper is plated on Zn electrode
Cu+2 + 2e- → Cu(s)
Zn(s) → Zn+2 + 2e-
0.3419 V
0.7618 V
Cu+2 + Zn(s) → Zn+2 + Cu(s) 1.1 V
E > 0, spontaneous
Note, no need for electron to flow external to cell for reaction to occur!!
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A Copper-Zinc battery – What Matters?
Consider reduction potentials:
Cu+2 + 2e- → Cu(s)
Zn+2 + 2e- → Zn(s)
0.3419 V
-0.7618 V
Place Cu electrode in zinc sulfate solution – What happens?
Zn doesn’t plate on copper electrode?!
Cu(s) → Cu+2 + 2eZn+2 + 2e- → Zn(s)
-0.3419 V
-0.7618 V
Zn+2 + Cu(s) → Cu+2 + Zn(s) -1.1 V
E < 0, not spontaneous
No reaction occurs !!
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Fig. 19-3, p.918
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What are the ½ reactions?
What is the overall reaction?
Identify the
oxidation, reduction,
anode, and cathode
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SHE: Standard Hydrogen Electrode
2 H3O+(aq, 1.00 M) + 2e- <->
H2(g, 1 atm) + 2H2O(l)
Eo = 0V
Standard conditions:
1M, 1atm, 25oC
Fig. 19-7, p.922
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Measuring Relative Potentials
Table of Standard
Reduction Potentials
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Standard
Reduction
Potentials
What is the standard
potential of a
Au+3/Au/Mg+2/Mg
cell?
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The half-reaction with the more positive standard reduction
potential occurs at the cathode as reduction.
The half-reaction with the more negative standard reduction
potential occurs at the anode as oxidation.
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Nernst
Picture from: www.corrosion-doctors.org/ Biographies/images/
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Is potential always the same?
Standard conditions: 1 atm, 25oC, 1 M
What will influence the potential of a cell?
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Mathematical Relationships:
Nernst
The Nernst Equation:
E = Eo - RT ln Q
nF
Eo = standard potential of the cell
R = Universal gas constant = 8.3145 J/mol*K
T = temperature in Kelvin
n = number of electrons transferred
F = Faraday’s constant = 96,483.4 C/mol
Q = reaction quotient (concentration of anode divided by
the concentration of the cathode)
Cu+2 + Zn(s) → Zn+2 + Cu(s) Q =
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Applying the Nernst Equation
Cu+2 + Zn(s) → Zn+2 + Cu(s)
This cell is operating at 25oC with 1.00x10-5M Zn2+ and
0.100M Cu2+?
Predict if the voltage will be higher or lower than the
standard potential
E = Eo - RT ln Q
nF
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Cu+2 + Zn(s) → Zn+2 + Cu(s)
E = Eo - RT ln Q
nF
Eo = standard potential of the cell
Zn+2 + 2e- -> Zn -0.76 V
Cu+2 + 2e- -> Cu 0.34 V
R = Universal gas constant = 8.3145 J/mol*K
T = temperature in Kelvin
n = number of electrons transferred
25oC + 273 = K
n =
F = Faraday’s constant = 96,483.4 C/mol
Q = reaction quotient (concentration of anode divided by the
concentration of the cathode)
Q = [Zn+2]/[Cu+2]
1.00x10-5M Zn2+ and 0.100M Cu2
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Were your predictions correct?
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Ampere
Picture from: musee-ampere.univ-lyon1.fr/
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The Units of Electrochemistry
• Coulomb
– 1 coulomb equals 2.998 x 109 electrostatic units (eu)
– eu is amount of charge needed to repel an identical charge 1
cm away with unit force
– Charge on one electron is -1.602 x 10-19 coulomb
Problem:
An aluminum ion has a +3 charge.
What is this value in coulombs?
magnitude of charge is same at that of e-, opposite sign
3 x 1.602 x10-19 = 4.806 x 10-19 coulomb
Key Point: electrons or ions charges can be measured in coloumbs
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The Units of Electrochemistry
• Ampere
– Amount of current flowing when 1 coulomb passes a given
point in 1 second
– Units of Amperes are Coulombs per second
– Current (I) x time (C/s x s) gives an amount of charge.
Problem:
How much current is flowing in a wire in which
5.0 x 1016 electrons are flowing per second?
The charge transferred each second
= (5.0 x 1016 electrons/sec) x (1.602 x 10-19 coulomb/electron)
= 8.0 x 10-3 coulombs/sec = amps
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The Units of Electrochemistry
• Ampere
– Amount of current flowing when 1 coulomb passes a given
point in 1 second
– Units of Amperes are Coulombs per second
– Current (I) x time (C/s x s) gives an amount of charge.
– We can express electron or ION flow in amps!
Problem:
If 1 mol Al+3 ions passes a given point in one hour, what
is the current flow?
1 hour = 3600 seconds; Al+3 charge is 4.806 x 10-19 coulomb
1 mol Al +3 ions
6.022 x 1023 Al +3 ions
4.806 x 10-19 coulomb
1 hour
Hour
1 mol Al +3 ions
1 Al+3 ion
3600 sec
= 80 C/s
= 80 A
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Electrolysis
The process of using electrical current to drive a redox
reaction that is not spontaneous. Balance the reaction in
acid.
At the anode (where oxidation occurs):
H2O (l) → O2 (g)
At the cathode (where reduction occurs):
H3O+ (aq) → 2 H2 (g)
Overall reaction:
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Overall reaction: 2 H2O (l) → 2 H2(g) + O2 (g)
Overall reaction: 2 H2O (l) → 2 H2(g) + O2 (g)
p.951
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Standard
Reduction
Potentials
What is the standard
potential for the
electrolysis of water?
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Electrolysis Reactions
How many coulombs of charge are required to electrolyze 1
mol of water?
Note that 6 water are consumed at the anode, but 4 are produced
at the cathode.
At the anode (where oxidation occurs): 6 H2O (l) → O2 (g) + 4 H3O+ (aq) + 4eAt the cathode (where reduction occurs): 4 H3O+ (aq) + 4 e- → 2 H2 (g) + 4 H2O (l)
The net amount of water consumed per every 4 electrons is 2
molecules.
1.0 mol H2O
4 mol e-
96,487 coulombs
= 190000 C
2 mol H2O 1 mol e-
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Electrolysis Reactions
A power supply puts out a maximum current of 10 amps.
How long will it take to
electrolyze 18 g (1 mole) water?
190,000
coulombs
sec
hour
= 5.4 hours
10 coulombs
3600 secs
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Electrolysis
What similarities do you find between voltaic
cells and electrolysis?
What differences?
Overvoltage: However, the potential that must be
applied to electrolysis cell is always greater than that
calculated from standard reduction potentials. This
excess potential is called an overvoltage. This additional
voltage is needed to overcome limitations in electron
transfer rate at the interface between electrode and
solution.