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Electrochemistry
Ch. 17
Electrochemistry
• Generate current from a reaction– Spontaneous reaction– Battery
• Use current to induce reaction– Nonspontaneous reaction– Electroplating
Oxidation-Reduction Reaction
• aka Redox
• Transfer of electrons
Donor + Acceptor(reducing agent) (oxidizing agent)
(is oxidized) (is reduced)
Mnemonics are cool!
Oxidation
Involves
Loss of electrons
Reduction
Involves
Gain of electrons
Loss of
Electrons is
Oxidation
says
Gain of
Electrons is
Reduction
Assigning Oxidation States(1) Covalent bond btw identicalatoms => Split electrons evenly
(2) Covalent bond btw differentatoms => All electrons given to moreelectronegative atom.
(3) For ionic compounds, oxidation states are equal to ionic charge.
(4) Oxidation state for an elemental atom is zero.
(5) Oxidation state for monatomic ionis the same as the charge.
(6) In compounds, fluorine alwayshas an O.S. of -1.
(7) Oxygen usually has an O.S. of -2,except when in a peroxide or when inOF2. H2O-2 H2O2
-1 +2OF2
(8) With a nonmetal, hydrogen has an O.S. of +1. With a metal, H is assigned an O.S. of -1. NH3
+1 LiH-1
(9) The sum of the oxidation states must add up to the overall charge.
Examples
Assign the oxidation states to eachatom of the following compounds.
CO2 CH4 K2Cr2O7
Redox Reactions
CH4 + O2 CO2 + H2O
Which species is oxidized?
Which species is reduced?
Which species is the oxidizing agent?
Which species is the reducing agent?
Balancing Redox Reaction
• Balance……# of atoms
…# of electrons transferred
…overall charge
• Types of reactions– Acidic conditions– Basic conditions
Redox in Acidic Solutions
Cr2O72- + C2H5OH Cr3+ + CO2
1. Assign oxidation states
2. Write half reactions
Red: Cr2O72- Cr3+
Ox: C2H5OH CO2
3. Balance elements except H and O
Cr2O72- 2Cr3+
C2H5OH 2CO2
4. Balance oxygen by adding H2O
Cr2O72- 2Cr3+ + 7H2O
3H2O + C2H5OH 2CO2
5. Balance hydrogen by adding H+
14H+ + Cr2O72- 2Cr3+ + 7H2O
3H2O + C2H5OH 2CO2 + 12H+
6. Balance charge by adding electrons
6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O
3H2O + C2H5OH 2CO2 + 12H+ + 12e-
7. Equalize the number of electrons
12e- + 28H+ + 2Cr2O72- 4Cr3+ + 14H2O
3H2O + C2H5OH 2CO2 + 12H+ + 12e-
8. Cancel like terms and add reactions
16H+ + 2Cr2O72- + C2H5OH
4Cr3+ + 2CO2 + 11H2O
9. Check your answer!
Balancing in basic solution
Following the same algorithm used for acidic solutions through step #8 then…
9. Add the same # of OH- to both sides of equation as there are H+ on one side
10. Combine H+ and OH- on same sides of equation to make H2O
11. Cancel any like terms and check
Galvanic Cells
• Spontaneous chemistry generating current
• Some terms– Reducing agent– Oxidizing agent– Half reactions– Anode– Cathode– Cell potential
Building a Galvanic CellOverall Reaction
8H+(aq) + MnO4-(aq) + 5Fe2+(aq)
Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)Half Reactions
Reduction:
8H+ + MnO4- + 5e- → Mn2+ + 4H2O
Oxidation:5(Fe2+ → Fe3+ + e-)
Reduction: 8H+ + MnO4- + 5e- Mn2+ + 4H2O
Oxidation: Fe2+ Fe3+ + e-
salt bridge
KNO3
Calculating Cell PotentialReduction: 8H+ + MnO4
- + 5e- Mn2+ + 4H2O
εº(reduction) = 1.51 V
Oxidation: 5(Fe2+ Fe3+ + e-)
εº(oxidation) = -0.77 V
εº(cell) = εº(red) + εº(ox) = 0.74 V
Comments on Cell Potential• Potential is an intensive property
• DO NOT multiply potential by balancing factor
• The º indicates standard conditions– 1.0 M and 1 atm
• Potentials references to standard H+ red.
2H+ + 2e- → H2 εº = 0.00 V
Fe3+(aq) + Cu(s) → Cu2+(aq) + Fe2+(aq)
Oxidation: Reduction:
salt bridgeCu
Fe or Pt
Cu2+ Fe3+
e- e- e-
← - +→
Cu Cu2+ + 2e- Fe3+ + e- Fe2+
Line Notation
2Al3+(aq) + 3Mg(s) → 2Al(s) + 3Mg2+(aq)
Line Notation II
MnO4-(aq) + H+(aq) + ClO3
-(aq) →
ClO4-(aq) + Mn2+(aq) + H2O(l)
Pt(s) │ ClO3-(aq), ClO4
-(aq) ║ MnO4
-(aq), Mn2+(aq) │ Pt(s)
To Review
• Full description of galvanic cell requires:– Composition of solutions– Composition of electrodes– Direction of electron flow– Direction of ion flow– Calculation of cell potential– Labels: “anode” and “cathode”
Cell Potential and Free Energy
arg
96485
o o
w
q
w q
w G q
Ch eof moleof electrons Faraday F
CF w G q nF
molee
G nF G nF
Reconsidering Cell Potential
Given:
Al3+ + 3e- Al ΔG1 and ε1 = -1.66V
Mg2+ + 2e- Mg ΔG2 and ε1 = -2.37 V
Find ε(cell) for:
2Al3+ + 3Mg 2Al + 3Mg2+
3 1 2
3 3 1 1 2 2
1 1 2 23
3
3
3
2 3
2 3
2 3
2(3)( 1.66) (3)(2)( 2.37)
6
1.66 2.37 0.71
o o o
o o o
o oo
o
o
G G G
n F n F n F
n n
n
V
Cell Potential and Spontaneity
• Can bromine oxidize iodide to iodine?
• Can Cr(II) reduce oxygen gas under acidic conditions to produce water?
• Can Ag(I) oxidize chloride to chlorine?
• Can hydrogen reduce Fe(II) to elemental iron?
Non-Standard Cell Potentialln
ln
ln
0.0592log @ 25
o
o o
o
o
o o
G G RT Q
G nF G nF
nF nF RT Q
RTQ
nF
Q Cn
Practice ProblemsDetermine ε for the following reaction and
conditions:
2Al(s) + 3Mn2+(aq) 2Al3+(aq) + 3Mn
(a) [Al3+] = 2.0 M; [Mn2+] = 1.0 M @ 25°C
(b) [Al3+] = 1.0 M; [Mn2+] = 3.0 M @ 25°C
2
3
( ) ln
0.48
(8.314)(298) (2.0)0.48 ln
(6)(96485) (1.0)
0.47 ( ) 0.49
o
o
RTa Q
nF
V
V b V
Do Worksheet
Potential and EquilibriumCell potential at equilibrium = 0.0 V
Q = K at equilibrium0 ln
ln
log @ 250.0592
o
o
oo
RTK
nF
nFK
RT
nK C
Types of Batteries• Lead storage (car battery)
• Dry cell battery– Acidic– Alkaline– Rechargeable
• Lithium ion battery
• Fuel cell
Lead Storage BatteryAnode:
Pb + HSO4- → PbSO4 + H+ + 2e-
Cathode:
PbO2 + HSO4- + 3H+ + 2e- → PbSO4 + 2H2O
Overall Potential: εº = 2.04 V
Dry Cell Battery (acidic)
Anode:
Zn → Zn2+ + 2e-
Cathode:
2NH4+ + 2MnO2 + 2e- → Mn2O3 + 2NH3 + H2O
Overall Potential: εº = 1.5 V
Dry Cell Battery (alkaline)
Anode:
Zn → Zn2+ + 2e-
Cathode:
2MnO2 + H2O + 2e- → Mn2O3 + 2OH-
Overall Potential: εº = 1.5 V
Lithium Ion Battery
Anode:
Li Li+ + e-
Cathode:
MnO2 + Li+ + e- LiMnO2
Cell Potential: εº = 3.6 V
Fuel Cell
Anode:
2H2(g) + 4OH-(aq) 4H2O + 4e-
Cathode:
O2 + 2H2O + 4e- 4OH-
Cell Potential: εº = 1.23 V
Corrosion• A significant portion of construction is
done to replace corroded materials.
Cathode:
O2 + 2H2O + 4e- 4OH- ε = 0.40 VAnode:
Fe Fe2+ + 2e- ε = 0.44 V
ε(cell) = 0.84 V
Corrosion and Acidic Conditions
Cathode:
O2 + 4H+ + 4e- 2H2O ε = 1.23 V
Anode:
Fe Fe2+ + 2e- ε = 0.44 V
ε(cell) = 1.67 V
Electrolysis
• Supply current to perform chemistry
• Performed in an electrolytic cell
• Stoichiometric relationship btw. charge and chemical amount
• Factor-label fun!
• Current measured in Ampere = 1 coulomb per second
Example Problem I
How long will it take to plate out 1.00 kg of aluminum from an aqueous solution of Al3+ using a current of 100.0 A?
Al3+ + 3e- Al
1 3 96485 1sec(1000 )
26.98 1 1 100
107285second = 29.8 h
mol mol e Cg Al
g Al mol Al mol e C
Example Problem II
What volume of F2 gas, at 25C and 1.00 atm, is produced when molten KF is electrolyzed by a current of 10.0 A for 2.00 hours? What mass of K metal is produced? At which electrode does each reaction occur?
Solution II
Molten KF contains K+ and F-
Cathode:
K+ + e- K
Anode:
F- 1/2F2 + e-
Volume of F2
2
2
0.560min 60 10.0 12.00
min 96485 1
0.373
(0.373)(0.0821)(298)9.12
1
mol Fs C mol eh
h s C mol e
mol F
nRTV L
P
Mass of K
K+ + e- K
F- 1/2F2 + e-
0.373 mol F2 0.746 mol K
(0.746 mol K)(39.10 g/mol) = 29.2 g K
Electrolysis in WaterAnode:
2H2O O2 + 4H+ + 4e- ε = -1.23V
Cathode:
4H2O + 4e- 2H2 + 4OH-ε = -0.83V
2H2O O2 + 2H2 ε(cell) = -2.06 V
Assuming [H+] = [OH-] = 1.0 M
Electrolysis in Pure Water
• In pure water: [H+] = [OH-] = 1.0 x 10-7 M• Use Nernst equation to determine ε
Anode:
Cathode:
47100.05920.83 log 0.42
4 1V
47100.05921.23 log 0.82
4 1V
Electrolysis in Pure Water• The overall potential for the electrolysis of
pure water is -1.24 V.
• Need to consider several possible oxidations and reductions when performing electrolysis of aqueous salt solutions.
• Consider the electrolysis of 1.0 M NaCl(aq)
1.0 M NaCl(aq) contains:
1.0 M Na+1.0 M Cl-
H2O 10-7 M H+ 10-7 M OH-
Reducible species: Na+ H+ H2O
Oxidizable species: Cl- OH- H2O
Possible Reductions
• Using pH = 7.00
Na+ + e- Na ε = -2.71 V
H+ + e- 1/2H2 ε = -0.414 V**
H2O + e- 1/2H2 + OH- ε = -0.416 V**
So, H2 produced at the cathode
**Potentials found using Nernst equation
Possible Oxidations
• Using pH = 7.00
Cl- 1/2Cl2 + e- ε = -1.36 V
2OH- 1/2O2 + H2O + 2e- ε = -0.814 V**
H2O 1/2O2 + 2H+ + 2e- ε = -0.816 V**
So, O2 expected to form at the anode.
But…