Electrochemistry Ch. 17. Electrochemistry Generate current from a reaction –Spontaneous reaction...
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Transcript of Electrochemistry Ch. 17. Electrochemistry Generate current from a reaction –Spontaneous reaction...
![Page 1: Electrochemistry Ch. 17. Electrochemistry Generate current from a reaction –Spontaneous reaction –Battery Use current to induce reaction –Nonspontaneous.](https://reader033.fdocuments.us/reader033/viewer/2022051216/56649e385503460f94b28570/html5/thumbnails/1.jpg)
Electrochemistry
Ch. 17
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Electrochemistry
• Generate current from a reaction– Spontaneous reaction– Battery
• Use current to induce reaction– Nonspontaneous reaction– Electroplating
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Oxidation-Reduction Reaction
• aka Redox
• Transfer of electrons
Donor + Acceptor(reducing agent) (oxidizing agent)
(is oxidized) (is reduced)
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Mnemonics are cool!
Oxidation
Involves
Loss of electrons
Reduction
Involves
Gain of electrons
Loss of
Electrons is
Oxidation
says
Gain of
Electrons is
Reduction
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Assigning Oxidation States(1) Covalent bond btw identicalatoms => Split electrons evenly
(2) Covalent bond btw differentatoms => All electrons given to moreelectronegative atom.
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(3) For ionic compounds, oxidation states are equal to ionic charge.
(4) Oxidation state for an elemental atom is zero.
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(5) Oxidation state for monatomic ionis the same as the charge.
(6) In compounds, fluorine alwayshas an O.S. of -1.
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(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.
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Examples
Assign the oxidation states to eachatom of the following compounds.
CO2 CH4 K2Cr2O7
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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?
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Balancing Redox Reaction
• Balance……# of atoms
…# of electrons transferred
…overall charge
• Types of reactions– Acidic conditions– Basic conditions
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Redox in Acidic Solutions
Cr2O72- + C2H5OH Cr3+ + CO2
1. Assign oxidation states
2. Write half reactions
Red: Cr2O72- Cr3+
Ox: C2H5OH CO2
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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+
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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!
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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
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Galvanic Cells
• Spontaneous chemistry generating current
• Some terms– Reducing agent– Oxidizing agent– Half reactions– Anode– Cathode– Cell potential
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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-)
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Reduction: 8H+ + MnO4- + 5e- Mn2+ + 4H2O
Oxidation: Fe2+ Fe3+ + e-
salt bridge
KNO3
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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
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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
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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+
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Line Notation
2Al3+(aq) + 3Mg(s) → 2Al(s) + 3Mg2+(aq)
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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)
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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”
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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
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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+
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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
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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?
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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
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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
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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
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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
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Types of Batteries• Lead storage (car battery)
• Dry cell battery– Acidic– Alkaline– Rechargeable
• Lithium ion battery
• Fuel cell
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Lead Storage BatteryAnode:
Pb + HSO4- → PbSO4 + H+ + 2e-
Cathode:
PbO2 + HSO4- + 3H+ + 2e- → PbSO4 + 2H2O
Overall Potential: εº = 2.04 V
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Dry Cell Battery (acidic)
Anode:
Zn → Zn2+ + 2e-
Cathode:
2NH4+ + 2MnO2 + 2e- → Mn2O3 + 2NH3 + H2O
Overall Potential: εº = 1.5 V
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Dry Cell Battery (alkaline)
Anode:
Zn → Zn2+ + 2e-
Cathode:
2MnO2 + H2O + 2e- → Mn2O3 + 2OH-
Overall Potential: εº = 1.5 V
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Lithium Ion Battery
Anode:
Li Li+ + e-
Cathode:
MnO2 + Li+ + e- LiMnO2
Cell Potential: εº = 3.6 V
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Fuel Cell
Anode:
2H2(g) + 4OH-(aq) 4H2O + 4e-
Cathode:
O2 + 2H2O + 4e- 4OH-
Cell Potential: εº = 1.23 V
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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
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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
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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
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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?
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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
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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?
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Solution II
Molten KF contains K+ and F-
Cathode:
K+ + e- K
Anode:
F- 1/2F2 + e-
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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
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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
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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
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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
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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)
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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
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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
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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…