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
+
(reducing agent)
(is oxidized)
Acceptor
(oxidizing agent)
(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 identical
atoms => Split electrons evenly
(2) Covalent bond btw different
atoms => All electrons given to more
electronegative 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 ion
is the same as the charge.
(6) In compounds, fluorine always
has an O.S. of -1.
(7) Oxygen usually has an O.S. of -2,
except when in a peroxide or when in
OF2. 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 each
atom 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+ + 12e7. Equalize the number of electrons
12e- + 28H+ + 2Cr2O72-  4Cr3+ + 14H2O
3H2O + C2H5OH  2CO2 + 12H+ + 12e8. 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 Cell
Overall 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 Potential
Reduction: 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)

e-
←
Cu
e- 
e-
-
Fe or Pt
+→
salt bridge
Cu2+
Oxidation:
Cu  Cu2+ + 2e-
Fe3+
Reduction:
Fe3+ + e-  Fe2+
Cu2+ + Zn  Zn2+ + Cu
Demo
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
w

q
 w  q
w  G   q
Ch arg e of mole of electrons  Faraday  F
96485 C
F
w  G   q  nF 

mole e
G  nF 
G o   nF  o
Reconsidering Cell Potential
Given:
Al3+ + 3e-  Al ΔG1 and ε1 = -1.66V
Mg2+ + 2e-  Mg ΔG2 and ε1 = -2.37 V
Find ε(cell) for:
2Al3+ + 3Mg  2Al + 3Mg2+
G  2G  3G
o
3
o
1
o
2
 n3 F   2n1 F   3n2 F 
o
3
o
1
o
2
2n   3n 
 
n3
o
3
o
1 1
o
2 2
2(3)(1.66)  (3)(2)(2.37)
 
6
o
 3  1.66  2.37  0.71V
o
3
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 Potential
G  G  RT ln Q
o
G   nF 
G   nF 
o
o
 nF    nF   RT ln Q
o
RT
  
ln Q
nF
o
0.0592
o
  
log Q @ 25 C
n
o
Practice Problems
Determine 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
RT
(a)    
ln Q
nF
o
  0.48V
o
2
(8.314)(298) (2.0)
  0.48 
ln
3
(6)(96485)
(1.0)
  0.47 V
(b) 0.49V
Do Worksheet
Potential and Equilibrium
Cell potential at equilibrium = 0.0 V
Q = K at equilibrium
RT
0 
ln K
nF
o
nF 
ln K 
RT
o
n
o
log K 
@ 25 C
0.0592
o
Types of Batteries
• Lead storage (car battery)
• Dry cell battery
– Acidic
– Alkaline
– Rechargeable
• Lithium ion battery
• Fuel cell
Lead Storage Battery
Anode:
Pb + HSO4- → PbSO4 + H+ + 2eCathode:
PbO2 + HSO4- + 3H+ + 2e- → PbSO4 + 2H2O
Overall Potential: εº = 2.04 V
Dry Cell Battery (acidic)
Anode:
Zn → Zn2+ + 2eCathode:
2NH4+ + 2MnO2 + 2e- → Mn2O3 + 2NH3 + H2O
Overall Potential: εº = 1.5 V
Dry Cell Battery (alkaline)
Anode:
Zn → Zn2+ + 2eCathode:
2MnO2 + H2O + 2e- → Mn2O3 + 2OHOverall Potential: εº = 1.5 V
Lithium Ion Battery
Anode:
Li  Li+ + eCathode:
MnO2 + Li+ + e-  LiMnO2
Cell Potential: εº = 3.6 V
Fuel Cell
Anode:
2H2(g) + 4OH-(aq)  4H2O + 4eCathode:
O2 + 2H2O + 4e-  4OHCell 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 V
Anode:
Fe  Fe2+ + 2eε = 0.44 V
ε(cell) = 0.84 V
Corrosion and Acidic Conditions
Cathode:
O2 + 4H+ + 4e-  2H2O
Anode:
Fe  Fe2+ + 2eε(cell) = 1.67 V
ε = 1.23 V
ε = 0.44 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 mol  3 mol e  96485 C  1sec 
(1000 g Al ) 




 26.98 g Al  1 mol Al  1 mol e  100 C 
 107285second = 29.8 h
Example Problem II
What volume of F2 gas, at 25C 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 FCathode:
K+ + e-  K
Anode:
F-  1/2F2 + e-
Volume of F2
 60 min  60 s  10.0C   1 mol e  0.5 mol F2 
 2.00h  





 h  min  s   96485 C  1 mol e 
 0.373 mol F2
nRT (0.373)(0.0821)(298)
V

 9.12 L
P
1
Mass of K
K+ + e-  K
F-  1/2F2 + e0.373 mol F2  0.746 mol K
(0.746 mol K)(39.10 g/mol) = 29.2 g K
Electrolysis in Water
Anode:
2H2O  O2 + 4H+ + 4eCathode:
4H2O + 4e-  2H2 + 4OH-
ε = -1.23V
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:
 107 4 
0.0592
  0.82V
  1.23 
log 
 1 
4
Cathode:


 107 4 
0.0592
  0.42V
  0.83 
log 
 1 
4


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 ClH2O
10-7 M H+
10-7 M OH-
Reducible species: Na+
H + H 2O
Oxidizable species: Cl-
OH-
H 2O
Possible Reductions
• Using pH = 7.00
Na+ + e-  Na
H+ + e-  1/2H2
H2O + e-  1/2H2 + OH-
ε = -2.71 V
ε = -0.414 V**
ε = -0.416 V**
So, H2 produced at the cathode
**Potentials found using Nernst equation
Possible Oxidations
• Using pH = 7.00
Cl-  1/2Cl2 + eε = -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…