Chapter 20
Overview
• Oxidation-Reduction reactions
» Balancing Redox Reactions
• Half-Reaction method
• Acidic Solution
• Basic Solution
• Voltaic Cells
• Cell EMF--standard reduction potentials
• Oxidizing & Reducing reagents
• Spontaneity of Redox reactions
• Effect of Concentration
» Nernst Equation
» Equilibrium Constants
• Commercial Voltaic Cells
• Electrolysis
» Quantitative Aspects
» Electrical Work
Redox Reactions
• Involve a transfer of electrons
» Oxidation  loss of one or more electron(s)
• oxidation state will increase
» Reduction  gain of one or more electron(s)
• oxidation state will decrease
» Must occur simultaneously
Zn(s) +
Cu2+(aq)

Zn2+(aq) + Cu(s)
Zn  Zn2+(aq) + 2e-
Cu2+(aq) + 2e-  Cu(s)
oxidation ½ rxn
reduction ½ rxn
You must know oxidation states:
(Review: Section 8.10)
• What are the oxidation states of each atom in
the following:
»
»
»
»
H2
CO
ClO2HC2H3O2
H 0
C +2,
O -2
Cl +3, O -2
H +1, C1 +3, C2 -3, O
Balancing Redox Reactions
• Mass balance must be observed
• e--transfer must be balanced
• Simple reactions:
» Sn2+ + Fe3+  Sn4+ + Fe2+
• Sn2+  Sn4+ + 2e• Fe3+ + e-  Fe2+ x 2
2Fe3+ + 2e-  2Fe2+
oxidation ½ rxn
reduction ½ rxn
Sn2+ + 2Fe3+  Sn4+ + 2Fe2+
• Reactions involving H & O in acid:
» MnO4- + C2O42-  Mn2+ + CO2
» write both ½ reactions
• MnO4• C2O42-


Mn2+
CO2
» mass balance (all except H & O)
• MnO4• C2O42-


Mn2+
2CO2
» add H2O & H+ to balance O & H
• 8H+ + MnO4
• C2O42- 
2CO2
Mn2+ + 4H2O
» balance charge by adding electrons
• 5e- + 8H+ + MnO4
• C2O42- 
2CO2 + 2e-
Mn2+ + 4H2O
» balance electrons transferred
• 10e- + 16H+ + 2MnO4
• 5C2O42- 
10CO2 + 10e-
2Mn2+ + 8H2O
» add half reactions
• 16H+ + 2MnO4-+ 5C2O42-  10CO2 + 2Mn2+ + 8H2O
»check the balance
• Reactions in base:
MnO4- + CN-  CNO- + MnO2
» use exactly the same process
• H2O + CN-  CNO- + 2H+ + 2e• 3e- + 4H+ + MnO4-  MnO2 + 2H2O
» since H+ cannot exist in basic solution, add OH• 2OH- + CN-  CNO- + H2O + 2e• 3e- + 2H2O + MnO4-  MnO2 + 4OH-
» balance electrons transferred & sum
• 6OH- + 3CN-  3CNO- + 3H2O + 6e• 6e- + 4H2O + 2MnO4-  2MnO2 + 8OH-
» 3CN- + H2O + 2MnO4-  2MnO2 + 3CNO- +2OH-
»check balance
Voltaic Cells
• A spontaneous redox reaction that does
work
• Anode
» electrode at which oxidation occurs
» loses mass
» electrons released, sign is negative
• Cathode
» electrode at which reduction occurs
» gains mass
» electrons consumed, sign is positive
Cell EMF
• Difference in potential energy of electrons at
the anode and cathode
» Diff. in potential energy per electrical charge
measured in volts
»1V = 1 J
C
• Potential difference = EMF, electromotive force
• Ecell = cell potential = cell voltage
» Eºcell = cell potential under std. conditions
• 1 M, 1 atm, 25 ºC
• Standard reduction potentials
» E ºred in tables
» E ºcell = E ºred (cathode) - E ºred (anode)
• Based on “standard hydrogen electrode”
» 2H+(aq, 1M) + 2e-  H2(g, 1atm)
E ºred = 0 V
» Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g) E ºcell = 0.76 V
» 0.76 V = 0 V - E ºred (anode)
• Zn2+(aq, 1M) + 2e-  Zn(s) E ºred (anode) = -0.76 V
Problem:
• Calculate Eºcell for
• 2Al(s) + 3I2(s)  2Al3+(aq) + 6I-(aq)
» Anode:
» Cathode:
2Al  2Al3+ + 6e-
3I2 + 6e-  6I-
• Eºcell = E ºred (cathode) - E ºred (anode)
• E ºcell = 0.54 V - (-1.66 V)
• E ºcell = 2.20 V
• Note: stoichiometric coefficient does not
affect the value of the E ºred
(it is an intensive property)
• E ºox = - E ºred
• 2Al(s) + 3I2(s)  2Al3+(aq) + 6I-(aq)
• 2Al  2Al3+ + 6e-
E ºox = +1.66 V
• 3I2 + 6e-  6I-
E ºred = +0.54 V
• E ºcell = E ºox + E ºred = 2.20V
• The more positive the E ºcell the more driving
force for the reaction
Oxidizing/Reducing Agents
• Oxidizing agents cause oxidation
» oxidizing agents are reduced
» the more (+) the E ºred the better the ox. agent
• Reducing agents cause reduction
» reducing agents are oxidized
» the more (-) the E ºred the better the red. agent
• Which is the better oxidizing agent?
• NO3- + 4H+ + 3e-  NO + 2H2O
E ºred 0.96 V
• Ag+ + e-  Ag
E ºred 0.80 V
• Cr2O72- + 14H+ + 6e-  2Cr3+ + H2O
E ºred 1.33 V
• Which is the strongest reducing agent?
• I2 + 2e-  2I-
Eºred +0.54 V
• Fe2+ + 2e- 
Eºred -0.44 V
Fe
• MnO4- + 8H+ + 5e-  Mn2+ + 4H2O
Eºred +1.51 V
Spontaneity of Redox Reactions
• Spontaneous redox rxns have positive
potentials
• Non-spontaneous redox rxns have negative
potentials
• Is this rxn spont. or non-spont.?
» MnO4- + 8H+ + 5Fe2+  5Fe3+ + Mn2+ + 4H2O
• Fe2+  Fe3+ + 1e• MnO4- + 8H+ + 5e-  Mn2+ + 4H2O
• E ºox + E ºred = + 0.74 v Yes
Eºox = -0.77 v
E ºred = +1.51 v
EMF & Free Energy
• If both DG & E are a measure of spontaneity,
they must be related
» DG = - nFE
• F is Faraday’s constant
1 F = 96,500 J/v mol e-
• remember: 1 C = 1 J/v
• n = mol e- transferred
» In the standard state DGº = - nFEº
• Calculate the standard free energy change for
Hg + 2Fe3+  Hg2+ + 2Fe2+
» n = 2 mol electrons transferred
• Hg  Hg2+ + 2e-
Eox = - 0.854 v
• 2Fe3+ +2e-  2Fe2+
Ered= + 0.771 v
• Ecell = - 0.083 v
» DG = - (2 mol e-)(-0.083 v)(96,500 J/v mol e-)
»
= + 16 kJ
Concentration & Cell EMF
• Nernst Equation
» relationship between DG & concentrations
• DG = DGº + RT ln Q
Q = [prod]x/[react]y
» substitute -nFE for DG
• E = Eº - (RT/nF) ln Q or
• E = Eº - (2.303 RT/nF) log Q
• 2.303 RT/F = 0.0592 v-mol e- at std. temp.
• E = Eº - (0.0592/n) log Q
• Calculate the emf that the following cell
generates when [Mn2+] = 0.10 M & [Al3+] =
1.5 M
2Al + 3Mn2+  2Al3+ + 3Mn
• Eº = + 0.48 v
• E = (+ 0.48 v) - (0.0592 v/ 6) log [(1.5)2/(0.10)3]
• E = + 0.45 v
• when [Mn2+] = 1.5 M & [Al3+] = 0.10 M
• E = (+ 0.48 v) - (0.0592 v/ 6) log [(0.10)2/(1.5)3]
• E = + 0.51 v
Equilibrium Constants
• Remember DG = DGº + RT ln Q, if Q = K,
then DG = 0, therefore -nFE = 0 and
• 0 = Eº - (RT/nF) ln K or
• 0 = Eº - (0.0592/n) log K
• K can be calculated from cell potentials
• log K = nE º/0.0592
• Calculate the equilibrium constant, K, for
2IO3- + 5Cu + 12H+  I2 + 5Cu2+ + 6H2O
• Eº = + 0.858 v
• n = 10 mol e- transferred
• log K = nEº/0.0592
• log K = 145
• K = 1 x 10145
Voltaic Cells
• Lead storage battery
» PbO2 + SO4-2 + 4H+ + 2e-  PbSO4 + H2O
Pb + SO42-  PbSO4 + 2e• Ecell = + 2.041 v
• Dry cell
» NH4+ + 2MnO2 + 2e-  Mn2O3 + 2NH3 + H2O
Zn  Zn2+ + 2e• In an alkaline cell the NH4Cl is replaced with KOH
• Ni-Cd
» NiO2 + 2H2O + 2e-  Ni(OH)2 + 2OH-
Cd + 2OH-  Cd(OH)2 + 2e-
• Fuel cells
» 4e- + O2 + 2H2O  4OH2H2 + 4OH- 
4H2O
Electrolytic Cells
• Redox reactions that are not spontaneous
• Must be driven by an outside source of
electrical energy
• Cathode
» reduction occurs
» by sign convention, is negative
• Anode
» oxidation occurs
» by sign convention, is positive
Quantitative Aspects
• Redox reactions occur in stoichiometric
relationship to the transfer of electrons
• Electrons put into a system through electrical
energy, can be quantized
» Coulomb = quantity of charge passing through
electrical circuit in 1 s at 1 ampere (A) current
• Coulomb = (amp) (seconds)
Problem:
Calculate the mass of Mg formed
upon passage of a current of 60.0 A for a period of
4.00 x 10 3 s.
• MgCl2  Mg + Cl2
» Mg2+ + 2e-  Mg
2Cl-  Cl2 + 2e-
• we are concerned with the reduction
• (60.0 A)(4 x 103s)(1C/1 A-s) = 2.4 x 105 C
• (2.4 x 105 C)(1 mol e-/ 96,500 C) = 2.49 mol e-
• (2.49 mol e-)(1 mol Mg/2 mol e-) = 1.24 mol Mg
• (1.24 mol Mg)(24.3 g/mol) = 30.1 Mg
Electrical Work
• DG = wmax
DG = - nFE
wmax = - nFE
• Max work proportional to potential
•
=
wmax
•J
=
-
n
F
E
(mol) (C/mol) (J/C)
• Electrical work = (watt) (time)
• 1 watt (W) = 1 J/s
• 1 kWh = 3.6 x 106 J
or watt-s = J