Electrochemistry

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Chapter 20
Electrochemistry
2
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

3

• Effect of Concentration
• Nernst Equation
• Equilibrium Constants
• Commercial Voltaic Cells
• Electrolysis
• Quantitative Aspects
• Electrical Work

4
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-
oxidation ½ rxn
oxidation
Cu2(aq) 2e- ? Cu(s)
reduction
reduction ½ rxn
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You must know oxidation states(Review Section
8.10)

• What are the oxidation states of each atom in the
  following
• H2
• CO
• ClO2-
• HC2H3O2

H 0 C 2, O -2 Cl 3, O -2 H 1, C1
3, C2 -3, O -2
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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

oxidation ½ rxn
reduction ½ rxn
x 2
2Fe3 2e- ? 2Fe2 Sn2 2Fe3 ?
Sn4 2Fe2
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• Reactions involving H O in acid
• MnO4- C2O42- ? Mn2 CO2
• write both ½ reactions
• MnO4- ? Mn2
• C2O42- ? CO2
• mass balance (all except H O)
• MnO4- ? Mn2
• C2O42- ? 2CO2
• add H2O H to balance O H
• 8H MnO4- ? Mn2 4H2O
• C2O42- ? 2CO2

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• balance charge by adding electrons
• 5e- 8H MnO4- ? Mn2 4H2O
• C2O42- ? 2CO2 2e-
• balance electrons transferred
• 10e- 16H 2MnO4- ? 2Mn2
  8H2O
• 5C2O42- ? 10CO2 10e-
• add half reactions
• 16H 2MnO4- 5C2O42- ? 10CO2 2Mn2
  8H2O
• check the balance

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• Reactions in base MnO4- CN- ? CNO-
  MnO2
• use exactly the same process
• CN- ? CNO-
• MnO4- ? MnO2

H2O
2H
2e-
3e-
2H2O
4H

• 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

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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

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Cell EMF

• Difference in potential energy of electrons at
  the anode and cathode
• Diff. in potential energy per electrical charge
  measured in volts
• 1 V 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

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• 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

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Problem

• Calculate Eºcell for
• 2Al(s) 3I2(s) ? 2Al3(aq) 6I-(aq)
• Anode 2Al ? 2Al3 6e-
• Cathode 3I2 6e- ? 6I-
• Eºcell E ºred (cathode) - E ºred (anode)
• E ºcell 0.54 V - (-1.66 V)
• E ºcell 2.20 V

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• 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

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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

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• 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- ? Fe Eºred -0.44 V
• MnO4- 8H 5e- ? Mn2 4H2O Eºred 1.51 V

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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- Eºox -0.77 v
• MnO4- 8H 5e- ? Mn2 4H2O E ºred 1.51 v
• E ºox E ºred 0.74 v

Yes
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EMF Free Energy

• If both DG E are a measure of spontaneity, they
  must be related
• DG - nFE
• F is Faradays 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

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• 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

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Concentration Cell EMF

• Nernst Equation
• relationship between DG concentrations
• DG DGº RT ln Q Q prodx/reacty
• 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

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• 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

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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

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• 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

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Voltaic Cells

• Lead storage battery
• PbO2 SO4-2 4H 2e- ? PbSO4 H2OPb
  SO42- ? PbSO4 2e-
• Ecell 2.041 v
• Dry cell
• NH4 2MnO2 2e- ? Mn2O3 2NH3 H2OZn ?
  Zn2 2e-
• In an alkaline cell the NH4Cl is replaced with KOH

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• Ni-Cd
• NiO2 2H2O 2e- ? Ni(OH)2 2OH- Cd 2OH-
  ? Cd(OH)2 2e-
• Fuel cells
• 4e- O2 2H2O ? 4OH- 2H2 4OH- ?
  4H2O

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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

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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)

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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

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Electrical Work

• DG wmax DG - nFE wmax - nFE
• Max work proportional to potential
• wmax - n F E
• J (mol) (C/mol) (J/C)
• Electrical work (watt) (time)
• 1 watt (W) 1 J/s or watt-s J
• 1 kWh 3.6 x 106 J