1. Chapter 20
Electrochemistry

2. Overview Oxidation-Reduction reactions Voltaic Cells
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) e-
oxidation
oxidation ½ rxn
Cu2+(aq) + 2e-  Cu(s)
reduction ½ rxn
reduction

5. 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

6. Balancing Redox Reactions
Mass balance must be observed
e--transfer must be balanced
Simple reactions:
Sn Fe3+  Sn Fe2+
Sn2+  Sn e-
Fe e-  Fe2+
oxidation ½ rxn
x 2
reduction ½ rxn
2Fe e-  2Fe2+
Sn Fe3+  Sn Fe2+

7. Reactions involving H & O in acid:
MnO C2O  Mn CO2
write both ½ reactions
MnO  Mn2+
C2O  CO2
mass balance (all except H & O)
C2O  CO2
add H2O & H+ to balance O & H
8H MnO  Mn H2O

8. check the balance balance charge by adding electrons
5e H MnO  Mn H2O
C2O  CO e-
balance electrons transferred
10e H MnO  2Mn H2O
5C2O  CO e-
add half reactions
16H+ + 2MnO4-+ 5C2O42-  10CO2 + 2Mn H2O
check the balance

9. check balance Reactions in base: MnO4- + CN-  CNO- + MnO2
use exactly the same process
CN-  CNO-
MnO4-  MnO2
H2O +
+ 2H+
+ 2e-
3e- +
4H+ +
+ 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 CN-  3CNO- + 3H2O + 6e-
6e- + 4H2O + 2MnO4-  2MnO2 + 8OH-
3CN- + H2O + 2MnO4-  2MnO2 + 3CNO- +2OH-
check balance

10. 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

12. 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

13. 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) = V

16. Problem: Calculate Eºcell for Anode: 2Al  2Al3+ + 6e-
2Al(s) + 3I2(s)  2Al3+(aq) I-(aq)
Anode: 2Al  2Al e-
Cathode: 3I e-  6I-
Eºcell = E ºred (cathode) - E ºred (anode)
E ºcell = V - (-1.66 V)
E ºcell = V

17. The more positive the E ºcell the more driving force for the reaction
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) I-(aq)
2Al  2Al e- E ºox = V
3I e-  6I- E ºred = V
E ºcell = E ºox + E ºred = 2.20V
The more positive the E ºcell the more driving force for the reaction

18. 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

19. Which is the better oxidizing agent?
NO H e-  NO + 2H2O E ºred V
Ag e-  Ag E ºred V
Cr2O H+ + 6e-  2Cr3+ + H2O E ºred V
Which is the strongest reducing agent?
I e-  2I Eºred V
Fe e-  Fe Eºred V
MnO4- + 8H+ + 5e-  Mn2+ + 4H2O Eºred V

20. 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 = v
MnO4- + 8H+ + 5e-  Mn2+ + 4H2O E ºred = v
E ºox + E ºred = v
Yes

21. 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

22. DG = - (2 mol e-)(-0.083 v)(96,500 J/v mol e-) = + 16 kJ
Calculate the standard free energy change for Hg + 2Fe3+  Hg2+ + 2Fe2+
n = 2 mol electrons transferred
Hg  Hg2+ + 2e- Eox = v
2Fe3+ +2e-  2Fe2+ Ered= v
Ecell = v
DG = - (2 mol e-)( v)(96,500 J/v mol e-)
= + 16 kJ

23. 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 = v-mol e- at std. temp.
E = Eº - (0.0592/n) log Q

24. Calculate the emf that the following cell generates when [Mn2+] = 0
Calculate the emf that the following cell generates when [Mn2+] = 0.10 M & [Al3+] = 1.5 M 2Al + 3Mn2+  2Al3+ + 3Mn
Eº = v
E = ( v) - ( v/ 6) log [(1.5)2/(0.10)3]
E = v
when [Mn2+] = 1.5 M & [Al3+] = 0.10 M
E = ( v) - ( v/ 6) log [(0.10)2/(1.5)3]
E = v

25. 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

26. Calculate the equilibrium constant, K, for 2IO3- + 5Cu + 12H+  I2 + 5Cu2+ + 6H2O
Eº = v
n = 10 mol e- transferred
log K = nEº/0.0592
log K = 145
K = 1 x 10145

27. Voltaic Cells Lead storage battery Dry cell
PbO2 + SO H+ + 2e-  PbSO4 + H2O Pb + SO  PbSO4 + 2e-
Ecell = v
Dry cell
NH4+ + 2MnO2 + 2e-  Mn2O3 + 2NH3 + H2O Zn  Zn2+ + 2e-
In an alkaline cell the NH4Cl is replaced with KOH

28. Ni-Cd
NiO2 + 2H2O + 2e-  Ni(OH)2 + 2OH- Cd + 2OH-  Cd(OH)2 + 2e-
Fuel cells
4e- + O H2O  4OH- 2H2 + 4OH-  H2O

29. 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

30. 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)

31. Problem: Calculate the mass of Mg formed upon passage of a current of 60.0 A for a period of 4.00 x s.
MgCl2  Mg Cl2
Mg e-  Mg 2Cl-  Cl e-
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

32. 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 = x 106 J