1. Unit 12 (Chp 20): Electrochemistry (E, ∆G, K)
Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Unit 12 (Chp 20): Electrochemistry (E, ∆G, K)
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.

2. Electrochemical Reactions
electrons are transferred from one species to another (redox).
assign oxidation numbers to identify which loses e– and which gains e– .

4. Oxidation and Reduction
A species is oxidized when it loses e– .
Zn loses 2 e– to from Zn metal to the Zn2+ ion.
A species is reduced when it gains e– .
H+ ions gain 1 e– and combine to form H2 .
Video Clips: OxidationReduction1 and OxidationReduction2 from Brown text resources Chp 20 emedia_Library
LEO says GER
2 video clips

5. Assigning Oxidation Numbers
All elements are 0. (all compounds are 0)
Monatomic ion is its charge. (Ex. Na+ ion)
Most nonmetals tend to be negative, but some are positive in certain compounds or ions. (Ex. SO3)
O is −2 (but in peroxide ion is −1 [ O2–2 ]
H is +1 with nonmetals (but −1 with metals)
F is −1 (always)
other halogens usually −1,
but are positive with O
Ex. ClO3– or NO3– or SO42–
HW p. 890 #11,16ab

6. Balancing Redox Reactions
by…
Half-Reactions
“O, he balance you.”
O H E BALANCE U
5 Steps:
comp – diss – cross – net – bal

7. Balance RedOx by Half-Rxns
5 Steps:
Balance RedOx by Half-Rxns +3 +2 1:
Ox #’s
Zn + Fe(NO3)3 
Zn(NO3)2 + Fe
(spectator ion: NO3–)
comp–diss–cross –net– balanced? 2:
Half rxns
RED: Fe+3  Fe
3 e− +
OX: Zn  Zn+2
+ 2 e− 3:
Electrons
RED: 6 e− + 2 Fe+3  2 Fe 4:
Bal.
same e–’s
OX: Zn  3 Zn e− 5:
Unite
3 Zn + 2 Fe+3  3 Zn Fe
(Balanced Overall)

8. Readily Oxidized or Reduced?
WS Aq. Soln’s & Chm Rxns II
Which region and group in the periodic table
shown contains elements that are:
most readily oxidized?
most readily reduced?
metals least readily oxidized? A
1 (alkali) D
17 (halogens) C
Ag, Au, Pt, Hg

9. WS Aq. Soln’s & Chm Rxns II #6
5 Steps:
WS Aq. Soln’s & Chm Rxns II #6 +2 +2 1:
Ox #’s
Fe + Pb(NO3)2 
Fe(NO3)2 + Pb
(NO3– is a spectator ion) 2:
Half rxns
RED: Pb+2  Pb
2 e− +
OX: Fe  Fe+2
+ 2 e− 3:
Electrons
RED: 2 e− + Pb+2  Pb 4:
Bal.
same e–’s
OX: Fe  Fe e− 5:
Unite
Fe + Pb+2  Fe+2 + Pb
(Balanced Overall)

10. WS Aq. Soln’s & Chm Rxns II #10
5 Steps:
WS Aq. Soln’s & Chm Rxns II #10 1:
Ox #’s +1 –1 +1 –1
Cl KI 
KCl I2
(K+ is a spectator ion) 2:
Half rxns
RED: Cl2  Cl–
2 e−
OX: I–  I2 e− 3:
Electrons
RED: e− + Cl2  2 Cl– 4:
Bal.
same e–’s
OX: I–  I e− 5:
Unite
Cl I–  2 Cl– + I2
(Balanced Overall)

11. WS Aq. Soln’s & Chm Rxns II #13
5 Steps:
WS Aq. Soln’s & Chm Rxns II #13 +1 –2 +2 –2 +1 1:
Ox #’s
Ca + H2O 
Ca(OH)2 + H2 2:
Half rxns
RED: H2O  OH– + H2
2 e−
OX: Ca  Ca+2
+ 2 e− 3:
Electrons
RED: 2 e− + 2 H2O  2 OH– + H2 4:
Bal.
same e–’s
OX: Ca  Ca e− 5:
Unite
Ca + 2 H2O  Ca OH– + H2
(Balanced Overall)

12. WS Aq. Soln’s & Chm Rxns II #23
5 Steps:
WS Aq. Soln’s & Chm Rxns II #23 +1 +2 1:
Ox #’s
Zn + H2SO4 
ZnSO4 + H2
(SO4–2 is a spectator ion) 2:
Half rxns
RED: H+  H2
2 e− + 2
OX: Zn  Zn+2
+ 2 e− 3:
Electrons
RED: 2 e− + 2 H+  2 H2 4:
Bal.
same e–’s
OX: Zn  Zn e− 5:
Unite
Zn + 2 H+  Zn+2 + H2
(Balanced Overall)

13. Redox Titration
The analytical technique used to calculate the moles in an unknown soln.
mass % (g /gtotal)
molarity (mol/L)
molar mass (g/mol)
buret
titrant
known vol. (V)
known conc. (M)
Stoich: L X  mol X  mol Y
analyte
M X
known vol. (V)
unknown conc. (M)
(or moles)

14. Redox Titration Consider a titration of H2O2 with KMnO4 :
2 MnO4− + 5 H2O2 + 6 H+  2 Mn O2 + 8 H2O +7 –1 +2
colorless
purple
excess MnO4– titrant
limitedMnO4– titrant
Why is it colorless?
Why is it purple?

15. Redox Titration equivalence point,(Veq): end point:
equal stoichiometric
amounts react completely
2 MnO4– + 5 H2O2 
mol X = 5/2 mol Y
end point:
permanently changes color

16. Redox Titration Titration of H2O2 with MnO4– : 16.8 mL 0.124 M 10.0 mL
HW p. 163 #103
Titration of H2O2 with MnO4– :
2 MnO4− + 5 H2O2 + 6 H+  2 Mn O2 + 8 H2O +7 –1 +2
16.8 mL
0.124 M
10.0 mL
? M
0.124 mol MnO4–
1 L MnO4–
5 mol H2O2
2 mol MnO4–
L MnO4– x x
= mol H2O2
mol H2O2
L H2O2 =
0.521 M H2O2

17. Voltaic Cells favorable redox reactions transfer e–’s release free
energy (–ΔG)
Zn + Cu+2  Zn2+ + Cu
Zn2+ 2+ 2+ 2+ 2+ Cu 2+ 2+ Zn Zn 2+ 2+ 2+ 2+
Cu2+
Cu2+

18. Voltaic Cells
the energy can be used to do work if the electrons flow through an external device.
We call this a voltaic cell.
(or galvanic cell, or electrochemical cell)

19. _________ at the cathode
Oxidation
________ at the anode
_________ at the cathode
Reduction
Voltaic Cells
RED
CAT AN OX
Zn  Zn e−
2 e− + Cu+2  Cu

20. salt bridge maintains charge balance
Voltaic Cells
As 1 e– flows from AN to CAT, the charges in each half-cell would be unbalanced and e– flow stopped.
(Anions to anode)
(Cations to cathode)
RED
CAT AN OX
salt bridge maintains charge balance

21. Voltaic Cells +cations form and dissolve at AN
HW p. 891 #22, 24
+cations form and dissolve at AN
(e– ’s flow from AN to CAT)
e– ’s reduce +cations to deposit solid metal on the CAT
animation: VoltaicCopperZincCell
Zn(s)  Zn+2 + 2e−
2e− + Cu+2  Cu(s) Zn Zn 2+
Zn2+ 2+
Cu2+
Cu2+ 2+ Cu 2+ Cu

22. Electric Potential Energy
Water only flows favorably in one direction.
(battery)
e–’s only favorably flow from:
higher to lower potential energy.
e– flow

23. Cell Potential (E) e– flow J 1 V = 1 C
The potential difference between anode and cathode in a cell is called the cell potential, Ecell (or E),
measured in volts (V).
(battery)
e–’s flow from:
higher to lower potential energy.
e– flow
1 V = 1 J C
(joules)
(coulomb)
(volts)
energy
charge
potential =

24. Cell Potential (E) OX: RED: Zn  Zn2+ + 2 e– Cu2+ + 2 e–  Cu
1.10 V
HOW?

25. BUT… Cell Potential (E) Ered = ? RED: Cu2+ + 2 e–  Cu
OX: Zn  Zn e–
Eox = ?
Overall: Zn + Cu+2  Zn2+ + Cu
Ecell = 1.10 V
E = Ered + Eox
BUT…

26. Standard Reduction Potentials
Cell Potential (E) o
Ered = ?
RED: Cu e–  Cu o
OX: Zn  Zn e–
Eox = ? o
Overall: Zn + Cu+2  Zn2+ + Cu
Ecell = 1.10 V o o o
E = Ered + Eox
Standard Cell Potential (E )
from
Standard Reduction Potentials
(Ered or SRP’s) o … o

27. Standard Hydrogen Electrode
(SHE) 
2 H+(aq, 1M) + 2 e−  H2(g, 1 atm)
Ered = 0 V
(defined)
1 atm
H2(g) or Zn Cu
SRP’s are measured against a SHE (0.00 V)

28. Standard Reduction Potentials
(Ered) 
SRP’s measured, tabulated.
(likely reduced) 
defined as Ered =
(SHE)
(NOT likely reduced)
(oxidized)

29. Standard Cell Potential (Eo)
Ered = ?
RED: Cu e–  Cu
OX: Zn  Zn e–
Eox = ?
Overall: Zn + Cu+2  Zn2+ + Cu
Ecell = 1.10 V
Ecell = Ered + Eox 
RED:
Cu e–  Cu
Ered =
+0.34 V 
OX:
Zn  Zn e–
Eox = ?
+0.76 V 
Ered =
–0.76 V
Ecell
= Ered(0.34) + Eox(????)   
+0.76
Ecell = V
(–Ered) 
HOW?
RED OX
Cu2+ Zn

30. Standard Cell Potential (Eo)
Ecell under standard conditions (1.0 M) using
Ered’s (SRP’s) is calculated with the equation: 
Ecell 
= Ered + Eox
CAT AN
NOT on equation sheet 
(–Ered)
(multiplying coefficients does not affect Eo)
1 V = 1 J C
Eo is potential energy per charge (J/C), amount of each substance does NOT matter.
(more C, more J)
(but J/C is unchanged)

31. Likely Oxidized or Reduced? (based on Ered) 
reduced easily (highest Ered)
F2(g) e–  2 F–(aq)
2 H+(aq) e–  H2(g)
Li+(aq) + e–  Li(s)
+2.87 V …
0 V
–3.05 V
(most nonmetals) (halogens)
Reduced Easily
2 H+(aq) e–  H2(g) V
Oxidized Easily
(most metals) (alkali) 
oxidized easily (lowest Ered)

32. Standard Cell Potential (Eo)
= Ered + Eox
CAT AN 
(–Ered)
+0.34 
greater difference in Ered’s, greater voltage (∆V)
(potential difference)
RED OX
Ecell
= (0.34) + (+0.76)  
Ecell = V
HW p. 892
#26,30,32,34,36,40abc
–0.76

33. Potential (Eo) & Free Energy (DGo)
Go for a redox reaction can be found by using the equation:
Go = −nFEo
n : moles of e– transferred
F : Faraday’s constant (96,485)
1 F = 96,485 C/mol e–
= 96,485 J/V∙mol e–
(in J)
on equation sheet
on equation sheet
1 V = 1 J C

34. (Eo) & (DGo) & ______________
(K) Equilibrium
G = −nFE
G = −RT ln K  
on equation sheet 
on equation sheet
–∆Go RT
= ln K
UNITS!!!
∆Go & R
both in
kJ or J
Solved for K :
–∆Go RT
K = e^
NOT on equation sheet

35. – + – + Eo, ΔGo , & K Go = −RT ln K Go = −nFEo K ∆Go = –RT(ln K)
HW p. 893 #48a,50,54
Go = −RT ln K
Go = −nFEo K
∆Go = –RT(ln K)
∆Go = –nF(Eo)
Fav or UNfav
therm.
fav. –
–RT ( + )
–nF( ) +
> 1 =
– =
–nF( ) =
–RT ( 0 )
0 =
= 1
neither
therm.
UNfav. – +
–RT ( – )
–nF( )
< 1 =
+ =

36. Ecell (nonstandard) (not Eo)
[P]y
[R]x
Q =
If Eo, then Q = __
x R  y P
Q = 1
[1.0]y
[1.0]x
NON-standard conditions :
[R] or [P] ≠ 1.0 M so Q ≠ 1
(if Q < 1 then E > Eo)
[R] & [P] = 1.0 M so Q = 1, & Eo= +(fav) but…
…as rxn proceeds  , [R] & [P], and…
Q > 1 so E < Eo until Q = K and E = 0
Q ≠ 1
or (if Q > 1 then E < Eo)
(Eo unchanged)
(equilibrium)
(“dead” cell, 0.00 V)

37. Example 1: Ecell (nonstandard) (not Eo)
System initially under standard conditions (1.0 M):
Add 2.0 M AgNO3, [Ag+] inc., [Ag+] > 1.0 M
Q < 1 (more reactant concentration)
E increases (E > Eo)
Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s)
(inc.)
[Cu2+]
[Ag+]2
Q =
Q =
(dec.)
System initially under standard conditions (1.0 M):
Add NaCl, [Ag+] dec. as AgCl(s), [Ag+] < 1.0 M
Q > 1 (less reactant concentration)
E decreases (E < Eo)

38. Example 2: Ecell (nonstandard) (not Eo)
System initially under standard conditions (1.0 M):
Add 2.0 M CuCl2, [Cu2+] inc., [Cu2+] > 1.0 M
Q > 1 (more product concentration)
E decreases (E < Eo)
Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s)
(inc.)
[Cu2+]
[Ag+]2
Q =
(dec.)
System initially under standard conditions (1.0 M):
Add H2O to Cu/Cu2+, [Cu2+] dec., [Cu2+] < 1.0 M
Q < 1 (less product concentration)
E increases (E > Eo)

39. Ecell (not Eo) (not 1.0 M) As Q inc↑, E ________. dec to 0 Q = ?
HW p. 894 #58, 60
2.00 V
As Q inc↑, E ________.
dec to 0 Al
Q = ?
___ + ___  ___ + ___ 2 Al 3
Cu2+ 2
Al3+ 3 Cu
[Al3+]2
[Cu2+]3
Q =
Al3+

40. Concentration Cells (E , not Eo)
0.00 V
+0.09 V
same electrodes, diff. conc’s Ni Ni Ni
Q < 1 AN OX
RED
CAT
[Ni2+]
Q = Ni
0.5 M Ni2+
0.5 M Ni2+
0.001 M Ni2+
1.0 M Ni2+
Ni + Ni+2  Ni2+ + Ni
Ni + Ni+2  Ni2+ + Ni
[0.5]
Q =
Q = 1
[0.001]
[1.0]
Q =
Q < 1
Eo = 0
E > Eo
Eo = 0
Eo = ( ) + ( ) –

41. The 1.0 M Ni2+ must decrease concentration, and
HW p. 894
#63
+0.09 V
Q =
[Ni2+]
Q =
[0.001]
[1.0]
Ni + Ni+2  Ni2+ + Ni
Q < 1
E > Eo
0.001 M Ni2+
1.0 M Ni2+
Eo = 0
RED:
Ni2+(1.0M) + 2e–  Ni(s)
OX:
Ni(s)  Ni2+(0.001M) + 2e–
Net Ionic:
Ni2+(1.0M)  Ni2+(0.001M) E = V
The 1.0 M Ni2+ must decrease concentration, and
the M Ni2+ must increase concentration.

42. Summary K = e^ E  = Ered + Eox  (–Ered) Go = −nFEo –∆Go RT
WS Electro II #1-6
“standard”
cell potential (Eo) from (Eored’s or SRPs) E 
= Ered + Eox
CAT AN 
(–Ered)
free energy change (∆Go) from Eo & n
Go = −nFEo
equilibrium constant (K) from ∆Go (from Eo & n)
UNITS!
∆Go & R
J or kJ
–∆Go RT
K = e^
Go = −RT ln K
( Q = K , E = 0 )
(Q = 1 , E = Eo)
non-“standard”
cell potential (E)
(≠ 1.0 M) from Eo & Q
[P]y
[R]x
Q =
(Q > 1 , E < Eo)
(Q < 1 , E > Eo)

43. Electrolysis

44. 
E = –4.07 V 
E = (–2.71) + (–1.36)  
Ered(Cl2)
= V
Ered(Na+)
= –2.71 V
Electrolytic Cell
OX of anions at AN
RED of cations at CAT

45. Electrolysis
electrical energy input used to cause an UNfavorable (E = –) REDOX rxn to plate out a solid mass of neutral metal from aq. ions.
Electrolytic Cell:
NOT electrochemical (voltaic/galvanic) cell b/c current is applied.
Can calculate:
time(s) to plate given mass
mass (g) plated in given time
e– (mol) or charge (C) transferred

46. recall Faraday’s constant, F = 96,485 C/mol e–
Electrolysis
current: amount of charge passing through an area per second. q t
I =
on equation sheet
I = current [Amperes (A)]
q = charge [Coulombs (C)]
t = time [seconds (s)]
recall Faraday’s constant, F = 96,485 C/mol e–

47. Electrolysis Faraday’s law is used to calculate the mass
of metal produced in electrolytic cells.
If I (current, in A or C/s) and t (in s) are known,
then moles or grams of substance can be
determined.
(use the UNITS as a guide)
(t)(I)(F–1)(mol-ratio–1)(M) = m
t (time)
m (mass)
? C
1 s
1 mol e–
96,485 C
_?_ g X
1 mol X
1 mol X
? mol e–
? s x x x x =
____ g X
time
current F
mol-to-mol ratio
molar mass
mass
(mol e– transferred per 1 mol X)

48. Electrolysis 60 s 1 min 0.987 C 1 s 12.4 min x x = 734 C 1 mol e–
Examples: (shorter)
(a) How much charge passes through the cross-section of a wire with A for 12.4 min?
q = ? C
current
time
60 s
1 min
0.987 C
1 s
12.4 min x x =
734 C
(b) How many mol of e– passed through?
1 mol e–
96,485 C
734 C x =
7.61 x 10–3 mol e–

49. Electrolysis Examples: (longer) t = ? s
(c) How long will it take to plate out 0.61 g Cu(s) from a solution of Cu2+ ions with 2.5 amps?
t = ? s
mass
current
1 mol Cu
63.55 g
2 mol e–
1 mol Cu2+
96,485 C
1 mol e–
1 s
2.5 C
0.61 g x x x x =
741 s
(d) How many grams of Ag(s) will be produced when 21.0 A flows through a solution of Ag+ ions for 45.0 min?
current
time
60 s
1 min
21.0 C
1 s
1 mol e–
96,485 C
1 mol Ag+
1 mol e– g
1 mol Ag
63.4 g
45.0 min x x x x x =

50. Electrolysis = 25.9 g 1 F = 96,485 C 1 mol e–
WS Electro II #12,16
Example: (MC no CALCULATOR)
(e) What mass of Pb(s) will be deposited by passing 0.50 F through a solution of Pb4+ ?
1 mol Pb4+
4 mol e–
207.2 g Pb
1 mol Pb
0.50 mol e– x x =
25.9 g
1 F = 96,485 C
1 mol e–
96,485 C
1 F
1 mol e–
96,485 C
1 mol Pb4+
4 mol e–
207.2 g Pb
1 mol Pb
0.50 F x x x x =

51. Electrolysis of Molten (l) SaltsP 
RED:
Na+ + e–  Na
Ered =
–2.71 V 
OX:
2 Cl–  Cl e–
Eox =
–1.36 V
Cl e–  2 Cl–
Na+
Cl– 
Ered =
+1.36 V
NaCl(l)
(molten)
Ecell
= (–2.71) + (–1.36) 
Ecell = –4.07 V 

52. Electrolysis Aqueous (aq) Salts P
RED:
Zn e–  Zn
Ered =
–0.76 V
2 Zn H2O  Zn + O2 + 4 H+ 
Ered =
–0.83 V
2 H2O + 2 e–  H2 + 2 OH–
H2O 
Eox =
OX:
2 Cl–  Cl e–
–1.36 V
Zn2+
Cl–
ZnCl2(aq) 
Eox =
–1.23 V 
RED: highest Ered
OX: highest Eox
2 H2O  O2 + 4 H+ + 4 e– 

53. Electrolysis Aqueous (aq) Salts
RED:
K+ + e–  K
Ered =
–2.92 V P 
2 H2O + 2 I– +  H2 + 2 OH– + I2
Ered =
–0.83 V
2 H2O + 2 e–  H2 + 2 OH–
H2O 
Eox =
OX:
2 I–  I e–
–0.53 V K+ I–
KI(aq) 
Eox =
–1.23 V 
RED: highest Ered
OX: highest Eox
2 H2O  O2 + 4 H+ + 4 e– 
WS Electro II #9,10

54. Hydrogen Fuel Cells
(product)
(reactant)
(reactant)

55. Reaction:
2 H2 + O2  2 H2O

56. 2H2(g) O2(g) 2H2O(l)   Ered = 0.00 V Ered = 1.23 V 1.23 V
2 H2  4 H+ + 4 e–
O2 + 4 H+ + 4 e–  2 H2O  
Ered = 0.00 V
Ered = 1.23 V
1.23 V
How do we get the H2 fuel?
2 H2 + O2  2 H2O
2H2(g)
O2(g) e–
fuel
ANOX
RED
CAT e–
Make H2 : electrolysis of H2O
E = –1.23 V e– e–
2H2O(l)
PEM

57. K = e^ E  = Ered + Eox  (–Ered) Go = −nFEo –∆Go RT Go = −RT ln K
CAT AN
“standard”
cell potential (Eo) 
(–Ered)
free energy change (∆Go)
Go = −nFEo
UNITS!
∆Go & R
J or kJ
equilibrium constant (K)
–∆Go RT
Go = −RT ln K
K = e^
(Q = K , E = 0)
(Q = 1 , E = Eo)
non-“standard”
cell potential (E)
(≠ 1.0 M) from Eo & Q
[P]y
[R]x
Q =
(Q > 1 , E < Eo)
(Q < 1 , E > Eo)
q (C)
t (s)
I (A) =
electrolysis (E = –)
(calculate: g, s, etc.)
96,485 C
1 mol e–
1 F =