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– .

3. Oxidation Numbers .

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:

7. Balance RedOx by Half-Rxns
5 Steps:
Balance RedOx by Half-Rxns 1:
Ox #’s +3 +2
Zn + Fe+3  Zn Fe
comp–diss–cross –net– balanced? 2:
Half rxns
RED: Fe+3  Fe
2 ( )
3 e− +
OX: Zn  Zn+2
3 ( )
+ 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. mass % (gsample /gtotal)
Redox Titration
The analytical technique used to calculate the moles in an unknown soln.
buret
molarity (mol/L)
molar mass (g/mol)
mass % (gsample /gtotal)
titrant
mol X = mol Y
known vol. (V)
known conc. (M)
MXVX = MYVY
x y
analyte
known vol. (V)
unknown conc. (M)
xX + yY  X– + Y+
(red) (ox)
(or moles)

14. Redox Titration Consider a titration of FeCl2 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 MXVX = MYVY x y mol X = mol Y equivalence point,(Veq):
equal stoichiometric
amounts react completely
MXVX = MYVY
x y
2 MnO4– + 5 H2O2 
MXVX = MYVY
mol X = mol Y
end point:
permanently changes color

16. Redox Titration MXVX = MYVY Titration of H2O2 with MnO4– : 16.8 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
mol X = mol Y
MXVX = MYVY
x y
( L)(0.124 M) = MY( L)
MY = 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. Standard Reduction Potentials
(Ered) 
SRP’s measured, tabulated.
(likely reduced) 
defined as Ered =
(SHE)
(NOT likely reduced)
(oxidized)

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 = Ecathode - Eanode
E = Ered + Eox or
BUT…

26. E =Ered of cathode – Ered of anode 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 o o o
E = Ered + Eox
E =Ered of cathode – Ered of anode
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 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
Ecell = Ecathode - Eanode 
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

29. Standard Cell Potential (Eo)
Ecell under standard conditions (1.0 M) using
Ered’s (SRP’s) is calculated with the equation: 
Ecell 
= Ered(CAT) + Eox(AN)
RED OX
NOT on equation sheet 
(–Ered)

30. 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)

31. Standard Cell Potential (Eo)
= Ered (CAT) + Eox (AN)
RED OX 
(–Ered)
+0.34 
greater difference in Ered’s, greater voltage (∆V)
(potential difference)
RED OX
Ecell
= (0.34) + (+0.76)  
Ecell = V
–0.76

32. Electrolysis
Note that the electric current here drives the direction of the reaction and not SRP

33. 
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

34. 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

35. 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–

36. Electrolysis of Molten (l ) Salts
RED:
Na+ + e–  Na 
Ered =
–2.71 V P
What drives the reaction is electric current and not SRP 
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 