1. Unit 8 (Chp 20): Electrochemistry
Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Unit 8 (Chp 20): Electrochemistry
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.

2. LEO says GER A species is oxidized when it loses e– .
A species is reduced when it gains e– .
LEO says GER
oxidized
reduced

3. Balancing Redox Half-Rxns+3 +2
Step 1:
Ox #’s
Zn + FeCl3  ZnCl2 + Fe
OX: Zn  Zn e−
Step 2:
Half, e–’s
RED: 3 e− + Fe+3  Fe
OX: 3 Zn  3 Zn e−
Step 3:
Balance
&
Unite
Overall
RED: 6 e− Fe+3  2 Fe
3 Zn Fe+3  3 Zn Fe

4. (maintain charge balance)
Voltaic Cells
oxidation occurs at the anode.
reduction occurs at the cathode.
RED
CAT OX AN
salt bridge
(maintain charge balance)

5. Voltaic Cells +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

6. 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

7. 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)

8. 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

9. (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 J or kJ
Solved for K :
–∆Go RT
K = e^
NOT on equation sheet

10. – + – + Eo, ΔGo , & K Go = −RT ln K Go = −nFEo K ∆Go = –RT(ln K)
Fav or UNfav
therm.
fav. –
–RT ( + )
–nF( ) +
> 1 =
– =
–nF( ) =
–RT ( 0 )
0 =
= 1
neither
therm.
UNfav. – +
–RT ( – )
–nF( )
< 1 =
+ =

11. 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)

12. Ecell (not Eo) (not 1.0 M) As Q inc↑, E ________. dec to 0 Q = ?
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+

13. recall Faraday’s constant, F = 96,485 C/mol e–
Electrolysis
electrical energy input used to cause an UNfavorable (E = –) REDOX rxn to plate out a solid mass of neutral metal from aq. ions.
on equation sheet
current: charge (C) per second (s). q t
I =
I = current [Amperes (A)]
q = charge [Coulombs (C)]
t = time [seconds (s)]
recall Faraday’s constant, F = 96,485 C/mol e–

14. 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 =

15. Electrolysis = 25.9 g 1 F = 96,485 C 1 mol e–
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 =

16. 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 

17. Electrolysis Aqueous (aq) Salts
RED:
K+ + e–  K
Ered =
–2.92 V P 
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– 

18. K = e^ E  = Ered (CAT) + Eox (AN) Go = −nFEo –∆Go RT Go = −RT ln K
RED OX
“standard”
cell potential (Eo)
free energy change (∆Go)
Go = −nFEo
equilibrium constant (K)
UNITS!
∆Go & R
J or kJ
–∆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 =