1. Voltaic (or Galvanic) Cells
In a voltaic (or galvanic) cell, e– transfer
occurs via an external pathway that
links the reactants.
e.g.,
a battery
anorexic ox
electrodes: the two solid metals in a
voltaic/galvanic cell
anode
cathode
fat,
red
cat
-- oxidation occurs here
-- reduction occurs here
-- written w/a (–) sign
-- written w/a (+) sign

2. WHY do the e– go the way they do?
Consider a solution of Zn(NO3)2(aq) and Cu(NO3)2(aq)
with electrodes as shown… e– e–
Zn anode
Cu cathode
NO3–
Na+
salt bridge containing
electrolyte
(e.g., NaNO3)
in a porous gel
Zn2+
Cu2+
Zn2+
NO3–
Cu2+
NO3–
(needed to neutralize
both solutions)
-- Zn anode dissolves into sol’n
-- Cu2+ plates out as Cu on the cathode
WHY do the e– go the way they do?

3. > Cell EMF PE of anode’s e– PE of cathode’s e– Thus…
anode’s e–s spontaneously flow
towards cathode, if given a path.
Difference in
is important; PE
charge
# of charges is NOT.
Voltage (or “potential”)
difference is also called the
electromotive force (emf).

4. Eocell = Eored,cath – Eored,an
For a particular cell, (i.e., a particular anode and
cathode), the cell’s emf is written Ecell and is called
the cell potential. --
it is + for spontaneous cell reactions --
it depends on materials, conc., and temp.
-- standard emfs occur at 25oC
-- To calculate Ecell, look up tabulated
standard reduction potentials for
each half-cell…
e.g., Ag+(aq) + e–  Ag(s)
Eored = V
(Look it up!)
…and then use the equation:
Eocell = Eored,cath – Eored,an

5. The reference point for reduction potentials is the
standard hydrogen electrode (SHE):
2 H+(aq, 1 M) e–  H2(g, 1 atm)
Eored = 0 V
This is analogous to the arbitrary ref. line when calc. PEg in physics.
V = J C
Multiples of coefficients don’t affect Eored.
e.g.,
Zn2+(aq, 1 M) e–  Zn(s)
2 Zn2+(aq, 1 M) e–  2 Zn(s)
(Eored = –0.76 V)
Because the same chemicals are
used for each, AAA-, AA-, C-, and
D-batteries all have an emf of 1.5 V.

6. A 9-volt battery is six AAA batteries wired together in series.

7. Eocell = Eored,cath – Eored,an
For Cr(s) + Cu2+(aq)  Cr2+(aq) + Cu(s), Eocell is
measured to be 1.25 V. Given that Eored for Cr2+ to Cr
is –0.91 V, find Eored for the reduction of Cu2+ to Cu.
Eocell = Eored,cath – Eored,an
1.25 V =
Eored, cath
–0.91 V –
–0.91 V
(Cu2+  Cu)
Cu2+ solution
Cu metal
Cr metal
Cr2+ solution
Eo =
Cu2+  Cu
+0.34 V

8. A galvanic cell has half-rxns:
(a) Al3+(aq) e–  Al(s) Eored = –1.66 V
(b) Ba2+(aq) e–  Ba(s) Eored = –2.90 V
Calculate Eocell and write the balanced equation.
For a galvanic cell, Eo must be > 0.
Thus, (a) represents the cathode
and (b) represents the anode.
Eocell = Eored,cath – Eored,an
Eocell =
–1.66 V – –2.90 V
= V 2
Al3+(aq) Ba(s)  Al(s) Ba2+(aq) 3 2 3

9. For a half-reaction, the more (+) the Eored value, the
greater the tendency for that reaction to “go” in that
direction (i.e., reduction). Strongest oxidizer is… F2
F2(g) e–  2 F–(aq)
Eored = V
Other strong oxidizers are… the other halogens
(Cl2, Br2, I2)
and oxyanions in which the central atom has a
large ___ charge.
(+)
e.g.,
MnO4–, Cr2O72–, ClO3–, etc.
Strong oxidizers LOVE their own e–s
(and they want everybody else’s, too).

10. Poorest oxidizer is… Li.
Li+(aq) + e–  Li(s) Eored = –3.05 V
(–) sign indicates poor tendency to “go” in this direction, but large magnitude (i.e., 3.05 V) shows strong tendency to “go” in other direction
(i.e., oxidation).
Poor oxidizers HATE their own e–s
(and have no interest in accepting
anyone else’s e– s, either.)
Lithium batteries take
advantage of lithium’s
strong tendency
to BE oxidized
(i.e., to REDUCE other stuff.)

11. In comparing the reduction potentials of two half-reactions, consider the scale shown. The “higher-up” reaction is the reduction half-cell; the “lower-down” reaction is the oxidation half-cell.
0 V
(+) V
(–) V A–
Red. A–
Ox. C–
Red. B–
Red. B–
Ox. C–
Ox.

12. Activity Series for MetalsLi Rb K Ba Sr Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb H2 Sb Bi Cu Hg Ag Pt Au
“e– haters”
“e– lovers”
Activity Series for Metals
Spontaneity of Redox Reactions
Eocell = Eored,cath – Eored,an
(same equation as before)
standard
conditions
nonstandard
conditions
If Eo (or E, or emf) is +…spontaneous.
–…nonspontaneous.
-- The Activity Series is based on
standard reduction potentials.

13. ( ) Relationship between E and DG… DG = –nFE In standard states…
DGo = –nFEo
n = # of mol of transferred e–
96,500 J
V . mol e–
F = Faraday’s constant = C
mol e–
( )
Michael Faraday
Josiah Gibbs
(1839–1903)
(1791–1867)

14. Eocell = Eored,cath – Eored,an
For…
5 Fe2+ + MnO4– + 8 H+  5 Fe3+ + Mn H2O
(a) What is n?
(b) Find DGo.
(a) n = 5
(b)
Fe3+ + e–  Fe Eored = 0.77 V
MnO4– + 8 H+ + 5 e–  Mn H2O
Eored = 1.51 V
Eocell = Eored,cath – Eored,an
Eocell =
1.51 V – 0.77 V =
0.74 V
DGo = –nFEo
= –5 mol e– (0.74 V)
96,500 J
V . mol e– ( )
= –357 kJ
(SPONTANEOUS)

15. Effect of Concentration on Cell EMF
Walther Nernst
(1864–1941)
Cell emf drops gradually due to
changing concentrations of
reactants and products. When
emf = 0 V, cell is “dead.”
Nernst equation:
At 25oC (298 K):

16. Fe(s) + Cd2+(aq)  Cd(s) + Fe2+(aq)
Find emf at 25oC when [Cd2+] = M
and [Fe2+] = 2.0 M.
Eo = V
Fe2+(aq) + 2 e–  Fe(s) Eored = –0.44 V
Cd2+(aq) + 2 e–  Cd(s) Eored = –0.40 V
nonspont.
= –0.01 V
If [Cd2+] = 2.0 M and [Fe2+] = M…
spont.
= V

17. At equilibrium, DG = ___, E = ___ and Q = ___. The
Nernst equation can be rearranged to give the
relationship between K and Eo. K K – =
log K =
At 25oC (298 K):
–DGo
5706 =
log K = n Eo

18. -- spontaneous redox reactions in which a metal
Corrosion
-- spontaneous redox reactions in which a metal
reacts with some substance in its environment
to form an unwanted compound
-- For some metals (e.g., Al and Mg)…
a protective
oxide coating (Al2O3, MgO) prevents further
corrosion of the underlying substrate.
-- Galvanized iron is coated
with a protective
layer of ____.
galvanized electrical conduit
zinc

19. Offshore oil rigs are often cathodically protected.
-- cathodic protection:
protecting a metal by making
it the cathode in an
electrochemical cell
 oxidized metal is called the ______________
sacrificial anode
EX.
Mg is used in the cathodic protection of
underground Fe pipe.
The Mg has to be
replaced
every so
often. e– Fe Mg
Mg2+
Offshore oil rigs are often
cathodically protected.

20. using an outside source of electrical energy to cause nonspontaneous
Electrolysis:
using an outside source of electrical
energy to cause nonspontaneous
redox reactions to “go”
-- Electrolysis occurs in electrolytic cells, which consist
of two electrodes in a molten salt or a solution.
Ag-plated
brass trumpet
reduction at cathode;
oxidation at anode
Cr-plated Fe pipe
Ni-plated
steel rotor
Cu-plated Pb shot

21. Consider plating chromium onto an iron pipe:
Fe2+/Fe  Eored = –0.44 V
Cr3+/Cr  Eored = –0.74 V
For Cr to plate out on the Fe pipe, the equation is:
2 Cr Fe  3 Fe Cr
and Eo (for the galvanic cell) would be:
Cathode – Anode
= “Cr3+/Cr”– “Fe2+/Fe”
= –0.74 V – –0.44 V = –0.30 V
The rxn is nonspontaneous in the forward direction. A
galvanic cell would run opposite the way we want. We
need to put an external “oomph” into the rxn to make
it go…which is the definition of an electrolytic cell.

22. electroplating:
using electrolysis to deposit a thin
layer of one metal onto another
Fe cathode to be plated with Ni
Ni anode
Ni2+(aq)
Vext
For electrolysis, current
in amperes (1 A = 1 C/s)
is passed through the
liquid. One mole of
transferred e– carries
with it 96,500 C of charge.
(Faraday’s constant)
1 e– =
1.60 x 10–19 C
6.02 x 1023 e– =
96,500 C ~

23. For how long must a 50.0 A current be passed through
molten BaBr2 in order to produce 500. g of barium?
**To solve these problems, use unit cancellation.
500. g Ba
= 14,100 s
Hall-Heroult process for purifying Al
from Al2O3 and Na3AlF6 (1886)
= h
Paul Heroult
(1863–1914)
Charles Martin Hall