A galvanic cell is made from two half cells. In the first, a
platinum electrode is immersed in a solution at pH = 2.00 that
is 0.100 M in both MnO4-(aq) and Mn2+(aq). In the second, a
zinc electrode is immersed in a 0.0100 M solution of Zn(NO3)2.
Calculate the cell voltage at 25oC
What is the overall cell reaction?
Diagram the galvanic cell
Ion-Selective Electrodes
pH or concentration of ions can be measured by using an
electrode that responds selectively to only one species of
ion.
In a pH meter, one electrode is sensitive to the H3O+(aq)
concentration, and the other electrode serves as a
reference.
A calomel electrode has a reduction half reaction
Hg2Cl2 (s) + 2 e- -> 2 Hg(l) + 2 Cl- (aq)
Eo = +0.27 V
When combined with the H+(aq)/H2(g) electrode, the overall
cell reaction is:
Hg2Cl2 (s) + H2 (g) -> 2 H+ (aq) + 2 Hg(l) + 2 Cl- (aq)
Q = [H+(aq)]2 [Cl- (aq)]2 / PH2
If PH2 is held at 1 atm then Q = [H+(aq)]2 [Cl- (aq)]2
DE = DEo - (RT/ n F ) ln [H+(aq)]2 [Cl- (aq)]2
The [Cl- (aq)] is held constant since the calomel electrode
consists of a saturated solution of KCl.
DE depends only on [H+(aq)].
Other electrodes are selectively sensitive to ions such as
Ca2+, NH4+, Na+, S2-.
A galvanic cell is constructed of a Cu electrode dipped into a
1 M solution of Cu2+ and a hydrogen gas electrode (PH2 = 1
atm) in a solution of unknown pH. The potential measured is
0.573 V at 25oC. What is the pH of the solution?
Cu2+(aq) + 2e-  Cu(s)
Eo = 0.337 V
2 H+(aq) + 2e-  H2 (g)
Eo = 0 V
Cell reaction
Cu2+(aq) + H2  Cu(s) + 2 H+(aq)
DEo = 0.337 V
DE = DEo - (R T / n F) ln Q
DE = DEo - (0.025693 / n) ln Q
= DEo - (0.025693 / n) 2.303 log Q = DEo - (0.0.05916 / n) log Q
Q = [H+(aq) ]2 / ([Cu2+(aq)] PH2 )
0.573 V = 0.337 V - (0.0591 / 2) log [H+(aq) ]2 / ([Cu2+(aq)] PH2 )
- log [H+]2 = - 2 log [H+] = 2(0.573 - 0.337)/0.0591
- log [H+] = 4.00 = pH
Equilibrium Constants
DGro = - n F DEo
DGro = - R T ln K
ln K = (n FDEo)/ (R T)
Determine the equilibrium constant at 25oC for
AgCl(s) -> Ag+(aq) + Cl- (aq) (not a redox reaction itself)
Write this reaction in terms of two half reactions; one for
oxidation and one for reduction
(1) AgCl(s) + e-  Ag(s) + Cl-(aq)
(2) Ag+(aq) + e-  Ag(s)
Eo = +0.22 V
Eo = +0.80 V
To get the desired overall reaction, reverse (2) and add to (1)
AgCl(s)  Ag+(aq) + Cl-(aq)
ln K = (n F DEo)/ (R T)
n = 1; DEo = -0.58 V
K = 1.6 x 10-10
DEo = 0.22 V - 0.80 V = - 0.58 V
Corrosion
2 H2O(l ) + 2 e-  H2 (g) + 2 OH- (aq)
Eo = - 0.83 V
The standard reduction potential is when [OH-(aq)] = 1 M; a
pH of 14.
At pH = 7, the reduction cell potential is E ≈ - 0.42 V
Any metal with a standard reduction potential more negative
than - 0.42 V can reduce water at pH = 7; at pH = 7 the metal
will be oxidized by water.
Fe2+(aq) + 2e+  Fe(s)
Eo = - 0.44 V
Fe has a very slight tendency to be oxidized by water at
pH = 7
Fe(s)  Fe2+(aq) + 2e+
When O2 is dissolved in water, the following half reaction is
important:
O2 (g) + 4 H+(aq) + 4 e-  2 H2O(l)
Eo = + 1.23 V
At pH = 7, E ≈ + 0.82 V for this reaction; greater than
Fe2+(aq) + 2e+  Fe(s) Eo = - 0.44 V
Fe(s) is readily oxidized to Fe2+(aq) in O2 containing water
Rust: Fe2O3. H2O(s)
To prevent corrosion: protect surface from exposure to O2
and H2O
Galvanize metal by coating with an unbroken film of Zn.
Electro-deposit Zn on metal; Zn lies below Fe and so is more
readily oxidized than Fe.
Another way is to use a “sacrificial”
anode
Magnesium is oxidized preferentially,
supplying electrons to the iron for the
reduction of O2.
Mg2+(aq) + 2e-  Mg(s)
Eo = -2.36 V
Fe2+(aq) + 2e-  Fe (s)
Eo = -0.44 V
The spontaneous cell reaction:
Mg(s) + Fe2+(aq)  Fe (s) + Mg2+(aq)
Metabolism
Metabolism – process by which living systems acquire and
use free energy to carry out vital processes
Oxidation-reduction reactions are responsible (directly or
indirectly) for all work done in living organisms
“Biological circuitry” - spontaneous, enzyme-catalyzed,
redox reactions to do biological work
break down of complex molecules
building complex molecules
A number of steps in metabolism use NAD+ (nicotinamide
adenine dinucleotide) as an electron acceptor or oxidizing
agent
Reduced metabolite + NAD+ 
oxidized metabolite + NADH + H+
The reduced NADH is then oxidized by oxygen to convert it
back to NAD+
NADH + 1/2O2 + H+  H2O + NAD+
DEo = 1.135 V
In the reduction of pyruvate by NADH:
Pyruvate + NADH + H+
lactate + NAD+
The reduction potentials for the half-reactions are
Pyruvate + 2H+ + 2elactate
E° = -.185 V
NAD+ + H+ + 2eNADH
E° = -.315 V
For the overall reaction involving the two half-reactions: DE°
= E°(cathode) – E°(anode)
So for the reduction of pyruvate by NADH:
DE° = -.185 V – (-.315 V) = +.13 V