Matter and Measurement

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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
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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- -gt 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) -gt 2 H (aq) 2 Hg(l) 2
  Cl- (aq)

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• 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-.

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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
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Q H(aq) 2 / (Cu2(aq) PH2 ) 0.573 V
0.337 V - (0.0591 / 2) log H(aq) 2 /
(Cu2(aq) PH2 ) - log H2 - 2 log H
2(0.573 - 0.337)/0.0591 - log H 4.00 pH
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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) -gt 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) Eo 0.22 V
• (2) Ag(aq) e- ? Ag(s) Eo 0.80 V
• To get the desired overall reaction, reverse (2)
  and add to (1)

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AgCl(s) ? Ag(aq) Cl-(aq) DEo 0.22 V - 0.80
V - 0.58 V ln K (n F DEo)/ (R T) n 1
DEo -0.58 V K 1.6 x 10-10
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Corrosion
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• 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

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• 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

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Rust Fe2O3. H2O(s)
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• 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.
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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)
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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
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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
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In the reduction of pyruvate by NADH Pyruvate
NADH H lactate NAD The reduction
potentials for the half-reactions are Pyruvate
2H 2e- lactate E -.185
V NAD H 2e- NADH
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