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Cell EMF
Eocell = Eored(cathode) - Eored(anode)
Example: Zn + Cu+2  Zn+2 + Cu
E0cell = 0.34 V - (-0.76 V) = + 1.10 V
A voltaic cell is based on two half-reactions:
Cd+2/Cd Sn+2/Sn
Which half-reaction takes place at the cathode?
Which half-reaction takes place at the anode?
What is the standard cell potential?
A voltaic cell is based on two half-reactions:
Cd+2/Cd Sn+2/Sn
Which half-reaction takes place at the cathode?
Sn+2 + 2e-  Sn
Which half-reaction takes place at the anode?
Cd  Cd+2 + 2eWhat is the standard cell potential?
0.267 V
Spontaneity of Redox Reactions
Eo = Eored(reduction) - Eored(oxidation)
E0 = (+) spontaneous
E0 = (-) nonspontaneous
Cu +
+
2H

+2
Cu
Calculate the value of
+ H2
0
E.
Cu +
+
2H

+2
Cu
Calculate the value of
+ H2
0
E.
E0 = -0.34 V
NOT SPONTANEOUS
EMF and Free-Energy Change
G = -nFE
n = a positive # (the # of electrons transferred)
F = Faraday’s constant. 1F = 96,500 J/V-mol
E = EMF
Use the standard reduction potentials to
calculate the standard free-energy change, Go,
for the following reaction:
4Ag + O2 + 4H+  4Ag+ + 2H2O
Use the standard reduction potentials to
calculate the standard free-energy change, Go,
for the following reaction:
4Ag + O2 + 4H+  4Ag+ + 2H2O
Go = -(4)(96,500J/V-mol)(+0.43V)
- 170 kJ/mol
The Nernst Equation
(at 298 K)
Calculate the emf at 298K generated by
-2
the following cell, when (Cr2O7 ) = 2.0
M, (H+) = 1.0 M, (I-) = 1.0 M, and
(Cr+3) = 1.0 x 10-5 M:
Cr2O7 (aq) +

(aq) +
+3
2Cr (aq) + 3I2(s) + 7H20(l)
-2
+
14H
6I (aq)
Calculate the emf at 298K generated by
-2
the following cell, when (Cr2O7 ) = 2.0
M, (H+) = 1.0 M, (I-) = 1.0 M, and
(Cr+3) = 1.0 x 10-5 M:
Cr2O7-2 + 14H+ + 6I-  2Cr+3 + 3I2 + 7H20
E = 0.79 V - 0.0592V/6 log(5.0 x 10-11)
E = 0.89 V