Section 10
Electrochemical Cells and Electrode
Potentials
Electrochemistry
Oxidation/Reduction Reactions
• “Redox” reactions involve electron transfer from
one species to another
• Ox1 + Red2  Red1 + Ox2
• Ox1 + ne-  Red1 (Reduction ½ reaction)
• Red2
 Ox2 + ne- (Oxidation ½ reaction)
• “Reducing agent” donates electrons (is oxidezed)
• “Oxidizing agent” accepts electrons (is reduced)
Electrochemistry
Oxidation/Reduction Reactions
• Typical oxidizing agents:
Standard Potentials,V
– O2 + 4H+ + 4e-  2H2O
+1.229
– Ce4+ + e-  Ce3+
+1.6 (acid)
– MnO4- + 8H+ + 5e-  Mn2+ + 4H2O +1.51
• Typical reducing agents:
– Zn2+ + 2e-  Zno
– Cr3+ + e-  Cr2+
– Na+ + e-  Nao
-0.763
-0.408
-2.714
The salt bridge allows charge transfer through the solution and prevents mixing.
The spontaneous cell reaction (Fe2+ + Ce4+ = Fe3+ + Ce4+) generates the cell potential.
The cell potential depends on the half-reaction potentials at each electrode.
The Nernst equation describes the concentration dependence.
A battery is a voltaic cell. It goes dead when the reaction is complete (Ecell = 0).
Fig. 12.1. Voltaic cell.
©Gary Christian, Analytical Chemistry,
6th Ed. (Wiley)
Electrochemistry
Standard Reduction Potentials
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Half-Reaction Potentials:
They are measured relative to each other
Reference reduction half-reaction:
standard hydrogen electrode (SHE)
normal hydrogen electrode (NHE)
2H+(a=1.0) + 2e-  H2(g 1atm) 0.0000 volts
The more positive the Eo, the better oxidizing agent is the oxidized form (e.g., MnO4-).
The more negative the Eo, the better reducing agent is the reduced form (e.g., Zn).
©Gary Christian, Analytical Chemistry,
6th Ed. (Wiley)
Electrochemistry
Reduction Potentials
• General Conclusions:
• 1. The more positive the electrode potential, the
stronger an oxidizing agent the oxidized form is
and the weaker a reducing agent the reduced form
is
• 2. The more negative the reduction potential, the
weaker the oxidizing agent is the oxidized formis
and the stronger the reducing agent the reduced
form is.
Electrochemistry
Oxidation/Reduction Reactions
• Typical oxidizing agents:
Standard Potentials,V
– O2 + 4H+ + 4e-  2H2O
+1.229
– Ce4+ + e-  Ce3+
+1.6 (acid)
– MnO4- + 8H+ + 5e-  Mn2+ + 4H2O +1.51
• Typical reducing agents:
– Zn2+ + 2e-  Zno
– Cr3+ + e-  Cr2+
– Na+ + e-  Nao
-0.763
-0.408
-2.714
Electrochemistry
Oxidation/Reduction Reactions
•
•
•
•
•
Net Redox Reactions:
Standard Potentials,V
MnO4-  Mn2+
MnO4- + 8H+ + 5e-  Mn2+ + 4H2O
+1.51
Sn4+ + 2e-  Sn2+
+0.154
Balanced Net Ionic Reaction:
• 2MnO4- + 16H+ + 5Sn2+  2Mn2+ + 5Sn4+ + 8H2O
Electrochemistry
Voltaic Cell
• The spontaneous (Voltaic) cell reaction is the one that
gives a positive cell voltage when subtracting one halfreaction from the other.
• Eocell = Eoright – Eoleft = Eocathode – Eoanode =Eo+ - Eo• Which is the Anode? The Cathode?
• Convention:
• The anode is the electrode where oxidation occurs 
the more negative half-reaction potential
• The cathode is the electrode where reduction occurs 
the more positive half-reaction potential
• anode  solution  cathode
Electrochemistry
Oxidation/Reduction Reactions
•
•
•
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•
Net Redox Reactions:
Standard Potentials,V
MnO4-  Mn2+
MnO4- + 8H+ + 5e-  Mn2+ + 4H2O
+1.51
Sn4+ + 2e-  Sn2+
+0.154
Balanced Net Ionic Reaction:
• 2MnO4- + 16H+ + 5Sn2+  2Mn2+ + 5Sn4+ + 8H2O
• Eocell = Eocat – Eoan = (+1.51 – (+0.154)) = +1.36 V
Electrochemistry
Nernst Equation
• Effects of Concentrations on Potentials:
• aOx + ne-  bRed
• E = Eo – (2.3026RT/nF) log([Red]b/[Ox]a
– Where E is the reduction at specific conc.,
– Eo is standard reduction potential, n is number of electrons
involved in the half reaction,
– R is the gas constant (8.3143 V coul deg-1mol-1),
– T is absolute temperature,
– and F is the Faraday constant (96487 coul eq-1).
• At 25oC(298.16K) the value of 2.3026RT/F is 0.05916
• Note: Concentrations should be activities
Electrochemistry
• Calculations:
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MnO4- + 8H+ + 5e-  Mn2+ + 4H2O Eo = +1.51 V
For [H+] = 1.0M, [MnO4-] = 0.10M, [Mn2+] = 0.010M
E = Eo – 0.05916/5 (log ([Mn2+]/[MnO4-][H+]8)
E = +1.51 – 0.1183(-1) = +1.63 V vs NHE
Note: This is more positive than Eo
Greater tendency to be reduced compared to standard
conditions.
Electrochemistry
• Calculations:
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Silver electrode/silver chloride deposit/0.010M NaCl
AgCl + 1e-  Ago + ClE=?
Ag+ + 1e-  Ago
Eo = +0.799 V
AgCl  Ag+ + ClKsp= 1.8 x 10-10
AgCl + e-  Ago + ClE = Eo - (0.05916/1) Log (1/[Ag+])
[Ag+] = Ksp/[Cl-] = 1.8 x 10-10/(0.010) = 1.8 x 10-8
E = +0.799 – (0.05916)(7.74) = +0.341 V