Oxidation & Reduction
Electrochemistry
BLB 11th Chapters 4, 20
Chapter Summary
 Oxidation and Reduction (redox) – introduced
in chapter 4
 Oxidation Numbers
 Electron-transfer
 Balancing redox reaction
 Electrochemical cells
 Corrosion
 Electrolysis
21.1, 4.4 Oxidation-Reduction Reactions
 Oxidation
 Loss of electrons
 Increase in oxidation number
 Gain of oxygen or loss of hydrogen
 Reduction
 Gain of electrons
 Decrease in oxidation number
 Loss of oxygen or gain of hydrogen
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
 Oxidizing agent or oxidant – reactant that
contains the element being reduced; is itself
reduced
 Reducing agent or reductant – reactant that
contains the element being oxidized; is itself
oxidized
Oxidation Numbers (p. 137)
Assign according to the following order:
 Atoms
zero (since neutral)
 Ions
equal to charge of the ion
 Nonmetals
1.
2.
3.
4.
O
H
F
X
−2
+1 (when bonded to other nonmetals)
−1 (when bonded to metals)
−1
−1 except when combined with oxygen
Sum of the oxidation numbers equals zero or the charge
of the polyatomic ion.
Oxidation numbers practice
1. O2
6. CH2O
2. CH4
7. Cu2+
3. NO3¯
8. OCl¯
4. CH3OH
5. Cr2O72-
Redox Reactions
 Combustion, corrosion, metal production,
bleaching, digestion, electrolysis
 Metal oxidation



Activity Series (Table 4.5, p. 141)
Some metals are more easily oxidized and
form compounds than other metals.
Displacement reaction – metal or metal ion is
replaced through oxidation
A + BX → AX + B
20.2 Balancing Redox Reactions
 Goal: Balance both the atoms and the electrons
 Examples:
Al(s) + Zn2+(aq) → Al3+(aq) + Zn(s)
MnO4¯(aq) + Cl¯(aq) → Mn2+(aq) + Cl2(g)
The Rules (p. 847)
In acidic solution:
1. Divide into half-reactions.
2. Balance all elements but H and O.
3. Balance O with H2O.
4. Balance H with H+.
5. Balance charge with electrons (e-).
6. Cancel out electrons by integer multiplication.
7. Add half reactions & cancel out.
8. Check balance of elements and charge.
MnO4¯(aq) + Cl¯(aq) → Mn2+(aq) + Cl2(g)
CH3OH(aq) + Cr2O72-(aq) → CH2O(aq) + Cr3+(g)
The Rules (p. 849-850)
In basic solution:
Proceed as for acidic solution through step 7.
8. Add OH¯ to neutralize the H+. (H+ + OH¯ → H2O)
9. Cancel out H2O.
10. Check balance of elements and charge.
Cr(s) + CrO4¯(aq) → Cr(OH)3(aq)
20.3 Voltaic Cells
 A spontaneous redox reaction can perform
electrical work.
 The half-reactions must be placed in separate
containers, but connected externally.
 This creates a potential for electrons to flow.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Line notation:
Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s)
Movement of Electrons
e¯
Zn(s) → Zn2+(aq) + 2 e¯
Cu2+(aq) + 2 e¯ → Cu(s)
Net reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
20.4 Cell EMF Under Standard Conditions
 EMF – electromotive force – the potential energy
difference between the two electrodes of a
voltaic cell; Ecell; measured in volts
 E°cell – standard cell potential (or standard emf)
 For the Zn/Cu cell, E°cell = 1.10 V
 electrical work = Coulombs x volts
J=CxV
J
V
C
Standard Reduction (Half-cell) Potentials
 E° - potential of each half-cell
 E°cell = E°cell(cathode) - E°cell(anode)
 For a product-favored reaction:
 ΔG°
<0
 E°cell > 0
 Measured against standard hydrogen
electrode (SHE); assigned E° = 0 V.
p. 857, more in Appendix E
Problem
Voltaic cell with: Al(s) in Al(NO3)3(aq) on one
side and a SHE on the other.
Problem
Voltaic cell with: Pb(s) in Pb(NO3)2(aq) on one
side and a Pt(s) electrode in NaCl(aq) with Cl2
bubbled around the electrode on the other.
20.5 Free Energy and Redox Reactions
n = # moles of e¯ transferred
ΔG° = wmax = −nFE° F = 96,485 C/mol (Faraday constant)
wmax = max. work
 ΔG° < 0
 E°cell > 0
ΔG° for previous problems
20.6 Cell EMF Under Nonstandard Conditions
 Concentrations change as a cell runs.
 When E = 0, the cell is dead and reaches equilibrium.
 Nernst equation allows us to calculate E under
nonstandard conditions:
J
R

8
.
3145
RT


mol K
E  E  
 ln Q
C
F  96,485 mol
 nF 
@ 298 K
 0.0257 
E  E  
 ln Q or
 n 
 0.0592 
E  E  
 log Q
 n 
Concentration Cells
 A cell potential can be created by using same half-
cell materials, but in different concentrations.
Problem 66
Problem 66
Cell EMF and Equilibrium
 When E = 0, no net change in flow of electrons and
cell reaches equilibrium.
0.0257
E 
ln K
n
and
nE 
ln K 
0.0257
or
0.0592
E 
log K
n
nE 
or log K 
0.0592
K of previous problems