Oxidation & Reduction Electrochemistry BLB 11 th Chapters 4, 20
Feb 23, 2016
Oxidation & ReductionElectrochemistry
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
20.1, 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. 132)
Assign according to the following order: Atoms zero (since neutral) Ions equal to charge of the ion Nonmetals
1. O −22. H +1 (when bonded to other nonmetals)
−1 (when bonded to metals)3. F −14. X −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
2. CH4
3. NO3¯
4. CH3OH
5. Cr2O72-
6. CH2O
7. Cu2+
8. OCl¯
Redox Reactions
Combustion, corrosion, metal production, bleaching, digestion, electrolysis
Metal oxidation Activity Series (Table 4.5, p. 136) Some metals are more easily oxidized and
form compounds than other metals. Displacement reaction – metal or metal ion is
replaced through oxidationA + 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. 830-1)
In acidic solution:1. Divide equation into two half-reactions (ox and red).2. Balance all elements but H and O.3. Balance O by adding H2O.4. Balance H by adding H+.5. Balance charge by adding 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. 833)
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. Reactant metal is the most reactive; product
metal the least.Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Line notation:Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s)
20.3 Voltaic Cell
Net reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Cu2+(aq) + 2 e¯ → Cu(s)Zn(s) → Zn2+(aq) + 2 e¯
Movement of Electrons
e¯
Net reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
20.4 Cell Potentials 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 = C x VCJV
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.
V 2.37- Mg(s) 2e )(Mg -2 aq
App. E, p. 1064 More E° values
ProblemVoltaic cell with: Al(s) in Al(NO3)3(aq) on one side and a SHE on the other. Sketch the cell, determine the balance equation, and calculate the cell potential.
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. Sketch the cell, determine the balance equation, and calculate the cell potential.
Problem
20.5 Free Energy and Redox Reactions
ΔG° < 0 E°cell > 0
ΔG° for previous problems
ΔG° = wmax = −nFE°
n = # moles of e¯ transferred
F = 96,485 C/mol (Faraday constant)
wmax = max. work
20.6 Cell Potentials 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:
Qn
EEorQn
EE
KFR
QnFRTEE
molCKmol
J
log0592.0ln0257.0298@
485,963145.8
ln
Concentration Cells
A cell potential can be created by using same half-cell materials, but in different concentrations.
Problem 69
Problem 69
Cell EMF and Equilibrium
When E = 0, no net change in flow of electrons and cell reaches equilibrium.
K of previous problems
0592.0log
0257.0ln
log0592.0ln0257.0
nEKornEK
and
Kn
EorKn
E
20.7 Batteries and Fuel Cells
Batteries self-contained electrochemical power source More cells produce higher potentials Primary – non-rechargeable (anode/cathode)
Alkaline: Zn in KOH/MnO2
Secondary – rechargeable (anode/cathode) Lead-acid: Pb/PbO2 in H2SO4
nicad: Cd/[NiO(OH)] NiMH: ZrNi2/[NiO(OH)] Li-ion: C(s,graphite)/LiCoO2
Hydrogen Fuel Cells Convert chemical energy directly into electricity Fuel and oxidant supplied externally continuously Products are only electricity and water
cathode: O2(g) + 4 H+(aq) + 4 e¯ → 2 H2O(l)anode: 2 H2(g) → 4 H+(aq) + 4 e¯ overall: 2 H2(g) + O2(g) → 2 H2O(l)
20.8 Corrosion
RUST! Anode: M(s) → Mn+(aq) + n e¯ Cathode: O2(g) + 4 H+(aq) + 4 e¯ → 2 H2O(l)
or: O2(g) + 2 H2O(l) + 4 e¯ → 4 OH¯ (aq)
Preventing Corrosion
Anionic inhibition painting oxide formation coating
Cathodic inhibition sacrificial anode – attach a metal (like Mg)
more easily oxidized galvanizing steel – coating with zinc
20.9 Electrolysis
Electrical energy chemical change
Hall-Héroult Process for Al Production
C(s) + 2 O2-(l) → CO2(g) + 4 e¯
3 e¯ + Al3+(l) → Al(l)