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)

PEM Fuel Cell

►

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)