1 chapter 20 principles of reactivity: electron transfer reactions

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Chapter 20 Principles of Reactivity: Electron TransferReactions

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I.Oxidation-Reduction ReactionsA.Balancing oxidation-reduction reactions

Balancing Redox Equations by the Oxidation Number Change Method (Other Stuff page)

II.Voltaic (Galvanic) CellsA.Cell constructionB.Cell potentialC.Effect of concentration on cell potentialD.Commercial voltaic cells

III.Electrolytic CellsA.ElectrolysisB.Quantitative aspects of electrolysis

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I. Oxidation-Reduction Reactions

Review:

2 Na + Cl2 2 NaCl(0) (0) (+1)(–1)

oxidation = increase in oxidation number (loss of electrons)reduction = decrease in oxidation number (gain of electrons)

e.g., Assign oxidation numbers to the species being oxidized and reduced in the following equation, and label the oxidizing agent and reducing agent.

NaI + 3 HOCl NaIO3 + 3 HCl

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I. Oxidation-Reduction ReactionsA. Balancing oxidation-reduction reactions

1. Determine the oxidation numbers of the species being oxidized and reduced (and make sure there are the same number of each on each side).

2. Balance the changes in oxidation numbers by multiplying each species by the appropriate coefficient (i.e., balance the electrons gained and lost).

3. Balance charges with: H+ in acidic solutionOH– in basic solution

4. Balance H (and O!) with H2O.

hard way: ion-electron (half-reaction) method (text)easy way: oxidation number change method (Web - Other Stuff)

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I. Oxidation-Reduction ReactionsA. Balancing oxidation-reduction reactions

e.g., PH3 + I2 H3PO2 + I– (acidic solution)

e.g., MnO4– + H2SO3 Mn2+ + SO4

2– (acidic)

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I. Oxidation-Reduction ReactionsA. Balancing oxidation-reduction reactions

e.g., Cl2 Cl– + ClO3– (basic)

e.g., CrO2– + S2O8

2– CrO42– + SO4

2– (basic)

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II. Voltaic (Galvanic) Cells

Produce electricity:chemical energy electrical energy

2Na+ + 2Cl– 2Na + Cl2 G >> 0• requires input of electrical energy (electrolysis)

2Ag+ + Ni 2Ag + Ni2+ G < 0• produces energy• but with Ag+ and Ni in contact, we can’t generate electricity; electrons just flow from Ni to Ag+

• have to use a voltaic (galvanic) cell…

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II. Voltaic (Galvanic) CellsA. Cell construction

Ni

Ni(NO3)2(aq)

Ag

AgNO3(aq)

Ni Ni2+ + 2e–

(oxidation)

Ni2+

Ag+ + e– Ag(reduction)

Ag+

–––

–––

+++

+++

e–

e–

salt bridge• KCl in gelatin• allows electrolytic conduction without mixing

K+Cl–

net: Ni + 2Ag+ Ni2+ + 2Ag

cathode(+)

anode(–)

electrical device

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II. Voltaic (Galvanic) CellsA. Cell construction

Ni Ag

Ni2+ Ag+

salt bridge or porous partition

Shorthand notation:

Ni | Ni2+ || Ag+ | Ag

Potentiometer (voltmeter)• measures cell potential (voltage or electromotive force)

• depends on:• species in redox equation• concentrations• temperature

• cell potential, E= actual cell voltage

• standard cell potential, Eº = voltage at standard state

25ºC1 atm pressure1 M concentration

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II. Voltaic (Galvanic) CellsB. Cell potential

1. standard reduction potential, Eº

a. Eº = tendency for a species to be reduced

Ag+ + e– Ag Eº(Ag+)Ni2+ + 2e– Ni Eº(Ni2+)

can’t measure directly

Can only measure difference in a voltaic cell:

2Ag+ + 2e– 2Ag Eº(Ag+)Ni Ni2+ + 2e– –Eº(Ni2+)

(not 2 x)

net: Ni + 2Ag+ Ni2+ + 2Ag Eºcell = Eº(Ag+) – Eº(Ni2+)

= 1.05 V

General: Eºcell = Eº(species reduced) – Eº(species oxidized)

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II. Voltaic (Galvanic) CellsB. Cell potential

1. standard reduction potential, Eº

b. standard: hydrogen electrode at standard state

1 atm H2

1 MH+

Pt electrode

by definition:

at standard state, the reduction

2H+ + 2e– H2 has Eº = 0.000 V (exactly)

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II. Voltaic (Galvanic) CellsB. Cell potential

1. standard reduction potential, Eº

c. standard reduction potentials- measure others against the standard hydrogen electrode:

DVM

Ag 1 MAg+

1 MH+

1 atmH2

Find:• cathode: 2Ag+ + 2e– 2Ag• anode: H2 2H+ + 2e– • Eºcell = 0.80 V

Since Eºcell = Eº(Ag+) – Eº(H+)

0.80 V = Eº(Ag+) – 0.00 V

Eº(Ag+) = +0.80 V(i.e., more easily reduced than H+)

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II. Voltaic (Galvanic) CellsB. Cell potential

1. standard reduction potential, Eº

c. standard reduction potentials

Find:• cathode: 2H+ + 2e– H2

• anode: Ni Ni2+ + 2e– •Eºcell = 0.25 V

Since Eºcell = Eº(H+) – Eº(Ni2+)

0.25 V = 0.00 V – Eº(Ni2+)

Eº(Ni2+) = –0.25 V(i.e., less easily reduced than H+ or Ag+)

Ni | Ni2+(1 M) || H+(1 M) | H2

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II. Voltaic (Galvanic) CellsB. Cell potential

1. standard reduction potential, Eº

d. determining cell potentials

Ag+ + e– Ag Eº = +0.80 VNi2+ + 2e– Ni Eº = –0.25 V

Ni | Ni2+(1 M) || Ag+(1 M) | Ag

more easily reduced

Eºcell = Eº(Ag+) – Eº(Ni2+)= +0.80 V – (–0.25 V)= +1.05 V

2Ag+ + 2e– 2Ag Ni Ni2+ + 2e–

Ni + 2Ag+ Ni2+ + 2Ag

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II. Voltaic (Galvanic) CellsB. Cell potential

2. spontaneity of redox reactions

F2 + 2e–

Ag+ + e–

2H+ + 2e–

Ni2+ + 2e–

Li+ + e–

2F–

AgH2

NiLi

Eº +2.87 V+0.80 V 0.00 V–0.25 V–3.05 V

easiest toreduce

(strongestoxidant)

hardest toreduce

(weakestoxidant)

hardest tooxidize

(weakestreductant)

easiest tooxidize

(strongestreductant)

A species on the left will react spontaneously with a species on the right that is below it in the table.

Or: The species with the more positive Eº will be reduced, and the species with the more negative Eº will be oxidized (Eºcell always > 0).

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II. Voltaic (Galvanic) CellsB. Cell potential

2. spontaneity of redox reactions

Given the two half reactions below, what is the net cell reaction? What is Eº? Draw a galvanic cell using these half cells and label the anode and cathode, their charges, and the direction electrons flow in the circuit.Fe3+ + 3e– Fe Eº = –0.04 V

Zn2+ + 2e– Zn Eº = –0.76 V

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II. Voltaic (Galvanic) CellsB. Cell potential

3. cell potential and free energy

Gsys = –wsurr

w = # e–s energy

e–

= coulombs joules

coulomb (= joules)

= coulombs volts

coulombs = nF (n = # of moles of e–s in redox reaction)(F = 96,500 coulombs/mol)

w = nFE

G = –nFEGº = –nFEº

Eºcell > 0, Gº < 0, spontaneous (voltaic)Eºcell < 0, Gº > 0, nonspontaneous (electrolytic)

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II. Voltaic (Galvanic) CellsB. Cell potential

3. cell potential and free energy

e.g., Zn | Zn2+(1 M) || Fe3+(1 M) | Fe

2Fe3+ + 3Zn 2Fe + 3Zn2+ Eºcell = 0.72 V

What is Gº for the cell?

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II. Voltaic (Galvanic) CellsC. Effect of concentration on cell potential

1. Nernst equation

G = Gº + RTlnQ

–nFE = –nFEº + RTlnQ

E = Eº – RTlnQnF

at 25ºC: E = Eº –0.0592

nlogQ

Nernstequation

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II. Voltaic (Galvanic) CellsC. Effect of concentration on cell potential

1. Nernst equation

e.g., Ni | Ni2+ (0.05 M) || Ag+ (0.01 M) | Ag

Ni + 2Ag+ Ni2+ + 2Ag Eº = 1.05 V

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II. Voltaic (Galvanic) CellsC. Effect of concentration on cell potential

2. applicationsa. measuring Ksp

e.g., AgCl(s) Ag+ + Cl– Ksp = [Ag+][Cl–]

Find: E = 0.53 V; What is Ksp for AgCl?

Ni + 2Ag+ Ni2+ + 2Ag Eº = 1.05 V

DVM

Ag Ni

AgCl(s)

0.10 MCl–

[Ag+] = ?

1.0 MNi2+

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II. Voltaic (Galvanic) CellsC. Effect of concentration on cell potential

2. applicationsb. measuring pH

Ag1.0 MAg+

1 atmH2

lemon juice

[H+] = ?

DVMFind E = 0.94 V; What is pH?

H2 + 2Ag+ 2H+ + 2Ag Eº = 0.80 V

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II. Voltaic (Galvanic) CellsC. Effect of concentration on cell potential

2. applications

A cell was constructed using the standard hydrogen electrode ([H+] = 1.0 M) in one compartment and a lead electrode in a 0.10 M K2CrO4 solution in contact with undissolved PbCrO4 in the other. The potential of the cell was measured to be 0.51 V with the Pb electrode as the anode. Determine the Ksp of PbCrO4 from this data. (Pb2+ + 2e– Pb Eº = –0.13 V)

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II. Voltaic (Galvanic) CellsC. Effect of concentration on cell potential

2. applications

A galvanic cell was constructed with a Cu electrode in a solution of 1.0 M Cu2+ in one compartment and a hydrogen electrode immersed in a sample of a soft drink. The cell potential was measured to be 0.523 V. What was the pH of the soft drink? (Cu2+ + 2e– Cu Eº = 0.337 V)

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II. Voltaic (Galvanic) CellsD. Commercial voltaic cells

anode: Pb + SO42– PbSO4 + 2e–

cathode: PbO2 + 4H+ + SO42– + 2e– PbSO4 + H2O

discharge: H2SO4 consumed, H2O produced• dilutes electrolyte solution• can measure with densitometer

discharge

chargecell: Pb + PbO2 + 4H+ + 2SO4

2–

2PbSO4 + 2H2O E ~ 2 V (6 cells in series)

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II. Voltaic (Galvanic) CellsD. Commercial voltaic cells

anode: Zn + 2OH– ZnO + H2O + 2e–

cathode: 2MnO2+ H2O + 2e– Mn2O3 + 2OH–

cell: Zn + 2MnO2 ZnO + Mn2O3

E ~ 1.5 V

Zinc cupanode

Graphitecathode

Moist pasteof MnO2,KOH andH2O Porouspartition

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II. Voltaic (Galvanic) CellsD. Commercial voltaic cells

Anodecap

Partition

Cathode:Ag2O paste

Anode:Zn and KOH

Gasket

Cellcan

anode: Zn + 2OH– ZnO + H2O + 2e–

cathode: Ag2O + H2O + 2e– 2Ag + 2OH–

cell: Zn + Ag2O ZnO + 2Ag

E ~ 1.5 V

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III. Electrolytic cellsA. Electrolysis

electrical energy chemical energy

e.g., NaCl(l) Na (l) + Cl2(g) G >> 0

cell: 2Na+ + 2Cl– 2Na + Cl2

Eº = Eº(Na+) - Eº(Cl2) = (-2.71) - (1.36) = -4.07 V

Gº = +786 kJ/mol

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III. Electrolytic cellsA. Electrolysis

e.g., NaCl(aq) ?

Eº cathode: 2H2O + 2e– H2 + OH– -0.83 V(reduction) Na+ + e– Na -2.71 V

H2O more easilyreduced than Na+

Eº anode: Cl2 + 2e– 2Cl– +1.36 V(oxidation) O2 + 4H+ + 4e– 2H2O +1.23 V H2O more easily

oxidized than Cl–

net: (2H2O + 2e– H2 + OH–) x 2 2H2O O2 + 4H+ + 4e– 2H2O 2H2 + O2

E

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III. Electrolytic cellsA. Electrolysis

e.g., CuCl2(aq) ?

Eº cathode: Cu2+ + 2e– Cu +0.34 V(reduction) 2H2O + 2e– H2 + OH– -0.83 V

Cu2+ more easilyreduced than H2O

net: (Cu2+ + 2e– Cu) x 22H2O O2 + 4H+ + 4e– 2Cu2+ + 2H2O 2Cu + O2 + 4H+

E

Eº anode: Cl2 + 2e– 2Cl– +1.36 V(oxidation) O2 + 4H+ + 4e– 2H2O +1.23 V H2O more easily

oxidized than Cl–

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IV. Electrolytic cellsB. Quantitative aspects of electrolysis

Units of charge: 1 faraday (F) = 1 mol e–s1 coulomb = 1 amp ·1 sec (A·s)

experimentally: 1 F = 96,500 C

e.g., How many moles of Na and Cl2 are produced in the electrolysis of NaCl(l) when a current of 25 A is applied for 8.0 hours?

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IV. Electrolytic cellsB. Quantitative aspects of electrolysis

e.g., How long would it take to deposit 21.4 g of Ag from a solution of AgNO3 using a current of 10.0 A?