Electrochemistry Ch. 17. Electrochemistry Generate current from a reaction –Spontaneous reaction...

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Electrochemistry Ch. 17

Transcript of Electrochemistry Ch. 17. Electrochemistry Generate current from a reaction –Spontaneous reaction...

Page 1: Electrochemistry Ch. 17. Electrochemistry Generate current from a reaction –Spontaneous reaction –Battery Use current to induce reaction –Nonspontaneous.

Electrochemistry

Ch. 17

Page 2: Electrochemistry Ch. 17. Electrochemistry Generate current from a reaction –Spontaneous reaction –Battery Use current to induce reaction –Nonspontaneous.

Electrochemistry

• Generate current from a reaction– Spontaneous reaction– Battery

• Use current to induce reaction– Nonspontaneous reaction– Electroplating

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Oxidation-Reduction Reaction

• aka Redox

• Transfer of electrons

Donor + Acceptor(reducing agent) (oxidizing agent)

(is oxidized) (is reduced)

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Mnemonics are cool!

Oxidation

Involves

Loss of electrons

Reduction

Involves

Gain of electrons

Loss of

Electrons is

Oxidation

says

Gain of

Electrons is

Reduction

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Assigning Oxidation States(1) Covalent bond btw identicalatoms => Split electrons evenly

(2) Covalent bond btw differentatoms => All electrons given to moreelectronegative atom.

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(3) For ionic compounds, oxidation states are equal to ionic charge.

(4) Oxidation state for an elemental atom is zero.

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(5) Oxidation state for monatomic ionis the same as the charge.

(6) In compounds, fluorine alwayshas an O.S. of -1.

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(7) Oxygen usually has an O.S. of -2,except when in a peroxide or when inOF2. H2O-2 H2O2

-1 +2OF2

(8) With a nonmetal, hydrogen has an O.S. of +1. With a metal, H is assigned an O.S. of -1. NH3

+1 LiH-1

(9) The sum of the oxidation states must add up to the overall charge.

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Examples

Assign the oxidation states to eachatom of the following compounds.

CO2 CH4 K2Cr2O7

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Redox Reactions

CH4 + O2 CO2 + H2O

Which species is oxidized?

Which species is reduced?

Which species is the oxidizing agent?

Which species is the reducing agent?

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Balancing Redox Reaction

• Balance……# of atoms

…# of electrons transferred

…overall charge

• Types of reactions– Acidic conditions– Basic conditions

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Redox in Acidic Solutions

Cr2O72- + C2H5OH Cr3+ + CO2

1. Assign oxidation states

2. Write half reactions

Red: Cr2O72- Cr3+

Ox: C2H5OH CO2

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3. Balance elements except H and O

Cr2O72- 2Cr3+

C2H5OH 2CO2

4. Balance oxygen by adding H2O

Cr2O72- 2Cr3+ + 7H2O

3H2O + C2H5OH 2CO2

5. Balance hydrogen by adding H+

14H+ + Cr2O72- 2Cr3+ + 7H2O

3H2O + C2H5OH 2CO2 + 12H+

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6. Balance charge by adding electrons

6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O

3H2O + C2H5OH 2CO2 + 12H+ + 12e-

7. Equalize the number of electrons

12e- + 28H+ + 2Cr2O72- 4Cr3+ + 14H2O

3H2O + C2H5OH 2CO2 + 12H+ + 12e-

8. Cancel like terms and add reactions

16H+ + 2Cr2O72- + C2H5OH

4Cr3+ + 2CO2 + 11H2O

9. Check your answer!

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Balancing in basic solution

Following the same algorithm used for acidic solutions through step #8 then…

9. Add the same # of OH- to both sides of equation as there are H+ on one side

10. Combine H+ and OH- on same sides of equation to make H2O

11. Cancel any like terms and check

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Galvanic Cells

• Spontaneous chemistry generating current

• Some terms– Reducing agent– Oxidizing agent– Half reactions– Anode– Cathode– Cell potential

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Building a Galvanic CellOverall Reaction

8H+(aq) + MnO4-(aq) + 5Fe2+(aq)

Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)Half Reactions

Reduction:

8H+ + MnO4- + 5e- → Mn2+ + 4H2O

Oxidation:5(Fe2+ → Fe3+ + e-)

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Reduction: 8H+ + MnO4- + 5e- Mn2+ + 4H2O

Oxidation: Fe2+ Fe3+ + e-

salt bridge

KNO3

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Calculating Cell PotentialReduction: 8H+ + MnO4

- + 5e- Mn2+ + 4H2O

εº(reduction) = 1.51 V

Oxidation: 5(Fe2+ Fe3+ + e-)

εº(oxidation) = -0.77 V

εº(cell) = εº(red) + εº(ox) = 0.74 V

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Comments on Cell Potential• Potential is an intensive property

• DO NOT multiply potential by balancing factor

• The º indicates standard conditions– 1.0 M and 1 atm

• Potentials references to standard H+ red.

2H+ + 2e- → H2 εº = 0.00 V

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Fe3+(aq) + Cu(s) → Cu2+(aq) + Fe2+(aq)

Oxidation: Reduction:

salt bridgeCu

Fe or Pt

Cu2+ Fe3+

e- e- e-

← - +→

Cu Cu2+ + 2e- Fe3+ + e- Fe2+

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Cu2+ + Zn Zn2+ + Cu

Demo

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Line Notation

2Al3+(aq) + 3Mg(s) → 2Al(s) + 3Mg2+(aq)

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Line Notation II

MnO4-(aq) + H+(aq) + ClO3

-(aq) →

ClO4-(aq) + Mn2+(aq) + H2O(l)

Pt(s) │ ClO3-(aq), ClO4

-(aq) ║ MnO4

-(aq), Mn2+(aq) │ Pt(s)

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To Review

• Full description of galvanic cell requires:– Composition of solutions– Composition of electrodes– Direction of electron flow– Direction of ion flow– Calculation of cell potential– Labels: “anode” and “cathode”

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Cell Potential and Free Energy

arg

96485

o o

w

q

w q

w G q

Ch eof moleof electrons Faraday F

CF w G q nF

molee

G nF G nF

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Reconsidering Cell Potential

Given:

Al3+ + 3e- Al ΔG1 and ε1 = -1.66V

Mg2+ + 2e- Mg ΔG2 and ε1 = -2.37 V

Find ε(cell) for:

2Al3+ + 3Mg 2Al + 3Mg2+

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3 1 2

3 3 1 1 2 2

1 1 2 23

3

3

3

2 3

2 3

2 3

2(3)( 1.66) (3)(2)( 2.37)

6

1.66 2.37 0.71

o o o

o o o

o oo

o

o

G G G

n F n F n F

n n

n

V

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Cell Potential and Spontaneity

• Can bromine oxidize iodide to iodine?

• Can Cr(II) reduce oxygen gas under acidic conditions to produce water?

• Can Ag(I) oxidize chloride to chlorine?

• Can hydrogen reduce Fe(II) to elemental iron?

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Non-Standard Cell Potentialln

ln

ln

0.0592log @ 25

o

o o

o

o

o o

G G RT Q

G nF G nF

nF nF RT Q

RTQ

nF

Q Cn

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Practice ProblemsDetermine ε for the following reaction and

conditions:

2Al(s) + 3Mn2+(aq) 2Al3+(aq) + 3Mn

(a) [Al3+] = 2.0 M; [Mn2+] = 1.0 M @ 25°C

(b) [Al3+] = 1.0 M; [Mn2+] = 3.0 M @ 25°C

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2

3

( ) ln

0.48

(8.314)(298) (2.0)0.48 ln

(6)(96485) (1.0)

0.47 ( ) 0.49

o

o

RTa Q

nF

V

V b V

Do Worksheet

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Potential and EquilibriumCell potential at equilibrium = 0.0 V

Q = K at equilibrium0 ln

ln

log @ 250.0592

o

o

oo

RTK

nF

nFK

RT

nK C

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Types of Batteries• Lead storage (car battery)

• Dry cell battery– Acidic– Alkaline– Rechargeable

• Lithium ion battery

• Fuel cell

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Lead Storage BatteryAnode:

Pb + HSO4- → PbSO4 + H+ + 2e-

Cathode:

PbO2 + HSO4- + 3H+ + 2e- → PbSO4 + 2H2O

Overall Potential: εº = 2.04 V

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Page 37: Electrochemistry Ch. 17. Electrochemistry Generate current from a reaction –Spontaneous reaction –Battery Use current to induce reaction –Nonspontaneous.

Dry Cell Battery (acidic)

Anode:

Zn → Zn2+ + 2e-

Cathode:

2NH4+ + 2MnO2 + 2e- → Mn2O3 + 2NH3 + H2O

Overall Potential: εº = 1.5 V

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Dry Cell Battery (alkaline)

Anode:

Zn → Zn2+ + 2e-

Cathode:

2MnO2 + H2O + 2e- → Mn2O3 + 2OH-

Overall Potential: εº = 1.5 V

Page 39: Electrochemistry Ch. 17. Electrochemistry Generate current from a reaction –Spontaneous reaction –Battery Use current to induce reaction –Nonspontaneous.
Page 40: Electrochemistry Ch. 17. Electrochemistry Generate current from a reaction –Spontaneous reaction –Battery Use current to induce reaction –Nonspontaneous.

Lithium Ion Battery

Anode:

Li Li+ + e-

Cathode:

MnO2 + Li+ + e- LiMnO2

Cell Potential: εº = 3.6 V

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Page 42: Electrochemistry Ch. 17. Electrochemistry Generate current from a reaction –Spontaneous reaction –Battery Use current to induce reaction –Nonspontaneous.

Fuel Cell

Anode:

2H2(g) + 4OH-(aq) 4H2O + 4e-

Cathode:

O2 + 2H2O + 4e- 4OH-

Cell Potential: εº = 1.23 V

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Page 44: Electrochemistry Ch. 17. Electrochemistry Generate current from a reaction –Spontaneous reaction –Battery Use current to induce reaction –Nonspontaneous.

Corrosion• A significant portion of construction is

done to replace corroded materials.

Cathode:

O2 + 2H2O + 4e- 4OH- ε = 0.40 VAnode:

Fe Fe2+ + 2e- ε = 0.44 V

ε(cell) = 0.84 V

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Corrosion and Acidic Conditions

Cathode:

O2 + 4H+ + 4e- 2H2O ε = 1.23 V

Anode:

Fe Fe2+ + 2e- ε = 0.44 V

ε(cell) = 1.67 V

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Electrolysis

• Supply current to perform chemistry

• Performed in an electrolytic cell

• Stoichiometric relationship btw. charge and chemical amount

• Factor-label fun!

• Current measured in Ampere = 1 coulomb per second

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Example Problem I

How long will it take to plate out 1.00 kg of aluminum from an aqueous solution of Al3+ using a current of 100.0 A?

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Al3+ + 3e- Al

1 3 96485 1sec(1000 )

26.98 1 1 100

107285second = 29.8 h

mol mol e Cg Al

g Al mol Al mol e C

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Example Problem II

What volume of F2 gas, at 25C and 1.00 atm, is produced when molten KF is electrolyzed by a current of 10.0 A for 2.00 hours? What mass of K metal is produced? At which electrode does each reaction occur?

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Solution II

Molten KF contains K+ and F-

Cathode:

K+ + e- K

Anode:

F- 1/2F2 + e-

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Volume of F2

2

2

0.560min 60 10.0 12.00

min 96485 1

0.373

(0.373)(0.0821)(298)9.12

1

mol Fs C mol eh

h s C mol e

mol F

nRTV L

P

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Mass of K

K+ + e- K

F- 1/2F2 + e-

0.373 mol F2 0.746 mol K

(0.746 mol K)(39.10 g/mol) = 29.2 g K

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Electrolysis in WaterAnode:

2H2O O2 + 4H+ + 4e- ε = -1.23V

Cathode:

4H2O + 4e- 2H2 + 4OH-ε = -0.83V

2H2O O2 + 2H2 ε(cell) = -2.06 V

Assuming [H+] = [OH-] = 1.0 M

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Electrolysis in Pure Water

• In pure water: [H+] = [OH-] = 1.0 x 10-7 M• Use Nernst equation to determine ε

Anode:

Cathode:

47100.05920.83 log 0.42

4 1V

47100.05921.23 log 0.82

4 1V

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Electrolysis in Pure Water• The overall potential for the electrolysis of

pure water is -1.24 V.

• Need to consider several possible oxidations and reductions when performing electrolysis of aqueous salt solutions.

• Consider the electrolysis of 1.0 M NaCl(aq)

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1.0 M NaCl(aq) contains:

1.0 M Na+1.0 M Cl-

H2O 10-7 M H+ 10-7 M OH-

Reducible species: Na+ H+ H2O

Oxidizable species: Cl- OH- H2O

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Possible Reductions

• Using pH = 7.00

Na+ + e- Na ε = -2.71 V

H+ + e- 1/2H2 ε = -0.414 V**

H2O + e- 1/2H2 + OH- ε = -0.416 V**

So, H2 produced at the cathode

**Potentials found using Nernst equation

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Possible Oxidations

• Using pH = 7.00

Cl- 1/2Cl2 + e- ε = -1.36 V

2OH- 1/2O2 + H2O + 2e- ε = -0.814 V**

H2O 1/2O2 + 2H+ + 2e- ε = -0.816 V**

So, O2 expected to form at the anode.

But…