Electrochemistry

Ch. 17

Electrochemistry

• Generate current from a reaction– Spontaneous reaction– Battery

• Use current to induce reaction– Nonspontaneous reaction– Electroplating

Oxidation-Reduction Reaction

• aka Redox

• Transfer of electrons

Donor + Acceptor(reducing agent) (oxidizing agent)

(is oxidized) (is reduced)

Mnemonics are cool!

Oxidation

Involves

Loss of electrons

Reduction

Involves

Gain of electrons

Loss of

Electrons is

Oxidation

says

Gain of

Electrons is

Reduction

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

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

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

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

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

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

(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.

Examples

Assign the oxidation states to eachatom of the following compounds.

CO2 CH4 K2Cr2O7

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?

Balancing Redox Reaction

• Balance……# of atoms

…# of electrons transferred

…overall charge

• Types of reactions– Acidic conditions– Basic conditions

Redox in Acidic Solutions

Cr2O72- + C2H5OH Cr3+ + CO2

1. Assign oxidation states

2. Write half reactions

Red: Cr2O72- Cr3+

Ox: C2H5OH CO2

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+

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!

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

Galvanic Cells

• Spontaneous chemistry generating current

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

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-)

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

Oxidation: Fe2+ Fe3+ + e-

salt bridge

KNO3

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

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

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+

Cu2+ + Zn Zn2+ + Cu

Demo

Line Notation

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

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)

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”

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

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+

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

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?

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

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

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

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

Types of Batteries• Lead storage (car battery)

• Dry cell battery– Acidic– Alkaline– Rechargeable

• Lithium ion battery

• Fuel cell

Lead Storage BatteryAnode:

Pb + HSO4- → PbSO4 + H+ + 2e-

Cathode:

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

Overall Potential: εº = 2.04 V

Dry Cell Battery (acidic)

Anode:

Zn → Zn2+ + 2e-

Cathode:

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

Overall Potential: εº = 1.5 V

Dry Cell Battery (alkaline)

Anode:

Zn → Zn2+ + 2e-

Cathode:

2MnO2 + H2O + 2e- → Mn2O3 + 2OH-

Overall Potential: εº = 1.5 V

Lithium Ion Battery

Anode:

Li Li+ + e-

Cathode:

MnO2 + Li+ + e- LiMnO2

Cell Potential: εº = 3.6 V

Fuel Cell

Anode:

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

Cathode:

O2 + 2H2O + 4e- 4OH-

Cell Potential: εº = 1.23 V

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

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

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

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?

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

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?

Solution II

Molten KF contains K+ and F-

Cathode:

K+ + e- K

Anode:

F- 1/2F2 + e-

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

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

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

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

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)

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

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

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…