Electrochemistry

Electrode Potentials and Their Measurement

Cu(s) + 2Ag+(aq)

Cu2+(aq) + 2 Ag(s)

Cu(s) + Zn2+(aq)

No reaction

Zn(s) + Cu2+(aq)

Cu(s) + Zn2+(aq)

In this reaction:Zn(s) g Zn2+(aq) OxidationCu2+(aq) g Cu(s) Reduction

•If Zn(s) and Cu2+(aq) is in the same solution, then the electron is a transferred directly between the Zn and Cu.

No useful work is obtained. However if the reactants areseparated and the electrons shuttle through an external path...

ΔG: the maximum non-p, V work at T, p = const.

An Electrochemical Cell/1

Anode (-)

Negative electrode generates electrons

Oxidation occurs

Cathode (+)

Positive electrode accepts electrons

Reduction occurs

Cu(s) | Cu2+(aq) || Ag+(aq) | Ag(s) ΔEcell = 0.460 V

Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) DEcell = 1.103 V

An Electrochemical Cell/2 (Daniell)

Anode (-)

Negative electrode generates electrons

Oxidation occurs

Cathode (+)

Positive electrode accepts electrons

Reduction occurs

Electron Transfer at the Electrodes

Anode

Cathode

Terminology• Electromotive force, DEcell.

– The cell voltage or cell potential.• Cell diagram.

– Shows the components of the cell in a symbolic way.– Anode (where oxidation occurs) on the left.– Cathode (where reduction occurs) on the right.

• Boundary between phases shown by |.• Boundary between half cells

(usually a salt bridge) shown by ||.

• Couple, M|Mn+A pair of species related by a change in number of e-.

Terminology

• Galvanic cells.– Produce electricity as a result of spontaneous

reactions.

• Electrolytic cells.– Non-spontaneous chemical change driven by

electricity.

Standard Electrode Potentials

• Cell voltages, the potential differences between electrodes, are among the most precise scientific measurements.

• The potential of an individual electrode is difficult to establish.

• Arbitrary zero is chosen.The Standard Hydrogen Electrode (SHE)

Standard Hydrogen Electrode

2 H+(a = 1) + 2 e- D H2(g, 1 bar) E° = 0 V

Pt|H2(g, 1 bar)|H+(a = 1)

Standard Electrode Potential, E°

• E° defined by international agreement.• The tendency for a reduction process to occur at

an electrode.– All ionic species present at a = 1 (approximately 1 M).– All gases are at 1 bar (approximately 1 atm).– Where no metallic substance is indicated, the potential is

established on an inert metallic electrode (ex. Pt).

Reduction CouplesCu2+(1M) + 2 e- D Cu(s) E°Cu2+/Cu = ?

Pt|H2(g, 1 bar)|H+(a = 1) || Cu2+(1 M)|Cu(s) ΔE°cell = 0.340 V

Standard cell potential: the potential difference of a cell formed from two standard electrodes.

ΔE°cell = E°cathode - E°anode

cathodeanode

Standard Cell PotentialPt|H2(g, 1 bar)|H+(a = 1) || Cu2+(1 M)|Cu(s) ΔE°cell = 0.340 V

ΔE°cell = E°cathode - E°anode

ΔE°cell = E°Cu2+/Cu - E°H+/H2

0.340 V = E°Cu2+/Cu - 0 V

E°Cu2+/Cu = +0.340 V

H2(g, 1 atm) + Cu2+(1 M) D 2H+(1 M) + Cu(s) ΔE°cell = 0.340 V

Measuring Standard Reduction Potential

cathode cathode anodeanode

Standard Reduction Potentials

Most spontaneous Oxidizing Agent

Most non-spontaneous Spontaneous in the reverse direction. Reducing Agent

ΔEcell, ΔG, and Keq

• Cells do electrical work.– Moving electric charge.

• Faraday constant, F = 96,488 C mol-1 = q´NA= 1.6022´10-19 C ´ 6.022045´1023 mol-1

= charge of one mole of electrons.

welec, rev = ΔG = -QΔE

ΔG = -nFΔEΔG° = -nFΔE°

Spontaneous Change• ΔG < 0 for spontaneous change.• Therefore ΔEcell > 0 because ΔGcell = -nFΔEcell• ΔEcell > 0– Reaction proceeds spontaneously as written.

• ΔEcell = 0– Reaction is at equilibrium.

• ΔEcell < 0– Reaction proceeds in the reverse direction

spontaneously.

The Behavior or Metals Toward AcidsM(s) D M2+(aq) + 2 e- E° = -E°M2+/M

2 H+(aq) + 2 e- D H2(g) E°H+/H2 = 0 V

2 H+(aq) + M(s) D H2(g) + M2+(aq)

ΔE°cell = E°H+/H2 - E°M2+/M = -E°M2+/M

When E°M2+/M < 0, E°cell > 0. Therefore ΔG° < 0.

Metals with negative reduction potentials react with acids

Combining Half-Cell Reactions/1

Reaction 1: Cu2+(aq) + 2e- D Cu(s) Reaction 2:

Cu+(aq) + e- D Cu(s)

Reaction 3: Cu2+(aq) + e- D Cu+(aq)

Since Reaction 3 = Reaction 1 - Reaction 2

NO!!

Reaction 1: Cu2+(aq) + 2e- D Cu(s) Reaction 2:

Cu+(aq) + e- D Cu(s)

Reaction 3: Cu2+(aq) + e- D Cu+(aq)

Combining Half Reactions/2Fe3+(aq) + 3e- D Fe(s) E°Fe3+/Fe = ?

Fe2+(aq) + 2e- D Fe(s) E°Fe2+/Fe = -0.440 V

Fe3+(aq) + e- D Fe2+(aq) E°Fe3+/Fe2+ = 0.771 V

Fe3+(aq) + 3e- D Fe(s) E°Fe3+/Fe = +0.331 V

Equation 3 = Equation 1 + Equation 2

1

2

3

Relationship Between DE°cell and Keq

ΔG° = -RT ln Keq = -nFDE°cell

DE°cell =nF

RTln Keq

Summary of Thermodynamic, Equilibrium and Electrochemical Relationships.

DEcell as a Function of ConcentrationΔG = ΔG° +RT ln Q

-nFΔEcell = -nFΔEcell° +RT ln Q

ΔEcell = ΔEcell° - ln QnF

RT

Convert to log10 and calculate constants

ΔEcell = ΔEcell° - log Qn

0.0592 VThe Nernst Equation:

DEcell as a Function of Concentration: an Alternative Route

Cathode: Ox1®Red1Anode: Red2®Ox2

Applying the Nernst Equation for Determining DEcell.

What is the value of DEcell for the voltaic cell pictured below and diagrammed as follows?

Example

Pt|Fe2+(0.10 M),Fe3+(0.20 M)||Ag+(1.0 M)|Ag(s)

ΔEcell = ΔEcell° - log Qn

0.0592 V

Pt|Fe2+(0.10 M),Fe3+(0.20 M)||Ag+(1.0 M)|Ag(s)

ΔEcell = ΔEcell° - logn

0.0592 V [Fe3+]

[Fe2+] [Ag+]

Fe2+(aq) + Ag+(aq) D Fe3+(aq) + Ag (s)

ΔEcell = 0.029 V – 0.018 V = 0.011 V

Alternative Route

C

A

Dismutation/1

Spontaneous

Dismutation/2

Non-spontaneous

Concentration CellsTwo half cells with identical electrodes

but different ion concentrations.

2 H+(1 M) D 2 H+(x M)

Pt|H2 (1 atm)|H+(x M)||H+(1.0 M)|H2(1 atm)|Pt(s)

2 H+(1 M) + 2 e-D H2(g, 1 atm)

H2(g, 1 atm) D 2 H+(x M) + 2 e-

Concentration Cells

ΔEcell = ΔEcell° - logn

0.0592 V x2

12

ΔEcell = 0 - log2

0.0592 V x2

1

ΔEcell = - 0.0592 V log x

ΔEcell = (0.0592 V) pH

2 H+(1 M) D 2 H+(x M)ΔEcell =Δ Ecell° - log Qn

0.0592 V

Measurement of Ksp

Ag+(0.100 M) D Ag+(sat’d M)

Ag|Ag+(sat’d AgI)||Ag+(0.10 M)|Ag(s)

Ag+(0.100 M) + e-D Ag(s)

Ag(s) D Ag+(sat’d) + e-

Using a Voltaic Cell to Determine Ksp of a Slightly Soluble Solute.

With the date given for the reaction on the previous slide, calculate Ksp for AgI.

Example

AgI(s) D Ag+(aq) + I-(aq)

Let [Ag+] in a saturated Ag+ solution be x:

Ag+(0.100 M) D Ag+(sat’d M)

ΔEcell = ΔEcell° - log Q = n

0.0592 VΔEcell° - log

n

0.0592 V

[Ag+]0.10 M soln

[Ag+]sat’d AgI

ΔEcell =Δ Ecell° - log n

0.0592 V

[Ag+]0.10 M soln

[Ag+]sat’d AgI

ΔEcell =Δ Ecell° - log n

0.0592 V

0.100

x

0.417 V= 0 - (log x – log 0.100) 1

0.0592 V

0.417log 0.100 -

0.0592log x = = -1 – 7.04 = -8.04

x = 10-8.04 = 9.1´10-9 Ksp = x2 = 8.3´10-17

Batteries: Producing Electricity Through Chemical Reactions

• Primary Cells (or batteries).– Cell reaction is not reversible.

• Secondary Cells.– Cell reaction can be reversed by passing

electricity through the cell (charging).• Flow Batteries and Fuel Cells.–Materials pass through the battery which

converts chemical energy into electric energy.

The Leclanché (Dry) Cell

Dry CellZn(s) D Zn2+(aq) + 2 e-Oxidation:

2 MnO2(s) + H2O(l) + 2 e- D Mn2O3(s) + 2 OH-Reduction:

NH4+ + OH-D NH3(g) + H2O(l) Acid-base reaction:

NH3 + Zn2+(aq) + Cl- → [Zn(NH3)2]Cl2(s)Precipitation reaction:

Alkaline Dry Cell

Zn2+(aq) + 2 OH-D Zn (OH)2(s)

Zn(s) D Zn2+(aq) + 2 e-

Oxidation reaction can be thought of in two steps:

2 MnO2(s) + H2O(l) + 2 e- D Mn2O3(s) + 2 OH-Reduction:

Zn (s) + 2 OH-D Zn (OH)2(s) + 2 e-

Lead-Acid (Storage) Battery• The most common secondary battery

PbO2(s) + 3 H+(aq) + HSO4-(aq) + 2 e- D PbSO4(s) + 2 H2O(l)

Lead-Acid Battery

Oxidation:

Reduction:

Pb (s) + HSO4-(aq) D PbSO4(s) + H+(aq) + 2 e-

PbO2(s) + Pb(s) + 2 H+(aq) + HSO4-(aq) D 2 PbSO4(s) + 2 H2O(l)

ΔE°cell = E°PbO2/PbSO4 - E°PbSO4/Pb = 1.74 V – (-0.28 V) = 2.02 V

The Silver-Zinc Cell: A Button Battery

Zn(s),ZnO(s)|KOH(sat’d)|Ag2O(s),Ag(s)

Zn(s) + Ag2O(s) D ZnO(s) + 2 Ag(s) Ecell = 1.8 V

The Nickel-Cadmium Cell

Cd(s) + 2 NiO(OH)(s) + 2 H2O(L) D 2 Ni(OH)2(s) + Cd(OH)2(s)

Fuel CellsO2(g) + 2 H2O(l) + 4 e- D 4 OH-(aq)

2{H2(g) + 2 OH-(aq) D 2 H2O(l) + 2 e-}

2H2(g) + O2(g)D 2 H2O(l)

ΔE°cell = ΔE°O2/OH- - ΔE°H2O/H2= 0.401 V – (-0.828 V) = 1.229 V

e = ΔG°/ ΔH° = 0.83

Corrosion: Unwanted Voltaic CellsO2(g) + 2 H2O(l) + 4 e- D 4 OH-(aq)

2 Fe(s) D 2 Fe2+(aq) + 4 e-

2 Fe(s) + O2(g) + 2 H2O(l) → 2 Fe2+(aq) + 4 OH-(aq)

Ecell = 0.841 V

EO2/OH- = 0.401 V

EFe2+/Fe = -0.440 V

In neutral solution:

In acidic solution:

O2(g) + 4 H+(aq) + 4 e- D 4 H2O (aq) EO2/OH- = 1.229 V

Corrosion Protection

Corrosion Protection

Bernard Moitessier(1925 - 1994)

Joshua

1968 - 69

Electrolysis: Causing Non-spontaneous Reactions to Occur

Galvanic Cell:

Zn(s) + Cu2+(aq) D Zn2+(aq) + Cu(s) DE0(dir)= 1.103 V

Electolytic Cell:

Zn2+(aq) + Cu(s) D Zn(s) + Cu2+(aq) DE0(inv) = -1.103 V

DED0 = - DE0(inv) Decomposition potential

Anode (+) Cathode (-)

Quantitative Aspects of ElectrolysisFaraday’s Laws

1. The amount of substance deposited at the electrodesis proportional to the amount of current passed throughthe circuit.

2. The amount of substance that reacts at an electrodeis proportional to its equivalent mass.

3. A current of 1 F (96488 C) makes 1 equivalent of substancereact at the electrodes.

Ag+ + e- D Ag(s)

1 mol of electrons (1 F) deposits 1 mol of Ag º 1 equivalent of Ag

Cu2+ + 2e- D Cu(s)

1 mol of electrons (1 F) deposits 1/2 mol of Cu º 1 equivalent of Cu

Al3+ + 3e- D Al(s)

1 mol of electrons (1 F) deposits 1/3 mol of Al º 1 equivalent of Al

1 mol e- = 96488 C

Charge (C) = current (C/s) × time (s)

ne- =I × t

F

Complications in Electrolytic Cells

• Overpotential.• Competing reactions.• Non-standard states.• Nature of electrodes.

Overpotential

εA, eC > O

εA - anodic overpotential

εC - cathodic overpotential

When a current I flows, the ohmic potential drop IR must beIncluded:

Industrial Electrolysis ProcessesElectro-refining

Electroplating

Chlor-Alkali Process