Prentice-Hall © 2007General Chemistry: Chapter 20Slide 1 of 54

Juana Mendenhall, Ph.D.

Assistant Professor

Lecture 2 March 15

Chapter 20: Electrochemistry

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Objectives

1. Describe and illustrate how a voltaic cell operates, with the concepts of electrodes, salt bridges, half-cell equations, net cell reaction and cell diagram.

2. Describe the standard hydrogen electrode and explain how other standard electrode potentials are related to it.

3. Using tabulated standard potentials, Eº, to determine E°cell for an oxidation-reduction reaction and predict whether the reaction is spontaneous.

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An Electrochemical Half Cell

Anode

Cathode

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An Electrochemical Cell

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Terminology

Electromotive force, Ecell.

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 ||.

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Terminology

Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)

Ecell = 1.103 VThis typical electrochemical cell was one of the first developed to produce

electrical energy (to run Morse code machines). It is called a Daniell cell.

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Terminology

Galvanic cells. Produce electricity as a result of spontaneous reactions.

Electrolytic cells. Non-spontaneous chemical change driven by electricity.

Couple, M|Mn+

A pair of species related by a change in number of e-.

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Example

A galvanic cell consists of a Mg electrode in a 1.0M Mg(NO3)

2 solution

and a Ag electrode in a 1.0M AgNO3 solution. Calculate the standard

emf of this electrochemical cell at 25º C.

Answer Mg2+(aq) + 2e- Mg(s) Eº = -2.37 V

Cathode: 2Ag+(aq) + 2e- 2Ag(s)

Ag+ will oxidize the Mg2+ (diagonal rule):

Anode: Mg(s) Mg2+(aq) + 2e-

Ag+(aq) + e- Ag(s) °E = 0.80 V

Overall: Mg (s) + 2Ag+(aq) Mg2+(aq) + 2Ag(s)

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Example cont.

Eº = Eºmg/mg2+ + EºAg+/Ag

= 2.37 V + 0.80 V

= 3.17 V

The sign of the emf of the cell allows us to predict the spontaneity of a

redox rxn. Recall that under steady-state conditions for rxns & products,

the redox rxn is spontaneous in the forward direction if the standard

emf of the cell is positive. If it is negative, Eºcell does not mean the rxn

will not occur, it means the the equilibrium will shift to the left.

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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)

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Standard Hydrogen Electrode2 H+(a = 1) + 2 e- H2(g, 1 bar) E° = 0 V

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

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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).

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Reduction Couples

Cu2+(1M) + 2 e- → Cu(s) E°Cu2+/Cu = ?

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

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

E°cell = E°cathode - E°anode

cathodeanode

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Standard Cell Potential

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

E°cell = E°cathode - E°anode

E°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) → H+(1 M) + Cu(s) E°cell = 0.340 V

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Ecell, ΔG, and Keq

Cells do electrical work. Moving electric charge.

Faraday constant, F = 96,485 C mol-1

elec = -nFE

ΔG = -nFE

ΔG° = -nFE°

Michael Faraday 1791-1867

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Example

Calculate the standard free-energy for the following reaction @ 25ºC:

2Au(s) + 3Ca2+ (aq) 2Au3+(aq) +3Ca(s)

Answer: First we break up the overall rxn into half-rxns:

Oxidation: 2Au(s) 2Au3+(aq) + 6e-

Reduction: 3Ca2+ (aq) + 6e- +3Ca(s)

Eº = EºAu/Au3+ + EºCa2+/Ca

= -1.50 V - 2.87 V

= -4.37 V

Gº = -nFEº

= - (6 mol)(96,500 J/V*mol)(-4.37 V)

= 2.53 x 103 kJ

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Spontaneous Change

ΔG < 0 for spontaneous change. Therefore E°cell > 0 because ΔGcell = -nFE°cell

E°cell > 0 Reaction proceeds spontaneously as written.

E°cell = 0 Reaction is at equilibrium.

E°cell < 0 Reaction proceeds in the reverse direction spontaneously.

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Combining Half Reactions

Fe3+(aq) + 3e- → Fe(s) E°Fe3+/Fe = ?

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

Fe3+(aq) + 3e- → Fe2+(aq) E°Fe3+/Fe2+ = 0.771 V

Fe3+(aq) + 3e- → Fe(s)

ΔG° = +0.880 J

ΔG° = -0.771 J

ΔG° = +0.109 VE°Fe3+/Fe = +0.331 V

ΔG° = +0.109 V = -nFE°

E°Fe3+/Fe = +0.109 V /(-3F) = -0.0363 V