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Chapter 20: Electrochemistry. Juana Mendenhall, Ph.D. Assistant Professor Lecture 2 March 15. Objectives. Describe and illustrate how a voltaic cell operates, with the concepts of electrodes, salt bridges, half-cell equations, net cell reaction and cell diagram. - PowerPoint PPT Presentation
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Prentice-Hall © 2007General Chemistry: Chapter 20Slide 1 of 54
Juana Mendenhall, Ph.D.
Assistant Professor
Lecture 2 March 15
Chapter 20: Electrochemistry
Prentice-Hall © 2007General Chemistry: Chapter 20Slide 2 of 54
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.
Prentice-Hall © 2007General Chemistry: Chapter 20Slide 3 of 54
An Electrochemical Half Cell
Anode
Cathode
Prentice-Hall © 2007General Chemistry: Chapter 20Slide 5 of 54
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 ||.
Prentice-Hall © 2007General Chemistry: Chapter 20Slide 6 of 54
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.
Prentice-Hall © 2007General Chemistry: Chapter 20Slide 7 of 54
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-.
Prentice-Hall © 2007General Chemistry: Chapter 20Slide 8 of 54
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)
Prentice-Hall © 2007General Chemistry: Chapter 20Slide 9 of 54
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.
Prentice-Hall © 2007General Chemistry: Chapter 20Slide 10 of 54
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)
Prentice-Hall © 2007General Chemistry: Chapter 20Slide 11 of 54
Standard Hydrogen Electrode2 H+(a = 1) + 2 e- H2(g, 1 bar) E° = 0 V
Pt|H2(g, 1 bar)|H+(a = 1)
Prentice-Hall © 2007General Chemistry: Chapter 20Slide 12 of 54
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).
Prentice-Hall © 2007General Chemistry: Chapter 20Slide 13 of 54
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
Prentice-Hall © 2007General Chemistry: Chapter 20Slide 14 of 54
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
Prentice-Hall © 2007General Chemistry: Chapter 20Slide 15 of 54
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
Prentice-Hall © 2007General Chemistry: Chapter 20Slide 16 of 54
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
Prentice-Hall © 2007General Chemistry: Chapter 20Slide 17 of 54
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.
Prentice-Hall © 2007General Chemistry: Chapter 20Slide 18 of 54
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