Lecture 38 Electrochemistry II

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Lecture 38 - Electrochemistry II
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Standard Reduction Potentials
Eo, V
Stronger Oxidizing Agent
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The Standard Cell Potential, Eocell

• Eocell Eoanode Eocathode
• Cu(s) Cu2(aq) 2 e- - 0.34 V
• 2 Ag(aq) 2 e- 2 Ag(s) 0.80 V
• Cu(s) 2 Ag(aq) Cu2(aq) 2 Ag(s) 0.46 V

(n2)
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voltmeter
electrons
0.46 V
KNO3(aq)
NO3-
K
Ag
Cu
NO3-
Ag
Anode
Cathode
Cu2
NO3-
1 M Cu(NO3)2(aq)
1 M AgNO3(aq)
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A More Complex Reaction...

• Fe(s) Fe2 2 e-
• MnO4- 8 H 5 e- Mn2 4 H2O

(n10)
Inert Electrode
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Electrodes involving gases

• e.g. 2 H(aq) 2 e- H2(g)

The Standard Hydrogen Electrode Eo 0.00 V
H2(g), 1 atm
Pt electrode
H(aq), 1 M
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Cell Potential and Free Energy
work (J)
Potential Difference (V)
charge (C)
-w
work done by cell
Ecell
q
charge transferred from cell
thus, w -qEcell
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Cell Potential and Free Energy

• wmax DG

free energy change in system (cell)
maximum work obtainable from cell
thus, DG -qEcell
or, DG -nFEcell
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DG -nFEcell
free energy change of reaction
number of moles of electrons transferred (mol
e-)
cell potential
(J/C)
Faraday constant
(J)
(C/mol e- )
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• under standard conditions,
• DGo -nFEocell

• solutes at 1.00 M, pure solids,25oC, 1 atm
• if Eocell gt 0, DGo lt0

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Will a reaction proceed?

• as usual, DGlt0 for spontaneity
• DG -nFEcell
• if Eocell gt 0, DGo lt 0
• or, if Eocell gt 0, reaction is spontaneous

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for example,

• Cu(s) 2 Ag(aq) Cu2(aq) 2 Ag(s)
• Eocell 0.46 V
• DGo -nFEocell
• - (2 mol e-) (96500 C (mol e-)-1)(0.46 J C-1)
• -88,800 J (i.e. spontaneous!)

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Will Gold Dissolve in 1.0 M Nitric Acid?

• 1. Set up the REDOX reaction
• 2. Calculate Eo
• 3. Calculate DGo

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2. Calculate Eocell

• Au(s) Au3(aq) 3 e- Eo -1.50 V
• NO3-(aq) 2 H(aq) e- NO2(g) H2O(l)
• Eo 0.79 V
• Au(s) 6 H(aq) 3 NO3-(aq)
• Au3(aq) 3 NO2(g) 3 H2O(l)
• Eocell -0.71 V

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3. Calculate DGo

• DGo - nFEocell
• - 3 mol e- (96500 C (mol e-) -1)(-0.71 J C-1)
• 205545 J
• 206 kJ

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Does ANY gold dissolve?
-DGo
Kc exp
RT
4.0 x 10-28
Au3 NO23

NO3-3 H6
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Does ANY gold dissolve?
Making a few assumptions,
Au3 lt 10-14 M