Electrochemical Potential, Work, and Energy
I.
Potential, Work, and Energy
A.
Units
1) Joule (J) = unit of energy, heat, or work (w) = kg•m2/s2
2) Coulomb (C) = unit of electrical charge (q). 1 e- = 1.6 x 10-19 C
3) Volt (V)  Work (J) = electrical potential (e)
Charge (C)
4)
1 J of work is produced when 1 C of charge is transferred between two points
differing by 1 V of electrical potential
5)
6)
Work flowing out of a system (Galvanic Cell) is taken to be negative work
Cell Potential is always positive
ε
7)
-w
q
or
- w  qε
From last chapter, wmax = DG
DG  w max  - qε max
B.
Electrochemical Problems
1) When current flows, we always waste some of the energy as heat instead of work
w < wmax
2) We can, however, measure emax with a potentiometer, so we can find the
hypothetical value of wmax
3)
4)
Example: eocell = 2.50 V 1.33 mole e- pass through the wire. eactual = 2.10 V
a) 1 Faraday (F) = the charge on 1 mole of electrons = 96,485 C
(6.022 x 1023 e-/mol)(1.6 x 10-19 C/e-) = 96,485 C/mol
b)
c)
w = -qe = -(1.33 mol e-)(96,485 C/mole e-)(2.10 J/C) = -2.69 x 105 J
wmax = -qemax = -(1.33 mol e-)(96,485 C/mole e-)(2.50 J/C) = -3.21 x 105 J
d)
Efficiency = w/wmax = -2.69 x10-5 J/-3.21 x 105 J = 0.838 or 83.8%
Free Energy (DG) DG  w max  - qε max
a) q = nF where n = number of moles, F = 96,485 C/mole
b) DG = -nFe (assuming the maximum e)
c) Maximum cell potential is directly related to DG between reactants and
products in the Galvanic Cell (This lets us directly measure DG)
 ε  - DG  Spontaneou s Process
5)
Example: Calculate DGo for the reaction
Cu2+(aq) + Fe(s)
Cu(s) + Fe2+(aq)
eo = 0.34 V o
E cell = +0.78 V
o
e = 0.44 V
a)
Half Reactions: Cu2+ + 2eCuo
Feo
Fe2+ + 2e-
b)
DGo = -nFeo = -(2 mol e-)(96,485 C/mol e-)(0.78 J/C) = -1.5 x 105 J
Example: Will 1 M HNO3 dissolve metallic gold to make 1 M Au3+?
a) Half Reaction: NO3- + 4H+ + 3eNO + 2H2O
eo = +0.96 V
Auo
Au3+ + 3eeo = -1.50 V
Au(s) + NO3-(aq) + 4H+(aq)
Au3+(aq) + NO(g) + 2H2O(l) eocell = -0.54V
b) Since e is negative (DG = +) the reaction will not occur spontaneously
6)
II.
Cell Potential and Concentration
A.
Concentration Cells
1) Up until now, concentration for all Galvanic solutions = 1 M (Gives eo)
2) What happens if we change these concentrations?
3)
Le Chatelier’s Principle
a) Cu(s) + 2Ce4+(aq)
b) Increase Ce4+ concentration,
c) Increase Cu2+ concentration,
d) Example
Cu2+(aq) + 2Ce3+(aq)
(e > eo)
(e < eo)
eocell = 1.36 V
4)
Concentration Cell = Galvanic Cell driven by the fact that concentrations of the
same reactants are different on the two sides of the cell.
5)
Example: Ag+ + eAgo
eo1/2 = +0.80 V
a) If both sides had [Ag+] = 1 M, then eocell = +0.80 V + (-0.80 V) = 0.00 V
b) If [Ag+]right = 1 M and [Ag+]left = 0.1 M then we should have a potential
i.
Diffusion would try to equalize Ag+ on the right side and the left side
(Entropy favors even distribution, like gas particles in two chambers)
Electrons would flow from left to right to even out [Ag+]
ii.
A very small voltage would be generated
i.
iii. Example
B.
The Nernst Equation
1) Derivation
a) DG = DGo + RTlnQ = -nFe
b) DGo = -nFeo
c) -nFe = -nFeo + RTlnQ
oC,
this simplifies to
RT
εε lnQ
nF
o
0.0592
εε logQ
n
o
2)
At 25
3)
Example: 2Al(s) + 3Mn2+(aq)
2Al3+(aq) + 3Mn(s) eocell = 0.48 V
a) Oxidation: 2Al(s)
2Al3+(aq) + 6eb) Reduction: 3Mn2+(aq) + 6e3Mn(s)
c) [Mn2+] = 0.5 M, [Al3+] = 1.5 M
d) Q = [Al3+]2 / [Mn2+]3 = (1.5)2 / (0.5)3 = 18
0.0592
ε cell  0.48 V log18  0.47 V
6
e)
f)
As the reaction proceeds, ecell
Calculating K:
0
(Q
K) = Dead Battery!
o
0.0592
nε
0  εo logK  logK 
n
0.0592
Example: [VO2+] = 2M, [H+] = 0.5M, [VO2+] = 0.01M, [Zn2+] = 0.1M
+ + 2H+ + e2+ + H O eo = 1.00 V
VO
VO
2
2
Find Ecell
2+
Zn + 2e
Zn
eo = -0.76 V
Ion-Selective Electrodes
1) Cell potential depends on concentration of an ion
2) pH meter
a) Standard electrode of known potential
b) Glass electrode filled with known [HCl] whose potential changes based on
external [H+]
c) Potentiometer measures the potential difference
g)
C.
3)
You can make similar Na+, K+, or NH4+, Cl-, F-, etc…selective electrodes
a) Glass “senses” the presence of H+ in open sites (pH meter)
b) Change the type of glass for sensing other ions
Line Notation for a typical pH electrode:
Ag | AgCl | Cl- || H+ outside | H+ inside, Cl- | AgCl | Ag
Outer ref elec.
sample
Known H+
H+ sensing glass membrane
Inner ref elec.
III. Batteries
A.
Battery Basics
1) Battery = galvanic cells used as a portable source of electrical potential
2) Batteries are a source of direct current only; not suitable for providing alternating
current like permanent outlets do
B.
Lead Storage Batteries
1) Highly rechargeable, durable batteries that can operate between –30 and 120 oF
2) Lead anode, Lead oxide cathode, Sulfuric Acid electrolyte
Anode: Pb + H2SO4
PbSO4 + H+ + 2eCathode: PbO2 + HSO4- + 3H+ + 2ePbSO4 + 2H2O
Cell: Pb(s) + PbO2(s) + 2H+(aq) + 2HSO4-(aq)
2PbSO4(s) + 2H2O eo = 2.0V
3)
4)
5)
6)
7)
For cars: 6 of these cells in series with grid electrodes provides 12 V (2
V each)
Sulfuric Acid is consumed; so density of the acid drops over its life
Water is also consumed; can “top off” the battery with water. New
Ca/Pb electrodes no longer use up water (sealed batteries)
Alternator recharges battery by forcing current in opposite direction
Physical Damage, not chemical depletion, usually “kills” the battery
C.
Other Batteries
1) Dry Cell Batteries = calculators, watches, etc…
a) Acid Version: Zn anode, C cathode, MnO2/NH4Cl/C paste as electrolyte 1.5V
Anode: Zn
Zn2+ + 2eCathode: 2NH4+ + 2MnO2 + 2eMn2O3 + 2NH3 + H2O
b) Alkaline Version has KOH or NaOH as electrolyte
Anode: Zn + 2OHZnO + H2O + 2eCathode: 2MnO2 + H2O + 2eMn2O3 + 2OHc) Rechargable Nickel—Cadmium Batteries
Anode: Cd + 2OHCd(OH)2 + 2eCathode: NiO2 + 2H2O + 2eNi(OH)2 + 2OHd) Nickel-Metal Hydride (NiMH) Batteries
Anode: M∙H + OHM + H2O + eCathode: NiO2 + 2H2O + 2eNi(OH)2 + 2OHe) Lithium Ion Batteries: flow of Li+ inside battery matched by e- in wire
D.
Fuel cells = galvanic cell with continuous source of reactants
1) The Hydrogen—Oxygen Fuel Cell is used for NASA spaceflights
2) The reactant gases can be stored as liquids in tanks
a) Anode: 2H2 + 4OH4H2O + 4ee1/2 = 0.83V
b) Cathode: 4e- + O2 + 2H2O
4OHe1/2 = 1.20V
c) Overall: 2H2(g) + O2(g) + catalyst
2H2O(l)
eo = 2.03V