Chem 340 Fall 2013 – Lecture Notes 12- Electrochemistry (Chap. 6)
Charged particle energies affected by applied electric fields,
similarly dissolution of metals from electrodes to create ions also
creates a potential difference between electrode and solution,
e.g. Zn0 (s)  Zn+2 (aq) + 2eSolution gets positive and electrode negative (e- left), but the
electrical potential that develops is dependent on the metal and
ion and affects chemical potential of charged species in solution
dwrev = (2 –1) dQ,
where dQ = - zF dn
where F is a Faraday = 96485 C/mol (charge on mol e-)
this is reversible, non-expansion work, so dG = dwrev
1
– 2z dn
where z is # charge transfer
zelectrochemical potential,
0
2
2z – 1z = 2z – 1z + z (2–1)F plug in def., drop dn
but 2 = 1 i.e. same thing (metal), different phases (diff. ), so
2 – 1 = + z(2 – 1)F can only measure difference in potential, let 1=0
2 = 1 + z 2 F  means charged particles differ in chemical potential by 
Charge will flow to reduce potential, negative particles toward more positive region
Chemist can control chemical potential of charged particles, even change sign of Grxn
Work:
Electrochemical cells
Example shown right, has two half-cells, each with
different metal electrode and a salt solution
(dissociated) with ions of the same metal. Electrodes
connected with meter to measure electrical potential,
s ts k w th “s t b
” – electrolyte (e.g. KCl) in a
gel (not moving much). Note, substitute power supply
c “p t out” m t s f om s t so ut o .
Metal—salt equilibrium makes different potential
each side, but need both to get measurement.
Establish standard state, at = 0 for ions in solution
or M+M+(as before), with field: M+M+ + zF
for e- in electrode e= 0, eF, and M M
At equil.: M M++ eM+ + zFzFM+ 
Add components, note - metal pure element so
MM+=0 and M+M+ + zF = zF
bottom line, chem pot. depend on electrical potential
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Still only measure difference in cells, reference to: Standard Hydrogen Electrode, SHE
H+(aq) + e-  ½ H2(g), at equilibrium: H+ + e- = ½ H2

f~a
For unit activities:
since H2 = 0 and Gfo(H+) = 0,
 reference electrode
Need standard concentration, aH+ = 1
Pt electrode catalyze H2  2H+ +2e-
SHE
Measure cell potential against SHE and get
Zn/Zn+2
potential for the half-cell reaction, in figure
potential is for the Zn/Zn+2 half-cell, collect
relative potentials to SHE and then can
determine cell potentials relative to each other  shorthand: (-) Zn|Zn+2||H+|H2|Pt (+)
cell previous page Left : Zn(s)  Zn+2(aq) + 2eoxidation  anode
+2
Right: Cu (aq) +2e  Cu(s)
reduction  cathode
+2
+2
e balance out
Sum: Zn(s) + Cu (aq)  Zn (aq) + Cu(s)
or: Zn|Zn+2||Cu+2|Cu
chem. potentials: Grxn = Zn2+ Cu - Cu2+ – Zn = Zn2+ - Cu2+ + RTln(aZn2+/aCu2+)
ionGion~
Grxn = Gorxn + RTln(aZn2+/aCu2+) = -nF
Grxn ~ emf
o


 – emf - electromotive force: G rxn = -nF generalize  RT/nF )RTln(aM1/aM2)
Nerst equation:
RT/nF) RTln(Q) or 0.05916/n) log10(Q) T=298K
If know activities and o then can calculate cell potential
Current? 1 mole of Zn0 Zn+2, then 2 mole e- flow, 1 mole Cu+2 Cu0 (not normal use)
Also works for half-cells,
Ox +ne-  Red:
Combining half-cells:
Tables list reduction potentials,
but ored = -oox
So cell potential is: ocell =ored + oox since Grxn = -nFthen Grxn = -nF
Clearly, if process, reaction spontaneous, then o > 0
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Examples Engel 9.4,9.5
Eo’ = -0.320V
Eo’ = -0.414V
Eo’ = 0.295V
Erxn’o = - 0.320 + (+ 0.414V) = 0.096 V
(same since 1H+ in both sides) also Eoxo’= - Eredo’
Change coupled half-cell can reverse the
NAD + H+ +e-  NADH reaction, flip order
Erxn’o = - 0.320 + (- 0.295V) = -0.615 V
Red. pot. for NAD+, so Eoxo’ = +0.615V
(sign same value differ, 2H+ for H2O)
Entropy determine from temperature variation of o:
Enthalpy can be obtained from G = H – TS H = G + TS
Combine K and  measurement: RT/nF) RTln(Q) =RT/nF) RTln(K/Q)
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Engel Example 9.5
S (-) could be expected for gas to solvated ion
At equilibrium,  = 0, rewrite Nernst: o = (RT/nF) ln K standard potential gives K, but
in principle need unit activities to determine o, alternate: combine half-cell values
+
-
Pt|NAD |NADH||HCOO |CO2|Pt
NAD+ +HCOO-  CO2 + NADH
NAD+ +H+ +2e-  NADH
o = -0.105 V
HCOO-  CO2 + H+ +2eo = 0.200 V
Grxn = -nF = - 2(96485 C/mol)(-0.105 + 0.200)V = -18.3 kJ/mol
ln K = -G/RT = 18.3 kJ/mol/(8.314 x 298 J/mol) = 7.4  K = exp(7.4) = 1640
reaction strongly favors products, challenging to measure equil. conc. using ordinary
methods (eg spectroscopy) but with electrochem, trivial to get value with voltmeter
Engel Example 9.6 – more dramatic
In this case, no reactants left, but
potential still easy to measure
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Solubility products (view salt dissolution as two steps):
AgBr(s) + e-  Ag(s) + Br -(aq)
Eo = 0.07133
Ag(s)  Ag+(aq) + eEo = -0.7996
AgBr(s)  Ag+(aq) + Br-(aq)
Eo = -0.7283
ln Ksp = (nF/RT )Eo = (1x96485 C/mol)(-0.7283 V)/(8.314J/molK)(298 K) = -28.4
Ksp = 4.9x10-13 here not measure conc. directly, except electrode and voltmeter easy
Determine activity coefficients, since Debye Hückel works at low
concentrations and since ±  0 at low concentrations then
measuring  at low concentrations will have less error from
activity coefficients.
 =  o + RT/F ln a± = o + RT/F ln b± +RT/F ln ±
As decrease concentration, Debye Hückel for 1:1, I = b±:
ln ± = -1.173 (b±)½ at T=298K:
 – 0.02568 ln b± = o – 0.03013 (b±)½ or
 – 0.05916 log10b± = o – 0.03013 (b±)½
or (m±)½ 
Plot: ( – 0.05916 log10b±) vs (b±)½ and extrapolate b0
This gives o – use this and concentration to calcuate ± - most accurate values
Biochemical standard state, since aoH+ = 10-7 standard instead of aoH+ = 1 will impact the
equilibrium constant, e.g. rR  pP + vH+ then normally: K = (aP/ao)p(aH+/aoH+)v/(aR/ao)r
If chemical standard state all the ao values are 1, K = (aP)p(aH+)v/(aR)r
but if use biochem standard state, K’ ( P)p(aH+/10-7)v/(aR)r so K’/K 107v
Relate to Go: Go’ -RT K’ - RTln K – 7vRT ln10 = Go – 7vRT ln10
Electrochemical cell same: o’ -Go’/ F (RT/ F) K’ (RT/ F)( K 7vRT 10)
o’ = o + 7v(ln10)(RT/nF)
tables in chemical reduction potentials or biochemical reduction potentials (below)
 ’(V)
o
Reaction
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Thermodynamic cycles and emf, cell potentials are intrinsic quantities (only depend on
concentration not how much you have), but G is extrinsic (Go’ f
s p mo )
o
o
G ’ - nF  ’
So if combining reactions in a thermodynamic cycle, need to account for quantity (n)
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CH3COO- + 3H+ + 2e-  CH3CHO + H2O
Eo’ -0.581 V
CH3CHO + 2H+ + 2e-  CH3CH2OH
Eo’ -0.197 V
Calculate Eo’ fo combined half-cell: CH3COO- + 5H+ + 4e-  CH3CH2OH + H2O
Example:
“Easy way” sum the Eo’ v u s but ov cou ts since overall rxn is 4e-, parts are 2eIf convert to Go’
F o’ fo
ch th sumGo’ s usu
b ck co v t to o’

G1o’ = -nF 1o’ -2x96485C/mol(-0.581V) = 112 kJ/mol

G2o’ -nF 2o’ -2x96485C/mol(-0.197V) = 38 kJ/mol
Go’ G1o’ G2o’ (112 38) kJ/mo 150kJ/mo  o’ Go’ / F - 0.388V
Note: this is actually (1o’ 2o’ )/2 (-0.581-0197)V/2 = -0.389 V
General: for A+nAe-  B+nBe-  C: (nA+nA)A/Co’ nAA/Bo’ nBB/Co’
Transmembrane effects
Cellular membranes (lipid bilayer) are relatively impervious to ions and transport of
charged species and can maintain a charge/potential difference
Transport of ions usually incorporates channels and proteins that act as pumps
Work in elect. transport: wE = Q, and GE = ZF, for moving ion of charge z
Difference in chemical potential then :  =E + C = zF + RT lnQ
At equilibrium, chem. pot. difference :  = 0 or zF = - RT lnK or  = - (RT/zF) lnK
Example: mammalian plasma membrane has  = in - out ~ -70mV. It has pores to
allow K+ to equilibrate. If extracellular [K+] = 5mM, what is concentration inside the cells
at 37oC, assume unit activity coefficients
For K+, z=1, RT/zF = 8.314x310/96485 = 0.0267 V
So lnQ = - /(RT/zF) = - (-0.07)/0.0267 = 2.62 and Q = 13.7 or
[K+in] =13.7 [K+out] = 13.7x0.005 M = 69 mM
Donnan effect: separate two solutions with a semipermeable membrane that will let
small ions, e.g. Na+ and Cl-, pass but will block macromolecules, e.g. Protein dialysis
put salt solution on one side, protein solution on other, and assume protein is z:1 salt,
i.e. dissociate to P-z + zNa+, and start with: bNazP on side 1, aNaCl side 2,
at equilibrium, get; (zb+x)Na+, xCl- and bP-z on side 1 and (a-x)Na+, (a-x)Cl- on side 2
also at equilibrium 1± = 2± and std. state also: 1± = 2±
so a1± = a2± or a1+a1- = a2+a2- if ± = 1, i.e. very dilute: c1+/c2+ = c1-/c2- =1
would work if no protein, since Na+ and Cl- would go through membrane and equalize
Donnan effect: equilibrium c1+/c2+ = c1-/c2- = rD ≠ 1 where rD is Donnan ratio
At equilibrium have (assume ± = 1): c1+c1- = c2+c2- or (zb+x)x = (a-x)(a-x)
The Na+ and Cl- both go through membrane to maintain charge neutrality, but
protein concentration is a constant (b), solve to get: x = a2/(2a+zb) and get all conc.
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rD = c1-/c2- = [a - a2/(2a+zb)]/[a2/(2a+zb)]= [(a(2a+zb)-a2)/(2a+zb)]/[a2/(2a+zb)] =
= [(a2+azb)/a2] = [(a+zb)/a] = rD
show it’s same for c1+/c2+ = rD
Put in electrodes, cell at equilibrium so  = 0, but can still be described by Nernst
 = D+(RT/F)ln(c1+/c2+) = D+(RT/F)ln rD = 0 
D = - (RT/F)ln rD =D + (RT/F)ln[(a+zb)/a]
where D is the Donnan potential
Physically this arises due to the difference in concentration of a given ion on
either side and manifests itself as a polarization in the membrane
Since concentrations differ on both sides of membrane, an osmotic pressure develops:
net = (c2-c1)RT = [(1+z)b + 2x -2(a-x)]RT = [(1+z)b-2a+4x]RT =
= [((1+z)b-2a)+4a2/(2a+zb)]RT = [(zb2+2ab+z2b2)/(zb+2a)]RT =net
Cells are permeable to small ions and not to proteins or nucleic acids in general.
Concentration of anionic proteins above described is passive, can be high in cell, would
cause a rupture except that sodium/potassium pump (in membrane) changes
concentration outside/inside to counter net – active transport balance it out
Problem need to transport against chem pot, so need to do work or consume energy
ATPase Na-K pump helps this: 3Nain+ +2Kout+ +ATP  3Naout+ +2Kin+ +ADP + Pi
Can be inhibited by cardiotonic steroids, e.g. digitoxin or other poisons, initially stimulate
heartbeat then can shut it down
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Practical note – this is a high single protein
concentration (100 mM), but not so high
for multiprotein containing cell
Batteries – opposite: consume chemical reagents to get energy (electrical work)
Wet cell, storage battery, 4 e- oxidation between Pb and PbO2, use PbSO4 as medium
Pb + HSO4-  PbSO4 + H+ + 2eox PbPb+2
o = -0.126
2e- + PbO2 + 3H+ + HSO4-  PbSO4 + 2H2O red Pb+4 Pb+2
o = 1.45
Pb + PbO2 + 2H+ + 2HSO4-  2PbSO4 + 2H2O
o = -0.126 -1.45 = -1.57V
Actual V depend on concentration and temperature, Nernst: RT/nF) RTln(Q)
Use PbSO4 to maintain high Pb+2 conc~const, and large excess H2SO4
Zn + xNH3  [Zn(NH3)x]+2 + 2e2MnO2 + 2NH4+ +2e-  Mn2O3 + H2O + 2NH3
Done with salts in paste (conduct) with Zn as outer can, C-electrode in center,  ~ 1.6 V
Dry cell:
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