1

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) + 2e-

Solution 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

Work: 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

zelectrochemical potential, 0

2z – 1z = 2z – 1z + z2(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

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, eF, and M M

At equil.: M M++ eM+ + zFzFM+

Add components, note - metal pure element so

MM+=0 and M+

M+ + zF = zF

bottom line, chem pot. depend on electrical potential

2

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-

Measure cell potential against SHE and get

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) + 2e- oxidation anode

Right: Cu+2(aq) +2e- Cu(s) reduction cathode

e- balance out Sum: Zn(s) + Cu+2(aq) Zn+2(aq) + Cu(s) or: Zn|Zn+2||Cu+2|Cu

chem. potentials: Grxn = Zn2+ Cu - Cu2+ – Zn = Zn2+ -

Cu2+ + RTln(aZn2+/aCu2+)

ionGion~ Grxn = Gorxn + RTln(aZn2+/aCu2+) = -nF Grxn ~ emf

– emf - electromotive force: Gorxn = -nFgeneralize

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 = -o

ox

So cell potential is: ocell =

ored + o

ox since Grxn = -nFthen Grxn = -nF

Clearly, if process, reaction spontaneous, then o > 0

SHE

Zn/Zn+2

3

Examples Engel 9.4,9.5

Entropy determine from temperature variation of o:

Enthalpy can be obtained from G = H – TS H = G + TS

Combine K and measurement: RT/nF) RTln(Q) =RT/nF) RTln(K/Q)

Change coupled half-cell can reverse the

NAD + H+ +e- NADH reaction, flip order

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’= - Ered

o’

Erxn’o = - 0.320 + (- 0.295V) = -0.615 V

Red. pot. for NAD+, so Eoxo’ = +0.615V

(sign same value differ, 2H+ for H2O)

4

Engel Example 9.5

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

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

S (-) could be expected for gas to solvated ion

In this case, no reactants left, but

potential still easy to measure

Pt|NAD+|NADH||HCOO-|CO2|Pt

5

Solubility products (view salt dissolution as two steps):

AgBr(s) + e- Ag(s) + Br -(aq) Eo = 0.07133

Ag(s) Ag+(aq) + e- Eo = -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 b0

This gives o – use this and concentration to calcuate ± - most accurate values

Biochemical standard state, since aoH+ = 10-7 standard instead of ao

H+ = 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)

o’(V) Reaction

6

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 )

Go’ - nF o’

So if combining reactions in a thermodynamic cycle, need to account for quantity (n)

7

Example: 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

“Easy way” sum the Eo’ v u s but ov cou ts since overall rxn is 4e-, parts are 2e-

If convert to Go’ F o’ fo ch th sumGo’ s usu b ck co v t to o’

G1o’ = -nF 1

o’ -2x96485C/mol(-0.581V) = 112 kJ/mol

G2o’ -nF 2

o’ -2x96485C/mol(-0.197V) = 38 kJ/mol

Go’ G1o’ G2

o’ (112 38) kJ/mo 150kJ/mo o’ Go’ / F - 0.388V

Note: this is actually (1o’ 2

o’ )/2 (-0.581-0197)V/2 = -0.389 V

General: for A+nAe- B+nBe- C: (nA+nA)A/Co’ nAA/B

o’ nBB/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.

8

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

9

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+ + 2e- ox PbPb+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

Dry cell: Zn + xNH3 [Zn(NH3)x]+2 + 2e-

2MnO2 + 2NH4+ +2e- Mn2O3 + H2O + 2NH3

Done with salts in paste (conduct) with Zn as outer can, C-electrode in center, ~ 1.6 V

Practical note – this is a high single protein

concentration (100 mM), but not so high

for multiprotein containing cell