Chapter 13:

Fundamentals of Electrochemistry

Chapter 13:

Fundamentals of Electrochemistry

Electrochemistry is the branch of chemistry concerned with the interrelation of electrical and chemical effects.

The study of chemical changes caused by passage of an electric current and production of electrical energy by chemical reactions.

• Makes use of electrochemistry at the surface of electrode for the purpose of analysis

• A voltage (potentiometry) or current (voltammetry) signal originating from an electrochemical cell is related to the activity or concentration of a particular species in the cell.

• Excellent detection limit (10-8 ~ 10-3 M):

• Inexpensive technique.

• Easily miniaturized : implantable and/or portable (biosensor, biochip)

Electroanalytical Chemistry

Adapted from Sawyer, Heineman & Beebe, Chemistry Experiments for Instrumental Methods, Wiley

AuPtcarbon

3 electrodeBAS, Inc.

I. Papautksy, University of Cincinnati

Electrochemical Cells for Voltammetric SensorsControl E – measure i

voltammetry – scan E amperometry – hold E

A -----A’

polyester substrateinsulation

enzyme pastenon-enzyme paste

Ag conductor

A

A’

connection pad

AgCl

WE1

WE2 RE

outer membrane

carbon

Silverconductionlayer

Carbonbaseelectrode

Dielectriclayer

Active enzymepastelayer

Outermembranelayer

Ag/AgCllayer

Non-activeenzymepastelayer

rPolyestesubstrate

RE

WE1

WE2

Biosensor DesignBiosensor Design Fabrication ProcessFabrication Process

Screen printing technology for thick-film electrode

A redox reaction: transfer of electrons from one species to another

Oxidation: loss of electronsReduction: gain of electrons

Fe3+ + V2+ -->Fe2+ + V3+ ---- (13.1)

electron

Fe3+: oxidantV2+: reductant

When electrons from a redox reaction flow through an electric circuit

Current ∝ the rate of the electrochemical reaction

Voltage ∝ the free energy change for the electrochemical reaction

13-1. Basic Concepts

Chemistry and Electricity

Electric ChargeElectric Charge

Unit: coulomb (C)

The magnitude of the charge of a single electron = 1.602 x 10-19 C

Charge of 1 mol of electrons = 9.649 x 104 C/mol ; Faraday constant (F)

Relation between charge and moles:

------- (13.2)

Q: If 5.585 g of Fe3+ were reduced in rxn 13-1,how many coulombs of charge must have been transferred from V2+ to Fe3+ ?

Sol: Atomic weight of Fe = 55.85 g/mol5.585g of Fe3+ = 0.100 mol of Fe3+

Each Fe3+ ion requires one electron in the reactionà 0.100 mol of electrons must have been transferred

(0.100 mol e-)(9.649x104 C/mol e-) = 9.649 x 103 C

Example: Relating Coulombs to Quantity of ReactionExample: Relating Coulombs to Quantity of Reaction

q = n · Fn= # moles of electron transferred

Charles Augustin de Coulomb (1736-1806)

Michael Faraday (1791-1867)

Current

Current: the quantity of charge flowing each second through a circuit

Unit: ampere, 1 A = 1 coulomb/sec

Example: Relating Current to Rate of ReactionExample: Relating Current to Rate of Reaction

Power Supply

KCl

Salt bridge Pt Cu

Anode Cathode

Zn2+

SO42-Sn4+Sn2+

2e-

Cu2+

Cl-

Cu2+

SO42-

SO42-

Cu2+

Cu

2e-

e-

(-) (+)

e-

K+

(-) (+)

Q: If Sn4+ is reduced to Sn2+ at a constant rate of 4.23 mmol/h, how much current flows ?

Sol:

Sn4+ + 2e- Sn2+

Sn4+ reduction rate = 4.23 mmol/h à Electron flows at a rate of 2 x (4.23 mmol/h )

(8.48 mmol/h)/(3600s/h) = 2.356 x10-3 mmol/s

Current = (2.356x10-6 mol/s)(9.649x104 C/mol)= 0.227C/s = 0.227 A

André-Marie Ampère(1775-1836)

Voltage, Work, and Free Energy

The difference in electric potential (E) between two points is a measure of the work

that is needed when an electric charge moves from one point to the other.

When a charge, q, moves through a potential difference, E, the work done is

Example: Electrical WorkExample: Electrical Work

Work = E x q(jouls) = (volts) x (coulomb)

------- (13.3)

1 volt = 1 J/C

Q: How much work is needed to move 2.36 mmol of electrons through a potential difference of 1.05 V ?

S: To get coulombs of charge, q = n x F = (2,36 x 10-3 mol)(9.649 x 104 C/mol) = 2.277 x 102 C

The work required = E x q = (1.05 V) x (2,277 x102 C) = 239 J

At constant T, P

The free energy change (ΔG) for a chemical reaction =

the maximum possible electrical work that can be done by the reaction on its surroundings

Work done on surroundings = - ΔG ------- (13.4)

From equations (13-2, 13-3, 13-4)

ΔG = - work = - E · q = - nFE

ΔG = - nFE ------- (13.5)

Relationship between Free Energy Difference and Electric Potential Difference

A galvanic cell : uses a spontaneous chemical reaction to generate electricity

To accomplish this: 1. One reagent must be oxidized2. The other must be reduced3. The two reagents must be physically separated

à electrons are forced to flow through external circuit to go from one reagent to the other

13-2. Galvanic Cells (Voltaic Cells)

Anode reaction : oxidation

Cathode reaction: reduction

Cd(s) Cd2+ (aq) + 2 e-

2 AgCl(s) + 2 e- 2Ag(s) + 2Cl-(aq)

When electrons flow from the left electrode tothe right electrode : positive voltage

When electrons flow from the right electrode to the left electrode : negative voltage

Cd(s) + 2 AgCl(s) Cd2+ (aq) +2Ag(s) + 2Cl-(aq)

High input impedance potentiometer (voltmeter)

(+)

Positive ions increase Positive ions decrease

The Ag+ ions in solution can directly reactat Cd(s) surface

à No flow of electrons through the external circuit

The purpose of salt bridge is to maintain electroneutrality (no charge buildup) through the cell

Physical separation

Cd(s) Cd(NO3)2(aq) AgNO3(aq) Ag(s)

A cell that will not work Salt Bridge: A cell that work

A quantitative description of the relative driving force for a half-cell reaction. A relative quantity vs standard hydrogen electrode [assigned to zero volt. E0(SHE)=0, 25 oC]

13-3. Standard (reduction) Potentials (activities of all species = 1)

H+ (aq, A = 1) + e- ½ H2 (g, A = 1)

E0(SHE)=0

SHE

Reduction :spontaneous

Oxidation:Spontaneous

Le Chatelier’s principle: increasing reactant concentrations drives the reacting to the right

The net driving force of the reaction is expressed by the Nernst equation

The Nernst equation tells us

the potential of a cell whose reagents are not all unit activity

13-4. Nernst Equation (activities of all species = 1)

Nernst Equation for a Half-Reaction

aA + ne- bB

------- (13.13)aA

bBo

AA

nFRTEE ln-=

ΔG = ΔGo + RT lnQ (Q; reaction quotient)

-nFE = -nFEo + RT lnQ (양변을 nF로나누어준다)

E = Eo –(RT/nF) lnQ

R: gas constant = 8.314 J/KmolT: temperature (K)

Nernst Equation at 25 oC

aA + ne- bB

------- (13.15))log(05916.0aA

bBo

AA

nVEE -=

QnFRTE

AA

nFRTEE o

aA

bBo lnln -=-=

QWhen all activities are unity, Q = 1 and ln Q = 0, thus E = Eo

If Q changes 10-fold, the potential changes by (59.16/n)mV

Example: Writing the Nernst Equation for a Half-Reaction

1/4 P4 (s, white) + 3H+ + 3e- PH3 (g) Eo = -0.046 V

)][

log(3

05916.0046.0 33

+--=HP

E PHSol:

Reduction of phosphorus to phosphine gas

Walter H. Nernst(1864 – 1941): 독일

1920년노벨화학상수상

Nernst Light Source(Infra Red)

Q: Find the voltage of the cell in Fig. 14-6

if the right cell contains 0.50M AgN03(aq) and the left cell contains 0.010M Cd(NO3)2(aq)

(Step 1)

Right electrode: 2Ag+ + 2e 2Ag(s) Eo+ = 0.799V

Left electrode: Cd2+ + 2e Cd(s) Eo- = – 0.402V

(Step 2)

Nernst equation for right electrode:

E+ = (Eo+) – (0.05916/2) log (1/[Ag+]2) = 0.799 – (0.05916/2) log(1/0.52) = 0.781 V

(Step 3)

Nernst equation for left electrode:

E- = (Eo-) – (0.05916/2) log (1/[Cd2+]) = – 0.402 – (0.05916/2) log(1/0.010) = – 0.461 V

(Step 4)

Cell voltage: E = E+ – E- = 0.781 – (– 0.461) = + 1.242 V

(Fig. 14-4) right electrode

(1) AgCl(s) + e- Ag(s) + Cl- Eo+ = 0.222 V

E+ = Eo+ - 0.05916 log [Cl-]

CdCl2 Cd2+ + 2Cl-

0.0167 M 0.0334 M

E+ = 0.222- 0.05916 log (0.0334) = 0.3093 V

(2) Ag+ + e- Ag(s) Eo+ = 0.799 V

E+ = 0.799 - 0.05916 log(1/ [Ag+])

AgCl Ag+ + Cl-

x x+0.0334

x(x+0.0334) = Ksp= 1.8 x 10-10, x = [Ag+] = 5.4 x 10-9 M

E+ = 0.799- 0.05916 log (1/5.4 x 10-9) = 0.3099 V

0.0334

Why different E+ values ?

1. Activity coefficient: neglected2. Inaccurate Ksp

Different Description of the Same Reaction

Your choice of reactions depend on whether concentrations of reactants is easier to figure out

It is nearly impossible to construct a standard electrochemical cell because we have no way to adjust the concentrations and ionic strength to give unit activities

In reality, activities less than unity are used in each half-cell, and the Nernst equation is used to extract the value of Eo from the cell voltage

Advice for Finding Relevant Half-Reactions

The Nernst Equation is Used in Measuring Standard Reduction Potentials

• A galvanic cell produces electricity because the cell reaction is not at equilibrium

• The potentiometer allows negligible current to flow

àThe concentration in each half-cell remains unchanged

13-5. Eo and Equilibrium Constant

0.412 VAt the outset

0 VAt equilibrium

[Cu2+] 증가[Ag+] 감소

Concentrations in the Operating Cell

A high-quality pH meter (voltmeter) à Resistance = 1013 Ω

If we measure 50 mV

Current = E/R = 0.05V/1013 Ω = 5 x 10-15 A

(5 x 10-15 C/s)/(9.649 x104 C/mol) = 5 x 10-20 mol e-/s

The production rate of Cd2+ = 2.5 x 10-20 mol/s ; negligible concentration change

If we replace the potentiometer with a wire, much more current would flow

à Concentrations would change until the cell reach equilibrium

E = E+ - E- = Eo+ - log( ) - [Eo- - log( )]n

0.05916aA

cC

AA

n0.05916

dD

bB

AA

E = Eo+ – Eo- - log( ) = Eo - log Q n0.05916

bB

aA

dD

cC

AAAA

Eo Q

n0.05916

Right electrode: aA + ne- cC : Eo+

Left electrode: dD + ne- bB : Eo-A + B C + D

When cell is at equilibrium, nothing would be driving the reaction,

and E = 0 and Q = K (DG = -nFE, DG = 0 à E = 0)

------- (13.23)Kn

E o log05916.0=

----- (13.22)

(14.14) becomes

Relationship between Eo and the Equilibrium Constant

Sol; 2Fe3+ + 2e- 2Fe2+ : Eo+ = 0.771 V

(-) Cu2+ + 2e- Cu(s) : Eo- = 0.339 V

Kn

E o log05916.0=

Example: Using Eo to Find the Equilibrium Constant

Q; Find the equilibrium constant for the reaction

Cu(s) + 2Fe3+ 2Fe2+ + Cu2+

Eo = Eo+ - Eo- = 0.771 – 0.339 = 0.432 V

Klog2

05916.0=

Therefore, K = 10(2x0.432)/0.05916 = 4 x1014

13-7. Formal Potential Eo’

Most important redox reactionsin living systems :Respiration(molecules of food is oxidized byO2 to yield energy and metabolicIntermediates)

If H+ is involved in the reaction,

Eo applies when pH = 0 (AH+ = 1)

pH inside a plant or animal cell:~ pH 7

The Formal Potential (Eo’): the reduction potential

under a specified set of conditions (pH, ionic strength, --)