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1 An Introduction to Electrochemistry Redox Reaction ox 1 + red 2 <=> red 1 + ox 2

An Introduction to Electrochemistry · An Introduction to Electrochemistry Redox Reaction ox 1 + red 2 red 1 + ox 2. 2 Outline 1. Review Concepts: Redox Reaction, Cell,

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Page 1: An Introduction to Electrochemistry · An Introduction to Electrochemistry Redox Reaction ox 1 + red 2  red 1 + ox 2. 2 Outline 1. Review Concepts: Redox Reaction, Cell,

1

An Introduction to

Electrochemistry

Redox Reaction

ox1 + red2 <=> red1 + ox2

Page 2: An Introduction to Electrochemistry · An Introduction to Electrochemistry Redox Reaction ox 1 + red 2  red 1 + ox 2. 2 Outline 1. Review Concepts: Redox Reaction, Cell,

2

Outline

1. Review Concepts: Redox Reaction, Cell, Potential

Balance an Equation

Nernst Equation

2. Redox Titration

3. Potentiometry

4. Voltammetry and Electrochemical Sensor

• Mass Balance: the amount of each element present at the

beginning of the reaction must be present at the end.

• Charge Balance: electrons are not lost in a chemical

reaction.

Half Reactions

• Half-reactions are a convenient way of separating

oxidation and reduction reactions.

12.1 Balancing Oxidation-12.1 Balancing Oxidation-

Reduction ReactionsReduction Reactions

Page 3: An Introduction to Electrochemistry · An Introduction to Electrochemistry Redox Reaction ox 1 + red 2  red 1 + ox 2. 2 Outline 1. Review Concepts: Redox Reaction, Cell,

3

Half Reactions

• The half-reactions for

Sn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe3+(aq)

are

Sn2+(aq) Sn4+(aq) +2e-

2Fe3+(aq) + 2e- 2Fe2+(aq)

• Oxidation: electrons are products.

• Reduction: electrons are reagents.

Balancing Equations by the Method of Half Reactions

• Consider the titration of an acidic solution of Na2C2O4

(sodium oxalate, colorless) with KMnO4 (deep purple).

Page 4: An Introduction to Electrochemistry · An Introduction to Electrochemistry Redox Reaction ox 1 + red 2  red 1 + ox 2. 2 Outline 1. Review Concepts: Redox Reaction, Cell,

4

How?

1. Write down the two half reactions.

2. Balance each half reaction:

a. First with elements other than H and O.

b. Then balance O by adding water.

c. Then balance H by adding H+.

d. Finish by balancing charge by adding electrons.

3. Multiply each half reaction to make the number of electrons equal.

4. Add the reactions and simplify.

5. Check!

Balancing Equations by the Method of Half Reactions

16H+(aq) + 2MnO4-(aq) + 5C2O4

2-(aq) 2Mn2+(aq) + 8H2O(l) + 10CO2(g)

Page 5: An Introduction to Electrochemistry · An Introduction to Electrochemistry Redox Reaction ox 1 + red 2  red 1 + ox 2. 2 Outline 1. Review Concepts: Redox Reaction, Cell,

5

Exercise:

Try to Balance:

Sn+2 + Fe+3 <=> Sn+4 + Fe+2

Fe+2 + MnO4- <=> Fe+3 + Mn+2

Sn+2 + Fe+3 <=> Sn+4 + Fe+2

Page 6: An Introduction to Electrochemistry · An Introduction to Electrochemistry Redox Reaction ox 1 + red 2  red 1 + ox 2. 2 Outline 1. Review Concepts: Redox Reaction, Cell,

6

Fe+2 + MnO4- <=> Fe+3 + Mn+2

Important Redox Titrants and the

Reactions

Oxidizing Reagents (Oxidants)

(1) Potassium Permanganate

MnO4

+ 8H+

+ 5e Mn2+

+ 4H2O

MnO

4+ 4H

++ 3e MnO

2(s) + 2H

2O

MnO4

+ e MnO4

2

Page 7: An Introduction to Electrochemistry · An Introduction to Electrochemistry Redox Reaction ox 1 + red 2  red 1 + ox 2. 2 Outline 1. Review Concepts: Redox Reaction, Cell,

7

Important Redox Titrants and the

Reactions

Oxidizing Reagents (Oxidants)

(2) Potassium Dichromate

Cr2O

7

2

+ 3U4 +

+ 2H+

2Cr3+

+ 3UO2

2++ H

2O

Cr2O

7

2

+14H+

+ 6e 2Cr3+

+ 7H2O

Important Redox Titrants and the

Reactions

Oxidizing Reagents (Oxidants)

(2) Potassium Iodate

IO3

+ 6H+

+ 5e1

2I

2+ 3H

2O

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8

Important Redox Titrants and the

Reactions

Reducing Reagent ( Reductants )

(2) Potassium Iodide

I1

2I

2+ e

Important Redox Titrants and the

Reactions

Reducing Reagent ( Reductants )

(2) Sodium Thiosulfate

2S2O32- S4O6

2- +2e

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9

12.2 Nernst Equation12.2 Nernst Equation

• Voltaic cells consist of

– Anode: Zn(s) Zn2+(aq) + 2e2

– Cathode: Cu2+(aq) + 2e- Cu(s)

– Salt bridge (used to complete the electrical circuit): cations

move from anode to cathode, anions move from cathode to

anode.

• The two solid metals are the electrodes (cathode and

anode).

Page 10: An Introduction to Electrochemistry · An Introduction to Electrochemistry Redox Reaction ox 1 + red 2  red 1 + ox 2. 2 Outline 1. Review Concepts: Redox Reaction, Cell,

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Fig. 12.1. Voltaic cell.

The salt bridge allows charge transfer through the solution and prevents mixing.

Page 11: An Introduction to Electrochemistry · An Introduction to Electrochemistry Redox Reaction ox 1 + red 2  red 1 + ox 2. 2 Outline 1. Review Concepts: Redox Reaction, Cell,

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Galvanic Cells

19.2

spontaneous

redox reaction

anode

oxidation

cathode

reduction

Page 12: An Introduction to Electrochemistry · An Introduction to Electrochemistry Redox Reaction ox 1 + red 2  red 1 + ox 2. 2 Outline 1. Review Concepts: Redox Reaction, Cell,

12

• This rearranges to give the Nernst equation:

• The Nernst equation can be simplified by collecting all

the constants together using a temperature of 298 K:

• (Note that change from natural logarithm to base-10 log.)

• Remember that n is number of moles of electrons.

E = E°RT

nFlnQ

E = E°0.0592

nlogQ

Nernst Equation

aA + bB + ne- <=> cC + dD

0.0592 [C]c[D]d

E = Eo - ---------- log ---------------- at 25oC n [A]a[B]b

E = E0 0.0592

nlog

[Re d]

[Ox]

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13

Standard Electrode Potentials

19.3

Standard reduction potential (E0) is the voltage associated

with a reduction reaction at an electrode when all solutes

are 1 M and all gases are at 1 atm.

E0 = 0 V

Standard hydrogen electrode (SHE)

2e- + 2H+ (1 M) H2 (1 atm)

Reduction Reaction

Standard Reduction (Half-Cell) Potentials

• The SHE is the cathode. It consists of a Pt electrode in a

tube placed in 1 M H+ solution. H2 is bubbled through the

tube.

• For the SHE, we assign

2H+(aq, 1M) + 2e- H2(g, 1 atm)

• E°red of zero.

• The potential of a cell can be calculated from standard

reduction potentials:

E°cell

= E°red

(cathode) E°red

(anode)

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Standard Reduction (Half-Cell) Potentials

• Consider Zn(s) Zn2+(aq) + 2e-. We measure Ecell

relative to the SHE (cathode):

E°cell = E°red(cathode) - E°Ox(anode)

0.76 V = 0 V - E°red(anode).

• Therefore, E°red(anode) = -0.76 V.

• Standard reduction potentials must be written as

reduction reactions:

Zn2+(aq) + 2e- Zn(s), E°red = -0.76 V.

Page 15: An Introduction to Electrochemistry · An Introduction to Electrochemistry Redox Reaction ox 1 + red 2  red 1 + ox 2. 2 Outline 1. Review Concepts: Redox Reaction, Cell,

15

12.3 Cell Diagram

anode//cathode

Zn/ZnSO4 (0.1 M)//CuSO4 (0.2

M)/Cu

If a cell,

Ecell = Ecathode - Eanode

if Ecell => positive, spontaneous as written

Page 16: An Introduction to Electrochemistry · An Introduction to Electrochemistry Redox Reaction ox 1 + red 2  red 1 + ox 2. 2 Outline 1. Review Concepts: Redox Reaction, Cell,

16

The more positive the Eo, the better oxidizing agent is the oxidized form (e.g., MnO4-).

The more negative the Eo, the better reducing agent is the reduced form (e.g., Zn).

©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)

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12.4 Redox Titration Curve

12.4 Redox Titration Curve

Page 18: An Introduction to Electrochemistry · An Introduction to Electrochemistry Redox Reaction ox 1 + red 2  red 1 + ox 2. 2 Outline 1. Review Concepts: Redox Reaction, Cell,

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12.4 Redox Titration Curve

Derivation of a titration curve

Fe+2 + Ce+4 <=> Ce+3 + Fe+3

EXAMPLE: Derive the titration curve for

50.00 mL of 0.0500 M Fe+2 with 0.1000 M

Ce+4 in a medium that is 1.0 M in H2SO4.

EXAMPLE: Derive the titration curve for 50.00

mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a

medium that is 1.0 M in H2SO4.

Fe+2 + Ce+4 <=> Ce+3 + Fe+3

E = Eo - (0.0592/n) log([red]/[ox])

At 0.00 mL of Ce+4 added, initial point

no Ce+4 present; minimal, unknown [Fe+3];

thus, insufficient information to calculate E

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Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with

0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4.

Fe+2 + Ce+4 <=> Ce+3 + Fe+3

At 15.00 mL of Ce+4 added, VFeMFe > VCeMCe

= 2.308 x 10-2 M

VCe MCe (15.00 mL)(0.1000 M)[Fe+3] = --------------- = ----------------------------

VFe + VCe (50.00 + 15.00)mL

EXAMPLE: Derive the titration

curve for 50.00 mL of 0.0500 M

Fe+2 with 0.1000 M Ce+4 in a medium

that is 1.0 M in H2SO4.

Fe+2 + Ce+4 <=> Ce+3 + Fe+3

At 15.00 mL of Ce+4 added, VFeMFe > VCeMCe

[Fe+3] = 2.308 x 10-2 M

VFe MFe - VCe MCe[Fe+2] = ------------------------

VFe + VCe

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EXAMPLE: Derive the titration

curve for 50.00 mL of 0.0500 M Fe+2

with 0.1000 M Ce+4 in a medium that

is 1.0 M in H2SO4.

Fe+2 + Ce+4 <=> Ce+3 + Fe+3

At 15.00 mL of Ce+4 added, VFeMFe > VCeMCe

[Fe+3] = 2.308 x 10-2 M

(50.00 mL)(0.0500 M) - (15.00 mL)(0.1000 M)[Fe+2] = --------------------------------------------------------

(50.00 + 15.00)mL

= 1.54 x 10-2 M

EXAMPLE: Derive the titration

curve for 50.00 mL of 0.0500 M Fe+2

with 0.1000 M Ce+4 in a medium that

is 1.0 M in H2SO4.

Fe+2 + Ce+4 <=> Ce+3 + Fe+3

At 15.00 mL of Ce+4 added, VFeMFe > VCeMCe

[Fe+3] = 2.308 x 10-2 M [Fe+2] = 1.54 x 10-2 M

for Fe+2 -> Fe+3 Eo = 0.69 v

0.0592 [Fe+2]Ecell = Ecell

o - ----------- log ----------- n [Fe+3]

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EXAMPLE: Derive the titration

curve for 50.00 mL of 0.0500 M Fe+2

with 0.1000 M Ce+4 in a medium that

is 1.0 M in H2SO4.

Fe+2 + Ce+4 <=> Ce+3 + Fe+3

At 15.00 mL of Ce+4 added, VFeMFe > VCeMCe

[Fe+3] = 2.308 x 10-2 M [Fe+2] = 1.54 x 10-2 M

for Fe+2 -> Fe+3 Eo = 0.69 v

0.0592 1.54 x 10-2

Ecell = 0.68 v - ---------- log ----------------- v = 0.69 v 1 2.31 x 10-2

EXAMPLE: Derive the titration

curve for 50.00 mL of 0.0500 M Fe+2

with 0.1000 M Ce+4 in a medium that

is 1.0 M in H2SO4.

Fe+2 + Ce+4 <=> Ce+3 + Fe+3

At 25.00 mL of Ce+4 added,

VFeMFe = VCeMCe, equivalence point

nFe EFeo + nCe ECe

o 1(1.44) + 1(0.68)Ecell = ----------------------- = ------------------------ v

nFe + nCe 1 + 1

= 1.06 v

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EXAMPLE: Derive the titration

curve for 50.00 mL of 0.0500 M Fe+2

with 0.1000 M Ce+4 in a medium that

is 1.0 M in H2SO4.

Fe+2 + Ce+4 <=> Ce+3 + Fe+3

At 26.00 mL of Ce+4 added, VCeMCe > VFeMFe

VFe MFe (50.00 mL)(0.0500 M)[Ce+3] = ---------------- = -------------------------------

VFe + VCe (50.00 + 26.00)mL

= 3.29 x 10-2 M

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EXAMPLE: Derive the titration

curve for 50.00 mL of 0.0500 M Fe+2

with 0.1000 M Ce+4 in a medium that

is 1.0 M in H2SO4.

Fe+2 + Ce+4 <=> Ce+3 + Fe+3

At 26.00 mL of Ce+4 added, VCeMCe > VFeMFe

[Ce+3] = 3.29 x 10-2 M

VCe MCe - VFe MFe[Ce+4] = ---------------------------

VFe + VCe

EXAMPLE: Derive the titration

curve for 50.00 mL of 0.0500 M Fe+2

with 0.1000 M Ce+4 in a medium that

is 1.0 M in H2SO4.

Fe+2 + Ce+4 <=> Ce+3 + Fe+3

At 26.00 mL of Ce+4 added, VCeMCe > VFeMFe

[Ce+3] = 3.29 x 10-2 M

(26.00 mL)(0.1000 M) - (50.00 mL)(0.0500 M)[Ce+4] = -------------------------------------------------------

(50.00 + 26.00)mL

= 1.32 x 10-3 M

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EXAMPLE: Derive the titration

curve for 50.00 mL of 0.0500 M Fe+2

with 0.1000 M Ce+4 in a medium that

is 1.0 M in H2SO4.

Fe+2 + Ce+4 <=> Ce+3 + Fe+3

At 26.00 mL of Ce+4 added, VCeMCe > VFeMFe

[Ce+3] = 3.29 x 10-2 M [Ce+4] = 1.32 x 10-3 M

for Ce+4 -> Ce+3 Eo = 1.44 v

0.0592 [Ce+3]Ecell = Ecell

o - ----------- log ----------n [Ce+4]

EXAMPLE: Derive the titration

curve for 50.00 mL of 0.0500 M Fe+2

with 0.1000 M Ce+4 in a medium that

is 1.0 M in H2SO4.

Fe+2 + Ce+4 <=> Ce+3 + Fe+3

At 26.00 mL of Ce+4 added, VCeMCe > VFeMFe

[Ce+3] = 3.29 x 10-2 M [Ce+4] = 1.32 x 10-3 M

for Ce+4 -> Ce+3 Eo = 1.44 v

0.0592 3.29 x 10-2

Ecell = 1.44 v - ------------ log --------------- v = 1.39 v 1 1.32 x 10-3

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Electrochemical Methods: Potentiometry•Reference Electrodes

•Indicator Electrodes

• pH Electrodes

• Ion-Selective Electrodes

Reference Electrodes

• A half-cell with a constant potential that is used as a

comparison for other measurements.

• Made of some stable chemical species, is readily

available and usually simple to use.

• Should be non-toxic if you are going to use it in a

biological system.

• Should be rugged and portable if you are going to use it

in the field!

• Several Varieties:

– S.H.E

– Ag/AgCl

– Calomel

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S.H.E. (again)• Cell made with hydrogen gas and hydrogen ion using a Pt

electrode

• Not practical for regular use due to the hydrogen gas

• Assigned a standard potential of zero by definition

Ag/AgCl• Constructed of a silver wire, coated with silver

chloride, in a solution containing silver chloride

– Sometimes AgCl is in solution saturated with KCl

– Saturated KCl keeps the chloride activity constant andhelps keep the electrode stable

AgCl(s)

+ e Ag(s) + Cl (aq )

E° = 222mV (Relative to S.H.E.)

E (sat. KCl) + 197 mV (relative to S.H.E.)

E = E°RT

nFln[Cl ]

RT

nF= 0.05916@25°C

IF we replace ln with base ten log!

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Page 30: An Introduction to Electrochemistry · An Introduction to Electrochemistry Redox Reaction ox 1 + red 2  red 1 + ox 2. 2 Outline 1. Review Concepts: Redox Reaction, Cell,

30

Calomel (S.C.E)• Calomel is mercury (I) chloride

• One of the most common reference

electrodes

Hg2Cl

2(s)+ 2e Hg

(1)+ 2Cl (aq )

E° = 268 mV (relative to S.H.E.)

E (sat. KCl) = 241 mV (relative to S.H. E.)

E = E°RT

nFln[Cl ]

2

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31

Conversions among different

scales (different electrodes)

• The reduction potential for iron (III)

under standard conditions is ? (look it

up)

– What is it if measured compared to the

Ag/AgCl reference electrode?

– A saturated calomel electrode?

Indicator Electrodes• Used for indicating

the potential caused

by some chemical

species as compared

to the reference

electrode.

• Usually connected to

the + (cathode) side

of the potentiometer

• Made of a variety of

often allegedly, but

never actually, inert

materials.

• Metal Electrodes

– Gold

– Platinum

– Carbon

• ?

• Solid carbon

• Carbon Paste

– Analyte of Interest

• Ion Selective

– Glass membrane

– Crystalline

• Solid State

– Liquid Membrane

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Typical Metal

Electrode

Arrangement

pH (Glass Membrane) Electrodes

• One of the simpler ion-selective electrodes (ISE)

• Hydrogen Ion imparts a charge across a hydrated

glass membrane

• Generally include an internal reference electrode

(Ag/AgCl) and a separate Ag/AgCl electrode for

sensing the charge imparted by the hydrogen ions

• Not as simple to use as you think!

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Basic Nernst Equation of a pH Electrode

(only the pH sensing cell, not the reference electrode)….

The constant is cancelled out by calibration in similar

Ionic strength solutions

is an efficiency factor that cancels out by calibration in similar

Ionic strength solutions.

So, what really matters are the activities of hydroben ions!

Don’t forget, electrodes respond to ACTIVITIES, even if we pretend

That they are concentrations to simplify our calculations.

E pH cell = constan t + x0.05916 x logA

H+ outside

AH

+inside

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Ionic Activity and Concentration

Craii

=

I

Izri

+=

1

51.0 log

2

Ionic Activity

Activity coefficient

Ionic Strength

Debye and Huckel in 1923

ISFET pH electrode

(Ion Selective Field Effect Transistor)

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What does 0.05916 mean?• It is a constant, if there is a one-

electron reaction

• It can be considered as the

equivalent of a constant of 59.16

mV

• A pH meter is a high-impedance

potentiometer (measures voltage)

• A pH change of “1” imparts a

change in 59.16 mV to the

potential recorded by the pH

meter!

• 1 pH unit change= 59.16 mV

7.00395.80

6.00336.64

5.00277.48

4.00218.32

3.00159.16

2.00100.00

pHmV (relative

readings)

Errors in pH Measurement….1. Junction potential due to the salt bridge and differences in junction

potentials over time due to contamination of the junction

• Overcome by regular recalibration

2. Sodium Error will result in high concentration of sodium solutions. The

sodium can also impart a charge across the glass membrane.

3. Acid error (strong acids) can saturate or contaminate the membrane

with hydrogen ion!

4. Equilibration Error is overcome by letting the electrode equilibrate with

the solution

5. Dried out glass membrane (ruins electrode)

6. Temperature. Since temperature affects activities, it is best to have all

solutions at the same, constant temperature!

7. Strong bases. Strongly basic solutions (>pH 12) will dissolve the glass

membrane!

8. Uncertainty in your buffer pH due to normal weighing, diluting errors.

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Ion Selective Electrodes……

• Selectivity Coefficient

– Defines how an ISE responds to the species of

interest versus some interfering species

• Interferences cause a signal (voltage) to be imparted on the

electrode that is NOT the result of the ion or chemical

species of interest

– You want the selectivity coefficient to be as SMALL

as possible

k(A ,X )

=electrode response to X

electrode response to A

Where X is the interfence and A is the analyte

E = constan t0.05916

nlog A

A+ k(A ,X ) A

X( )( )

Examples of Electrodes / Interferences

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Solid State Electrodes

• Uses a small amount of a doped crystal to transport charge from

the solution to an inner electrode

• The inner (sensing) electrode can be a Ag/AgCl electrode, and a

separate Ag/AgCl electrode can be present

– Combination electrode

• You can use a separate reference electrode also.

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• Typical equation (Fluoride Electrode):

• Use:

– Prepare calibration and sample solutions to

similar ionic strengths and temperatures

– Connect ISE and reference electrode to

potentiometer (pH meter in mV mode)

– Record potentials of calibration solutions

– Prepare calibration curve

– Measure potential of sample solutions and

calculate fluoride activity!

E = constan t 0.05916 log AF outside( )

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Liquid Membrane ISE’s

• Replace the solid

state crystal

with a liquid

ion-exchanger

filled membrane

• Ions impart a

charge across

the membrane

• The membrane

is designed to be

SELECTIVE for

the ion of

interest……

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VoltammetryVoltammetry can be used to

• Study electroactivity of ions and molecules at the

electrode/solution interface

• Probe coupled chemical reactions and measure

electron transfer rates

• Examine electrode surfaces

An electrochemical cell consists of a working (analyzing) electrode, an auxiliary

(counter) electrode, and a reference electrode. The control device is a potentiostat.

Clarifications of disk electrodes:

In quiescent solutions,

Conventional-sized Microelectrode Ultramicroelectrode Nanode

(>mm) (>25 mm) (0.1-25 μm) (nm)

In flowing or agitated solutions,

Rotating disk electrode, channel electrode, wall-jet electrode, and flow-through

tubular electrode.

Disk Working Electrode

Imbedded In Teflon or Kel-F

Pt/Au wire or coil auxiliary electrode

Page 42: An Introduction to Electrochemistry · An Introduction to Electrochemistry Redox Reaction ox 1 + red 2  red 1 + ox 2. 2 Outline 1. Review Concepts: Redox Reaction, Cell,

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In anodic stripping voltammetry (analysis), analytes are reduced and deposited into (onto) an

electrode. They are reoxidized during the stripping step.

e.g. Cd2+(aq) + 2e = Cd(Hg) Deposition Step

Cd(Hg) – 2e = Cd2+ (aq) Stripping

In the deposition step, solution is stirred or the electrode is rotated.

In cathodic stripping voltammetry, typically anions are oxidized and deposited onto an

electrode with subsequent stripping via a negative potential scan.

2I- + 2Hg -2e = Hg2I2 at a Hg electrode (deposition step)

Hg2I2 + 2e = 2 Hg + 2I- (cathodic stripping, reduction)

Because sample analytes are preconcentrated from a large-volume dilute solution into (onto)

a small-volume electrode under forced convection, trace analysis (enhanced sensitivity) can

be realized.

Stripping Analysis

Tissue-based biosensor

(voltammetric)

PPO: polyphenol oxidase

catechol