week 2 electrochemistry (full size)

MTE4599

Materials for energy technologies

So, which technologies ?• Wind turbines and thermal solar power plants; most main components are made from well-known materials like concrete and steal. Crucial materials engineering in the wind-turbine blades made from reinforced polymers – please refer to e.g. MTE4572.

Mainly the emerging technologies *Photo-voltaic cells (power producing solar-cells)

* Materials for water splitting (hydrogen production)

* Materials for storing and converting energy

* Fuel-cell materials

* Need electrochemistry !

Electrochemistry 1.0

Electrode potential

(batteries, fuel-cells and corrosion)

Electrochemical reactions

(water-splitting, dye sensitized solar-cells and reversible batteries)

Electrochemical analytical methods

Electrode potential

V

PART 1

Electrode potential

Electrode potential

Two equilibriums

Mg2+(aq) + 2e- � Mg

Cu2+(aq) + 2e- � Cu

(Standard) Electrode potential

Equilibrium E° (volts)

-3.03

-2.92

-2.87

-2.71

-2.37

-1.66

-0.76

-0.44

-0.13

0

+0.34

+0.80

+1.50

Standard conditions ? - 1M

(Standard) Electrode potential- If you can write the equilibrium equation…

Standard conditions ? - 1M and 1atm

Non metalsCl2 + 2e- � 2Cl- (+1.36V)

O2 + 4e- + 4H+ � 2H2O (+1.23V)

SO42+ + 2e- + 4H+ � SO2 + 2H2O (+0.17V)

S + 2e- + 2H+ � H2S (+0.14V)

CO2 + 2e- + 2H+ � HCOOH (-0.20V)

Carbon (?)

(Standard) Electrode potential- If you can write the equilibrium equation…

Standard conditions ? - 1M and 1atm

Non metalsCl2 + 2e- � 2Cl- (+1.36V)

O2 + 4e- + 4H+ � 2H2O (+1.23V)

SO42+ + 2e- + 4H+ � SO2 + 2H2O (+0.17V)

S + 2e- + 2H+ � H2S (+0.14V)

CO2 + 2e- + 2H+ � HCOOH (-0.20V)

Carbon (?)

“Redox couples”Fe3+ + e- � Fe2+ (+0.77V)

I3- + 3e- � 3I- (+0.53V)

Hydroquinone

Etc.

Electrode potential =>

Battery cell

Using standard reduction potentials:

0.34V – (- 0.76V) = 1.1V (?)

V

Salt bridge

Standard conditions ? And if not, call a friend !

A friend called Nernst

The free energy of the electrode reaction:

ΔG° = –nFE°

F

F = 96,485 C/mol


ΔG° = –nFE° (and ΔG = –nFE)



And

ΔG = ΔG° + RT ln Q (Q = the reaction quotient)

(http://en.wikipedia.org/wiki/Gibbs_free_energy)

Gives

–nFE = –nFE° + RT ln Q



And

ΔG = ΔG° + RT ln Q (Q = the reaction quotient)

Gives

–nFE = –nFE° + RT ln Q

Rearranged it becomes the Nernst Equation:

or

(at 25°C)

R = Gas constant = 8.314 J/(K*mol) F = 96,485 C/mol

Nernst in the lab

What is the electrode potential of Ag+ + e- � Ag at

various concentrations ?

E = 0.80 – 0.059/1*log(1/[Ag+])

(at 25°C)

Nernst in the lab

What is the electrode potential of Ag+ + e- � Ag at

various concentrations ?

E = 0.80 – 0.059/1*log(1/[Ag+])

(at 25°C)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

-10 -8 -6 -4 -2 0 2

V v

s S

HE

Log[Ag+]

E(Ag/Ag+)

Nernst in the lab

What is the electrode potential of Ag+ + e- � Ag ?

E = 0.80 – 0.059/1*log(1/[Ag+])

(at 25°C)

Concentration

batteries !- Any problems in

practical

applications ?

V

Nernst in the lab

What is the electrode potential of Ag+ + e- � Ag ?

- At different temperatures (1mM solution) ?

E = 0.80 -8.314*T

* ln(1/0.001)1*96,485

This temperature dependence is the

underlying principle for the “thermo-cell”

And also the problem of using

reference electrodes at

different temperatures !

Nernst Nernst Nernst Nernst and Water

Oxidation: H2(g) → 2H+ + 2e–

(at 25°C)

E = E° - (.059/2) × log (PH2/[H+]2)

Reduction: O2(g) + 4H+ + 4e– → H2O

E = 1.23 – (.059/4) x 4pH =

1.23 – 0.059 pH

And then what ?

E = 0 - (.059/2) × 2 pH = – 0.059 pH

Overall: 2H2(g ) + O2(g) → 2H2O

E = E° - (.059/4) × log (1/([H+]4 x PO2))

Nernst Nernst Nernst Nernst and Water

(at 25°C)

And then what ?

E = 1.23 – 0.059 pH

E = – 0.059 pH

Electrode potential => Fuel cell

A “continuous” battery…


1.23V – 0V = 1.23V

Splitting Water

Reduction: 2H+ + 2e– → H2(g)

E = – 0.059 pH

Oxidation H2O → O2(g) + 4 H+ + 4e–

E = 1.23 – 0.059 pH

Eh, what just happened ?

We went from measuring potentials on electrodes

to applying potentials to electrodes

– much more about that later under

Electrolytic cells and electrolysis(Next lecture or so)

The driving force in science:

What happens when…

Applying potential to

a half frog ?

Luigi Galvani, circa 1786,

electrochemistry and

electrophysiology began with

frogs' legs.

PART 2

Applying potential to an electrode

- in an electrolyte containing ions.

- Just a small potential though !

-

-

+

-

-

--

-

-

-

-

-

+

+

+

+

+ +

++

+




Double layer capacity

-

-

+

-

-

--

-

--

-

- +

+

+ +

+

++

+

+

++

+

++

+

+

+++

+

+

-

-

+

-

-

--

-

-

-

-

-

+

+

+

+

+ +

++

+

Applying potential to an electrode- in an electrolyte containing ions.


Double layer capacity• Surface area

• Salt and concentration

• Potential (!)

• Characteristic charging/discharging

curve at constant |currents|-

-

+

-

-

--

-

--

-

- +

+

+ +

+

++

+

+

++

+

++

+

+

+++

+

+

V

Time

+I -I

Applying potential to an electrode- in an electrolyte containing ions.

Higher potentials ?

-

-

+

-

-

--

-

--

-

- +

+

+ +

+

++

+

+

++

+

++

+

+

+++

+

+

V

Time

+I -I

????

????

Applying potential to an electrode- in water containing ions.

-

-

+

-

-

--

-

--

-

- +

+

+ +

+

++

+

+

++

+

++

+

+

+++

+

+

V

Time

+I -I

????

????

Applying potential to an electrode- in water containing ions.

-

-

+

-

-

--

-

--

-

- +

+

+ +

+

++

+

+

++

+

++

+

+

+++

+

+

V

Time

1.23V (?)

Water splitting-

-

+

-

-

--

-

-

-

-

-+

+

+ +

+

++

+

+

--

-

--

-

---

-

+

+

Applying potential to an electrode- in water containing ions,

constant current.

-

-

+

-

-

--

-

--

-

- +

+

+ +

+

++

+

+

++

+

++

+

+

+++

+

+ V vs SHE

Time

> 1.23V (?)

Water splitting

-

-

+

-

-

--

-

-

-

-

-+

+

+ +

+

++

+

+

--

-

--

-

---

-

+

+

Oxidation: H2O → O2(g) + 4 H+ + 4e–


Applying potential to an electrode(- in water containing ions and ?)

What else could happen ?

ANY reaction where the equilibrium

(“standard“) electrode potential is within

the potential range where water is stable.

Metal deposition:

Mz+ + ze- � M

Metal dissolution:

M � Mz+ + ze-

Gas reactions

e.g. SO2 + 2H2O � SO42+

+ 2e- + 4H+ (+0.17V)

“Redox” couples e.g.

Ferocene/ferocenium and

Iodine/iodide

(Oxidative) polymerization

e.g. conducting polymers

Not water ?

- same logic, but different potentials possible


Not water ?

- same logic, but different potentials possible

- “Electrochemical window”

E.g. Propylene carbonate

NB electrolyte salt (ions)

also need to stable !

> 4V


How much product are we getting from

electrochemical reactions ?

- Meet our second friend: M. Faraday

Faraday’s 1st law links the amount (mass/mol) of

material produced with the charge used (Q).

At constant current the number of moles is:

Importance in batteries ?


How much product are we getting ?

- Meet our second friend: M. Faraday

Please note that Faraday’s law is NOT taking the

potential needed to drive the reaction into account

=>

Faraday’s law can be used to estimate conversion

efficiency (“mole per coulomb”), but NOT to

evaluate ENERGY efficiency of the reaction.

(Remember: Power = Current x Voltage)

Applying potential to an electrode- in water containing ions,

constant current.

-

-

+

-

-

--

-

--

-

- +

+

+ +

+

++

+

+

++

+

++

+

+

+++

+

+ V vs SHE

Time

> 1.23V (?)

Water splitting

-

-

+

-

-

--

-

-

-

-

-+

+

+ +

+

++

+

+

--

-

--

-

---

-

+

+



A closer look;Equilibrium, over-potential and our next friends…

Cu2+ + 2e- � Cu

+0.34 V is the equilibrium potential.

This means that at this potential the reaction:

Cu2+ + 2e- �� Cu (cathodic exchange current density ic)

(charge transfer coefficient: αc) and

Cu2+ + 2e- Cu (anodic exchange current density ia)

(charge transfer coefficient (symmetry factor): αa)

The overall electrode current i is zero,

but the exchange current density i0 is not and

i0 = ||||ic|||| = ia at equilibrium

(NB. In one-step, one-electron reactions αc + αa = 1)


Defining the over-potential ηηηη = E - Eeq

The anodic and catodic exchange currents

can be written as a function of

• the exchange current density (io),

• the overpotential (η) and

• the symmetry factor (α)

ia = io[e(1-α)ηf] and

ic = io[-e(-α)ηf],

where f =nF/RT

We will not go into the

deeper reasoning

behind this math !

The main job for you

is to

accept/agree/imagine

that an exchange

current density exists !


Defining the over-potential ηηηη = E - Eeq

i(η) = ia + ic = io[ e (1 - α) η f – e -α η f ], where f =nF/RT

Known as the Butler-Volmer equation

A closer look;Overall current depending on exchange current density i0

i = ia + ic = io[ e (1 - α) η f – e -α η f ], where f =nF/RTHere for one-electron process n=1. Also α = 0.5

over-potential ηηηη = E - Eeq

A closer look;Overall current depending on charge transfer coefficient αααα

i = ia + ic = io[ e (1 - α) η f – e -α η f ], where f =nF/RTHere for one-electron process n=1


Can αααα be 1 ?At least close…

SO2 � SO42-

A good electro-catalyst ?

i = ia + ic = io[ e (1 - α) η f – e -α η f ], where f =nF/RT

High io for all steps in the reaction

e.g.

O2 + 4e- + 4H+ � 2H2O

Contains multiple steps

one for each electron transferred + adsorption and desorption

- And no α “against” you…

Take one of the currents in the Butler-Volmer equation

ln i = ln i0 + αηFn/RT Solving for η:

ηc = -b*log i0 + b*log i, (where b is 2.3RT/αnF)

This is the Tafel equation

Measuring io and α - meet our last friend: Julius Julius Julius Julius TafelTafelTafelTafel

i = ia + ic = io[ e (1 - α) η f – e -α η f ], where f =nF/RT


Tafel Slope is 2.3RT/αnF

Intercept “gives” log i0

Overall two possibilities:

1) Controlling current and measuring potential

2) Controlling potential and measuring current

But we can do it in many different ways depending on what you

want to get out of the measurement

A couple of examples…

- Chronovoltammetry (1)

- Chronoamperometry (2)

- Cyclic Voltametry (CV) (2)

Electrochemical Measurements

PART 3

Test Cell set-up: Two or three electrodes ?

Where the 3rd electrode is a reference electrode

- Chronovoltammetry - Chronoamperometry - Cyclic Voltametry (CV)

Electrochemical MeasurementsPART 3

Counter electrode



- Chronovoltammetry -Chronoamperometry -Cyclic Voltametry (CV)


Counter electrode

Ref CounterWorking

V

A

Potentiostat



- Chronovoltammetry -Chronoamperometry -Cyclic Voltametry (CV)


Ref CounterWorking

V

A

Potentiostat

Applying a constant current and measuring the potential

Typical use: Battery charge and discharge experiments

Capacitor characterization

Fuel-cell characterization

Metal deposition


Chronovoltammetry

PART 3



Capacitor characterisation


Metal deposition


Chronovoltammetry (1)

PART 3

C-rate

1C: charge or discharge in one hour

10C: charge or discharge in six minutes



Capacitor characterization


Metal deposition


Chronovoltammetry

PART 3

Applying a constant voltage and measuring the current

Typical use: Electro-catalytic measurements (over-potential)

Diffusion measurements (e.g. conducting polymers)

Stability test

Electro deposition


Chronoamperometry

PART 3

-2.5

-2.0

-1.5

-1.0

-0.5

0

-600 -400 -200 0 200

Ewe (mV vs. SCE)

I (m

A/c

m2 )

Steady state values

(after 1 hour) for

oxygen reduction

on

PEDOT/Au/Goretex

cathode

Applying a potential sweep and measuring the current

Typical use: Redox activity on electrodes

Diffusion measurements (!)

Determining limiting reaction (surface or bulk)

De-convoluting capacity form resistance and catalysis

Solar-cell characterisation

With rotation disk

Determining reaction (number of electrons involved)


Cyclic Voltametry (CV)

PART 3




Sweep- and Cyclic Voltametry (CV)

PART 3

V

I

Resistor R

I = 1/R*V





PART 3

V

I

Capacitor C

Q = C × V

I x t =C x V

I = C x V/t


Typical use: Redox activity on electrodes (e.g. Fe2+ � Fe3++ e-)



PART 3

Why this current response ?





PART 3





PART 3

Why the peak

current ?

Electrochemistry 1.0

• This was just the fast introduction !

• USE IT in the rest of the course !

• Next week: Materials for Fuel-Cells

Electrode potentials exists !

From last lecture:

Applying potential to an electrode- in acidic water containing ions,

constant current.

V vs SHE

Time

> 1.23V (?)

Water splitting


Reduction: 4H+ + 4e– → 2H2(g)

How much product are we getting ?





1.23V – 0V = 1.23V

Double layer

-

+++

++

+

+

++++

+

--

--

--

--

--

-

week 2 electrochemistry (full size)

Documents

emgcu2 aq

electrode reaction

redox couplesfe3 e fe2

mstandard electrode

electrode potentialbatteries

h 2h2o

nernst nernst nernst

h hcooh