chapter 15 electrochemical engineering

8/19/2019 Chapter 15 Electrochemical Engineering

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Chapter 15

Electrochemical Engineering


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What we will study

Electrochemical principlesAnodes and cathodes

Half cells and simple electrochemical cells

Fuels CellsRagone plot and battery capacities

Exploring Engineering


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Why electrochemical engineering?

Batteries and fuel cells are deeply embedded in “GreenEnergy” because solar and wind energy systems need tostore electrical energy

It’s also a hot topic because some vehicles use batteriesfor propulsion such as in hybrid cars and trucks

Electrochemical engineering is the study of whathappens inside batteries, fuel cells and ‘ultracapacitors’

So be prepared for a little more chemistry!



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Green Energy

“Green energy” refers to renewable energy suppliesthat do not spew forth greenhouse gases nor toxicimpurities

Wind and solar energy are two favored sources

Big disadvantage #1: Only works when the wind blowsor the sun shines

Big disadvantage #2: May make too much electricityexactly when you don’t need it.

Solution: Store the electrical energy until you do needit.



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Load leveling

Electricity not used whengenerated, nor available

when needed if the sun orwind go down

Solution: Battery storage

These may be very

large batteries!

Mismatch of green electric power and use



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Solar and Wind Need Equipment!

Batteries



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Why Do Batteries Work?

Matter is inherently electrically charged. Simplestcase is ionic bonding (i.e., attraction) in compoundssuch as common table salt: Na+Cl- in which Coulombicforces hold together positively charged sodium ionsand negatively charged chlorine ions. The forcebetween these ions is:

where e is the charge on an electron and r is theinterionic distance. k is the dielectric constant, whichis about 80 for water.

2

2

kr

eF =



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Electrochemistry

When Na+Cl- dissolves in water with k ≈ 80, the forcesbetween ions lessen allowing free ions as Na+ and asCl-

Ion-containing solutions are called “Electrolytes”

Overall ion-containing solutions are electricallyneutral

Locally ion-containing solutions have both charged

species at short distances to each otherBatteries use these ions when they can beseparated



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Electrolytes, Anodes and Cathodes

Electrodes are classifiedas anodes and cathodes

Anodes are “sources” ofelectrons, and cathodesare “sinks” for electrons

e- = electrons

E = Electrolyte

C = Cathode

A = Anode

e- flow

E

A C

- +



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Anodes are “sources”of electrons, andcathodes are “sinks”

for electronse- = electrons

E = Electrolyte

C = Cathode

A = Anode


e- flow

E

A C

+-


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Beware Franklin’serror!

e- = electrons

E = ElectrolyteC = Cathode

A = Anode

Conventional Current flow

E

A C

- +



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Lead-Acid Batteries

These are the batteries you find in a car

Both electrodes are based on lead, Pb one with aPbO2 coating

The electrolyte is sulfuric acid written H2SO4,

which dissolved in water is 2H+

/SO=

4 (the sulfateion has two electrons/molecule)

The principle anodic reaction is: Pb → Pb++ + 2e-

The two electrons flow through the external

circuit to the cathode on which:PbO2 + 4H

+ + SO4= + 2e- = PbSO4 + 2H2O



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Lead-Acid Batteries

Product of reaction is PbSO4

which precipitates duringdischarge and dissolves during charging.

The anodic voltage at the anode is 0.36V above areference cell and the cathodic is 1.69 V below.

Overall cell voltage = ~2.0 VA C

E

Anodic

0.36 V Reference

cell

Cathodic

1.69V



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Can You Power a Car Using Batteries ?

Property Lead-acidbattery

Gasoline

Mass 25. kg 25. kg

Energy 3,000 kJ 1.2 105 kJ

Mass energy storage density 120 kJ/kg 46,500 kJ/kg

Volumetric energy storagedensity

250 kJ/liter 34,400 kJ/liter

Power 5 kW Typically >100 kW

This battery is too heavy, contains too little energy, anddelivers too little power – that’s why hybrids are a

popular substitute.



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The Electrochemical SeriesCell voltage set by the tendency to transfer

electrons

Half Cell Chemistry Potential in volts

Li+ + e

-Li(s) –3.05 V

Na+ + e

- Na(s) –2.71 V

Mg++ + 2e- Mg(s) –2.37 VZn

++ + 2 e

- Zn(s) –0.76 V

Fe++

+ 2 e- Fe(s) –0.44 V

Ni++

+ 2e- Ni(s) –0.25 V

2H+ + 2 e

- H2(g) 0.00V (Hydrogen ½ cell is defined as zero)

Cu

++

+ 2 e

-

Cu(s) 0.34 VCu

+ + e

- Cu(s) 0.52 V

Ag+ + e

- Ag(s) 0.80 V

Pd++

+ 2e- Pd(s) 0.95 V



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The Electrochemical Series

These half-cell reactions in principle arereversible. The more negative the more theywant to lose electrons; the more positive themore to gain them

This determines how cells will behave



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Daniell cell

The electrolytes areZnSO4(aq) with a Zn anodeand CuSO

4

(aq) with a Cucathode. Write down thereactions in each half celland explain what happensin the salt bridge. What is

the voltage?



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Daniell cell

In bulk aqueous solution , Zn

++

and SO4=

must be inbalance with each other.

( )4 4ZnSO Aq Zn SO++ =↔ +

But the anode is also dissolving and thus yields

some locally extra Zn++ ions according to:

Zn(s) Zn 2e (V 0.76V)

++ −↔ + = +



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Daniell cell

Electrons from the anode flow through the externalcircuit precipitating copper at the cathode:

( )Cu 2e Cu s 0.34V++ −+ ↔ +We have removed copper ions from solution; therefore

there must be a corresponding reduction in SO4= ions

from the electrolyte. They must move into the salt

bridge to exactly counteract the Zn++ ions from the

anodic side.

The cell potential is equal to:

(+0.76 V) + (+0.34 V) = +1.10 V.



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Electroplating

An electroplater wants to coat a 10.0 cm by10.0 cm copper plate with 12.5 micrometersof silver. How many electrons must pass in

the external circuit? How many coulombsare passed? If the plating takes 1,200. swhat’s the electrical current in amperes in

the external circuit?



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Electroplating

Know: Atomic mass of Ag is108 kg/kmol. Its density is10,500 kg/m3. Avogadro’snumber (NAv) is 6.02 × 10

26

atoms/kmol. What we callcurrent is nothing but therate of flow of electrons, so1.00 A = 1.00 coulomb/s and

one electron carries –1.60 ×10-19 coulombs.

Ag+, NO3- , H2O

Ag Cu

+ -

Electron

flow

Conventional

current flow



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Electroplating

The reaction at the anode is Ag(s) → Ag+ + e-

and the reaction at the cathode is Ag+ + e- →Ag(s); hence one atom of silver dissolves at

the anode and one atom of silver isdeposited at the cathode. For each atom ofsilver dissolving at the anode and depositingat the cathode, one electron must circulate in

the external circuit.



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Electroplating

[ ]-6

22 324

-3

12.5×10 ×100.×10,500 cmmMass =μm cm kg/m /μm m1.00×10

=1.31×10 kg

• Next convert to kmols:

[ ] kmols1022.1kg

kmolskg

108

1031.1kmols 5

3−

−

×=

×=



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Electroplating

atoms1032.7

]/][[1032.7

1002.61022.1

deposited atomsAg

21

21

265

×=

×=×××

=

−

kgatomskg

Next convert to atoms of Ag(s):



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Electroplating

A975.0]C/s[

.1200

1017.1currentHence

coulombs1017.1

]coulombs/e][[6010.11032.7

3

3

-1921

=×

=

×=

×× −− e

Next convert to mass of Ag(s):



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Fuel CellsFuel cells are just continuously refueled batteries.

They will not discharge while electrochemicalfuel is being fed to them.

Most fuel cells depend on “Proton Exchange

Membrane” or “PEM” to catalyze electrodereactions

RxnCathodic 022244

RxnAnodic442

22

2

H OH H Oe H

e H H

⇒+⇒++

+⇒−+−+

−+



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Fuel Cells

http://aq48.dnraq.state.ia.us/prairie/images /fuelcell.jpg


http://aq48.dnraq.state.ia.us/prairie/imageshttp://aq48.dnraq.state.ia.us/prairie/imageshttp://aq48.dnraq.state.ia.us/prairie/images


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Fuels cells

Note: Only H2 and O2(i.e.,air) in and only H2Oout.

Cell voltage is 1.23 V for overall

rxn H2 + O2 = H2O

Apparently no green house gas pollution!

Unfortunately to make H2 needs copious CO2



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The Ragone ChartBatteries must supply both energy and power

Typically batteries supply current measured at mA/cm2 ofelectrode area at a few volts

The more electrode area, the greater the current; this maybe internal area packing or simply more cells placed in

parallelThe more cells in series the higher the voltage

But it’s the application that demands whether the cellcan deliver both enough energy (say mileage between

charges on an electric car) and power (say to passanother car)



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The Ragone Chart

The best measures of energy and power efficiency aretheir mass densities: e = E /Wt (Whr/kg) and p = P/Wt(W/kg)

The energy density delivered by a power source for a

time t is simply e = p × t.Take log base 10 of this equation:

log10 e = log10 p + log10 t

Plot the log10 energy density of a battery vs. log10power density for the same battery and you getthe Ragone Plot



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The Ragone Chart

Modified from a graphic of Maxwell Technologies: http://www.maxwell.com.



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The Ragone Chart

Very convenient way to compare differentelectrochemical sources

Ideally you want to have your cake and eat it by

being in the upper right cornerReality shows what can be achieved by competingelectrochemical sources

“Ultracapacitors” are storage devices that can

store thousands of times the energy capabilityof an electrical capacitor.



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Ragone Chart

That that the form oflog10 e = log10 p + log10t is y = mx + c and has a

slope of m =1 giventhe log10 scaling inchart.

The battery’s discharge

time is given by e/ p

1,000

100

10

1

0.1

0.01

E n e r g y , W

h r / k g

10 100 1,000 10,000

Power, W/kg

1 h o u

r

3 6 0 s

3 6 s

3. 6 s

0. 3 6 s

3 6 m s

1 0 h o

u r s



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Summary

Green energy and load storage and leveling

Electrochemical series

Simple electrical cells

Simple electrochemistry

Principles of fuel cells

Ragone chart to characterize

E l i E i i

chapter 15 electrochemical engineering

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