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Through this course, we have learned that chemical processes involve energy transformations.

Can we apply what we have learned to design chemical systems to harness energy?

Unit 8How do we use chemical systems

to harness energy?

Chemical systems can thus be used to store

energy or to transform it into usable forms.

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M2. Inducing Electron Transitions.

M1. Controlling Electron TransferAnalyze electron transfer

between coupled systems.

Explore the effect of electron transitions in solid systems.

The central goal of this unit is to apply and extend central concepts and ideas discussed in this course

to design chemical systems to harness energy.

Unit 8How do use chemical systems to

harness energy?

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ContextTo illustrate the power of the concepts, ideas, and ways of thinking discussed in the course, we will

focus our attention on understanding how to design cleaner devices for energy transformation.

How would your chemistry

knowledge be useful in

identifying/designing alternative, cleaner sources of energy?

Why do we care?

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The Problem

There are a wide variety of chemical processes that release energy (exothermic) or transform energy

from one form to another.

The central question is how to apply chemical concepts, ideas, and ways of thinking to control

these energy transformations:

Amount, rate, and type of energy production

Reversibility

Nature of the products

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Module 1: Controlling Electron Transfer

Central goal:

To analyze electron transfer between

coupled chemical systems.

Unit 8How do we use chemical

systems to harness energy?

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The Challenge

In many chemical processes the redistribution of charge between reacting atoms or ions results in

the transformation of potential energy of the reactants into other forms of energy (heat, light).

TransformationHow do I change it?

How can we control the amount, rate, and type of new forms of energy

produced?

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The energy released in exothermic chemical reactions essentially results from charge

redistribution among reacting atoms.

Energy Changes

In these processes, electrons move from

states of higher to lower potential energy.

Ep

0

If we want to control the amount and rate of energy released, we need to control charge redistribution.

How can we do it?

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Basic ComponentsMany systems, natural and artificial, used to control

the rate of production of chemical energy include three basic parts:

I. A chemical system that undergoes an oxidation:

A – ne- An+

II. A chemical system that undergoes a reduction:

Bm+ + me- B

III. A mechanism to a) allow, b) control e- transfer:

Wire Molecule (e.g. protein) Reaction Sequence

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ILet’s Think

Analyze this galvanic cell:

Identify the three basic parts for energy production and harnessing;

Describe in detail what is happening in this system.

I. Oxidation: Zn(s) – 2e- Zn2+(aq)

II. Reduction: Cu2+(aq) + 2e- Cu(s)

III. Electrodes, wire, switch, electrolyte.

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Galvanic Cell

Batteries are a combination of one or more electrochemical cells using liquid (wet)

or solid (dry) electrolytes.

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ILet’s Think

Analyze this nano device:

Identify the three basic parts for energy production and harnessing;

Describe in detail what is happening in this system.

I. Oxidation: Asc – 2e- DhAsc

II. Reduction: 2H+(aq) + 2e- H2(g)

III. Transport proteins, light.

ON(+2) ON(+1)

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For these devices to be useful, the Gt for the overall process should be negative

(thermodynamically favored):

and the activation energy Ea for each half-reaction, as well as for the e- transfer process,

should be low (~RT) (kinetically favored).

Requirements

mA + nBm+ mAn+ + nB Gt = mG1 + nG2 < 0

A – ne- An+ G1

Bm+ + me- B G2

x m

x n

Same # of e- lost and

gained. [mA – mxn e-] + [nBm+ + mxn e-] mAn+ + nB

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Given the interest in processes for which Gt < 0, chemists have devised a procedure to determine

Gi for many half REDOX reactions using a common standard system (H+/H2) as a reference:

Reference System

Hydrogen Reference Electrode

Electric Potential Difference (Eo) = 0.34 V

To understand how this works, let’s analyze this case.

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Cu2+/Cu

Cu2+(aq) + H2(g) Cu(s) + 2H+(aq)

The measured electric potential difference across this standard cell is Eo

reduction(Cu2+/Cu) = +0.34 V.

H2 – 2e- 2H+

Cu2+ + 2e- Cu

}

Go (J) = - 2 x 9.6485x104 x 0.34 = -6.6x10 kJ (Favored; Cu2+ is a stronger oxidizer than H+)

Go for the process is a measure of the energy needed to transport electrons across this electric potential:

Go (J) = - n F Eocell

n- # of e- transferred in the unit reaction.F- Electric charge of 1 mol of e- (NAqe) = 9.6485x104 C/mol

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Zn2+/Zn

Now, let’s imagine we couple these processes:

Zn2+(aq) + H2(g) Zn(s) + 2H+(aq) H2 – 2e- 2H+

Zn2+ + 2e- Zn

}In this case we find Eo

reduction (Zn2+/Zn) = -0.76 V

Go (J) = -2 x 9.6485x104 x (-0.76) = +1.5x102 kJ (Not Favored; H+ is a stronger oxidizer than Zn2+)

Based on these results, would you expect this reaction to be favored or not favored:

Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq)Let′s think!

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Cu2+(aq) + H2(g) Cu(s) + 2H+(aq) Go1 = -6.6x10 kJ

Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) Go2 = -1.5x102 kJ

Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq) Got = -2.2x102 kJ

Certainly favored!!

Cell Potentials

We can simplify this analysis by simply looking at the Eo

reduction values:

Zn2+ + 2e- Zn Eo2 = -0.76

Implies H+ is a stronger oxidizer than Zn2+

Cu2+ + 2e- Cu Eo1 = +0.34

Implies Cu2+ is a stronger oxidizer than H+

Thus, Cu2+ should be able to oxidize Zn.

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Zn - 2e- Zn2+ Eoox = -Eo

red = +0.76

Cu2+ + 2e- Cu Eored = +0.34

Cell Potentials

Cu Zn

If Eocell > 0 then

Go = -nFEocell < 0

FAVORED!!!!

Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq) Eocell = Eo

red + Eoox

Eocell = +1.10 V

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Activity Series

Eored

Na

Na+

-2.71

Zn

Zn2+

-0.76

Fe

Fe2+

-0.40

H2

H+

0

Cu

Cu2+

+0.34

Cl-

Cl2

+1.36

Oxidizers

Reducers

The farther away in the activity series, the more thermodynamically favored the

(counterclockwise) reaction between the REDOX pairs.

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ILet’s Think

Given the available information:

Build the voltaic cell that will generate the maximum Eo

cell;

Predict the direction of e- transfer.

Write the chemical equation for the overall process;

Calculate its Go;

Eocell

(V)Eo

cell (V)

Cu2+ + 2e- Cu 0.34 Zn2+ + 2e- Zn -0.76

2H+ + 2e- H2 0.0 Ag+ + e- Ag 0.80

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Let’s Think

2Ag+(aq) + Zn(g) 2Ag(s) + Zn2+(aq) Zn2+ + 2e- Zn

2Ag + 2e- 2Ag+

}Eo

cell = 0.76 + 0.80 = 1.56 V Go = -nFEocell = 3.0x102 kJ

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Alternative Energy Sources?

Batteries and related electrochemical devices are seen as a very promising route for cleaner

portable energy sources.

Make a list of actual and potential advantages and disadvantages of

these types of devices.Let′s think!

Limited power supply

Limited durability

Need to be disposed or recharged

Heavy

Expensive

Smaller C footprint

Portable, Noiseless

Rechargable

Small, diverse sizes

Sealed products

DisadvantagesAdvantages

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Two of the most critical issues in battery technology are discharge time and rechargeability.

Critical Issues

Why does Ecell decreases with time of use?Why is it not rechargable?

Consider a prototypical alkaline battery:

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As the reaction proceeds, electrodes dissolve and cannot be regenerated by reversing the reaction.

Alkaline Battery

Einitial = 1.54 V

With use, the concentration of

reactants decreases and the reactions moves towards chemical

equilibrium (Ecell 0).

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Nernst Equation

For a REDOX process of the form:

mA + nBm+ mAn+ + nB

The value of Ecell is determined, in a first approximation, by Nernst equation:

nnm

nmnocellcell BA

BA

nF

RTEE

][][

][][ln

As the reaction

proceeds, Ecell decreases.

At equilibrium:

KnF

RTEE ocellcell ln 0)ln(

RT

Go o

enF

RT

nF

G

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Depending on their size (amount of reactants) and composition, batteries discharge at different rates.

Discharge Rates

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Rechargable Batteries

In common rechargable batteries, the products of the reaction tend to get attached to the electrode,

so the overall process may be reversed by providing energy.

Lead-Acid Battery

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ILet’s Think

Given the half-reactions that occur during discharge of a lead-acid battery:

PbO2(s) + 2 H+(aq) + H2SO4(aq) + 2 e- PbSO4(s) + 2 H2O(l) Eo

red = +1.685 V

Pb(s) + H2SO4(aq) - 2 e- PbSO4(s) + 2 H+(l)Eo

ox = +0.356 V

Write the overall reaction for this process and calculate the cell potential Eo

cell.

PbO2(s) + Pb(s) + 2 H2SO4(aq) 2 PbSO4(s) + 2 H2O(l)

Eocell = Eo

red + Eoox = 2.041 V

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ILet’s Think

PbO2(s) + Pb(s) + 2 H2SO4(aq) 2 PbSO4(s) + 2 H2O(l)

Eocell = 2.041 V

Write Nernst equation for this system and calculate Ecell at 25 oC for [H2SO4] = 4.5 M

(Typical concentration in a fully charged car battery).

242 ][

1ln

2041.2

SOHF

RTEcell

Ecell = 2.08 V

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Other OptionsIn general, the issue of rechargeability has been

addressed by looking for systems in which:

a) The electrocehmical process is easily reverted (e.g. Li-ion batteries in laptops, iPhones, etc.)

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Other Options

b) The reactants can be replenished when needed (e.g. H2 fuel cells for cars)

O2(g) + 4H+ + 4e- 2H2O(l)

2H2(g) – 4e- 4H+

Eocell = +1.23 V

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I Assess what you know

Let′s apply!

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Let′s apply! Calculate

Write the overall favored reaction for this electrochemical cell and calculate Eo

cell.

The battery of modern hybrid cars are nickel metal hydride (NiMH).

The other is based on nickel alone:

NiO(OH)(s) + H2O(l) + e- Ni(OH)2(s) + OH-(aq)Eo

red = +0.490 V

M(s) + H2O(l) + e- MH(s) + OH-(aq) Eored = -0.828 V

One electrode uses alloys (M) that can soak up hydrogen atoms:

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ILet′s apply! Estimate

What is the total voltage generate by this type of car battery?

Toyota Prius: 28 nickel metal hydride

modules—each containing six cells—connected in series.

1.32 x 6 x 28 = 221.8 V

NiO(OH)(s) + H2O(l) + e- Ni(OH)2(s) + OH-(aq)

MH(s) + OH-(aq) – e- M(s) + H2O(l)

NiO(OH)(s) + MH(s) Ni(OH)2(s) + M(s)Eo

cell = Eored + Eo

ox = +1.318 V

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Write the Nernst Equation for this cell. Analyze this equation and discuss its implications

for the properties of this battery.

Let′s apply! Analyze

NiO(OH)(s) + MH(s) Ni(OH)2(s) + M(s)

Eocell = +1.318 V

NiO(OH)(s) + H2O(l) + e- Ni(OH)2(s) + OH-(aq)

MH(s) + OH-(aq) – e- M(s) + H2O(l)

Ecell = Eocell

The cell potential does not depend on the concentration of the electrolyte, which may extend

the life f the battery and its rechargeability.

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I Working in pairs, identify one central idea introduced in this

module.

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Controlling Electron TransferSummary

System used to control the rate of production of chemical energy include three basic parts:

I. A chemical system that undergoes an oxidation:

A – ne- An+

II. A chemical system that undergoes a reduction:

Bm+ + me- B

III. A mechanism to a) allow, b) control e- transfer:

Wire Molecule (e.g. protein) Reaction Sequence

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Standard half-cell potentials Eored, measured in

regerence with a standard H+/H2 electrode, can be used to make predictions about the directionality

of a REDOX reaction.

Standard Potentials

Eored

A

An+

-2.71

H+

0

B-

Bm+

+1.36

Eocell = Eo

red + EooxmA + nBm+ mAn+ + nB

H2

Go (J) = - n F Eocell nnm

nmnocellcell BA

BA

nF

RTEE

][][

][][ln

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For next class,

Investigate what are the basic properties of semiconducting materials.

What chemical systems tend to be semiconducting?