Brdy 6Ed Ch20 Electrochemistry

8/9/2019 Brdy 6Ed Ch20 Electrochemistry

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Chapter 20:Electrochemistry

Chemistry: The Molecular Natureof Matter, 6E

Jespersen/Brady/Hyslop


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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E

Electrochemistry

Study of chemical reactions that can produceelectricity or use electricity to produce desired

product. Study of interchange of chemical and electrical

energy

Electrochemical reaction always involvesoxidation-reduction reactions

Electron transfer reactions

Electrons transferred from one substance toanother

Also called redox reactions

2


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Oxidation-Reduction Reactions

Oxidation and reduction reactions occur inmany chemical and biochemical systems

Combustion reactions

Photosynthesis

6CO2 + 6H2O C6H12O6 + 6O2

Mitochondrial Respiration NADH NAD+

Methane monoxygenase

CH4 + NADH + H+ + O2 CH3OH + NAD

+ + H2O


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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 4

Oxidation Reduction Reactions

Involves two processes:

Oxidation loss of electrons from one reactant

Na (s) Na+ + e Oxidation Half Reaction

Reduction gain of electrons from another

reactantCl2 (g) + 2 e

2 Cl Reduction Half Reaction

Net reaction

2 Na (s) + Cl2 (g) 2 Na+ + 2 Cl

Oxidation and reductionALWAYS occur together

Can't have one without the other


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Oxidation-Reduction Reaction

Oxidizing Agent Substance that accepts the e's

Takes e's from another substance

is itself reduced

e acceptor

Cl2 (g) + 2 e

2 Cl

Reducing Agent

Substance that donates e's

Releases e's to another substance is itself oxidized

e donor

Na (s) Na+ + e


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Oxidation Numbers Way to keep track of e'sRules for Assigning Oxidation Numbers

1. Sum of all oxidation numbers of atoms in moleculeor polyatomic ion must equal charge on particle

2. Oxidation number:

1. of any free element is zero (0)2. of any simple, monoatomic ion is equal to charge on ion

3. of fluorine in its compounds is 1

4. of hydrogen in its compounds is +15. of oxygen in its compounds is 2

3. If there is a conflict between 2 rules, apply the

rule with lower number and ignore conflicting rule.


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Define Oxidation-Reduction in Terms

of Oxidation Number Oxidation

in oxidation number or more + oxidationnumber

Leo (Loss of es)

Oil (oxidation is e loss) Reduction

in oxidation number or more oxidationnumber

Ger (Gain of es)

Rig (Reduction is e gain)


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Your Turn!

What is the oxidation number of Cl in HClO4?

A. +1B. +3

C. +5

D. +7

H is +1 and O is -2. There are 4 O atoms. +1 +(4)(-2) + x = 0

x = +78


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Spontaneous Redox Reactions When a copper wire is placed in a solution of

silver nitrate Silver metal spontaneously precipitates

Copper ion spontaneously forms as evidenced byblue color of solution


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Electrochemistry Possible to separate oxidation and reduction

processes and cause to occur in two differentlocations

Can use spontaneous redox reaction to produceelectricity

Can use electricity to make nonspontaneous redoxreactions happen

Biology does this by coupling nonspontaneous

redox reactions with spontaneous reactions thatprovide the driving force.

Can harness electrical energy to do work


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Galvanic (or Voltaic) Cell

Electrochemical cell in which spontaneousredox reactions occur

Produce electricity spontaneously Energy released by this spontaneous redox

reaction can be used to perform electrical

work Transfer of electrons takes place through an

external pathway (wire) rather than directly

between reactants Ex. Batteries used to power laptops, cell phones,

camera, etc.


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Spontaneous Redox Reactions

Ex 1. Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s)

Ex 2. Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq)

Lets look at Ex.1 in detail Imagine dividing up the reaction in Ex. 1 into

individual oxidation and reduction half reactions

Cu(s) Cu2+(aq) + 2e 2Ag+(aq) + 2e 2Ag(s)

Physically this can be accomplished by having a

strip of the given metal (called an electrode) ina solution of the corresponding ion


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Example of Spontaneous Redox

Silver metal placed in a solution of AgNO3 Copper metal placed in solution of Cu(NO3)2 Each compartment is called a half-cell


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Half-Cells

When metal ion collides withelectrode and gainselectrons, ion becomesreduced

If metal atom on surface ofelectrode loseselectrons,becomes oxidized

Left on their own, each individual cell quickly establishes an

equilibrium between metal and ions in solution.

M(s) Mn+(aq) + ne


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Electrode Names By combining 2 different half-cells we can

cause es to flow from 1 cell to the other

One half-cell undergoes oxidation = anode

Other half-cell undergoes reduction = cathode

Anode = electrode at which oxidation (e

loss) occurs Cathode = electrode at which reduction (e

gain) occurs An Ox and a Red Cat


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Learning Check:

Identify cathode and anode in these acidicsolutions

CrO3 (s) + MnO2 (s) MnO4 (aq) + Cr3+ (aq)

CrO3

(s) + 6H+ + 3 e Cr3+ (aq) + 3H2

O (redn)

MnO2(s) + 2H2O MnO4(aq) + 4H+ + 3e (oxidn)

Cathode = reduction = CrO3 Cr3+ (aq) half rxn

Anode = oxidation = MnO2(s) MnO4(aq) half rxn


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Learning Check:

Identify the cathode and anode in these acidicsolutions

H2O2(aq) + CO2(g) H2C2O4(aq) + O2(g)

1H2

O2

(aq) O2

(g) + 2H+ (aq) + 2e (oxidn)

2CO2 (g) + 2H+ + 2e H2C2O4 (aq) (redn)

Cathode = reduction =

2CO2 (g) H2C2O4 (aq) half rxn

Anode = oxidation = H2O2 (aq) O2 (g) half rxn


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Your Turn!

Identify the half-reaction that occurs at the anode forthe reaction:

4H+(aq) + 3S(s) + 4NO3-(aq) 3SO2(g) + 4NO(g) + 2H2O

A. 3S(s) + 6H2O 3SO2(g) + 12H+(aq) + 12e-

B. 12e- + 4NO3-(aq) + 16H+(aq) 4NO(g) + 8H2OC. 3S(s) + 6H2O 3SO2(g) + 12H

+(aq) + 8e-

D. 8e- + 4NO3-(aq) + 16H+(aq) 4NO(g) + 8H2O

Oxidation occurs at the anode and oxidation is the loss ofelectrons. Answer C is not a balanced equation.

18


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How can we get electrons to migrate intothe other solution?

Connect the two solutions with a metal wire that willallow es to pass from one cell to another

Now es can flow, but the reaction still will not initiate.

Why? Consider what would happen to solution if es did flow

from one cell to the other


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At copper end, 2 es are given up Cu(s) Cu2+(aq) + 2e

Travel across with wire to the silver half-cell

2Ag+

(aq) + 2e

2Ag(s) What happens to the charge in each half-cell?

Each time the reaction occurs Net charge at copper end by 2

Net charge at silver end by 2

Violates principle of electroneutrality Cannot have solution with a net charge

We need to balance the charge in

order for the es to flow

Electron Flow Between Half-Cells


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Two Types of Charge Conduction

Metallic Conduction External to the cell

Electrical charge is transported from one electrodeto the other by movement of electrons through thewires

How metals conduct electricity Electrolytic Conduction

Inside electrochemical cells

Electrical charge is carried through the liquid bymovement of ions

Transport of electrical charge by ions


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Needed to complete circuit Tube filled with solution of an electrolyte

Salt composed of ions not involved in cell reaction

KNO3 and KCl often used

Porous plugs at each end of tube Prevent solution from pouring out

Enable ions from salt bridge to migrate betweenhalf-cells to neutralize charges in cell compartments Anions always migrate toward anode

Cations always migrate toward cathode

Salt Bridge

Salt bridge


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Summary of a Galvanic Cell

Half-cells (compartments containing reactantsfor each half-reaction)

Electrodes to conduct current through solution Wire connecting two half-cells

Salt bridge to offset ionmovement

Resistance to electrical flow

Supporting electrolyte


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Your Turn!

Which of the following species would beappropriate for a salt bridge solution?

A. AgCl

B. C12H22O11 sucrose (sugar)

C. NaCl

D. C6H6

The solution needs to be an electrolyte. AgCl isnot soluble and the organic compounds are not

ionic.25


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Cell Reaction

Net overall reaction in the cell To get add individual half-reactions, after make

sure that number of es gained in reduction = number of es

lost in oxidation

Cu(s) Cu2+(aq) + 2e

(oxidation)2Ag+(aq) + 2e 2Ag(s) (reduction)

2Ag+(aq) + Cu(s) 2Ag(s) + Cu2+(aq)

Use ion electron method to balance half-reactions (see Ch 6)


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Standard Cell Notation

Drawing sketches of electrochemical cells canbe cumbersome

Simpler representation called standard cellnotation used instead

Cu(s)|Cu2+(aq)||Ag+(aq)|Ag(s) Anode electrode where oxidation occurs is

placed on left

Cathode electrode where reduction occurs inplaced on right

anode cathode


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Cell Notation

Cu(s)|Cu2+(aq)||Ag+(aq)|Ag(s)

Single slash = boundary between phases (solidelectrode and aqueous solution of ions)

Double slash represents salt bridge Separates cell reactions

In each half (half-cell)

Electrodes appear at outsides Reaction electrolytes in inner section

Species in same state separated with ;

Concentrations shown in ( )

anode cathode

anode

electrode

anode

electrolyte

cathode

electrolyte

cathode

electrode

Salt Bridge


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Learning Check

Write the standard cell notation for thefollowing electrochemical cells:

Fe (s) + Cd2+

(aq)

Cd(s) + Fe2+

(aq)Anode = ox = Fe(s)

Cathode = red = Cd2+(aq)

Fe(s)|Fe2+(aq)||Cd2+(aq)|Cd(s)

Al(s) + Au3+(aq) Al3+(aq) + Au(s)

Anode = ox = Al(s)

Cathode = red = Au3+(aq)

Al(s)|Al3+

(aq)||Au3+

(aq)|Au(s)


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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E30

Learning Check

Where there are no conductive metals involvedin a process, an inert electrode is used. C(gr)or Pt(s) are often used. Write the Standard

Cell notation for the reactions

H2O2(aq) + CO2(g) H2C2O4(aq) + O2(g)

Anode = ox = H2O2(aq) | O2(g)Cathode = an = CO2(g) | H2C2O4(aq)

C(gr)|H2O2(aq); H+(aq)|O2(g)||CO2(g)|H2C2O4(aq); H+(aq)|C(gr)

Pt(s)|H2O2(aq); H+(aq)|O2(g)||CO2(g)|H2C2O4(aq); H

+(aq)|Pt(s)


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Your Turn!

Write the standard cell notation (Pt electrodes) for thefollowing reaction:

2Mn3+(aq) + 2I-(aq) Mn2+(aq) + I2(s)

A. Pt(s)|Mn3+(aq); Mn2+(aq)||I-(aq)|I2(s)|Pt(s)

B. Pt(s)|I-(aq)|I2(s)||Mn3+(aq); Mn2+(aq)|Pt(s)

C. Mn3+(aq)|Pt(s); Mn2+(aq)||I-(aq)|I2(s)|Pt(s)

D. Pt(s)|Mn3+

(aq); I-

(aq)||Mn2+

(aq)|I2(s)|Pt(s)

Oxidation reaction is on the right and reduction

reaction is on the left of the salt bridge (||).31


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Measuring Cell Potential

Instead of just a wire we can connect avoltmeter between 2 half-cells

Voltmeter Measures potential difference between 2 half-cells

Unit of Potential Volt (V)

Measure of amount of energy(Joules, J) that can bedelivered per SI unit ofcharge (coulomb, C)

1 V = 1 J/C

l


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Voltage

Voltage is driving force for reactions Greater voltage = greater driving force

Voltage or potential of galvanic cell varies withamount of current flowing through the circuit

Voltage is always positive in direction of aspontaneous reaction

Cell Potential (Ecell)

Maximum voltage that a given cell can generate

Depends on Composition of electrodes

[Ions] in half-cells

Temperature


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Standard Cell Potential, Ecell

To compare potentials for different cells, weneed a set of standard conditions

E

cell means at standard conditions T = 25C = 298 K

All gases at P = 1 atm

All [Ions] = 1 M

Rarely larger than a few volts Ex. 2Ag+(aq) + Cu(s) 2Ag(s) + Cu2+(aq)

If you need batteries with higher voltages,arrange several cells in series Ex. car batteries


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Electrical Potential Cell potential

Competition between two half-cells

Every half-cell has tendency to gain electrons

and proceed as reduction half-reaction Reduction Potential, Ered

Relative ease of gaining electrons

Standard Reduction Potential, Ered Reduction potential measured under standard

conditions All [solutes] = 1M

All gases at P = 1 atm

T = 25C

St d d C ll P t ti l E


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Standard Cell Potential, Ecell When two half-cells are connected The one with larger (more +) Ered acquires electrons

Undergoes reduction

The one with lower (more ) Ered donates electrons Undergoes oxidation

Measure cell potential, Ecell

Represents magnitude of difference between Ered of

one half-cell and Ered of other half-cell Is always taken as positive number for spontaneous

redox reactions Ecell = E

red E

red

Of substance

reduced

Of substance

oxidized

l l i


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Ex: Calculating Ecell Look at reaction in zinc-copper cell Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s)

Zinc is oxidized

Copper is reduced

Half reduction reactions

Zn2+

(aq) + 2 e

Zn (s) Cu2+ (aq) + 2 e Cu (s)

Reaction for Cu2+ must have greatertendency to proceed than Zn2+

as Cu2+ is the one reduced

Ecell = E

Cu2+ E

Zn2+

Refe ence Elect ode


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Reference Electrode

No way to measure reduction potential of isolatedhalf-reaction Can only measure difference in potential between 2 half-cells

To assign Ered to half reactions

Must arbitrarily choose a reference electrode (half-reaction)

Set its potential to exactly 0.00V

Standard Hydrogen Electrode H2 gas at 1 atm bubbled over

Pt electrode coated with finely divided Pt to provide large surface area

T = 25C

1.00 M [H+]

Standard Hydrogen Electrode


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Standard Hydrogen Electrode

2H+(aq, 1.00M) +2e H2(g, 1 atm) EH+ = 0.00 V

Double arrow indicates only that reaction is

reversible Not that there is true equilibrium

Whether oxidation or reduction occurs depends on

what half-reaction this is paired with How to list in cell notation

Pt(s), H2(g)|H+(aq)||Cu2+(aq)|Cu (s)

Zn(s)|Zn2+(aq)|| H+(aq)|Pt(s), H2(g)

Ex. Galvanic cell of Cu/Cu2+ with Hydrogen


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Ex. Galvanic cell of Cu/Cu with HydrogenElectrode

Need to know which half-cell undergoes reduction andwhich undergoes oxidation

When measuring potential in galvanic cell Cathode (reduction) carries + charge Anode (oxidation carries charge

Use voltmeter to determine potential

Shows that Cu carries + charge (cathode)

Ex Galvanic cell of Cu/Cu2+ with


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Ex. Galvanic cell of Cu/Cu2+ with

Hydrogen ElectrodeCu2+(aq) + 2e Cu(s) (cathode)

H2(g) 2H+(aq) + 2e (anode)

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

E

cell = E

reduced E

oxidized

Ecell = E

cathode E

anode

Ecell = ECu2+ E

H+

0.34 V = ECu2+ 0V

ECu2+ = 0.34 V

Ex Galvanic cell of Zn/Zn2+ with


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Ex. Galvanic cell of Zn/Zn2+ with

Hydrogen Electrode In this case the hydrogen electrode is positive so it

is the cathode

Zn(s) Zn2+(aq) + 2e (anode)2H+(aq) + 2e H2(g) (cathode)

Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) (cell reaction)

Ecell = E

reduced E

oxidized

E

cell = E

cathode E

anode Ecell = E

H+ E

Zn2+

0.76 V = 0.00V EZn2+

EZn2+ = 0.76 V

Standard Reduction Potentials (E )


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Standard Reduction Potentials (E red)

See Table 20.1 in the Jespersen textAll tabulated as reduction potentials

oxidized form + electrons reduced form

Can be ions, elements or compounds

Arranged from top to bottom by Ered

High values of E

red (E

red > 0) meanssubstance is easily reduced

More positive value of Ered = morelikely to

undergo reduction

F2(g, 1atm) + 2 e 2 F(aq, 1M) E = 2.87 V

Most easily reduced

Standard Reduction Potentials (E )


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Standard Reduction Potentials (E red)

Low value of Ered (Ered < 0) means substanceis easily oxidized More negative value of Ered = less likely to

undergo reduction

Li+ (aq, 1M) + e Li(s) E = 3.05 V More likely the reverse reaction occurs

Li(s) Li+ (aq, 1M) + e E = +3.05 V

When reaction reversed, sign of E reversed

All substances are compared to H+

, which has aEred of 0.00 V. Sign of Ered is sign of electrode when attached to

H+

/H2

Using E to Calculate E


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Using E cell to Calculate E red Ex. Go back to galvanic cell with

2Ag+(aq) + Cu(s) 2Ag(s) + Cu2+(aq)

Ecell = 0.46V Cu(s) undergoes oxidation

ECu

= + 0.34 V

Ag+(aq) undergoes reduction

What is Ered for the half-reaction of Ag+?

Ecell = EAg+ ECu +0.46 V = EAg+ (+0.34 V)

E

Ag+ = 0.46 V + 0.34 V = +0.80 V

Can we Use E to Predict


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Can we Use E red to Predict

Spontaneous Reactions?Yes!!

For substances in two half-reactions

Substance with more positive Eredalways occurs as written

Reduction Other half-reaction with less positive Ered

always forces to run in reverse

Oxidation

Predicting Reaction Spontaneity


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When have list of Ered as given in Table 20.1 Half-reaction with more positive potentials

(higher up in table) and half-reaction occurs as

written Other half-reaction (lower in table) is reversed and

occurs as oxidation

For spontaneous reaction Reactants found in left side on higher half-reaction

and on right side of lower half-reaction



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Ex. What spontaneous reaction occurs when Crand Au are added to a solution of Au3+ andCr3+ ?

Cr3+(aq) + 3e Cr(s) E = 0.74 VAu3+(aq) + 3e Au(s) E = +1.50 V

Au is more + so reduction

Cr (oxidations) so half reaction is reversed

Au3+(aq) + 3e Au(s)

Cr(s) Cr3+(aq) + 3e

Au3+(aq) + Cr(s)Au(s) + Cr3+(aq)

Is this Answer Reasonable?


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Is this Answer Reasonable?

Use reduction potentials to predict Ecell for thereaction.

Cr3+(aq) + 3e Cr(s) E = 0.74 V

Au3+(aq) + 3e Au(s) E = +1.50 V

Net Reaction =

Au3+(aq) + Cr(s)Au(s) + Cr3+(aq)

Ecell = E

reduced E

oxidized

Ecell = E

Au3+ E

Cr3+

Ecell = +1.50 V (0.74 V)

Ecell = +2.24 V

Calculating Cell Potentials E


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Calculating Cell Potentials, E cell Predict the reaction that will occur when Mg

and Cd are added to a solution containingMg2+ and Cd2+ ion.

Cd2+(aq) + 2e Cd(s) E = 0.40 V

Mg2+

(aq) + 2e

Mg(s) E

= 2.37 V Since Cd2+ higher in Table (more positive), it

is reduced and Mg is oxidized

Cd2+(aq) + 2e Cd(s) (reduction)Mg(s) Mg2+(aq) + 2e (oxidation)

Cd2+(aq) + Mg(s) Cd(s) + Mg2+(aq)

Calculating Cell Potentials E ll


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Calculating Cell Potentials, E cell

What would be the cell reaction and cellpotential for the galvanic cell employing the

following half-reactions? Co2+(aq) + 2e Co(s) E = 0.28 V

Au3+

(aq) + 3e

Au(s) E

= +1.50 VAu3+ has more positive Ered so

Au3+(aq) + 3e Au(s) reduction

Co(s) Co2+(aq) + 2e oxidation

2Au3+(aq) + 3 Co(s) 2Au(s) + 3Co2+(aq) net

2[

3[ ]

]

Calculating Cell Potentials, E ll


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Calculating Cell Potentials, E cellEcell = Ereduced EoxidizedEcell = E

Au3+ E

Co2+

Ecell

= +1.50 V (0.28 V)

Ecell = +1.78 V

Note:Although multiplying half-reaction by factors

to make electrons cancel, Do NOT multiply Ered by these factors

Ered are intensive quantities, do not depend on

amount Units = V

Same number of joules available for each coulomb of

charge regardless of total number of electrons shown

Predicting if Reaction is Spontaneous


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Predicting if Reaction is Spontaneous

Given any redox reaction

Calculate Ecell for reaction as written

If E

cell > 0 V (positive), then reaction isspontaneous

Galvanic cell = spontaneous

Ecell always positive

If Ecell < 0 V (negative), then reaction is

nonspontaneous Electrolytic cell = nonspontaneous

Ecell always negative

Using Ecell to Predict Spontaneity


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Using E cell to Predict Spontaneity

Determine whether the following reaction isspontaneous as written

6I

(aq) + BrO3

(aq) + 6H+

(aq) 3I2(s) + Br

(aq) + 6H2O I2(s) + 2e

2I(aq) Ered = + 0.54 V

BrO3

(aq) + 6H+(aq) + 6e Br(aq) + 6H2O

Ered = + 1.44 V BrO3

(aq) + 6H+(aq) + 6e Br(aq) + 6H2O redn

3 x [2I(aq) I2(s) + 2e ] oxidn

Ecell = EBrO3 E

I2

Ecell = 1.44 V (+0.54 V)

E

cell = +0.90 V spontaneous

Using Ecell to Predict Spontaneity


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Using E cell to Predict Spontaneity

Determine whether the following reaction isspontaneous as written

Au(s) + Al3+(aq) Au3+(aq) + Al(s)

Al3+(aq) + 3e Al(s) E = 1.66 V

Au3+(aq) + 3e Au(s) E = +1.50 V

Au(s) Au3+(aq) + 3e oxidation

Al3+(aq) + 3e Al(s) reduction

Ecell = EAl3+ EAu3+

Ecell = 1.66 V (+1.50 V)

Ecell

= 3.16 V nonspontaneous

Your Turn!


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Is the following reaction spontaneous and what is itsstandard cell potential?

2H2O2(aq) + S(s) SO2(g) + 2H2O

H2O2(aq) + 2H+(aq) + 2e- 2H2O Eo = +1.77 VSO2(g) + 4H

+(aq) +4e- S(s) + 2H2O Eo = +.45 V

A. spontaneous; +2.22V

B. spontaneous; +1.32V

C. spontaneous; +3.09V

D. non-spontaneous; -1.32V Eocell = (+1.77 0.45) V = +1.32 V

spontaneous since Eocell > 0

56

Are Ecell and G Related?


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cell

Yes, both let us predict spontaneity of reaction For chemical reaction

G = maximum work system can do In electrical system

Maximum work= nFEcell Where n= number of moles of electrons transferred

F= Faradays constant = number of Coulombs of

charge equivalent to 1 mole of e

1 F= 96,486 C

Ecell = potential of cell in Volts

Relationship between Ecell and G


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p cell

( ) Je

e =

=

coulomboules

molcoulombsmolWorkMax

Equating the 2 expressions for maximum work gives

G = nF

Ecell

If using standard potentials, you can calculate

standard free energy changes

G = nFEcell

n F Ecell

Calculating Cell Potentials, Ecell


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cell

VC

JV

e

CeG

11

1)16.3(

mol1

485,96)mol3(o

59

Ex. Calculate G for the following reaction

Au3+(aq) + Al(s) Au(s) + Al3+(aq)

Ecell = +3.16 V

Solving

Std conditions so T = 25C = 298 K

G = 914,678 J/mol = -915 kJ/mol

Applications of Electrochemistry


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Determination of Equilibrium Constant, K If we combine

G = RTlnKC

with

G = nFEcell

We get nFEcell = RTlnKC

Rearranging gives

Ccell KnRTE lnF

=o

Determine KC from E

cell


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C cell

Ex. Calculate KC for the redox reaction6I(aq) + BrO3

(aq) + 6H+(aq) 3I2(s) + Br

(aq) + 6H2O

E

cell = +0.90 V Determine n

BrO3

(aq) + 6H+(aq) + 6e Br(aq) + 6H

2O redn

6I(aq) 3I2(s) + 6e oxidn

So n= 6 T = 298 K

R= 8.3145 J/(molK)

Determine KC from E

cell


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C cell

Next Rearrange eqn to get lnKc

Plug values into equation

lnKc = 210.3

Ecell = +0.90 V, G = 521 kJ/mol, Kc = 2.1 10

90

Reaction spontaneous; goes to completion

RTnEK cellC

F

o

=ln

KKmolJ

molCCJKC 298/3145.8

485,96690.0ln

=

903.210

101.2 == eKC

Ccell

Kn

RTE ln

F

=o

Your Turn!


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Determine the equilibrium constant, KC , for areaction that has a standard cell potential of

+2.55 V and transfers four electrons in the

reaction process.

A. infiniteB. 1.00

C. 1.6 x 10397

D. 3.2 x 10172

63

Your Turn! - Solution


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172

2.55 / 4 96,485 /ln

8.3145 / 298ln 397.2

3.2 x 10

C

C

C

J C C mol K

J mol K K K

K

=

=

=

Relationship between Ecell and Ecell


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Ecell = cell potential at standard conditions All gases at 1 atm pressure

All ion concentrations = 1.00M

25C

When concentrations, pressures or temperature

changes, Ecell changes Just as G changes

Ex. In operating battery

Potential gradually as reactants are used up andcell reaction approaches equilibrium

When reaches equilibrium, Ecell = 0

Relationship between Ecell and Ecell


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To determine effect of [ion] on Ecell look athow G affected by [ion]

G =

G

+ RT lnQ Since G = nFEcell and G = nFE

cell

Then

nFEcell = nFEcell + RT lnQ

Dividing both sides by nF gives

Nernst equation

QnRTEE cellcell lnF

= o

Using the Nernst Eqn


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Need to construct mass action expression forredox reaction in either

[ ] in M for solutions

P in atm for gases

Ex. For the reaction,

Zn(s) + 2 H+(aq)

Zn2+(aq) + H2 (g)

2

2

]H[

]Zn[

ln2

+

+

=

H

cellcell

P

n

RTEE

F

o

2

2

]H[

]Zn[

ln2

+

+

=

HP

Q

Ex. Using the Nernst Eqn Calculate the cell potential for a galvanic cell


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Calculate the cell potential for a galvanic cellemploying the following half-reactions if[Fe3+]=0.0100 M, [Fe2+]=0.250 M,[Co2+]=0.050 M,

Fe3+(aq) + e Fe2+(aq) Ered = +0.77 V

Co2+(aq) + 2e Co(s) Ered = 0.28 V

Solution Fe3+ has more + Ered , so reduction

Fe3+(aq) + e Fe2+(aq) reduction

Co(s) Co2+(aq) + 2e oxidation

2Fe3+(aq) + Co(s) 2Fe2+(aq) + Co2+(aq) net rxn

]2 [

Ex. Using the Nernst Eqn n = total number of e transferred = 2


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n= total number of e transferred = 2

2Fe3+(aq) + Co(s) 2Fe2+(aq) + Co2+(aq)

Ecell = EFe

3+ ECo2+

Ecell = 0.77 V (0.28V) = +1.05 V

Plug in R, T, nand concentrations

E

cell = +1.05 V 0.0738 V = 0.98 V

23

222

]Fe[

]Co[]Fe[ln

2 +++

=F

RTEE cellcell

o

2

2

]0100.0[

]050.0[]250.0[ln

/485,962

298)/(3145.8V1.05

molC

KKmolJEcell

=

)100.1(

)03125.0(ln)V8012.0(V1.05

4=cellE

Ex. Using the Nernst Eqn


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In order to determine H+ concentration, agalvanic cell was constructed by connecting aCu electrode dipped in 0.50 M CuSO4 to a H2electrode at gas pressure of 1 atm dipped inthe H+ solution of unknown concentration. A

value of 0.49 V is recorded for Ecell at 298K.What is the pH of hydrogen half-cell?

Step 1: First write balanced redox reaction

and determine Ecell. H2 (g) + Cu

2+ (aq) 2H+ (aq) + Cu (s)

E

cell = ECu2+ EH+ = 0.34 V 0.00 V = 0.34 V

Ex. Determination of pH from Ecell


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Step 2: Determine the form of the Nernstequation for this reaction

n = 2 as 2 electrons are transferred

Step 3: Solve for [H+]

][

][ln

2 2

2

+

+

=Cu

HRTEE cellcell

F

o

)050.0(

][

ln)/500,962

298)/(314.8

34.049.0

2+

=

H

molC

KKmolJ

VV

)050.0(

][lnV01284.0V34.0V49.0

2+

= H


72/126

Your Turn!


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What is the pOH of the following reaction when thepressure of O2 is 1 atm, the temperature is 25

oC, the

concentration of Al3+ is 1.00M, and the cell potential

reads 1.341V?3O2(g) + 6H2O + 4Al(s) 4Al

3+(aq) + 12OH-(aq)

O2(g) + 2H2O + 4e- 4OH-(aq) Eo = +0.401V

Al3+(aq) Al(s) + 3e- Eo = -1.66V

A. 12.99

B. 7.00C. 1.01

D. Cannot be determined from the data given

73


74/126

Electrolytic Cell


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Nonspontaneous Use electrical energy to force nonspontaneous

reaction to occur

Electrodes switch relative to galvanic cells What was cathode becomes anode where oxidation occurs

What was anode becomes cathode where reduction occurs

Type of reaction that occurs when you rechargebatteries

Substance undergoing electrolysis must be molten or

in solution so ions can move freely and conduction canoccur

Example Electrolysis Cell


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Molten NaCl cell (801C) Inert electrodes placed into

molten liquid and connected

to electrical source Cathode: Na+() + e Na()

Anode: 2Cl() Cl2 (g) + 2e

Multiply first half-cell by 2to balance electrons

2Na+() + 2e 2Na()

2Na+() + 2Cl() 2Na() + Cl2(g) (Net) Since nonspontaneous

2Na+() + 2Cl() 2Na() + Cl2

(g)

cathode anode

electrolysis

Electrolytic vs. Galvanic Cells


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Electrolysis means application of electricity Electrolytic means that a particular reaction is not

spontaneous

Electrolytic Cell Galvanic CellCathode is positive (reduction)

Anode is negative (oxidation)

Non-spontaneous

Requires a battery

Cathode is positive (reduction)

Anode is negative (oxidation)

Spontaneous

Is a Battery

In both galvanic and electrolytic cells, positive ionsmove to cathode and negative ions move to anode Attracted there in electrolytic cell by charges on electrodes

Diffuse there in galvanic cell to balance developing charges

Electrical Conduction


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Only occurs because reactions take place at surface ofelectrode

Consider molten NaCl electrolytic cell When charged anode placed in molten NaCl Positive charge attracts Cl ions positive charge on electrode pulls e away from Cl to form Cl

atoms that combine to form Cl2


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Electrolysis Reactions in Aq. Solution


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More complex May have competingreactions Must consider possible

oxidation and reduction ofsolute

Must also consider redox

reactions of water Ex. K2SO4 in H2O Expected products

K and S2O82 Actual products H2 and O2 gases

cathode anode

SO42

Electrolysis of Aqueous K2SO4


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Why this difference in products? Must examine reduction potential data

Cathode

K+ (aq) + e K(s) E = 2.92 V

2H2O + 2e H2(g) + 2OH

(aq) E = 0.83 V

H2O much easier to reduce than K+

Anode

S2O82(aq) + 2e 2SO4

2 E = +2.01 V

O2(g) + 4H+(aq) + 4e 2H2OE = +1.23 V

S2O82 easier to reduce than O2

So H2O easier to oxidize than SO42

Electrolysis of Aqueous K2SO4 Alternatively calculate E


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ycell

1st for K+ and S2O82 reaction

Then for H2O to H2 and O2

K+

(aq) + e

K(s) 2SO4

2 S2O82(aq) + 2e

Ecell = EK+ ES2O82 = 2.92 V (+2.01 V)

Ecell = 4.93 V 2H2O + 2e

H2(g) + 2OH(aq)

2H2O O2(g) + 4H+(aq) + 4e

Ecell = EH2O EO2= 0.83 V (+1.23 V)

Ecell = = 2.06 V

So water reaction costs much less energy

Electrolysis of Aqueous K2SO4


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Net reaction

2H2O + 2e H2(g) + 2OH

(aq)

2H2O O2(g) + 4H+(aq) + 4e

6H2O O2(g) + 2H2(g) + 4OH(aq) + 4H+(aq)

Net = 2H2O

H2(g) + O2(g) So what is the role of K2SO4?

If not present, no electrolysis occurs

K2SO4 is the electrolyte K+ and SO4

2 carry charges through solution

Without it no current can pass through solution

2 [ ]

4H2O

2


84/126

Using Reduction Potentials to Predict


85/126


Electrolysis Products Possible anode reactions Oxidation of Br or H2O

Br2(aq) + 2e 2Br(aq) E = +1.07 V O2(g) + 4H

+(aq) + 4e 2H2OE = +1.23 V

EO2

more + so

O2 easier to reduce and Br easier to oxidize

Expected reaction

Cu2+(aq) + 2e Cu(s) (cathode) 2Br(aq) Br2(aq) + 2e

(anode)

Cu2+(aq) + 2Br(aq) Cu(s) + Br2(aq) (net)

Electrolysis of CuBr2 in H2)


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Anode Cathode

Blue color is Cu2+

Black bits on electrode

are Cu metal

Yellow-orange

color is Br2

Learning Check

h h d d f h


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What are the expected products for theelectrolysis of an aqueous solution of Na2S withtwo platinum electrodes connected to anexternal electrical source?

Possible cathode reactions

Reduction of Na+ or H2O

Na+(aq) + e Na(s) E = 2.71 V

2H2O + 2e H2(g) + 2OH

(aq) E = 0.83 V

H2O easier to reduce

Using Reduction Potentials to Predict

El t l i P d t


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Electrolysis Products Possible anode reactions Oxidation of S2 or H2O

S(s) + 2e S2(aq) E = 0.48 V O2(g) + 4H

+(aq) + 4e 2H2OE = +1.23 V

EO2

more + so

O2 easier to reduce and S2 easier to oxidize

Expected reaction

2H2O + 2e H2(g) + 2OH(aq) (cathode)S2(aq) S(s) + 2e (anode)

2H2O + S2(aq) H2(g) + S(s) + 2OH

(aq) (net)

Your Turn!

Whi h f th f ll i ti i t lik l


89/126


Which of the following reactions is most likelyto occur at the anode?

A. Cl2 + 2e- Cl- Eo = +1.36V

B. I2 + 2e-I- Eo = +0.54

C. Br2 + 2e-

Br- Eo = +1.07D. Li+ + e- Li Eo = -3.05V

Lithium is the easiest to oxidize

89

Michael Faraday

D t i d th t t f h i l h


90/126


Determined that amount of chemical changeoccurring during electrolysis is directlyproportional to amount of electrical charge

passed through cellAmpere (A)

SI unit of electrical current

Coulomb (C) SI unit of charge

1 coulomb = 1 ampere 1 second

1 C = 1A s

Faraday (F) 1F = 96,485 C/mol e

Kinetics of Electrolysis

F d E ti It


91/126


Faradays Equation q = charge (coulombs, C)

I = current (Amperes, A, or C/s)

t = time (s) n = moles of electrons transferred in process

F= Faradays constant (96,485 C/mol)

Units tell us how these quantities are related

This equation addresses kinetic aspects of

electrochemical cells

FnItq ==

F)()(mole

)(time)(current)(coulombs

Faradaysne

sAC

=

=

Electroplating

In Faradays equation use number of moles


92/126


In Faradays equation, use number of molesof electrons transferred, ne Because we cant see electrons, we gauge

this by the amount of metal deposited or lost Using the half-reaction and stoichiometry, we

can relate the metal to the number of molesof electrons

entcoefficien

tcoefficien

MM

molesmass

m

e

metal

metalmetal =

Calculations Related to Electrolysis

In the electrolysis of a solution containing Ni2+

( ) t lli Ni ( ) d it th th d


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(aq), metallic Ni (s) deposits on the cathode.Using a current of 0.150 A for 12.2 min, what

mass of nickel will form?Faradays

(mole e)

Current

& time

Charge (C)

Mass (g) reactant

(or product)

A s = C

F

mol en

mol reactant1

Mol reactant

(or product)

MM

Solution

1 Calculate charge passing through cell in


94/126


1. Calculate charge passing through cell in12.2 min

Coulombs (C) = current (A) time (s)

C = 0.150 A 12.2 min 60 s/min

C = 109.8 C

2. Calculate moles of electrons

== emol

C

emolCemol 10138.1

485,96

18.109 3

Solution (cont.)

3 Calculate amount of Ni in g


95/126


3. Calculate amount of Ni in g

= 0.033 g Ni

Could also do the entire calculation in one

step:

Ni1

Ni69.58

2

Ni110138.1 3

mol

g

emol

molemol

Ni033.0Ni1

Ni69.58

2

Ni1

1

485,96

1min/60min2.12150.0

gmol

g

emol

mol

sA

C

C

emolsA

=

Calculations Related to Electrolysis How long must a current of 0.800 A flow to

form 2 50 g of silver metal in an electroplating


96/126


form 2.50 g of silver metal in an electroplatingexperiment? The cathode reaction is Ag+(aq) +e Ag(s)

Faradays

(mole e)

Current

& time

Charge (C)

Mass (g) reactant

(or product)

A s = C

Fmol en

mol

reactant1

Mol reactant

(or product)

MM


97/126

Solution3. Calculate time required


98/126


Could also do the entire calculation in onestep:

min6.4660

min1

1

1

0.800

2236

)(current

)(Charge=

==

sC

sA

A

C

A

Ctime

min6.4660

min1

0.800

/1

1

485,96

Ag1

1

Ag07.91

Ag1Ag500.2

=

sA

Cs

emol

C

mol

emol

g

molg

Your Turn!60.3 minutes are required to deposit 1.00 g of

ld th th d f l ti f 1 00M


99/126


gold on the cathode from a solution of 1.00M

Au2+ solution. How many amps are used in the

electroplating process?

A. 0.250 AB. 0.125 A

C. 15.0 A

D. 0.0625 A

99

Your Turn! - Solution


100/126


1 Au 21.00 Au

196.97 Au 1 Au96,485 1 1min

i65.3min 601

i = 0.250 amps

mol mol e g

g mol C

smol e

=

Applications of Electrochemistry Batteries

G l i ll i ti


101/126


Galvanic cells in action

Electroplating

Electrolysis in action

101

Batteries Galvanic cells that generate portable electrical

energy


102/126


energy Usually a collection of several linked in series to

get higher voltages

Ex. Car batteries,usually 6 cellseach capable ofproducing 2 Vso Net = 12 V

Two Major Classes of BatteriesPrimary Cell Secondary Cell

Non rechargeable Rechargeable


103/126


Non-rechargeable

Ex. Alkaline dry cell

Rechargeable

Pb storage Battery

Alkaline Battery Zn/MnO

2battery 1.5 V

Uses basic or alkaline electrolyte


104/126


Uses basic or alkaline electrolyte Not rechargeable Compared to dry cell

Longer shelf life Delivers higher currents Less expensive

Reactions: Anode

Zn (s) + 2OH (aq) ZnO (s) + H2O + 2e

Cathode2MnO2 (s) + H2O + 2e

Mn2O3 (s) + 2OH (aq)

NetZn (s) + 2MnO2 (s) ZnO (s) + Mn2O3 (s)

Nicad Battery 1.4 V

Nickel Cadmium storage cellRecha geable


105/126


g Rechargeable

High energy density

Release energy quickly

Rapidly recharged

Used in portable power tools,

CD players, electric cars Reactions:

Anode: Cd(s) + 2OH(aq) Cd(OH)2(s) + 2e

Cathode: NiO2(s) + 2H2O + 2e Ni(OH)2(s) + 2OH (aq)

Net: Cd(s) + NiO2(s) Cd(OH)2(s) + Ni(OH)2(s)

Disadvantage = Cd toxicity, so disposal a problem

Properties that modern battery designers consider for

applications such as

Important Properties of Batteries


106/126


pp Cell phones, laptops, digital cameras, cordless tools and

pacemakers

Shelf-life How long do batteries hold their charge when not in use?

Rate of energy output

High currents Energy density

Ratio of available energy to battery volume

Specific energy Ratio of available energy to weight

Ni-MH Batteries Nickel Metal Hydride

Rechargeable


107/126


a g ab

1.35V

Some metals and alloys can

absorb H2 gas and effectively

store it, release to redox reaction

Advantages: 50% more power/volume than NiCad

Useful longer

Reactions:Anode: MH(s) + OH(aq) M(s) + H2O + e

Cathode: NiO(OH)(s) + H2O + e Ni(OH)2(s) + OH

(aq)

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

Lithium Ion Batteries

Rechargeable high specific energy (due tolow mass)


108/126


g g p gy (low mass)

High energy density (due to very () E0red)

Dont actually involve true oxidation andreduction

Li+

ions can slip between layers of atoms insolids such as graphite or LiCoO2 Process called intercalation

Li+ ion battery is based on transport of Li+

ions


109/126

Fuel Cells

Galvanic cells in which reactants are continuouslysupplied with reagents


110/126


supplied with reagents

Able to operate as long as supply of reactants is

maintained Attractive power source for long-term generation of

electricity

Major advantages: Clean burning

No electrode loss

Easily replenished High operational temperature

Highly efficient

Hydrogen-Oxygen Fuel Cell

Early design Cathode: O2(g) + 2H2O + 4e

4OH(aq)


111/126


Cathode: O2(g) 2H2O 4e 4OH (aq) Anode: H2(g) + 2OH

(aq) 2H2O + 2e

Net: 2H2(g) + O2(g) 2H2O Electrolyte

Hot (~200C) concentratedKOH in center compartment

In contact with 2 porouselectrodes containing Ptcatalyst that facilitatesreactions

Advances led to loweroperating temperatures anduse of H+ transfer membranes

Application of Electrolysis

Electroplating Put fork and bar of silver in AgCN bath


112/126


Set potential so that cell runs electrolytically

Ag dissolves from Ag electrode

Ag deposited on metal utensil

cathode anode

Bar of Silver

Uses of Electrolysis Separating metals in aqueous solution

If you have several metals in solution each with


113/126


ydifferent reduction potentials, you can oftenseparate by electrolysis

Start at very low voltage and gradually turn up

Ag+ + e Ag E = 0.80 V

Cu2+ + 2e Cu E = 0.34 V Zn2+ + 2e Zn E = 0.76 V

Order of oxidizing ability: Ag+ > Cu2+ > Zn2+

So Ag plates out first, then Cu, then Zn as voltage

113


114/126

Application of Electrolysis

Electrorefining ofMetals


115/126


Purifying metals such as

Cu Ultrapure Cu sheets =

cathodes

Lower between slabs ofimpure Cu = anodes

Aqueous CuSO4 electrolytesolution

4 weeks for anodes todissolve and pure Cu toplate onto cathodes

115

Refining Copper


116/126


Corrosion of Iron Steel mostly iron

Fe2O3 coating NOT impervious to further corrosion

Ph i l i l i i l


117/126


Physical stress points leave strain in metal More susceptible to corrosion

Creates anodic and cathodic regions

Electrochemical process

Anode region: Fe Fe2+ + 2e

Electrons flow through steel to cathode region Cathode region: O2 + 2H2O + 4e

4OH

117

Prevention of Corrosion

How to conserve our natural resources Primaryapply coating


118/126


y pp y g

Paint

Metal plating Use chromium and tin to plate steel

Galvanizingusing zinc to coat steel

Zinc more active metal so oxidized first

Zn(s) = sacrificial coating

Alloyingmixture of metals

Stainless steel contains chromium and nickel

118


119/126

Your Turn!Which metal would provide the best cathodic

protection for an iron tank?


120/126


A. CrB. Pb

C. CuD. Sn

Cr is a better reducing agent than the otherchoices.

120

Extra Practice


121/126

121

Learning CheckCalculate E

cell. Which are spontaneous?

Cu(s) + Ag+(aq) Cu2+(aq) + Ag(s)


122/126


Cr2O72(aq) + MnO2 (s) MnO4

(aq) + Cr(s)

Cu2+/Cu 0.337V Cr 2O72/Cr

1.33V

Ag+/Ag 0.799V MnO4/MnO2 1.695V

Pb2+/Pb 0.126V

[0.799 0.337]V = 0.462V

[1.33 1.695]V = 0.365V

Learning Check Calculate G0 in kJ.. Which reactions are spontaneous

under standard conditions? Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s)


123/126


Cr2O72(aq) + MnO2 (s) MnO4

(aq) + Cr(s)

Cu2+/Cu 0.337V Cr 2O72/Cr 1.33V

Ag+

/Ag 0.799V MnO4

/MnO2 1.695VPb2+/Pb -0.126V

G = 2 mol x 96,485 C/mol x 0.462 J/C

G = 89.1 kJ

G = 6 mol x 96,485 C/mol x 0.365 J/CG = 423 kJ

Learning Check: Calculate Kfor the following cells:

Cu(s) + Ag+(aq) Cu2+(aq) + Ag(s) E= 0.462 V


124/126


K= e35.967 = 4.18 1015

Cr2O72(aq) + MnO2 (s) MnO4(aq) + Cr(s)

E= 0.365 V

K= e170.486 = 9.10 10-75

967.3515298)(3148

4620485962ln =

= = K.KmolJ/.

J/C.C/mol,

RT

EnK

oF

486.17015298)(314836504859612ln =

==

K.KmolJ/.J/C.C/mol,

RTEnK

o

F

Learning Check: Calculate Ecell

Al|Al3+ (aq, 0.5M)||Zn2+(aq, 0.2M)|Zn

Al3+/Al 1.662V Zn2+/Zn 0.763V


125/126


3Zn2+(aq) + 2Al(s) 3Zn(s) + 2Al3+(aq)

E = [0.763 (1.662)]V = 0.899V

E = 0.899 V 0.015 V = 0.884 V

Al|Al3+ (aq, 0.5M)||Zn2+ (aq, 1M)|Zn

E

= 0.899 V + 0.006 V = 0.905 V

3

2

]2.0[

]5.0[ln

6

02569.0899.0

VVE =

3

2

]0.1[]5.0[ln

602569.0899.0 VVE =

Learning Check: What current must be supplied to deposit

3.00 g Au from a solution of AuCl3 in 200.0 s?


126/126


How much time (in s) does it take to plate10.2 g of Ag+ using a 0.1mA power source?

22.0 A

9 107 s

Documents

Brdy 6Ed Ch20 Electrochemistry