Electrochemistry Ch20b

8/11/2019 Electrochemistry Ch20b

1/13

Slide 50

The End of Chem1051 Classes!

Now, Review Classes

Practice lots of examples, a few each day, do not

try and study all of 1051 in a couple of days, it will

not work!

Read the questions! And recognise what the asker

wants!

Get help if you need it; Resource Room,

Tutors/TAs, Tutorial, Me (e-mail: [email protected])

Career advice - I am happy to provide this,

arrange an appointment or just try and find me.

Slide 49

2 more examples.

Gold can be plated out of a solution containing

Au3+. What mass of gold (in grams) will be plated

by the flow of 5.5 A of current for 25 minutes? Silver can be plated out of a solution containing

Ag+. How much time (in minutes) would it take to

plate 12 g of silver using a current of 3.0 A?

Slide 48

Problem Calculating quantities in electrolysis

PROBLEM: A technician is plating a faucet (tap) with 0.86 g of Cr from an

electrolytic bath containing aqueous Cr2(SO4)3. If 12.5 min is

allowed for the plating, what current is needed?

PLAN: SOLUTION:

mol of e- transferred

divide by M

9.65x104C/mol e-

3mol e-/mol Cr

divide by time

mass of Cr needed

mol of Cr needed

charge (C)

current (A)

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

0.86g (mol Cr) (3 mol e-)

(52.00 gCr) (mol Cr)

= 0.050 mol e-

0.050 mol e- (9.65x104 C/mol e-) = 4.8x103 C

4.8x103 C

12.5 min

(min)

(60s) = 6.4C/s = 6.4 A

Slide 47

Quantitative Aspects of Electrolysis

1 mol e- = 96485 C Charge (C) = current (C/s) x time (s)

Mass of product?

C/sCmol e-mol productg product

MASS (g)of substance

oxidized or

reduced

MASS (g)of substance

oxidized or

reduced

AMOUNT (MOL)

of electrons

transferred

AMOUNT (MOL)

of electrons

transferred

AMOUNT (MOL)

of substance

oxidized or

reduced

AMOUNT (MOL)

of substance

oxidized or

reduced

CHARGE (C)

CHARGE (C)

CURRENT (A)

CURRENT (A)

balanced

half-reaction

Faraday

constant

(C/mol e-)

M(g/mol)

time(s)


2/13

Slide 46

Problem Making qualitative predictions

Predict the half-reaction occurring at the anode and

the cathode for electrolysis of each of the following: A mixture of molten AlBr3 and MgBr2

An aqueous solution of LiI

Slide 45

continued

E0

= -0.42V2H2O(l) + 2e-

H2(g) + 2OH-

(aq)

E0 = -0.80V(b) Ag+(aq) + e- Ag(s)

Ag+ is the cation of an inactive metal and therefore will be reduced to Ag

at the cathode. Ag+(aq) + e- Ag(s)

The N in NO3- is already in its most oxidized form so water will have to be

oxidized to produce O2 at the anode. 2H2O(l) O2(g) + 4H+(aq) + 4e-

Mg is an active metal and its cation cannot be reduced in the presence of

water. So as in (a) water is reduced and H2(g) is produced at the cathode.

The S in SO42- is in its highest oxidation state; therefore water must be

oxidized and O2(g) will be produced at the anode.

E0 = -2.37V(c) Mg2+(aq) + 2e- Mg(s)

Slide 44

Problem Predicting half- and overall reactions in

electrolysis

PROBLEM: What products form during electrolysis of aqueous solution of the

following salts: (a) KBr; (b) AgNO3; (c) MgSO4?

SOLUTION:

PLAN: Compare the potentials of the reacting ions with those of water,

remembering to consider the 0.4 to 0.6V overvoltage.

The reduction half-reaction with the less negative potential, and the oxidation half-reaction with the less positive potential will occur at their respective electrodes.

E0 = -2.93V(a) K+(aq) + e- K(s)

E0 = -0.42V2H2O(l) + 2e- H2(g) + 2OH

-(aq)

The overvoltage would make the water reduction -0.82 to -1.02 but the

reduction of K+ is still a higher potential so H2(g) is produced at the cathode.

The overvoltage would give the water half-cell more potential than

the Br-, so the Br- will be oxidized. Br2(g) forms at the anode.

E0 = 1.07V2Br-(aq) Br2(g) + 2e-

2H2O(l) O2(g) + 4H+(aq) + 4e- E0 = 0.82V

Slide 43

Complications in Electrolytic Cells

Overpotential.(1) Competing reactions.(2)

Non-standard states.(3)

Nature of electrodes.(4)


3/13

Slide 42

Comparison of Voltaic and Electrolytic Cells

Cell Type G Ecell

Electrode

Name Process Sign

Voltaic

Voltaic

Electrolytic

Electrolytic

< 0

< 0

> 0

> 0

> 0

> 0

< 0

< 0

Anode

Anode

Cathode

Cathode

Oxidation

Oxidation

Reduction

Reduction

-

-

+

+

Slide 41

The tin-copper reaction as the basis of a voltaic (galvanic) and an

electrolytic cell.

Oxidation half-reactionSn(s) Sn2+(aq) + 2e-

Reduction half-reaction

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

Oxidation half-reactionCu(s) Cu2+(aq) + 2e-

Reduction half-reaction

Sn2+(aq) + 2e- Sn(s)

Overall (cell) reaction

Sn(s) + Cu2+(aq) Sn2+(aq) + Cu(s)


Sn(s) + Cu2+(aq) Sn2+(aq) + Cu(s)

voltaic cell electrolytic cell

Slide 40

20-7 Electrolysis: Causing

Non-spontaneous Reactions to Occur

Galvanic Cell: also known as voltaic

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Eocell= 1.103 V

Electrolytic Cell:

Zn2+(aq) + Cu(s) Zn(s) + Cu2+(aq) Eocell = -1.103 V

Batteries and Corrosion (Sections 20-5 and 20-6) omitted

Slide 39

Example 20-10

Ecell = Ecell - logn

0.0592 V

[Ag+]0.10 M soln

[Ag+]satd AgI


0.0592 V

0.100

x

0.417 =0 - (logx log 0.100)1

0.0592 V

0.417log 0.100-

0.0592logx= = -1 7.04 = -8.04

x= 10-8.04 = 9.1 x 10-9 Ksp = x2 = 8.3 x 10-17


4/13

Slide 38

Example 20-10

Using a Voltaic Cell to Determine Ksp of a Slightly

Soluble Solute.

With the data given for the reaction on the previous slide,

calculate Ksp for AgI.

AgI(s) Ag+(aq) + I-(aq)

Let[Ag+] in a saturated Ag+ solution be x:

Ag+(0.100 M) Ag+(satd M)

Ecell = Ecell - log Q =n

0.0592 VEcell - log

n

0.0592 V

[Ag+]0.10 M soln

[Ag+]satd AgI

Slide 37

Measurement of Ksp

Ag+(0.100 M) Ag+(satd M)

Ag|Ag+(satd AgI)||Ag+(0.10 M)|Ag(s)

Ag+(0.100 M) + e-Ag(s)

Ag(s) Ag+(satd) + e-

Slide 36

PROBLEM:A concentration cell consists of two Ag/Ag+ half-cells. In

half-cell A, electrode A dips into 0.0100M AgNO3; in half-cell

B, electrode B dips into 4.0x10-4M AgNO3. What is the cell

potential at 298 K? Which electrode has a positive charge?

PLAN: E0cell will be zero since the half-cell potentials are equal. Ecellis calculated from the Nernst equation with half-cell A (higher

[Ag+]) having Ag+ being reduced and plating out, and in half-cell B Ag(s) will be oxidized to Ag+.

SOLUTION: Ag+(aq, 0.010M) half-cell A Ag+(aq, 4.0x10-4M) half-cell B

Ecell = E0cell -

0.0592V

1log

[Ag+]dilute

[Ag+]concentrated

Ecell = 0 V - 0.0592 log 4.0x10-2 = 0.0828V

Half-cell A is the cathode and has the positive electrode.

Slide 35

Concentration Cells

Ecell = Ecell- logn

0.0592 V x2

12

Ecell = 0 - log2

0.0592 V x2

1

Ecell = - 0.0592 V logx

Ecell = (0.0592 V) pH

2 H+(1 M) 2 H+(xM)Ecell = Ecell - log Qn

0.0592 V


5/13

Slide 34

Concentration Cells

Two half cells with identical electrodes

but different ion concentrations.

2 H+(1 M) 2 H+(xM)

Pt|H2 (1 atm)|H+(xM)||H+(1.0 M)|H2(1 atm)|Pt(s)

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

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

Slide 33

Problem Predicting spontaneous reactions for

non-standard conditions

Will the cell reaction proceed spontaneously as

written for the following cell?

Sn(s)|Sn2+(0.50 M)||Pb2+(0.0010 M)|Pb(s)

Slide 32

PROBLEM: In a test of a new reference electrode, a chemist constructs a

voltaic cell consisting of a Zn/Zn2+ half-cell and an H2/H+ half-cell under the

following conditions:

PLAN:

SOLUTION:

[Zn2+] = 0.010M [H+] = 2.5M P = 0.30atmH2

Calculate Ecell at 298 K.

Find E0cell

and Q in order to use the Nernst equation.

Determining E0cell :

E0 = 0.00V2H+(aq) + 2e- H2(g)

E0 = -0.76VZn2+(aq) + 2e- Zn(s)

Zn(s) Zn2+(aq) + 2e- E0 = +0.76V

Q =P x [Zn2+]

H2

[H+]2

Q = 4.8x10-4

Q =(0.30)(0.010)

(2.5)2

Ecell = E0cell - 0.0592V

nlog Q

Ecell = 0.76 - (0.0592/2)log(4.8x10-4) = 0.86V

Slide 31

Example 20-8

Ecell = Ecell - log Qn

0.0592 V

Pt|Fe2+(0.10 M),Fe3+(0.20 M)||Ag+(1.0 M)|Ag(s)


0.0592 V [Fe3+

][Fe2+] [Ag+]

Fe2+(aq) + Ag+(aq) Fe3+(aq) + Ag (s)

Ecell = 0.029 V 0.018 V = 0.011 V


6/13

Slide 30

Example 20-8

Pt|Fe2+(0.10 M),Fe3+(0.20 M)||Ag+(1.0 M)|Ag(s)

Applying the Nernst Equation for Determining Ecell.

What is the value of Ecell for the voltaic cell pictured

below and diagrammed as follows?

Slide 29

20-4 Ecell as a Function of Concentration

G = G + RTln Q

-nFEcell = -nFEcell + RTln Q

Ecell = Ecell - ln QnF

RT

Convert to log10 and calculate constants

Ecell = Ecell - log Qn

0.0592 VThe Nernst Equation:

Slide 28

Summary of Important Thermodynamic, Equilibrium

and Electrochemistry Relationships

Slide 27

Problem Relating Keq to Eocell for a redox reaction

PLAN:

SOLUTION:

PROBLEM: Lead can displace silver from solution:

As a consequence, silver is a valuable by-product in the industrial extraction

of lead from its ore. Calculate K and G0 at 298 K for this reaction.

Pb(s) + 2Ag+(aq) Pb2+(aq) + 2Ag(s)

Break the reaction into half-reactions, find the E0 for each half-reaction

and then the E0cell.

E0 = -0.13V

E0 = 0.80V

2X

E0 = 0.13V

E0 = 0.80V

E0cell = 0.93V

Ag+(aq) + e- Ag(s)

Pb2+(aq) + 2e- Pb(s)

Ag+(aq) + e- Ag(s)

Pb(s) Pb2+(aq) + 2e-

E0cell = ln Keq0.025693V

n

lnK = K = 2.75x1031n x E0cell

0.025693V

(2)(0.93V)

0.025693V=

G0

= -nFE0

cell = -(2)(96.5kJ/mol*V)(0.93V)

G0 = -1.8x102kJ


7/13

Slide 26

Relationship Between Ecell and Keq

G = -RTln Keq = -nFEcell

Ecell =nF

RTln Keq

Slide 25

Relative Reactivities (Activities) of Metals

1. Metals that can displace H from acid

2. Metals that cannot displace H from acid

3. Metals that can displace H from water

4. Metals that can displace other metals fromsolution

Slide 24

The Behavior of Metals Toward Acids

M(s) M2+(aq) + 2 e- E = -EM2+/M

2 H+(aq) + 2 e- H2(g) EH+/H2 = 0 V

2 H+(aq) + M(s) H2(g) + M2+(aq)

Ecell = EH+/H2 - EM2+/M = 0 -EM2+/M = -EM2+/M

When EM2+/M < 0, Ecell > 0. Therefore G < 0.

Metals with negative reduction potentials react with acids

All metals belowhydrogen in table of Eo should displace H2(g)

from acidic solutions (Pb through Li)Slide 23

Problem Making qualitative predictions

Peroxodisulfate salts such as Na2S2O8 are good

oxidizing agents used in bleaching. Dichromates

such as K2Cr2O7 have been used as laboratory

oxidizing agents. Which is the better oxidizing

agent in acidic solution under standard conditions,

Cr2O72- or S2O8

2-?


8/13

Slide 22

Problem Applying the criterion for spontaneous

change in a redox reaction

PROBLEM: (a) Combine the following three half-reactions into three balanced

equations (A, B, and C) for spontaneous reactions, and

calculate E0cell for each.

PLAN:

(b) Rank the relative strengths of the oxidizing and reducing agents:

E0 = 0.96V(1) NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l)

E0 = -0.23V(2) N2(g) + 5H+(aq) + 4e- N2H5

+(aq)

E0 = 1.23V(3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l)

Put the equations together in varying combinations so as to produce

(+) E0cell for the combination. Since the reactions are written as

reductions, remember that as you reverse one reaction for an

oxidation, reverse the sign of E0. Balance the number of electrons

gained and lost without changing the E0.

In ranking the strengths, compare the combinations in terms of E0cell.Slide 21

Spontaneous Change

G < 0 for spontaneous change.

Therefore:

Ecell > 0, (Ecell +ve)

Reaction proceeds spontaneously as written.

Ecell = 0

Reaction is at equilibrium.

Ecell < 0, (Ecell -ve)

Reaction proceeds in the reverse direction spontaneously.

Slide 20

Combining Half Reactions

Fe3+(aq) + 3e- Fe(s) EFe3+/Fe = ?

Fe2+(aq) + 2e- Fe(s) EFe2+/Fe = -0.440 V

Fe3+(aq) + 3e- Fe2+(aq) EFe3+/Fe2+ = 0.771 V

Fe3+(aq) + 3e- Fe(s)

G = +0.880 J

G = -0.771 J

G = +0.109 VEFe3+/Fe = +0.331 V

G = +0.109 V = -nFE

EFe3+/Fe = +0.109 V /(-3F) = -0.0363 V

Slide 19

20-3 Ecell, G, and Keq

Cells do electrical work.

Moving electric charge.

Faraday constant, F = 96,485 C mol-1elec = nFE

G = -nFE

G = -nFE


9/13

Slide 18

Problem Determining Eo from an Eocellmeasurement

PROBLEM:A voltaic cell houses the reaction between aqueous bromine

and zinc metal:

PLAN:

SOLUTION:

Br2(aq) + Zn(s) Zn2+

(aq) + 2Br-

(aq) E0

cell = 1.83VCalculate E0bromine given E

0zinc = -0.76V

The reaction is spontaneous as written since the E0cell is (+).

Zinc is being oxidized and is the anode. Therefore the

E0bromine can be found using E0

cell = E0cathode - E

0anode.

anode: Zn(s) Zn2+(aq) + 2e- E = +0.76

E0Zn as Zn2+(aq) + 2e- Zn(s) is -0.76V

E0cell = E0cathode - E

0anode = 1.83 = E

0bromine - (-0.76)

E0bromine = 1.83 - 0.76 = +1.07 V

Slide 17

Problem Calculating Eocell for a reaction

A new battery system currently under study for possible use in

electric vehicles is the zinc-chlorine battery. The overall reaction

producing electricity in this cell is Zn(s) + Cl2(g) ZnCl2(aq).

What is Eocell of this voltaic cell?

Slide 16

Standard Reduction Potentials

Slide 15

Measuring Standard Reduction Potential

cathode cathode anodeanode


10/13

Slide 14

Standard Cell Potential

Pt|H2(g, 1 bar)|H+(a = 1) || Cu2+(1 M)|Cu(s) Ecell = 0.340 V

Ecell

= Eright

- Eleft

Ecell = ECu2+/Cu - EH+/H2

0.340 V = ECu2+/Cu - 0 V

ECu2+/Cu = +0.340 V

H2(g, 1 atm) + Cu2+(1 M) H+(1 M) + Cu(s) Ecell = 0.340 V

Slide 13

Reduction Couples

Cu2+(1M) + 2 e- Cu(s) ECu2+/Cu = ?

Pt|H2(g, 1 bar)|H+(a = 1) || Cu2+(1 M)|Cu(s) Ecell = 0.340 V

Standard cell potential: the potential difference of a

cell formed from two standardelectrodes.

Ecell = E(right)cathode - E(left)anode

cathodeanode

Slide 12

Standard Electrode Potential, E

E defined by international agreement.

The tendency for a reduction process to occur at

an electrode.

All ionic species present at a=1 (approximately 1 M).

All gases are at 1 bar (approximately 1 atm).

Where no metallic substance is indicated, the potential is

established on an inert metallic electrode (ex. Pt).

Slide 11

Standard Hydrogen Electrode

2 H+(a = 1) + 2 e- H2(g, 1 bar) E = 0 V

Pt|H2(g, 1 bar)|H+(a = 1)


11/13

Slide 10

20-2 Standard Electrode Potentials

Cell voltages, the potential differences between

electrodes, are among the most precise scientific

measurements. The potential of an individual electrode is difficult to

establish.

Arbitrary zero is chosen.

The Standard Hydrogen Electrode (SHE)

Slide 9

Problem - Representing a Redox Reaction by

Means of a Cell Diagram

PROBLEM: Diagram, show balanced equations, and write the notation for a

voltaic cell that consists of one half-cell with a Cr bar in a Cr(NO3)3solution, another half-cell with an Ag bar in an AgNO3 solution, and

a KNO3 salt bridge. Measurement indicates that the Cr electrode is

negative relative to the Ag electrode.

PLAN:

SOLUTION:

Identify the oxidation and reduction reactions and write each half-

reaction. Associate the (-)(Cr) pole with the anode (oxidation) and

the (+) pole with the cathode (reduction).

Voltmeter

Oxidation half-reaction

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

Reduction half-reactionAg+(aq) + e- Ag(s)


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

Cr

Cr3+

Ag

Ag+

K+

NO3-

salt bridge

e-

Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s)

Slide 8

A voltaic (galvanic) cell based on the zinc-copper reaction.

Oxidation half-reaction

Zn(s) Zn2+(aq) + 2e-Reduction half-reaction

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


Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

Slide 7

Terminology

Galvanic cells.

Produce electricity as a result of spontaneous

reactions, e.g. Cells on slides 4 & 6

Electrolytic cells.

Non-spontaneous chemical change driven by

electricity.

Couple, M|Mn+

A pair of species related by a change in number of e-.


12/13

Slide 6

Terminology

Ecell = 1.103 V

Slide 5

Terminology

Electromotive force, Ecell.

The cell voltage or cell potential.

Cell diagram. Shows the components of the cell in a symbolic way.

Anode (where oxidation occurs) on the left.

Cathode (where reduction occurs) on the right.

Boundary between phases shown by |.

Boundary between half cells

(usually a salt bridge) shown by ||.

Slide 4

An Electrochemical Cell

Slide 3

An Electrochemical Half-Cell

Anode

Cathode


13/13

Slide 2

Electrode Potentials and Their Measurement

Cu(s) + 2Ag+(aq)

Cu2+(aq) + 2 Ag(s)

Cu(s) + Zn2+(aq)

No reaction

General ChemistryPrinciples and Modern Applications

Petrucci Harwood Herring Madura

9th Edition

Chapter 20: Electrochemistry

Fran Kerton

Memorial University of Newfoundland,St. Johns, Canada

A1B 3X7

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Electrochemistry Ch20b