Electrochemistry

Chapter 17

Contents Galvanic cells

Standard reduction potentials

Cell potential, electrical work, and free energy

Dependence of cell potential on concentration

Batteries

Corrosion

Electrolysis

Commercial electrolytic processes

Oxidation reduction reactions

Oxidation reduction reactions

involve a transfer of electrons.

Oxidation Involves Loss of electrons

– Increase in the oxidation number

Reduction Involves Gain electrons

– Decrease in the oxidation number

17.1 Galvanic Cells

8H++MnO4-+ 5Fe+2

Mn+2 + 5Fe+3 +4H2O

If we break the reactions into half reactions.

8H++MnO4-+5e- Mn+2 +4H2O (Red)

5(Fe+2 Fe+3 + e- ) (Ox)

Electrons are transferred directly.

This process takes place without doing useful work

Example

H+ MnO4

- Fe+2

When the compartments of the two beakers are connected as shown the reaction starts

Current flows for an instant then stops

No flow of electrons in the wire, Why?

Current stops immediately because charge builds

up.

Oxidant

reductant

H+

MnO4-

Fe+2

Galvanic Cell

Salt Bridge allows

ions to flow without

extensive mixing in

order to keep net

charge zero.

Electrons flow

through the wire

from reductant to

oxidant

Solutions must be connected so ions can flow to

keep the net charge in each compartment zero

H+

MnO4- Fe+2

Porous

Disk

Fe2+

Reducing

Agent

Oxidizing

Agent

MnO4-

e-

e-

e- e-

e-

e-

Anode Cathode

Electrochemical Cells

19.2

Spontaneous redox reaction

_______

__________ _______

__________

Thus a Galvanic cell is a device in which a chemical energy is changed to electrical energy

The electrochemical reactions occur at the interface between electrode and solution where the electron transfer occurs

Anode: the electrode compartment at which oxidation occurs

Cathode: the electrode compartment at which reduction occurs

Cell Potential Oxidizing agent pulls the electrons

Reducing agent pushes the electrons

The total push or pull (“driving force”) is called the cell potential, E

cell

Also called the electromotive force (emf)

Unit is the volt(V)

= 1 joule of work/coulomb of charge

Measured with a voltmeter

Measuring the cell potential

Can we measure the total cell potential??

A galvanic cell is made where one of the two electrodes is a reference electrode whose potential is known.

Standard hydrogen electrode (H+ = 1M

and the H2 (g) is at 1 atm) is used as a

reference electrode and its potential was assigned to be zero at 25

0C.

1 M HCl

H+

Cl-

H2

Standard Hydrogen Electrode

This is the reference all other oxidations are compared to

Eº = 0

(º) indicates standard states of 25ºC, 1 atm, 1 M

solutions.

1 atm

Zn+2

SO4-2

1 M HCl

Anode

0.76

1 M ZnSO4

H+

Cl-

H2

Cathode

Standard Electrode Potentials

Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)

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

Zn (s) Zn2+ (1 M) + 2e- Anode (oxidation):

Cathode (reduction):

17.2 Standard Reduction Potentials, E

The E values corresponding to reduction half-

reactions with all solutes at 1M and all gases at 1

atm.

E can be measured by making a galvanic cell in

which one of the two electrodes is the Standard

Hydrogen electrode, SHE, whose E = 0 V

The total potential of this cell can be measured

experimentally

However, the individual electrode potential can

not be measured experimentally.

Why?

If the cathode compartment of the cell is SHE, then the half reaction would be

2H+ + 2e H2 (g); Eo = 0V

And the anode compartment is Zn metal in Zn2+, (1 M) then the half reaction would be

Zn Zn2+ + 2e

The total cell potential measured experimentally was found to be + 0.76 V

Thus, +0.76 V was obtained as a result of this calculation:

Eº cell = EºZn Zn2+ + Eº H

+ H2

0.76 V 0.76 V 0 V

Standard Reduction Potentials

The E values corresponding to reduction half-

reactions with all solutes at 1M and all gases at

1 atm. can be determined by making them half

cells where the other half is the SHE.

E0 values for all species were determined as

reduction half potentials and tabulated. For

example:

– Cu2+ + 2e Cu E = 0.34 V

– SO42 + 4H+ + 2e H2SO3 + H2O E = 0.20 V

– Li+ + e- Li E = -3.05 V

Some Standard Reduction Potentials

Li+ + e- ---> Li -3.045 v

Zn+2 + 2 e- ---> Zn -0.763v

Fe+2 + 2 e- ---> Fe -0.44v

2 H+(aq) + 2 e- ---> H2(g) 0.00v

Cu+2 + 2 e- ---> Cu +0.337v

O2(g) + 4 H+(aq) + 4 e- ---> 2 H2O(l) +1.229v

F2 + 2e- ---> 2 F- +2.87v

Standard Reduction Potentials at 25°C

• E0 is for the reaction as

written

• The more positive E0 the

greater the tendency for the

substance to be reduced

• The more negative E0 the

greater the tendency for the

substance to be oxidized

• Under standard-state

conditions, any species on

the left of a given half-

reaction will react

spontaneously with a

species that appears on the

right of any half-reaction

located below it in the table

• The half-cell

reactions are

reversible

• The sign of E0 changes when the

reaction is

reversed

• Changing the

stoichiometric

coefficients of a

half-cell reaction

does not change

the value of E0

Can Sn reduce Zn2+ under standard-state conditions?

•How do we find the answer?

•Look up the Eº values in in the table of reduction

potentials

•\Which reactions in the table will reduce Zn2+(aq)?

Zn+2 + 2 e- ---> Zn(s) -0.763v

Sn+2 + 2 e- ---> Sn -0.143v

Look up the Eº values in in the table of reduction potentials

Standard cell potential

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

The total standard cell potential is the sum of the potential at each electrode.

Eº cell = EºZn Zn2+ + Eº Cu

+2 Cu

We can look up reduction potentials in a table.

One of the reactions must be reversed, in order to change its sign.

Standard Cell Potential

Determine the cell potential for a galvanic cell based on the redox reaction.

Cu(s) + Fe+3(aq) Cu+2(aq) + Fe+2(aq)

Fe+3(aq) + e- Fe+2(aq) Eº = 0.77 V

Cu+2(aq)+2e- Cu(s) Eº = 0.34 V

Cu(s) Cu+2(aq)+2e- Eº = -0.34 V

2Fe+3(aq) + 2e- 2Fe+2(aq) Eº = 0.77 V

Eo cell = EoFe

3+ Eo Fe

2+ + EoCuEo

Cu2+

Eo cell = 0.77 + (-0.34) = o.43 V

The total reaction:

Cu(s) Cu+2(aq)+2e- Eº = -0.34 V

2Fe+3(aq) + 2e- 2Fe+2(aq) Eº = 0.77 V

Cu(s) + 2Fe+3(aq) Cu2+ + 2Fe2+

Eºcell = +0.43 V

Line Notation

SolidAqueousAqueoussolid

Anode on the leftCathode on the right

Single line different phases.

Double line porous disk or salt bridge.

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

If all the substances on one side are aqueous, a platinum electrode is indicated.

For the last reaction

Cu(s)Cu+2(aq)Fe+2(aq),Fe+3(aq)Pt(s)

Complete description of a Galvanic Cell

The reaction always runs spontaneously in the direction that produces a positive cell potential.

Four parameters are needed for a complete description:

1. Cell Potential

2. Direction of flow

3. Designation of anode and cathode

4. Nature of all the components- electrodes and ions

Exercise

Describe completely the galvanic cell based on the following half-reactions under standard conditions.

MnO4- + 8 H+ +5e- Mn+2 + 4H2O Eº=1.51

Fe+3 +3e- Fe(s) Eº=0.036V

1. Write the total cell reaction

2. Calculate Eo cell

3. Define the cathode and anode

4. Draw the line notation for this cell

17.3 Cell potential, electrical work and free energy

The work accomplished when electrons are transferred through a wire depends on the “push” (thermodynamic driving force) behind the electrons

The driving force (emf) is defined in terms of potential difference (in volts) between two points in the circuit

emf = potential difference (V)

= work (J) / Charge(C) =

q

w

The work done by the system has a

–ve sign

Potential produced as a result of doing a work should have a +ve sign

The cell potential, E, and the work, w, have opposite signs.

Relationship between E and w can be expressed as follows:

E = work done by system / charge

( )

q

wE

=

Charge is measured in coulombs. Thus,

-w = qE

Faraday = 96,485 C/mol e-

q = nF = moles of e- x charge/mole e-

w = -qE = -nFE = DG

Thus, DG = -nFE and

DGo = -nFEo

q

wE

=

Potential, Work, DG and spontaneity

DGº = -nFE º

if E º > 0, then DGº < 0 spontaneous

if E º < 0, then DGº > 0 nonspontaneous

In fact, the reverse process is spontaneous.

Spontaneity of Redox Reactions

DG = -nFEcell

DG0 = -nFEcell 0

n = number of moles of electrons in reaction

F = 96,500 J

V • mol = 96,500 C/mol

DG0 = -RT ln K = -nFEcell 0

Ecell 0 =

RT

nF ln K

(8.314 J/K•mol)(298 K)

n (96,500 J/V•mol) ln K =

= 0.0257 V

n ln K Ecell

0

= 0.0592 V

n log K Ecell

0

Spontaneity of Redox Reactions

If you know one, you can

calculate the other…

If you know K, you can

calculate DEº and DGº

If you know DEº, you can

calculate DGº

Spontaneity of Redox Reactions

Relationships among DG º, K, and Eºcell

2(3e- + Al3+ Al)

3 (Mg Mg2+ + 2e-) Oxidation:

Reduction:

Calculate DG0 for the following reaction at 250C.

2Al3+(aq) + 3Mg(s) 2Al(s) + 3Mg+2(aq)

n = ?

DG0 = -nFEcell 0

E0 = Ered + Eox cell 0 0

DG0 = -nFEcell 0 = ___ X (96,500 J/V mol) X ___ V

DG0 = _______ kJ/mol

17.4 Dependence of Cell Potential on Concentration

Qualitatively: we can predict direction of change in E from LeChâtelier pinciple

2Al(s) + 3Mn+2(aq) 2Al+3(aq) + 3Mn(s); Eo cell = 0.48 V

Predict if Ecell will be greater or less than Eºcell for the following cases:

if [Al+3] = 1.5 M and [Mn+2] = 1.0 M

if [Al+3] = 1.0 M and [Mn+2] = 1.5M

An increase in conc. of reactants would favor forward reaction thus increasing the driving force

for electrons; i.e. Ecell becomes > Eo cell

Concentration Cell: both compartments contain same

components but at different concentrations

Half cell potential are not

identical

Because the Ag+ Conc.

On both sides are not

same

Eright > Eleft

• To make them equal, [Ag+]

On both sides should same

• Electrons move from left to

right

The Nernst Equation

Effect of Concentration on Cell Emf

DG = DG0 + RT ln Q DG = -nFE DG0 = -nFE 0

-nFE = -nFE0 + RT ln Q

E = E0 - ln Q RT

nF Nernst equation

At 298K

- 0.0257 V

n ln Q E 0 E = -

0.0592 V n

log Q E 0 E =

The Nernst Equation

As reactions proceed concentrations of products increase and reactants decrease.

When equilibrium is reached

Q = K ; Ecell = 0

and DG = 0 (the cell no longer has

the ability to do work)

Qualitatively: we can predict the direction of change in E from Lechatelier principle

Find Q

Calculate E

E > 0; the reaction is spontaneous to the

right

E < 0; the reaction is spontaneous to the

left

Predicting spontaneity using Nernst equation

Will the following reaction occur spontaneously at 250C if

[Fe2+] = 0.60 M and [Cd2+] = 0.010 M?

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

2e- + Fe2+ 2Fe

Cd Cd2+ + 2e- Oxidation:

Reduction: n = 2

E0 = -0.44 + (+0.40)

E0 = -0.04 V

E0 = EFe /Fe + ECd /Cd 2+ 0 0

2+

- 0.0257 V

n ln Q E 0 E =

- 0.0257 V

2 ln -0.04 V E =

0.010

0.60

E = ____________

E ___ 0 ________________

Exercise- p. 843

Determine the cell potential at 25oC for the following cell, given that

2Al(s) + 3Mn+2(aq) 2Al+3(aq) + 3Mn(s)

[Mn2+] = 0.50 M; [Al3+]=1.50 M; E0 cell = 0.4

Always we have to figure out n from the balanced equation

2(Al(s)+ Al+3(aq) + 3e-)

3(Mn+2(aq) + 2e- Mn(s))

n = 6

- 0.0592 V

n log Q E 0 E =

Calculation of Equilibrium Constants

for redox reactions

At equilibrium, Ecell = 0 and Q = K.

Qn

EE o log059.0

=Then,

Kn

Eo log0591.0

0 =

0591.0log

onEK = at 25 oC

2e- + Fe2+ Fe

2Ag 2Ag+ + 2e- Oxidation:

Reduction:

What is the equilibrium constant for the following reaction

at 250C? Fe2+ (aq) + 2Ag (s) Fe (s) + 2Ag+ (aq)

= 0.0257 V

n ln K Ecell

0

E0 = -0.44 –0.80= -1.24 V

E0 = -1.24 V 0.0257 V

x n E0 cell exp K =

n = ___

0.0257 V

x 2 -1.24 V = exp

K = ________________

E0 = EFe /Fe+ EAg /Ag 0 0

2+ +

17.5 Batteries

Lead-Storage Battery

A 12 V car battery consists of 6 cathode/anode

pairs each producing 2 V.

Cathode: PbO2 on a metal grid in sulfuric acid:

PbO2(s) + SO42-(aq) + 4H+(aq) + 2e-

PbSO4(s) + 2H2O(l)

Anode: Pb:

Pb(s) + SO42-(aq) PbSO4(s) + 2e-

Batteries are Galvanic Cells

Anode:

Cathode:

Lead storage battery

PbO2 (s) + 4H+ (aq) + SO2- (aq) + 2e- PbSO4 (s) + 2H2O (l) 4

Pb (s) + SO2- (aq) PbSO4 (s) + 2e- 4

Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2- (aq) 2PbSO4 (s) + 2H2O (l) 4

Lead-Storage Battery

The overall electrochemical reaction is

PbO2(s) + Pb(s) + 2SO42-(aq) + 4H+(aq)

2PbSO4(s) + 2H2O(l)

for which

Ecell = Ered(cathode) - Ered(anode)

= (+1.685 V) - (-0.356 V)

= +2.041 V.

H2SO4 is consumed while the battery is

discharging

H2SO4 is 1.28g/ml and must be kept

Water is depleted thus the battery should be

topped off always

Dry cell Batteries

Zn (s) Zn2+ (aq) + 2e- Anode:

Cathode: 2NH4 (aq) + 2MnO2 (s) + 2e- Mn2O3 (s) + 2NH3 (aq) + H2O (l) +

Zn (s) + 2NH4 (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s)

Dry Cell Battery

Anode: Zn cap:

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

Cathode: MnO2, NH4Cl and C paste:

2NH4+(aq) + 2MnO2(s) + 2e- Mn2O3(s) + 2NH3(aq)

+ 2H2O(l)

Total reaction:

Zn + NH4+ +MnO2 Zn2+ + NH3 + H2O

This cell produces a potential of about 1.5 V.

The graphite rod in the center is an inert

cathode.

For an alkaline battery, NH4Cl is replaced with

KOH.

Anode: oxidation of Zn

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

Cathode: reduction of MnO2.

2MnO2 + H2O + 2e- Mn2O3 + 2OH-

• Total reaction

Zn(s) + 2 MnO2(s) ---> ZnO(s) + Mn2O3(s)

It lasts longer because Zn anode corrodes

less rapidly than under acidic conditions.

Alkaline Cell Battery

Alkaline Battery

Nickel-Cadmium (Ni-Cad) Battery

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

Cathode: NiO2 + 2H2O + 2e- ` Ni(OH)2 + 2OH-

NiO2 + Cd + 2H2O Cd(OH)2 +Ni(OH)2

NiCad 1.25 v/cell

The products adhere to the

electrodes thus the battery can

be recharged indefinite number

of times.

Fuel Cells

A fuel cell is a galvanic cell that requires a continuous

supply of reactants to keep functioning

Anode:

Cathode: O2 (g) + 2H2O (l) + 4e- 4OH- (aq)

2H2 (g) + 4OH- (aq) 4H2O (l) + 4e-

2H2 (g) + O2 (g) 2H2O (l)

17.6 Corrosion

Rusting - spontaneous oxidation of metals.

Most metals used for structural purposes have reduction potentials that are less positive than O2 . (They are readily oxidized by O2)

Fe+2 +2e- Fe Eº= - 0.44 V

O2 + 2H2O + 4e- 4OH- Eº= 0.40 V

When a cell is formed from these two half reactions a cell with +ve potential will be obtained

Au, Pt, Cu, Ag are difficult to be oxidized (noble metals)

Most metals are readily oxidized by O2 however, this process develops a thin oxide coating that protect the internal atoms from being further oxidized.

Al that has Eo = -1,7V is easily oxidized. Thus, it is used for making the body of the airplane.

Water

Rust

Iron dissolves forming a pit e-

Salt speeds up process by increasing conductivity

Anodic area

Cathodic area

Fe Fe+2 + 2e-

Anodic reaction

O2 + 2H2O + 4e- 4OH-

cathodic reaction

Electrochemical corrosion of iron

Fe2+ (aq) + O2(g) + (4-2n) H2O(l)

2F2O3(s).nH2O (s)+ 8H+(aq)

Fe on the steel surface is oxidized (anodic regions)

Fe Fe+2 +2e- Eº=- 0.44 V

e-’s released flow through the steel to the areas that have O2 and moisture (cathodic regions). Oxygen is reduced

O2 + 2H2O + 4e- 4OH- Eº= 0.40 V

Thus, in the cathodic region Fe+2 will react with O2

The total reaction is:

Fe2+ (aq) + O2(g) + (4-2n) H2O(l)

2F2O3(s).nH2O (s)+ 8H+(aq)

Thus, iron is dissolved to form pits in steel

Moisture must be present to act as the salt bridge

Steel does not rust in the dry air

Salts accelerates the process due to the increase in conductivity on the surface

Preventing of Corrosion

Coating to keep out air and water.

Galvanizing - Putting on a zinc coat

Fe Fe2+ + 2e- Eoox = 0.44V

Zn Zn2+ + 2e- Eoox = 0.76 V

Zn has a more positive oxidation potential than Fe, so it is more easily oxidized.

Any oxidation dissolves Zn rather than Fe

Alloying is also used to prevent corrosion. stainless steel contains Cr and Ni that make make steel as a noble metal

Cathodic Protection - Attaching large pieces of an active metal like magnesium by wire to the pipeline that get oxidized instead. By time Mg must be replaced since it dissolves by time

Cathodic Protection of an Underground Pipe

Cathodic Protection of an Iron Storage Tank

Running a galvanic cell backwards.

Put a voltage bigger than the galvanic potential and reverse the direction of the redox reaction.

Electrolysis: Forcing a current through a cell to produce a chemical change for which the cell potential is negative.

That is causing a nonspontaneous reaction to occur

It is used for electroplating.

17.7 Electrolysis

1.0 M

Zn+2

e- e-

Anode Cathode

1.10

Zn Cu 1.0 M

Cu+2

Galvanic cell based on spontaneous reaction:

Zn + Cu2+ Zn2+ + Cu

1.0 M

Zn+2

e- e-

Anode Cathode

A battery

>1.10V

Zn Cu 1.0 M

Cu+2

Electrolytic cell

Zn2+ + Cu Zn + Cu2+

Galvanic Cell Electrolytic Cell

Calculating plating

How much chemical change occurs with the flow of a given current for a specified time?

Determine quantity of electrical charge in coulombs

Measure current, I (in amperes) per a period of time

1 amp = 1 coulomb of charge per second

coulomb of charge = amps X seconds =

q = I x t

q/nF = moles of metal

Mass of plated metal can then be calculated

x ss

C

Stoichiometry of Electrolysis

• How much chemical change occurs with the flow of a given current for a given time?

• Current and time quantity of charge

moles of electrons moles of

analyte grams of analyte

Exercise

Calculate mass of Cu that is plated out when a current of 10.0 amps is passed for 30.0 min through a solution of Cu2+

Excercise

How long must 5.00 amp current be applied to produce 10.5 g of Ag from Ag+?

Electroplating

How many grams of chromium can be plated from a Cr+6 solution in 45 minutes at a 25 amp current?

How many grams of chromium can be plated from a Cr+6 solution in 45 minutes at a 25 amp current?

(45 min) #g Cr = ------------

(45 min)(60 sec) #g Cr = --------------------- (1 min)

How many grams of chromium can be plated from a Cr+6 solution in 45 minutes at a 25 amp current?

(45) (60 sec) (25 amp) #g Cr = --------------------------- (1)

(45)(60 sec)(25 amp)(1 C) #g Cr = ----------------------------- (1) (1 amp sec)

Faraday’s constant

(45)(25)(60)(1 C)(1 mol e-) #g Cr = ---------------------------------- (1)(1)(96,500 C)

(45)(60)(25)(1)(1 mol e-)(52 g Cr) #g Cr = ------------------------------------------- (1)(1)(96,500) (6 mol e-)

Electroplating

How many grams of chromium can be plated from a Cr+6 solution in 45 minutes at a 25 amp current?

(45)(60)(25)(1)(1 mol e-)(52 g Cr) #g Cr = ------------------------------------------- (1)(1)(96,500)(6 mol e-)

= 58 g Cr

Michael Faraday Lecturing at the Royal Institution Before Prince Albert and Others (1855)

Electrolysis of water

Electrolysis of Water

The Electrolysis of Water Produces Hydrogen Gas at the Cathode (on the Right) and Oxygen Gas at the Anode (on the Left)

Other uses of electrolysis

Separating mixtures of ions.

More positive reduction potential means the reaction proceeds forward.

We want the reverse.

Most negative reduction potential is easiest to plate out of solution.

17.8 Commercial electrolytic processes

A Schematic Diagram of an Electrolytic Cell for

Producing Aluminum by the Hall-Heroult Process.

The Downs Cell for the Electrolysis of Molten Sodium Chloride

The Mercury Cell for Production of Chlorine and

Sodium Hydroxide