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

17.1Galvanic 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)

55(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.

Oxidantreductant

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 PotentialOxidizing agent pullspulls the electronsReducing agent pushes the electrons The total push or pull (“driving force”)

is called the cell potential, EcellAlso called the electromotive force

(emf) Unit is the volt(V) = 1 joule of work/coulomb of chargeMeasured 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. Ecan be measured by making a galvanic cell in

which one of the two electrodes is the Standard

Hydrogen electrode, SHE, whose E0 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 = Eo

Fe3+ Eo

Fe2+ + Eo

CuEoCu

2+

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 direction that produces a positive cell potentialproduces a positive cell potential.

Four parameters are needed for a complete description:

1. Cell Potential2. Direction of flow3. Designation of anode and cathode4. 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=0.036V1. Write the total cell reaction2. Calculate Eo cell3. Define the cathode and anode4. 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 = G Thus, G = -nFE and

Go = -nFEo

q

wE

Potential, Work, G and spontaneity Gº = -nFE º

if E º > 0, then Gº < 0 spontaneous if E º < 0, then Gº > 0 nonspontaneous

In fact, the reverse process is spontaneous.

Spontaneity of Redox Reactions

G = -nFEcell

G0 = -nFEcell0

n = number of moles of electrons in reaction

F = 96,500J

V • mol = 96,500 C/mol

G0 = -RT ln K = -nFEcell0

Ecell0 =

RTnF

ln K(8.314 J/K•mol)(298 K)

n (96,500 J/V•mol)ln K=

=0.0257 V

nln KEcell

0

=0.0592 V

nlog KEcell

0

Spontaneity of Redox Reactions

If you know one, you can calculate the other…

If you know K, you can calculate Eº and Gº

If you know Eº, you can calculate Gº

Spontaneity of Redox Reactions

Relationships among G º, K, and Eºcell

2(3e- + Al3+ Al)

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

Reduction:

Calculate G0 for the following reaction at 250C. 2Al3+(aq) + 3Mg(s) 2Al(s) + 3Mg+2(aq)

n = ?

G0 = -nFEcell0

E0 = Ered + Eoxcell0 0

G0 = -nFEcell0 = ___ X (96,500 J/V mol) X ___ V

G0 = _______ 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 VPredict if Ecell will be greater or less than Eºcell for

the following cases:if [Al+3] = 1.5 M and [Mn+2] = 1.0 Mif [Al+3] = 1.0 M and [Mn+2] = 1.5MAn 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

G = G0 + RT ln Q G = -nFE G0 = -nFE 0

-nFE = -nFE0 + RT ln Q

E = E0 - ln QRTnF

Nernst equation

At 298K

-0.0257 V

nln QE0E = -

0.0592 Vn

log QE0E =

The Nernst Equation

As reactions proceed concentrations of products increase and reactants decrease.

When equilibrium is reached

Q = K ; Ecell = 0

and G = 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

nln QE0E =

-0.0257 V

2ln -0.04 VE =

0.0100.60

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

nlog QE0E =

Calculation of Equilibrium Constants for redox reactions

At equilibrium, Ecell = 0 and Q = K.

Qn

EE o log059.0

Then,

Kn

E o 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

nln KEcell

0

E0 = -0.44 –0.80= -1.24 V

E0 = -1.24 V 0.0257 Vx nE0 cellexpK =

n = ___

0.0257 Vx 2-1.24 V

= exp

K = ________________

E0 = EFe /Fe+ EAg /Ag0 0

2+ +

17.5 Batteries17.5 Batteries

Lead-Storage BatteryLead-Storage Battery A 12 V car battery consists of 6 cathode/anode A 12 V car battery consists of 6 cathode/anode

pairs each producing 2 V.pairs each producing 2 V. CathodeCathode: : PbOPbO22 on a metal grid in sulfuric acid: on a metal grid in sulfuric acid:

PbOPbO22((ss) + SO) + SO442-2-((aqaq) + 4H) + 4H++((aqaq) + 2e) + 2e--

PbSOPbSO44((ss) + 2H) + 2H22O(O(ll)) Anode: Pb:Anode: Pb:

Pb(Pb(ss) + SO) + SO442-2-((aqaq) ) PbSO PbSO44((ss) + 2e) + 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 BatteryLead-Storage Battery The overall electrochemical reaction is The overall electrochemical reaction is

PbOPbO22((ss) + Pb() + Pb(ss) + 2SO) + 2SO442-2-((aqaq) + 4H) + 4H++((aqaq) )

2PbSO2PbSO44((ss) + 2H) + 2H22O(O(ll))

for whichfor which

EEcellcell = = EEredred(cathode) - (cathode) - EEredred(anode) (anode)

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

= +2.041 V.= +2.041 V. HH22SOSO44 is consumed while the battery is discharging is consumed while the battery is discharging

HH22SOSO4 4 is 1.28g/ml and must be kept is 1.28g/ml and must be kept Water is depleted thus the battery should be Water is depleted thus the battery should be

topped off alwaystopped 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 BatteryDry Cell Battery Anode: Zn cap:Anode: Zn cap:

Zn(Zn(ss) ) Zn Zn2+2+((aqaq) + 2e) + 2e-- Cathode: MnOCathode: MnO22, NH, NH44Cl and C paste:Cl and C paste:

2NH2NH44++((aqaq) + 2MnO) + 2MnO22((ss) + 2e) + 2e-- Mn Mn22OO33((ss) + 2NH) + 2NH33((aqaq) + ) +

2H2H22O(O(ll)) Total reaction: Total reaction:

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

This cell produces a potential of about 1.5 V. This cell produces a potential of about 1.5 V. The graphite rod in the center is an inert The graphite rod in the center is an inert

cathode.cathode.

For an alkaline battery, NHFor an alkaline battery, NH44Cl is replaced with Cl is replaced with

KOH.KOH.Anode: oxidation of Zn Anode: oxidation of Zn

Zn(Zn(ss) + 2OH) + 2OH-- ZnO + H ZnO + H22O + 2eO + 2e--

Cathode: reduction of MnOCathode: reduction of MnO22..

2MnO2 + HH22O + 2eO + 2e-- Mn2O3 + 2OH-

• Total reaction

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

It lasts longer because Zn anode corrodes It lasts longer because Zn anode corrodes

less rapidly than under acidic conditions. less rapidly than under acidic conditions.

Alkaline Cell BatteryAlkaline Cell Battery

Alkaline BatteryAlkaline Battery

Nickel-Cadmium (Ni-Cad) BatteryNickel-Cadmium (Ni-Cad) Battery

Anode: CdAnode: Cd(s)(s) + 2OH + 2OH-- Cd(OH) Cd(OH)22 + 2e + 2e--

Cathode:Cathode: NiONiO22 + 2H + 2H22O + O + 2e2e- ` - ` Ni(OH)Ni(OH)22 + +

2OH2OH--

NiONiO22 + Cd + 2H + Cd + 2H22O Cd(OH)O Cd(OH)2 2 +Ni(OH)+Ni(OH)22 NiCad 1.25 v/cellNiCad 1.25 v/cell

The products adhere to theThe products adhere to theelectrodes thus the battery canelectrodes thus the battery can be recharged indefinite numberbe recharged indefinite number

of timesof 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.

AlAl 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- Eo

ox = 0.44V Zn Zn2+ + 2e- Eo

ox = 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 Cu1.0 M

Cu+2

Galvanic cell based on spontaneous reaction:Zn + Cu2+ Zn2+ + Cu

1.0 M

Zn+2

e- e-

AnodeCathode

A battery >1.10V

Zn Cu1.0 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

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-)(1 mol Cr)#g Cr = -------------------------------------------------

(1)(1)(96,500) (6 mol e-)

(45)(60)(25)(1)(1 mol e-)(52 g Cr)

#g Cr = ------------------------------------------- (1)(1)(96,500) (1 mol Cr)

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

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

Electrolysis of water

Other uses of electrolysis

Separating mixtures of ions.More positive reduction potential

means the reduction 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.

To reduce mp of AlFrom 2000 to 1000

The Downs Cell for the Electrolysis of Molten Sodium Chloride

The Mercury Cell for Production of Chlorine and

Sodium Hydroxide

Metal Plating