Oxydation-Reduction Reactions

Balancing Oxydation-Reduction Reactions

Voltaic Cells

Cell EMF

Spontaneity of Redox Reactions Spontaneity of Redox Reactions

Effect of Concentration of Cell EMF

Batteries

Corrosion

Electrolysis

OxidationOxidationOxidationOxidation----reductionreductionreductionreduction reactionsreactionsreactionsreactions thetransfer of electrons from one species toanother.

The oxidation state of one or moresubstances in the reaction changes.substances in the reaction changes.

The transfer of electrons can be used toproduce energy in the form of electricity.

ElectrochemistryElectrochemistryElectrochemistryElectrochemistry is the study of therelationships between chemical reactionsand electrical energy.

We identify a reaction as oxidation-reduction by comparing the oxidationnumbers of atoms in the reactants andproducts.

If the oxidation numbers change, the If the oxidation numbers change, thereaction is an oxidation-reduction.

zinc metal loses electrons to become acation, and thus zinc has been oxidized.

Hydrogen gains electrons to becomehydrogen gas.

Hydrogen has been reduced. Hydrogen has been reduced.

Zinc is the reducingreducingreducingreducing agentagentagentagent, or reductant,reductant,reductant,reductant,and hydrogen ion is the oxidizingoxidizingoxidizingoxidizing agentagentagentagent, oroxidantoxidantoxidantoxidant.

The amount of each element must be thesame on both sides of the equation.

Balancing the number of electronstransferred. In the equation

Balancing the mass automatically balancesthe total charge on each side.

In many reactions, though, balancing themass does not result in balancing charge.

In the next equation the law of conservationof mass appears to have been obeyed.

However, note that the total charge on theleft side is +3, while the total charge on theright side is only +2.right side is only +2.

The two processes, oxidation of manganesemetal and reduction of chromium(III) ion, donot correspond to the transfer of the samenumber of electrons.

If we multiply the manganese on each side by If we multiply the manganese on each side by3 and the chromium on each side by 2, weget

It is not always convenient to balance oxidation-reduction reactions by inspection.

The equation

would be impossible to balance by inspection.

To balance such an equation, we use a techniqueknown as the method of halfhalfhalfhalf----reactionsreactionsreactionsreactions.

A half-reaction is an equation that shows eitheroxidation or reduction alone.

The electrical energy produced by a spontaneousoxidation-reduction reaction can be "harnessed"using a voltaic cell (also called a galvanic cell).

In a voltaic cell the two half-reactions are madeto occur in separate compartments (half-cells).to occur in separate compartments (half-cells).

The electrons can be transferred only throughthe wire. oxidation and reduction half-reactions occur at separate

electrodes

electric current flows through the wire

Anode - the electrode at which oxidationtakes place. the negative (-) electrode

produces electrons

Cathode - the electrode at which reduction Cathode - the electrode at which reductiontakes place. the positive (+) electrode

consumes electrons

Oxidation-reduction reactions part IOxidation-reduction reactions part I

Galvanic cells I: the copper-zinc cellGalvanic cells I: the copper-zinc cell

Galvanic cells II: the zinc-hydrogen cellGalvanic cells II: the zinc-hydrogen cell

EXAMPLE:EXAMPLE:EXAMPLE:EXAMPLE:

Describe how you would construct a galvanic cell based on the following reaction:

Pb2+ (aq) + Zn (s) Pb (s) + Zn2+ (aq)Pb2+ (aq) + Zn (s) Pb (s) + Zn2+ (aq)

Pb2+ (aq) + 2 e- Pb (s) Zn (s) Zn2+ (aq) + 2 e- Looking at the two half-reactions, we find that

the Pb2+ is being reduced, and the Zn is beingoxidized.

Therefore, the anode compartment of our cell Therefore, the anode compartment of our cellwould consist of a strip of zinc metal immersedin a solution containing Zn2+ ions (such as zincnitrate).

The cathode compartment would consist of astrip of lead immersed in a solution containingPb2+ ions (such as lead (II) nitrate).

Pb2+ (aq) + 2 e- Pb (s) Zn (s) Zn2+ (aq) + 2 e- The two half-cells would be connected to

each other with a salt bridge and an externalwire.wire.

Electrons flow through the wire from the zincanode to the lead cathode.

Anions move from the cathode compartmenttowards the anode while cations migrate fromthe anode compartment toward the cathode.

Redox chemistry of iron and copperRedox chemistry of iron and copper

Single vertical line, , represents a phaseboundary.

Double vertical line, , represents a saltbridge.

Shorthand for the anode half-cell is alwayswritten on the left of the salt-bridge symbol,followed on the right of the symbol by theshorthand for the cathode half-cell. Reactants in each half cell are written first, followed Reactants in each half cell are written first, followed

by products.

Electrons move through the external circuit fromleft to right.

For Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s):Zn (s)Zn2+ (aq) Cu2+ (aq)Cu (s).

Cell involving a gas. Additional vertical line due to presence of

additional phase.

List the gas immediately adjacent to the appropriateelectrode.

Detailed notation includes ion concentrations Detailed notation includes ion concentrationsand gas pressures.

EXAMPLE:EXAMPLE:EXAMPLE:EXAMPLE:

Give the shorthand notation for a galvanic cell that employs the overall reaction

Pb(NO3)2(aq) + Ni (s) Pb(s) + Ni(NO3)2(aq)Pb(NO3)2(aq) + Ni (s) Pb(s) + Ni(NO3)2(aq)

Give a brief description of the cell.

SOLUTION: The two half-reactions for this overall reaction are:

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

Ni (s) Ni2+ (aq) + 2 e-

From these half-reactions, we know that lead is being reduced and nickel is being oxidized.

Therefore, Ni is the anode and Pb is thecathode. The cell notation is:

Ni(s)Ni2+(aq)Pb2+(aq)Pb(s) This cell would consist of a strip of nickel as

the anode dipping into an aqueous solutionthe anode dipping into an aqueous solutionof Ni(NO3)2 and a strip of Pb as the cathodedipping into an aqueous solution of Pb(NO3)2.

The two half-cells would be connected by asalt bridge and a wire.

Electrons spontaneously flow from onespecies to anotherand through the wirefrom the anode to the cathode of a voltaiccellbecause of a difference in potentialenergy, or a potential difference.energy, or a potential difference.

The potential difference between twoelectrodes is measured in volts.

One volt is equal to one joule per coulomb. 1

The potential difference that drives electrons through the wire in a voltaic cell is called the electromotive force or electromotive force or emf. For a voltaic cell the emf is denoted Ecell and referred to as the cell potential.

Cell potential - measured with a voltmeter. Gives a positive reading when the + and - terminals

of the voltmeter are connected to cathode (+) andanode (-), respectively.

can use voltmeter-cell connections to determine whichelectrode is the anode and which is the cathodeelectrode is the anode and which is the cathode

The value of a cell potential depends on whathalf-reactions are taking place in the twocompartments of the cell.

The cell potential measured under standardconditions, Ecell (25C, 1 M concentrations,conditions, Ecell (25C, 1 M concentrations,and 1 atm pressures), is the standardstandardstandardstandard cellcellcellcellpotentialpotentialpotentialpotential or standardstandardstandardstandard emfemfemfemf.

For the zinc and copper voltaic cell in Figure20.5, Ecell is 1.10 V.

That is for the reaction

at 25C, where the concentrations of copperand zinc ions are both 1 M.

Such potentials can be measuredexperimentally, but many of them can becalculated from tabulated standardstandardstandardstandard reductionreductionreductionreductionpotentialspotentialspotentialspotentials, Ered values.

The more positive the standard cell potential,the greater the driving force for electrons toflow from the anode to the cathode.

Because the cathode of a voltaic cell is alwaysthe half-reaction with the more positive (orthe half-reaction with the more positive (orless negative) standard reduction potential,the standard cell potential of a voltaic cell isalways positive.

The standard reduction potentials for the varioushalf-reactions are measured against a standardstandardstandardstandardhydrogenhydrogenhydrogenhydrogen electrodeelectrodeelectrodeelectrode (SHE).

The half-reaction of interest and the SHE, bothunder standard conditions, are made into aunder standard conditions, are made into avoltaic cell, as shown in Figure 20.11, and thecell potential is measured experimentally.

The standard potential of the standard hydrogenelectrode's half-reaction is arbitrarily assigned avalue of zero, so the measured potentialcorresponds to the half-reaction being evaluated.

The standard reduction potentials in Table20.1 can be used to compare the oxidizingpower or reducing power of a substance.

The more positive the value of Ered for aspecies, the more readily it undergoesspecies, the more readily it undergoesreduction and the better oxidizing agent it is.

As Ered becomes more negative, the specieson the right side of the arrow becomes astronger reducing agent.

Standard reduction potentialsStandard reduction potentials

It is possible to use standard reductionpotentials to predict whether or not a givenoxidation-reduction reaction will bespontaneous.

For instance, will copper solid be oxidized by For instance, will copper solid be oxidized bya solution containing iron(II) ions?

To determine the answer, we calculate thestandard cell potential for the cell as we havedescribed it.

In our description, copper is being oxidizedso the copper electrode is the anode.so the copper electrode is the anode.

Iron is being reduced, making the ironelectrode the cathode.

Ered values for the two half-reactions are

Ered = 0.34 V

Ered = - 0.44 V

The negative Ecell tells us that this reaction, aswritten, will not occur spontaneously. (Its reversereaction will be spontaneous.)

Our ability to predict the spontaneity of achemical reaction by calculating a standardcell potential points to a relationship betweenthe sign of Ecell and the sign of G , thestandard change in Gibbs free energy.standard change in Gibbs free energy.

Quantitatively, this relationship is expressedas

n is the number of moles of electronstransferred in the reaction,

F, called Faraday's constant, is the quantity ofelectrical charge on a mole of electrons.

Both n and F are always positive numbers.

Therefore, a positive value for E will alwayscorrespond to a negative value for G , bothdenoting a spontaneous reaction.

The relationship between cell potential and The relationship between cell potential andfree energy change holds at conditions otherthan standard, as well.

E = E o 0.0592

nlogQ

Nernst equation:

(in volts at 25oC). Enables us to calculate cell potentials under

nonstandard-state conditions.

E = E o 0.0592

nlogQ

nonstandard-state conditions.

Calculate Ecell for the following cell reaction:

2 Cr(s) + 3Pb2+(aq) 2 Cr3+ (aq) + 3Pb(s) [Pb2+] = 0.15 M;

[Cr3+] = 0.50 M

2 Cr (s) 2 Cr3+ + 6 e- Eo =+0.74 V

3 Pb2+ (aq) + 6 e- 3 Pb (s) Eo=-0.13V

( )Ecell Vo = + + = +0 74 013 0 61. . .

E = E o 0.0592

nlogQ

E = E n

logQ

Ecell = Ecello

0.0592

6log

Cr 3+[ ]2Pb2 +[ ]3

Ecell = +0.610.0592

6log

0.5( )2

0.15( )3= 0.59

Three different ways to determine the value of an equilibrium constant K:

K =C[ ]c D[ ]d

K =C[ ] D[ ]A[ ]a B[ ]b

ln K =Go

RT

ln K =nFEo

RT

Equilibrium constants for redox reactionstend to be either very large or very small incomparison with equilibrium constants foracid-base reactions. Positive value of Eo corresponds to K > 1. Positive value of E corresponds to K > 1.

Negative value of Eo corresponds to K < 1.

Important application of Nernst equation -electrochemical determination of pH using apH meter.

Consider a cell with a hydrogen electrode asthe anode and a second reference electrodethe anode and a second reference electrodeas the cathode. Pt (s)H2 (1 atm)H+ (? M)reference cathode.

Ecell = 0.0592pH + Eref

can measure the pH of a solution by

Ecell = 0.0592pH + Eref

pH =Ecell Eref

0.0592

can measure the pH of a solution bymeasuring Ecell

Actual pH measurements use a glasselectrode with a calomel electrode as thereference.

EXAMPLEEXAMPLEEXAMPLEEXAMPLE::::

The following cell has a potential of 0.49 V.Calculate the pH of the solution in the anodecompartment.

Pt(s) H2(g) (1 atm)H+(pH = ?)Cl-(aq) (1M)Hg2Cl2 Pt(s) H2(g) (1 atm)H+(pH = ?)Cl-(aq) (1M)Hg2Cl2

(s) Hg (l)

The cell reaction is

Hg2Cl2 (s) + H2 (g) 2 Hg (l) + 2 Cl- (aq) + 2H+ (aq)

The cell reaction is

Hg2Cl2 (s) + H2 (g) 2 Hg (l) + 2 Cl- (aq) + 2H+ (aq)

V 28.0V 28.0V 00.0ooo =+=+= EEE V 28.0V 28.0V 00.0oClHg,ClHg

o

HH

o-

22+

2=+=+=

EEE

pH =Ecell Eref

0.0592

55.30592.0

V 28.0V 49.0pH ==

EXAMPLEEXAMPLEEXAMPLEEXAMPLE::::

Calculate the equilibrium constant for thefollowing reaction at 25oC.

5 S O 2- (aq) + I (s) + 6 H O (l) 10 SO 2- (aq) + 2 IO3- (aq) + 5 S2O82- (aq) + I2 (s) + 6 H2O (l) 10 SO42- (aq) + 2 IO3- (aq) +12 H+ (aq)

S2O82- (aq) + 2 e- 2 SO42- (aq) Eo = +2.01

V

I2(s) + 6H2O (l) 2IO3- (aq) + 12 H+(aq) +10e- Eo= -1.20 V

Ecell

o = 2.01+ 1.20( )= +0.81 V

The value of n for this reaction is 10.

log K =(10)(0.81)

(0.0592)= 137

K = 10137

Eo =0.0592

nlogK

Most important practical application ofgalvanic cells is their use as batteries.

Features required in a battery depend on theapplication.

C. General features. C. General features. Compact and lightweight. Physically rugged and inexpensive. Provide a stable source of power for relatively long

periods of time.

A battery is a self-contained source ofelectrochemical energy made from one ormore voltaic cells.

Ordinary flashlight batteries consist of asingle voltaic cell, while car batteries are sixsingle voltaic cell, while car batteries are sixidentical voltaic cells connected in series.

It is worth noting that electrochemistry is oneof only a very few commercially viablemethods of generating electricity.

Car Battery

Cathode:

Anhode:

Car Battery

Cathode:

Anhode:

Corrosion - the oxidative deterioration of ametal.

Well-known example of corrosion -conversion of iron to rust. Requires both oxygen and water. Requires both oxygen and water.

Involves pitting of the metal surface.

rust is deposited at a location physically separatedfrom the pits

Corrosion is the undesirable oxidation of a metal.

A familiar example of corrosion is the rusting ofiron.

Iron metal is oxidized to Fe2+ by oxygen.

The Fe2+ is then further oxidized to Fe3+ in ahydrated form of Fe2O3, what we know as rust.

Corrosion of iron can be prevented by coatingit with paint or with another metal such as tinor zinc.

Coating with paint or with a less readilyoxidized metal such as tin protects the ironoxidized metal such as tin protects the ironsimply by preventing oxygen and water fromreaching the iron surface.

If such a coating is damaged and the iron isexposed, corrosion of iron will occur in theexposed area.

Some metals create their own sealant byoxidation.

Aluminum, for example, oxidizes fairlyreadily.

Ered = - 1.66 V

Coating iron with a more easily oxidizedmetal such as zinc, also prevents oxygen andwater from reaching the iron surface.

But unlike tin, zinc protects the iron even ifthe zinc coating is damaged.the zinc coating is damaged.

It does so by making the iron serve as thecathode in an voltaic cell.

When the zinc coating is damaged and iron isexposed, the zinc itself is oxidized ratherthan the iron.

Ered = - 0.44 V

Ered = - 0.76 V

Iron, with the more positive (less negative)Ered is more easily reduced and therefore,less easily oxidized.

Zinc will be oxidized, serving as the sacrificialanode.anode.

This type of corrosion prevention is calledcathodiccathodiccathodiccathodic protectionprotectionprotectionprotection.

Ered = - 0.44 V

Ered = - 0.76 V

One of the many uses of cathodic protection is toprevent corrosion of underground pipes andstorage tanksmany of which are iron.

A more easily oxidized metal is placed in electricalcontact with the object to be protected by ancontact with the object to be protected by aninsulated copper wire.

The iron object becomes the cathode (and the moreeasily oxidized metal the sacrificial anode) in avoltaic cell.

Oxidation eventually consumes the sacrificialanode, and it must be replaced periodically.

A voltaic cell is one in which a spontaneouschemical reaction is used to generate avoltage.

Electrolysis is the use of a voltage to drive anonspontaneous reaction.nonspontaneous reaction.

Reactions that are driven by an externallysupplied voltage are called electrolysiselectrolysiselectrolysiselectrolysisreactionsreactionsreactionsreactions, and electrochemical cells designedfor the purpose of carrying out electrolysisreactions are called electrolyticelectrolyticelectrolyticelectrolytic cellscellscellscells

Sodium metal is produced commercially byelectrolysis of molten sodium chloride.

Electrodes are immersed in molten sodiumchloride, and a voltage source driveselectrons from the anode to the cathode.electrons from the anode to the cathode.

Sodium is reduced at the cathode to moltensodium metal.

Chloride ions are oxidized to chlorine gas atthe anode.

Cathode:

Anode:

Ered = - 2.71 V

Ered = 1.36 V

Ecell = - 4.07 V

A negative cell potential means thatthe reaction is nonspontaneous.The value of the cell potential tells usthat a minimum of 4.07 volts must beapplied to the cell to drive the reactionin the desired direction.

Ered = - 2.71 VCathode:

Anode: Ered = 1.36 V

Ecell = - 4.07 V

Electrolysis of molten salts is usedcommercially to produce a number ofactive metals.

Ered = - 2.71 VCathode:

Anode: Ered = 1.36 V

Ecell = - 4.07 V

Electrolysis of molten salts is usedcommercially to produce a number ofactive metals.

Electrolysis can also be done with aqueoussolutions.

Although chlorine gas can be produced froma solution of sodium chloride, sodium metalcannot be produced.cannot be produced.

This has to do with the relative ease ofreduction of water versus sodium ions.

Ered = - 2.71 V

Ered = - 0.83 V

With a far less negative reduction potential,waterrather than sodium will be reducedat the cathode of an electrolytic cellcontaining aqueous sodium chloride.

Ered = - 2.71 V

Ered = - 0.83 V

Ered = - 2.71 V

Ered = - 0.83 V

In a comparison of reduction potentials, wemight also predict that water, and notchloride ions, would be oxidized at the anodeof such an electrolytic cell.

Ered = 1.36 V

Ered = 1.23 V

The first half-reaction, with the more positivereduction potential, is more apt to occur as areduction rather than an oxidation.

Experimentally, however, chlorine gas isproduced at the anode during electrolysis ofproduced at the anode during electrolysis ofaqueous sodium chloride.

Ered = 1.36 V

Ered = 1.23 V

This is explained by kinetics.

Although the oxidation of water isthermodynamically favored, the oxidation ofchloride ions is kinetically favored because ofa lower activation energy.a lower activation energy.

Ered = 1.36 V

Ered = 1.23 V

Electrolysis of water

Electroplating is an electrolytic process usedto deposit a thin layer of one metal onanother.

In such a process the metal to be deposited isused as the anode, and the metal on whichused as the anode, and the metal on whichthe deposit is to be made is the cathode.

Electroplating is the process by which silver-plated flatware is finished.

It is also the process used to coat iron carbodies with zinc to protect them from rust.

When the electrode in an electrolytic cell isinvolved in the reaction, it is called an activeelectrode.

When the same metal is to be oxidized fromthe anode and reduced (deposited) at thethe anode and reduced (deposited) at thecathode, the standard cell potential is zero.

It therefore requires only a very small voltageto drive such a process.

Electroplating

Using stoichiometry, we can relate theamount of metal deposited in anelectroplating process to the current and thelength of time for which the current isapplied.applied.

The charge passing through an electrolyticcell is measured in coulombs.

How many grams of copper would be reducedby the application of 5.00 amps to a solutionof copper sulfate for 35.0 minutes?

EXAMPLEEXAMPLEEXAMPLEEXAMPLE::::

How many grams of Cl2 would be produced inthe electrolysis of molten NaCl by a current of4.25 A for 35.0 min?

Remember that a coulomb is an A.s or that anampere is C/s.

2 Cl- Cl2 + 2 e- moles of electrons = 2

4.25 C

s 35.0 min

60 s

1 min

1 mol e -

96,500 C

1 mol Cl 22 mol e -

70.9 g Cl 21 mol Cl -

= 3.28 g Cl2