electrochemistry chapter 20

Chapter 20 1

ElectrochemistryElectrochemistry

Chapter 20Chapter 20

Chapter 20 2

Oxidation-Reduction ReactionsOxidation-Reduction Reactions• Zn added to HCl yields the spontaneous reaction

Zn(s) + 2H+(aq) Zn2+(aq) + H2(g).• The oxidation number of Zn has increased from 0 to 2+.• The oxidation number of H has reduced from 1+ to 0.• Therefore, Zn is oxidized to Zn2+ while H+ is reduced to

H2.• H+ causes Zn to be oxidized and is the oxidizing agent.• Zn causes H+ to be reduced and is the reducing agent.• Note that the reducing agent is oxidized and the

oxidizing agent is reduced.

Chapter 20 3

Balancing Oxidation-Reduction EquationsBalancing Oxidation-Reduction Equations

• Law of conservation of mass: the amount of each element present at the beginning of the reaction must be present at the end.

• Conservation of charge: electrons are not lost in a chemical reaction.

• In complicated redox reactions, we need to look at the transfer of electrons carefully.

Half-ReactionsHalf-Reactions• Half-reactions are a convenient way of separating

oxidation and reduction reactions.

Chapter 20 4


Half-ReactionsHalf-Reactions• The half-reactions for

Sn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe3+(aq)are

Sn2+(aq) Sn4+(aq) +2e-

2Fe3+(aq) + 2e- 2Fe2+(aq)• Oxidation: electrons are products.• Reduction: electrons are reagents.

Chapter 20 5


Balancing Equations by the Method of Half-Balancing Equations by the Method of Half-ReactionsReactions

• Consider the titration of an acidic solution of Na2C2O4 (sodium oxalate, colorless) with KMnO4 (deep purple).

• MnO4- is reduced to Mn2+ (pale pink) while the C2O4

2- is oxidized to CO2.

• The equivalence point is given by the presence of a pale pink color.

• If more KMnO4 is added, the solution turns purple due to the excess KMnO4.

Chapter 20 6



Chapter 20 7



What is the balanced chemical equation?1. Write down the two half reactions.2. Balance each half reaction:

a. First with elements other than H and O.b. Then balance O by adding water.c. Then balance H by adding H+.d. Finish by balancing charge by adding electrons.

Chapter 20 8



3. Multiply each half reaction to make the number of electrons equal.

4. Add the reactions and simplify.5. Check!

For KMnO4 + Na2C2O4:

Chapter 20 9



1. The two incomplete half reactions areMnO4

-(aq) Mn2+(aq)

C2O42-(aq) 2CO2(g)

2. Adding water and H+ yields 8H+ + MnO4

-(aq) Mn2+(aq) + 4H2O• There is a charge of 7+ on the left and 2+ on the right.

Therefore, 5 electrons need to be added to the left:5e- + 8H+ + MnO4

-(aq) Mn2+(aq) + 4H2O

Chapter 20 10



• In the oxalate reaction, there is a 2- charge on the left and a 0 charge on the right, so we need to add two electrons:

C2O42-(aq) 2CO2(g) + 2e-

3. To balance the 5 electrons for permanganate and 2 electrons for oxalate, we need a common multiple of 10 electrons for both. Multiplying gives:

Chapter 20 11



10e- + 16H+ + 2MnO4-(aq) 2Mn2+(aq) + 8H2O

5C2O42-(aq) 10CO2(g) + 10e-

4. Adding gives:16H+(aq) + 2MnO4

-(aq) + 5C2O42-(aq) 2Mn2+(aq) + 8H2O(l)

+ 10CO2(g)

5. Which is balanced!

Chapter 20 12

Balancing Oxidation-Reduction EquationsBalancing Oxidation-Reduction EquationsPractice Examples1) Balance the following reaction in acidic conditions:

Fe2+(aq) + MnO4-(aq) Fe3+(aq) + Mn2+(aq)

Ans:

5Fe2+(aq) + MnO4-(aq) + 8H+(aq) 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

•Balance the following reaction in acidic conditions:

Cr2O72-(aq) + I-(aq) Cr3+(aq) + I2(s)

Ans:

Cr2O72-(aq) + 6I-(aq) +14H+(aq) Cr3+(aq) + 3I2(s) + 7H2O(l)

Chapter 20 13


Balancing Equations for Reactions Occurring in Balancing Equations for Reactions Occurring in Basic SolutionBasic Solution•We use OH- and H2O rather than H+ and H2O.•The same method as above is used, but OH- is added to “neutralize” the H+ used (generally easier than using OH- and H2O)

•Let’s consider the following reaction run in base:

CN-(aq) + MnO4-(aq) CNO-(aq) + MnO2(s)

Chapter 20 14

CN-(aq) + MnO4-(aq) CNO-(aq) + MnO2(s)

Separating into half-reactions and balancing in acid, we get:

H2O(l) + CN-(aq) CNO-(aq) + 2H+(aq) + 2e-

and

3e- + 4H+(aq) + MnO4-(aq) MnO2(s) + 2H20(l)

After adjusting each equation to 6 electrons, and combining, we get:

2H+(aq) + 3CN-(aq) + 2MnO4-(aq) 3CNO-(aq) + 2MnO2(s) + H2O(l)


Chapter 20 15

The following equation was balanced in acid. We will convert it to basic conditions by adding OH- ions to both sides of the equation to neutralize the H+ ions present. When the OH- ions and the H+ ions are combined, they produce water. See the conversion below:

2H+(aq) + 3CN-(aq) + 2MnO4-(aq) 3CNO-(aq) + 2MnO2(s) + H2O(l)

H2O(aq) + 3CN-(aq) + 2MnO4-(aq) 3CNO-(aq) + 2MnO2(s) + 2OH-(aq)

This method for balancing in base is simpler because it avoids the complications of adding OH- and H2O, which both contain O and H


+ 2OH-+ 2OH-2H2O(l)

Chapter 20 16

Balancing Oxidation-Reduction EquationsBalancing Oxidation-Reduction EquationsPractice Examples1) Balance the following reaction in basic conditions:

MnO4-(aq) + SO3

2-(aq) MnO2(s) + SO42-(aq)

Ans:

2MnO4-(aq) + 3SO3

2-(aq) + H2O(l) 2MnO2(s) + 3SO42-(aq) + 2OH-(aq)

•Balance the following reaction in basic conditions:

MnO4-(aq) + I-(aq) MnO4

2-(aq) + IO3-(aq)

Ans:

6MnO4-(aq) + I-(aq) + 6OH-(aq) 6MnO4

2-(aq) + IO3-(aq) + 3H2O(l)

Chapter 20 17

Voltaic CellsVoltaic Cells• The energy released in a spontaneous redox reaction is

used to perform electrical work.• Voltaic or galvanic cells are devices in which electron

transfer occurs via an external circuit.• Voltaic cells are spontaneous.• If a strip of Zn is placed in a solution of CuSO4, Cu is

deposited on the Zn and the Zn dissolves by forming Zn2+.

• Zn is spontaneously oxidized to Zn2+ by Cu2+.• The Cu2+ is spontaneously reduced to Cu0 by Zn.• The entire process is spontaneous.

Chapter 20 18

Voltaic CellsVoltaic Cells

Chapter 20 19

Voltaic CellsVoltaic Cells• Voltaic cells consist of

– Anode: Zn(s) Zn2+(aq) + 2e2

– Cathode: Cu2+(aq) + 2e- Cu(s) – Salt bridge (used to complete the electrical circuit): cations move from anode to

cathode, anions move from cathode to anode.

• The two solid metals are the electrodes (cathode and anode).

• As oxidation occurs, Zn is converted to Zn2+ and 2e-. The electrons flow towards the anode where they are used in the reduction reaction.

Chapter 20 20

Voltaic CellsVoltaic Cells• We expect the Zn electrode to lose mass and the Cu

electrode to gain mass.• “Rules” of voltaic cells:

1. At the anode electrons are products. (Oxidation)2. At the cathode electrons are reagents. (Reduction)3. Electrons cannot swim.

• Electrons flow from the anode to the cathode.• Therefore, the anode is negative and the cathode is

positive.• Electrons cannot flow through the solution, they have to

be transported through an external wire. (Rule 3.)

Chapter 20 21

Voltaic CellsVoltaic Cells

Chapter 20 22

Voltaic CellsVoltaic Cells• Anions and cations move through a porous barrier or salt

bridge.• Cations move into the cathodic compartment to

neutralize the excess negatively charged ions (Cathode: Cu2+ + 2e- Cu, so the counterion of Cu is in excess).

• Anions move into the anodic compartment to neutralize the excess Zn2+ ions formed by oxidation.

Chapter 20 23

Voltaic CellsVoltaic CellsA Molecular View of Electrode ProcessesA Molecular View of Electrode Processes• Consider the spontaneous redox reaction between Zn(s)

and Cu2+(aq).• During the reaction, Zn(s) is oxidized to Zn2+(aq) and Cu2+

(aq) is reduced to Cu(s). • On the atomic level, a Cu2+(aq) ion comes into contact

with a Zn(s) atom on the surface of the electrode.• Two electrons are directly transferred from the Zn(s)

(forming Zn2+(aq)) to the Cu2+(aq) (forming Cu(s)).

Chapter 20 24

Cell EMFCell EMF• The flow of electrons from anode to cathode is

spontaneous.• Electrons flow from anode to cathode because the

cathode has a lower electrical potential energy than the anode.

• Potential difference: difference in electrical potential. Measured in volts.

• One volt is the potential difference required to impart one joule of energy to a charge of one coulomb:

C 1J 1V 1

Chapter 20 25

Cell EMFCell EMF• Electromotive force (emf) is the force required to push

electrons through the external circuit.• Cell potential: Ecell is the emf of a cell.• For 1M solutions at 25 C (standard conditions), the

standard emf (standard cell potential) is called Ecell.

Standard Reduction PotentialsStandard Reduction Potentials• Convenient tabulation of electrochemical data.• Standard reduction potentials, Ered are measured

relative to the standard hydrogen electrode (SHE).

Chapter 20 26

Cell EMFCell EMFStandard Reduction PotentialsStandard Reduction Potentials

Chapter 20 27

Cell EMFCell EMFStandard Reduction PotentialsStandard Reduction Potentials• The SHE is the cathode. It consists of a Pt electrode in a

tube placed in 1 M H+ solution. H2 is bubbled through the tube.

• For the SHE, we assign2H+(aq, 1M) + 2e- H2(g, 1 atm)

• Ered of zero.• The emf of a cell can be calculated from standard

reduction potentials:Ecell = Ered(cathode) - Ered(anode)

Chapter 20 28


Chapter 20 29

Cell EMFCell EMFStandard Reduction PotentialsStandard Reduction Potentials• Consider Zn(s) Zn2+(aq) + 2e-. We measure Ecell relative

to the SHE (cathode): Ecell = Ered(cathode) - Ered(anode)

0.76 V = 0 V - Ered(anode).• Therefore, Ered(anode) = -0.76 V.• Standard reduction potentials must be written as

reduction reactions:Zn2+(aq) + 2e- Zn(s), Ered = -0.76 V.

• Since Ered = -0.76 V we conclude that the reduction of Zn2+ in the presence of the SHE is not spontaneous.

Chapter 20 30

Cell EMFCell EMFStandard Reduction PotentialsStandard Reduction Potentials• The oxidation of Zn with the SHE is spontaneous.• Changing the stoichiometric coefficient does not affect

Ered.• Therefore,

2Zn2+(aq) + 4e- 2Zn(s), Ered = -0.76 V.• Reactions with Ered > 0 are spontaneous reductions

relative to the SHE.• Reactions with Ered < 0 are spontaneous oxidations

relative to the SHE.

Chapter 20 31


• The larger the difference between Ered values, the larger Ecell.

• In a voltaic (galvanic) cell (spontaneous) Ered(cathode) is more positive than Ered(anode).

• RecallEcell = Ered(cathode) - Ered(anode)

Chapter 20 32

Cell EMFCell EMFOxidizing and Reducing AgentsOxidizing and Reducing Agents• The more positive Ered the stronger the oxidizing agent

on the left.• The more negative Ered the stronger the reducing agent

on the right.• A species on the higher to the left of the table of

standard reduction potentials will spontaneously oxidize a species that is lower to the right in the table.

• That is, F2 will oxidize H2 or Li; Ni2+ will oxidize Al(s).• Any species on the right will spontaneously reduce

anything that is higher to the left in the series.

Chapter 20 33

Cell EMFCell EMFOxidizing and Reducing AgentsOxidizing and Reducing Agents

Chapter 20 34

Spontaneity of Redox ReactionsSpontaneity of Redox Reactions• In a voltaic (galvanic) cell (spontaneous) Ered(cathode) is

more positive than Ered(anode) since

Ecell = Ered(cathode) - Ered(anode)• More generally, for any electrochemical process

E = Ered(reduction process) - Ered(oxidation process).• A positive E indicates a spontaneous process (galvanic

cell).• A negative E indicates a nonspontaneous process.• The above equation is used to understand the activity

series.

Chapter 20 35

Spontaneity of Redox ReactionsSpontaneity of Redox Reactions• Consider the displacement of silver by nickel:

Ni(s) + 2Ag+(aq) Ni2+(aq) + 2Ag(s)has

E = Ered(Ag+/Ag) - Ered(Ni2+/Ni)

= (0.80 V) - (-0.28 V)= 1.08 V,

which indicates a spontaneous process.

EMF and Free-Energy ChangeEMF and Free-Energy Change• We can show that

G = -nFE

Chapter 20 36

Spontaneity of Redox ReactionsSpontaneity of Redox ReactionsEMF and Free-Energy ChangeEMF and Free-Energy Change G is the change in free-energy, n is the number of

moles of electrons transferred, F is Faraday’s constant, and E is the emf of the cell.

• We define

• Since n and F are positive, if G > 0 then E < 0.mol-J/V 96,500C/mol 500,961 F

Chapter 20 37

Effect of Concentration on Cell EMFEffect of Concentration on Cell EMF• A voltaic cell is functional until E = 0 at which point

equilibrium has been reached.• The point at which E = 0 is determined by the

concentrations of the species involved in the redox reaction.

The Nernst EquationThe Nernst Equation• The Nernst equation relates emf to concentration using

and noting that QRTGG ln

QRTnFEnFE ln

Chapter 20 38

Effect of Concentration on Cell EMFEffect of Concentration on Cell EMFThe Nernst EquationThe Nernst Equation• This rearranges to give the Nernst equation:

• The Nernst equation can be simplified by collecting all the constants together using a temperature of 298 K:

• (Note the change from natural logarithm to base-10 log.)• Remember that n is number of moles of electrons.

QnFRTEE ln

Qn

EE log0592.0

Chapter 20 39

Effect of Concentration on Cell EMFEffect of Concentration on Cell EMFConcentration CellsConcentration Cells• We can use the Nernst equation to generate a cell that

has an emf based solely on difference in concentration.• One compartment will consist of a concentrated

solution, while the other has a dilute solution.• Example: 1.00 M Ni2+(aq) and 1.00 10-3 M Ni2+(aq).• The cell tends to equalize the concentrations of Ni2+(aq)

in each compartment.• The concentrated solution has to reduce the amount of

Ni2+(aq) (to Ni(s)), so must be the cathode.

Chapter 20 40

Effect of Concentration on Cell EMFEffect of Concentration on Cell EMFConcentration CellsConcentration Cells• Since the two half-reactions are the same, Eº will be

zero.

V 0888.0 1.00

101.00log2

0592.0V 0

]Ni[

]Ni[log2

0592.0V 0

log0592.0

3-edconcentrat

2dilute

2

MM

Qn

EE

Chapter 20 41

Effect of Concentration on Cell EMFEffect of Concentration on Cell EMFCell EMF and Chemical EquilibriumCell EMF and Chemical Equilibrium• A system is at equilibrium when G = 0.• From the Nernst equation, at equilibrium:

.0592.0

log

.log0592.00

: and 0 mequilibriuAt

.ln0592.0

nEK

Kn

E

KQE

Qn

EE

Chapter 20 42

BatteriesBatteriesLead-Acid BatteryLead-Acid 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) + SO4

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

Chapter 20 43

BatteriesBatteriesLead-Acid BatteryLead-Acid Battery• The overall electrochemical reaction is PbO2(s) + Pb(s) + 2SO4

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

for whichEcell = Ered(cathode) - Ered(anode)

= (+1.685 V) - (-0.356 V)= +2.041 V.

• Wood or glass-fiber spacers are used to prevent the electrodes form touching.

Chapter 20 44

BatteriesBatteriesLead-Acid BatteryLead-Acid Battery

Chapter 20 45

BatteriesBatteriesAlkaline BatteryAlkaline 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)

• The graphite rod in the center is an inert cathode.• For an alkaline battery, NH4Cl is replaced with KOH.• Anode: Zn powder mixed in a gel:

Zn(s) Zn2+(aq) + 2e- • Cathode: reduction of MnO2.

Chapter 20 46

BatteriesBatteriesAlkaline BatteryAlkaline Battery

Chapter 20 47

BatteriesBatteriesFuel CellsFuel Cells• Direct production of electricity from fuels occurs in a fuel

cell.• On Apollo moon flights, the H2 - O2 fuel cell was the

primary source of electricity.

• Cathode: reduction of oxygen:O2(g) + 4H+(aq) + 4e- 2H2O(l)

• Anode: oxidation of hydrogen:H2(g) 2H+(aq) + 2e-

Chapter 20 48

BatteriesBatteriesFuel CellsFuel Cells

Chapter 20 49

CorrosionCorrosionCorrosion of IronCorrosion of Iron• Since Ered(Fe2+) < Ered(O2) iron can be oxidized by oxygen.• Cathode: O2(g) + 4H+(aq) + 4e- 2H2O(l).• Anode: Fe(s) Fe2+(aq) + 2e-.• Dissolved oxygen in water usually causes the oxidation

of iron.• Fe2+ initially formed can be further oxidized to Fe3+ which

forms rust, Fe2O3.xH2O(s).• Oxidation occurs at the site with the greatest

concentration of O2.

Chapter 20 50

CorrosionCorrosionCorrosion of IronCorrosion of Iron

Chapter 20 51

CorrosionCorrosionPreventing the Corrosion of IronPreventing the Corrosion of Iron• Corrosion can be prevented by coating the iron with

paint or another metal.• Galvanized iron is coated with a thin layer of zinc.• Zinc protects the iron since Zn is the anode and Fe the

cathode:Zn2+(aq) +2e- Zn(s), Ered = -0.76 V

Fe2+(aq) + 2e- Fe(s), Ered = -0.44 V• With the above standard reduction potentials, Zn is

easier to oxidize than Fe.

Chapter 20 52

CorrosionCorrosionPreventing the Corrosion of IronPreventing the Corrosion of Iron

Chapter 20 53

CorrosionCorrosionPreventing the Corrosion of IronPreventing the Corrosion of Iron• To protect underground pipelines, a sacrificial anode is

added.• The water pipe is turned into the cathode and an active

metal is used as the anode.• Often, Mg is used as the sacrificial anode:

Mg2+(aq) +2e- Mg(s), Ered = -2.37 V

Fe2+(aq) + 2e- Fe(s), Ered = -0.44 V

Chapter 20 54

CorrosionCorrosionPreventing the Corrosion of IronPreventing the Corrosion of Iron

Chapter 20 55

ElectrolysisElectrolysisElectrolysis of Aqueous SolutionsElectrolysis of Aqueous Solutions• Nonspontaneous reactions require an external current in

order to force the reaction to proceed.• Electrolysis reactions are nonspontaneous.• In voltaic and electrolytic cells:

– reduction occurs at the cathode, and– oxidation occurs at the anode.– However, in electrolytic cells, electrons are forced to flow from the anode to

cathode.– In electrolytic cells the anode is positive and the cathode is negative. (In galvanic

cells the anode is negative and the cathode is positive.)

Chapter 20 56

ElectrolysisElectrolysisElectrolysis of Aqueous SolutionsElectrolysis of Aqueous Solutions

Chapter 20 57

ElectrolysisElectrolysisElectrolysis of Aqueous SolutionsElectrolysis of Aqueous Solutions• Example, decomposition of molten NaCl.• Cathode: 2Na+(l) + 2e- 2Na(l)• Anode: 2Cl-(l) Cl2(g) + 2e-.• Industrially, electrolysis is used to produce metals like Al.

Electrolysis with Active ElectrodesElectrolysis with Active Electrodes• Active electrodes: electrodes that take part in

electrolysis.• Example: electrolytic plating.

Chapter 20 58

ElectrolysisElectrolysisElectrolysis with Active ElectrodesElectrolysis with Active Electrodes

Chapter 20 59

ElectrolysisElectrolysisElectrolysis with Active ElectrodesElectrolysis with Active Electrodes• Consider an active Ni electrode and another metallic

electrode placed in an aqueous solution of NiSO4:• Anode: Ni(s) Ni2+(aq) + 2e-

• Cathode: Ni2+(aq) + 2e- Ni(s).• Ni plates on the inert electrode.• Electroplating is important in protecting objects from

corrosion.

Chapter 20 60

ElectrolysisElectrolysisQuantitative Aspects of ElectrolysisQuantitative Aspects of Electrolysis• We want to know how much material we obtain with

electrolysis.• Consider the reduction of Cu2+ to Cu.

– Cu2+(aq) + 2e- Cu(s).– 2 mol of electrons will plate 1 mol of Cu.– The charge of 1 mol of electrons is 96,500 C (1 F).– Since Q = It, the amount of Cu can be calculated from the current (I) and time (t)

taken to plate.

Chapter 20 61

ElectrolysisElectrolysisElectrical WorkElectrical Work• Free-energy is a measure of the maximum amount of

useful work that can be obtained from a system.• We know

• If work is negative, then work is performed by the system and E is positive.

• The emf can be thought about as a measure of the driving force for a redox process.

...

max

max

nFEwnFEGwG

Chapter 20 62

ElectrolysisElectrolysisElectrical WorkElectrical Work• In an electrolytic cell an external source of energy is

required to force the reaction to proceed.• In order to drive the nonspontaneous reaction the

external emf must be greater than Ecell.• From physics: work has units watts:

1 W = 1 J/s.• Electric utilities use units of kilowatt-hours:

J. 106.3

W1J/s 1

h 1s 3600h 1 W1000kWh 1

6

electrochemistry chapter 20

Documents

h aq zn2 aq h2g

e2fe3 aq

balancing charge

h yields

transfer of electrons

oxidation number of

h causes zn

zn causes h