Electrochemistry

Chapter 20Electrochemistry

Chemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

John D. BookstaverSt. Charles Community College

St. Peters, MO 2006, Prentice Hall, Inc.

Electrochemistry

Electrochemical Reactions

In electrochemical reactions, electrons are transferred from one species to another.

Metals tend to lose electrons and are oxidized, non metals tend to gain electrons and are reduced.

Electrochemistry

LEOLEO GERGER

LLosingosing EElectrons islectrons is OOxidation.xidation.

GGainingainingEElectrons islectrons isRReductioneduction

Electrochemistry

OIL RIGOIL RIG

• OOxidationxidation• IIss• LLoss.oss.

• RReduction eduction • IIss GGain.ain.

Electrochemistry

REDOX REACTIONSREDOX REACTIONS

• Are reduction – Are reduction – oxidation reactions.oxidation reactions.

• Electrons are Electrons are transferred.transferred.

• When an atom is When an atom is losing electrons its losing electrons its O.N. increases. It is O.N. increases. It is being oxidized.being oxidized.

• When an atom gains When an atom gains electrons its O.N. electrons its O.N. decreases. It is being decreases. It is being reducedreduced

Electrochemistry

Oxidation Numbers

In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.

Electrochemistry

Assigning Oxidation Numbers1. Elements in their elemental form have an ON= 0.2. The oxidation number of a monatomic ion is its

charge.3. Nonmetals tend to have negative oxidation

numbers, although some are positive in certain compounds or ions.

Oxygen has an oxidation number of −2, except in the peroxide ion in which it has an oxidation number of −1.

Hydrogen is +1 except in metal hydrides when is −1 .

Fluorine always has an oxidation number of −1.

Electrochemistry

Assigning Oxidation Numbers The other halogens have an oxidation

number of −1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions

4. The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.

5. The sum of the oxidation numbers in a neutral compound is 0.

Electrochemistry

Oxidation and Reduction

• A species is oxidized when it loses electrons.Here, zinc loses two electrons to go from neutral

zinc metal to the Zn2+ ion.

Electrochemistry

Oxidation and Reduction

• A species is reduced when it gains electrons.Here, each of the H+ gains an electron and they

combine to form H2.

Electrochemistry

Oxidation and Reduction

• What is reduced is the oxidizing agent.H+ oxidizes Zn by taking electrons from it.

• What is oxidized is the reducing agent.Zn reduces H+ by giving it electrons.

Electrochemistry

• 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.

• 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.

Oxidation-Reduction Oxidation-Reduction ReactionsReactions

Electrochemistry

• 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.

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

oxidation and reduction reactions.

Balancing Oxidation-Balancing Oxidation-Reduction ReactionsReduction Reactions

Electrochemistry

Half Reactions• The half-reactions for

Sn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe2+(aq)

are

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

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

Loss of Gain of

Electrons is Electrons is

Oxidation Reduction

Electrochemistry

Balancing Equations by the Method of Half Reactions

• 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.

Electrochemistry

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. If it is in basic solution, remove H+ by adding OH-

e. Finish by balancing charge by adding electrons.3. Multiply each half reaction to make the number of electrons

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

Electrochemistry

Half-Reaction Method

Consider the reaction between MnO4− and C2O4

2− :

MnO4−(aq) + C2O4

2−(aq) Mn2+(aq) + CO2(aq)

Electrochemistry

Balancing Equations by the Method of Half Reactions

• 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.

Electrochemistry

Half-Reaction Method

First, we assign oxidation numbers.

MnO4− + C2O4

2- Mn2+ + CO2

+7 +3 +4+2

Since the manganese goes from +7 to +2, it is reduced.

Since the carbon goes from +3 to +4, it is oxidized.

Electrochemistry

For KMnO4 + Na2C2O4:

• The two incomplete half reactions are

MnO4-(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

Electrochemistry

• 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 10 electrons for both. Multiplying gives:

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

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

Electrochemistry

4. Adding gives:

16H+(aq) + 2MnO4-(aq) + 5C2O4

2-(aq) 2Mn2+(aq) + 8H2O(l) + 10CO2(g)

5. Which is balanced!

Electrochemistry

Balancing in Basic Solution

• If a reaction occurs in basic solution, one can balance it as if it occurred in acid.

• Once the equation is balanced, add OH− to each side to “neutralize” the H+ in the equation and create water in its place.

• If this produces water on both sides, you might have to subtract water from each side.

Electrochemistry

Examples – Balance the following oxidation-reduction reactions:

1. Cr (s) + NO3- (aq) Cr3+ (aq) + NO (g) (acidic)

2. Al (s) + MnO4- (aq) Al3+ (aq) + Mn2+ (aq) (acidic)

3. PO33- (aq) + Mn4

- (aq) PO43- (aq) + MnO2 (s)

(basic)

4. H2CO (aq) + Ag(NH3)2+ (aq) HCO3

- (aq) + Ag (s) + NH3 (aq) (basic)

Electrochemistry

Section 20-3 20-4

• Voltaic Cells – Spontaneous reactions

• ELECTRODES – POLARITIES

• SALT BRIDGE

• DRIVING FORCE- EMF

• STANDARD REDUCTION POTENTIALS

• HW Q 23, 25, 31, 33 (a, b ) , 35

Electrochemistry

Voltaic Cells

In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released.

Electrochemistry

Voltaic Cells

• We can use that energy to do work if we make the electrons flow through an external device.

• We call such a setup a voltaic cell.

Electrochemistry

Voltaic Cells

• A typical cell looks like this.

• The oxidation occurs at the anode.

• The reduction occurs at the cathode.

Electrochemistry

Voltaic Cells

Once even one electron flows from the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons would stop.

Electrochemistry

Voltaic Cells

• Therefore, we use a salt bridge, usually a U-shaped tube that contains a salt solution, to keep the charges balanced.Cations move toward

the cathode.Anions move toward

the anode.

Electrochemistry

Voltaic Cells• In the cell, then,

electrons leave the anode and flow through the wire to the cathode.

• As the electrons leave the anode, the cations formed dissolve into the solution in the anode compartment.

Electrochemistry

Voltaic Cells• As the electrons

reach the cathode, cations in the cathode are attracted to the now negative cathode.

• The electrons are taken by the cation, and the neutral metal is deposited on the cathode.

Electrochemistry

Electromotive Force (emf)• Water only

spontaneously flows one way in a

waterfall.• Likewise, electrons

only spontaneously flow one way in a

redox reaction—from higher to lower

potential energy.

Electrochemistry

Electromotive Force (emf)

• The potential difference between the anode and cathode in a cell is called the electromotive force (emf).

• It is also called the cell potential, and is designated Ecell.

Electrochemistry

Cell Potential

Cell potential is measured in volts (V).

One volt is the potential difference required to impart 1J of energy to a charge of 1 coulomb. (1 electron has a charge of 1.6 x 10-19 Coulombs). The potential difference between the 2 electrode provides the driving force that pushes the electron through the external circuit.

Electrochemistry

Cell Potential or Electromotive Force (emf)

• The “pull” or driving force on the electrons.

• ELECTROMOTIVE FORCE EMF

• CAUSES THE ELECTRON MOTION!

1 V = 1 JC

Electrochemistry

Standard Reduction Potentials

Reduction potentials for

many electrodes have been

measured and tabulated.

Electrochemistry

Standard Hydrogen Electrode

• Their values are referenced to a standard hydrogen electrode (SHE).

• By definition, the reduction potential for hydrogen is 0 V:

2 H+ (aq, 1M) + 2 e− H2 (g, 1 atm)

Electrochemistry

Standard Cell Potentials

The cell potential at standard conditions can be found through this equation:

Ecell = Ered (cathode) − Ered (anode)

Because cell potential is based on the potential energy per unit of charge, it is an intensive property.

Electrochemistry

For the reaction to be SPONTANEOUS

E has to be positive

Electrochemistry

Cell Potentials

• For the oxidation in this cell,

• For the reduction,

Ered = −0.76 V

Ered = +0.34 V

Electrochemistry

Cell Potentials

Ecell = Ered (cathode) − Ered (anode)

= +0.34 V − (−0.76 V)

= +1.10 V

Electrochemistry

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

• 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.

Electrochemistry

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

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

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

• Recall anodecathode oxredcell EEE

Electrochemistry

PNEUMONICS

• LEO GER

• OIL RIG

• RED CAT

• AND ALWAYS THE SOURCE OF ELECTRONS IS THE NEGATIVE ELECTRODE!!!

Electrochemistry

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

Or

• More generally, for any electrochemical process

• A positive E indicates a spontaneous process (galvanic cell).

• A negative E indicates a nonspontaneous process.

Spontaneity of Redox Spontaneity of Redox ReactionsReactions

anodecathode oxredcell EEE

processoxidation processreduction oxredcell EEE

anodecathode redredcell EEE

Electrochemistry

Cell Potential Calculations

• To Calculate cell potential using Standard Reduction Potentials:

• 1. One reaction and its cell potential’s sign must be reversed--it must be chosen such that the overall cell potential is positive.

• 2. The half-reactions must often be multiplied by an integer to balance electrons--this is not done for the cell potentials.

Electrochemistry

Cell Potential Calculations Continued

• Fe3+(aq) + Cu(s) ----> Cu2+

(aq) + Fe2+(aq)

• Fe3+(aq) + e- ----> Fe2+

(aq) Eo = 0.77 V

• Cu2+(aq) + 2 e- ----> Cu(s) Eo = 0.34 V

• Reaction # 2 must be reversed.

Electrochemistry

Cell Potential Calculations Continued

• 2 (Fe3+(aq) + e- ----> Fe2+

(aq)) Eo = 0.77 V

• Cu(s) ----> Cu2+(aq) + 2 e- Eo = - 0.34 V

• 2Fe3+(aq) + Cu(s) ----> Cu2+

(aq) + 2Fe2+(aq)

• Eo = 0.43 V

Electrochemistry

Oxidizing and Reducing Agents

The greater the difference between the two, the greater the voltage of the cell.

REMEMBER TO REVERSE THE SIGN OF THE SPECIE THAT GETS OXIDIZED (THE ONE BELOW!!!!!)

Electrochemistry

• Section 4.5 – 4.6

• Free energy and redox rx

• Cell EMF under nonstandard conditions

• Nerst Equation- Concentration cells

• Redox Applications - Batteries and fuel cells – Corrosion prevention

• HW 47, 51 a, 59, 63

Electrochemistry

Oxidizing and Reducing Agents• The strongest

oxidizers have the most positive reduction potentials.

• O.A. PULL electrons• The strongest

reducers have the most negative reduction potentials.

• R.A. PUSH their electrons

Electrochemistry

Examples – Sketch the cells containing the following reactions. Include E°cell, the direction of electron flow, direction of ion migration through salt bridge, and identify the anode and cathode:

1. Cr3+ + Cl2 (g) Cr2O72- + Cl-

2. Cu2+ + Mg (s) Mg2+ + Cu (s)

3. IO3- + Fe2+ Fe3+ + I2

4. Zn (s) + Ag+ Zn2+ + Ag

Electrochemistry

Free Energy

G for a redox reaction can be found by using the equation

G = −nFE

where n is the number of moles of electrons transferred, and F is a constant, the Faraday.

1 F = 96,485 C/mol = 96,485 J/V-mol

Electrochemistry

Free Energy and Cell Potential

Under standard conditions G = nFE

• n = number of moles of electrons• F = Faraday = 96,485 coulombs per

mole of electrons = 96,485 J/Vmol• E = V

Electrochemistry

Find the free energy for the reaction of

Zn + Cu2+ ---> Zn2+ + Cu

Electrochemistry

The Nernst Equation• 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 equation relates emf to concentration using

and noting that

Effect of Concentration Effect of Concentration on Cell EMFon Cell EMF

QRTGG ln

QRTnFEnFE ln

Electrochemistry

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

QnFRT

EE ln

Qn

EE log0592.0

Electrochemistry

Cell EMF and Chemical Equilibrium• A system is at equilibrium when G = 0.• From the Nernst equation, at equilibrium and 298 K (E =

0 V and Q = Keq):

0592.0log

log0592.0

0

nEK

Kn

E

eq

eq

Electrochemistry

Examples

1. Calculate E°cell for the following reaction:

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

2. Calculate Ecell if [Zn2+] = 1.88 M and [Cu2+] = 0.020 M

3. Calculate E°cell for the following reaction:

IO3- (aq) + Fe2+ (aq) Fe3+ (aq) + I2 (aq)

4. Calculate Ecell if [IO3-] = 0.20 M, [Fe2+] = 0.65

M, [Fe3+] = 1.0 M, and [I2] = 0.75 M

Electrochemistry

Concentration Cells

• Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes.

• For such a cell, would be 0, but Q would not.Ecell

• Therefore, as long as the concentrations are different, E will not be 0.

Electrochemistry

Concentration 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.

Electrochemistry

Concentration Cells

Electrochemistry

PROBLEM

• Use the Nernst equation to determine the (emf) at 25oC of the cell. Zn(s)│Zn2+(0.00100M ║Cu2+(10.0M)│Cu(s).

• The standard emf of this cell is 1.10 V.

Electrochemistry

Applications of Oxidation-Reduction

Reactions

Electrochemistry

Batteries

•A battery is a galvanic cell or, more commonly, a group of galvanic cells connected in series.

Electrochemistry

Batteries

Electrochemistry

17_370

H2SO4

electrolytesolution

Anode (leadgrid filled withspongy lead) Cathode (lead

grid filled withspongy PbO2)

A lead storage battery consists of a lead anode, lead dioxide cathode, and an electrolyte of 38% sulfuric acid.

Electrochemistry

Lead Storage Battery

• Anode reaction:

Pb(s) +H2SO4(aq) ---> PbSO4(aq) + 2H+(aq) + 2e-

• Cathode reaction:

PbO2(s)+H2SO4(aq)+ 2e-+2H+(aq)->PbSO4(aq)+2H2O(l)

• Overall reaction:

Pb(s)+ PbO2(s) + 2H2SO4(aq)-->PbSO4(aq)+ 2H2O (l)

Electrochemistry

17_371

Cathode(graphite rod)

Anode(zinc inner case)

Paste of MnO2,NH4CL, andcarbon

Common dry cell and its components.

Electrochemistry

Alkaline 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.

Electrochemistry

• Anode: Zn powder mixed in a gel:

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

• Cathode: reduction of MnO2.

Electrochemistry

Electrochemistry

Fuel 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:

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

• Anode:

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

Electrochemistry

17_372

Cathode (steel)

Anode (zinc container)

Solution of HgO (oxidizingagent) in a basic medium (KOHand Zn(OH)2)

Insulation

Mercury battery used in calculators.

Electrochemistry

Fuel Cells

• . . . galvanic cells for which the reactants are continuously supplied.

• 2H2(g) + O2(g) 2H2O(l)

• anode: 2H2 + 4OH 4H2O + 4e

• cathode: 4e + O2 + 2H2O 4OH

Electrochemistry

Hydrogen Fuel Cells

Electrochemistry

17_369

Reference solution ofdilute hydrochloric acid

Silver wire coated withsilver chloride

Thin-walled membrane

Ion selective electrodes are glass electrodes that measures a change in potential when [H+] varies. Used to measure pH.

Electrochemistry

Corrosion 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).

CorrosionCorrosion

Electrochemistry

• Oxidation occurs at the site with the greatest concentration of O2.

Preventing 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.

Electrochemistry

Electrochemistry

• 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.

Electrochemistry

Electrochemistry

Preventing 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

Electrochemistry

Electrochemistry

Corrosion•Some metals, such as copper, gold, silver and platinum, are relatively difficult to oxidize. These are often called noble metals.

•About 1/5 of all iron and steel produced each year is used to replace rusted metal.

Electrochemistry

Self-protecting Metals

• Some metals such as aluminum, copper, and silver form a protective coating that keeps them from corroding further.

• The protective coating for iron and steel flakes away opening new layers of metal to corrosion.

Electrochemistry

Prevention of Corrosion• Coating--painting or applying oil to

keep out oxygen and moisture.• Galvanizing--dipping a metal in a more

active metal -- galvanized steel bucket.• Alloying -- mixing metals with iron to

prevent corrosion -- stainless steel.• Cathodic protection -- attaching a more

active metal. Serves as sacrificial metal--used to protect ships, gas lines, and gas tanks.

Electrochemistry

Redox Titrations

• Same as any other titration.• the permanganate ion is used often because

it is its own indicator. MnO4- is purple, Mn+2 is

colorless. When reaction solution remains clear, MnO4

- is gone.

• Chromate ion is also useful, but color change, orangish yellow to green, is harder to detect.

Electrochemistry

Example• The iron content of iron ore can be

determined by titration with standard KMnO4 solution. The iron ore is dissolved in excess HCl, and the iron reduced to Fe+2 ions. This solution is then titrated with KMnO4 solution, producing Fe+3 and Mn+2 ions in acidic solution. If it requires 41.95 mL of 0.205 M KMnO4 to titrate a solution made with 10.613 g of iron ore, what percent of the ore was iron?