electrochemistry terminology oxidation oxidation – a process in which an element attains a more...

Electrochemistry Terminology Electrochemistry Terminology OxidationOxidation – A process in which an

element attains a more positive oxidation state

Na(s) Na+ + e-

ReductionReduction – A process in which an element attains a more negative oxidation state

Cl2 + 2e- 2Cl- Oxidizing agent Oxidizing agent -The substance that is

reduced is the oxidizing agent Reducing agent Reducing agent - The substance that is

oxidized is the reducing agentLEO says GER

Electrochemistry TerminologyElectrochemistry Terminology

Anode -Anode -The electrode where oxidation occurs

Cathode - Cathode - The electrode where reduction occurs

RedReduction uction at the at the

CatCathodehodeLeo is aLeo is a

Balancing EquationsBalancing Equations

Balancing EquationsBalancing Equations

Balancing EquationsBalancing EquationsStep 4:Step 4: Multiply each half-reaction by a factor Multiply each half-reaction by a factor

so that the reducing agent supplies as many so that the reducing agent supplies as many electrons as the oxidizing agent requires.electrons as the oxidizing agent requires.

Reducing agentReducing agent Cu ---> Cu Cu ---> Cu2+2+ + 2e- + 2e-Oxidizing agentOxidizing agent 22 Ag Ag++ + + 22 e- ---> e- ---> 22 Ag AgStep 5:Step 5: Add half-reactions to give the overall Add half-reactions to give the overall

equation.equation.Cu + 2 AgCu + 2 Ag++ ---> Cu ---> Cu2+2+ + 2Ag + 2AgThe equation is now balanced for both charge The equation is now balanced for both charge

and mass.and mass.

Balancing EquationsBalancing EquationsBalance the following in acid solution—Balance the following in acid solution— VOVO22

++ + Zn ---> VO + Zn ---> VO2+ 2+ + Zn+ Zn2+2+

Step 1:Step 1: Write the half-reactionsWrite the half-reactions

OxOx Zn ---> ZnZn ---> Zn2+2+

RedRed VOVO22++ ---> VO ---> VO2+2+

Step 2:Step 2: Balance each half-reaction for mass.Balance each half-reaction for mass.

OxOx Zn ---> ZnZn ---> Zn2+2+

RedRedVOVO22

++ ---> VO ---> VO2+2+ + + HH22OO2 H2 H++ ++

Add HAdd H22O on O-deficient side and add HO on O-deficient side and add H++ on on other side for H-balance.other side for H-balance.

Balancing EquationsBalancing EquationsStep 3:Step 3: Balance half-reactions for Balance half-reactions for

charge.charge.OxOx Zn ---> ZnZn ---> Zn2+2+ + + 2e- 2e-RedRed e- e- + 2 H+ 2 H++ + VO + VO22

++ --> VO --> VO2+2+ + H + H22OOStep 4:Step 4: Multiply by an appropriate factor.Multiply by an appropriate factor.OxOx Zn ---> ZnZn ---> Zn2+2+ + + 2e-2e-RedRed 22e-e- + + 44 H H++ + + 22 VO VO22

++ ---> ---> 22 VO VO2+2+ + + 22 H H22OO

Step 5:Step 5: Add Add balancedbalanced half-reactions half-reactionsZn + 4 HZn + 4 H++ + 2 VO + 2 VO22

++ ---> Zn ---> Zn2+2+ + 2 VO + 2 VO2+2+ + 2 H + 2 H22OO

Balancing Equations for Redox Balancing Equations for Redox ReactionsReactions

A great example of a thermodynamically spontaneous reaction is the thermite reaction. Here, iron oxide (Fe2O3 = rust) and aluminum metal powder undergo a redox (reduction-oxidation) reaction to form iron metal and aluminum oxide (Al2O3 = alumina):

Fe2O3(s) + Al(s) ↔ Al2O3(s) + Fe(l)Al = 0Al = 0Fe = +3Fe = +3 Fe = 0Fe = 0Al = +3Al = +3

Tips on Balancing Tips on Balancing EquationsEquations

How many electrons are transferred in the How many electrons are transferred in the following reaction?following reaction?

2ClO2ClO33–– + 12H + 12H++ + 10I + 10I–– → 5I → 5I22 + Cl + Cl22 + 6H + 6H22OO

12 5 2 30 10

0% 0% 0%0%0%

1) 122) 53) 24) 305) 10

0of5

Which of the following reactions is possible Which of the following reactions is possible at the anode of a galvanic cell?at the anode of a galvanic cell?

two of t

hese

0% 0% 0%0%0%

1. Zn → Zn2+ + 2e–

2. Zn2+ + 2e– → Zn3. Zn2+ + Cu → Zn + Cu2+ 4. Zn + Cu2+ → Zn2+ + Cu5. two of these

0of5

10SecondsRemainin

g

Which of the following species cannot Which of the following species cannot function as an oxidizing agent?function as an oxidizing agent?

S(s)

NO3–(ag)

Cr2O72–(aq)

I– (a

q)

MnO4– (a

q)

0% 0% 0%0%0%

1. S(s)2. NO3

–(ag)

3. Cr2O72–(aq)

4. I– (aq)5. MnO4

– (aq)0of5 15

Electrochemical CellsElectrochemical CellsElectrochemical CellsElectrochemical Cells• An apparatus that allows a An apparatus that allows a

redox reaction to occur by redox reaction to occur by transferring electrons through transferring electrons through an external connector.an external connector.

• Product favored reaction ---> Product favored reaction ---> voltaic or galvanic cellvoltaic or galvanic cell ----> ----> electric currentelectric current

• Reactant favored reaction ---> Reactant favored reaction ---> electrolytic cellelectrolytic cell ---> electric ---> electric current used to cause chemical current used to cause chemical change.change.

Batteries are voltaic cellsBatteries are voltaic cells

AnodeAnode CathodeCathode

Basic Concepts Basic Concepts of Electrochemical Cellsof Electrochemical Cells

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons

Terms Used for Voltaic CellsTerms Used for Voltaic Cells

CELL POTENTIAL, ECELL POTENTIAL, EZn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons

Calculating Cell VoltageCalculating Cell Voltage

Zn(s) ---> ZnZn(s) ---> Zn2+2+(aq) + 2e-(aq) + 2e-CuCu2+2+(aq) + 2e- ---> Cu(s)(aq) + 2e- ---> Cu(s)----------------------------------------------------------------------------------------CuCu2+2+(aq) + Zn(s) ---> Zn(aq) + Zn(s) ---> Zn2+2+(aq) + Cu(s)(aq) + Cu(s)

Zn(s) ---> ZnZn(s) ---> Zn2+2+(aq) + 2e-(aq) + 2e-CuCu2+2+(aq) + 2e- ---> Cu(s)(aq) + 2e- ---> Cu(s)----------------------------------------------------------------------------------------CuCu2+2+(aq) + Zn(s) ---> Zn(aq) + Zn(s) ---> Zn2+2+(aq) + Cu(s)(aq) + Cu(s)

Measuring Measuring Standard Standard Electrode Electrode PotentialPotential

Potentials are measured against a Potentials are measured against a hydrogen ion reduction reaction, which hydrogen ion reduction reaction, which is arbitrarily assigned a potential of is arbitrarily assigned a potential of zero zero voltsvolts..

Table of Table of Reduction Reduction PotentialsPotentials

Measured Measured against against

the the StandardStandardHydrogenHydrogenElectrodeElectrode

TABLE OF STANDARD TABLE OF STANDARD REDUCTION POTENTIALSREDUCTION POTENTIALS

TABLE OF STANDARD TABLE OF STANDARD REDUCTION POTENTIALSREDUCTION POTENTIALS

2

Eo (V)

Cu2+ + 2e- Cu +0.34

2 H+ + 2e- H 0.00

Zn 2+ + 2e- Zn -0.76

oxidizingability of ion

reducing abilityof element

To determine an oxidation from a To determine an oxidation from a reduction table, just take the opposite reduction table, just take the opposite sign of the reduction!sign of the reduction!

Zn/Cu Electrochemical CellZn/Cu Electrochemical Cell

Zn(s) ---> ZnZn(s) ---> Zn2+2+(aq) + 2e-(aq) + 2e- EEoo = +0.76 V = +0.76 VCuCu2+2+(aq) + 2e- ---> Cu(s)(aq) + 2e- ---> Cu(s) EEoo = +0.34 V = +0.34 V------------------------------------------------------------------------------------------------------------------------------CuCu2+2+(aq) + Zn(s) ---> Zn(aq) + Zn(s) ---> Zn2+2+(aq) + Cu(s) (aq) + Cu(s) EEoo = +1.10 V = +1.10 V

Cathode, Cathode, positive, positive, sink for sink for electronselectrons

Anode, Anode, negative, negative, source of source of electronselectrons

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons ++

Volts

Cd Salt Bridge

Cd2+

Fe

Fe2+

Volts

Cd Salt Bridge

Cd2+

Fe

Fe2+

EEoo for a Voltaic Cell for a Voltaic Cell

All ingredients are present. Which way does reaction proceed?All ingredients are present. Which way does reaction proceed?

Volts

Cd Salt Bridge

Cd2+

Fe

Fe2+

Volts

Cd Salt Bridge

Cd2+

Fe

Fe2+

EEoo for a Voltaic Cell for a Voltaic Cell

Day 2 (it’s electric!)

More About More About Calculating Cell VoltageCalculating Cell VoltageAssume IAssume I-- ion can reduce water. ion can reduce water.

2 H2O + 2e- ---> H2 + 2 OH-

2 I- ---> I2 + 2e--------------------------------------------------2 I- + 2 H2O --> I2 + 2 OH- + H2

2 H2O + 2e- ---> H2 + 2 OH-

2 I- ---> I2 + 2e--------------------------------------------------2 I- + 2 H2O --> I2 + 2 OH- + H2

Cathode

Anode

Galvanic (Electrochemical) Galvanic (Electrochemical) CellsCells

Spontaneous redox processes

have: A positive cell potential, E0

A negative free energy change, (-G)

Zn - Cu Zn - Cu Galvanic Galvanic

CellCell

Zn2+ + 2e- Zn E = -0.76V

Cu2+ + 2e- Cu E = +0.34V

From a From a table of table of reduction reduction potentials:potentials:

Zn - Cu Zn - Cu Galvanic Galvanic

CellCell

Cu2+ + 2e- Cu E =

+0.34V

The less positive, The less positive, or more negative or more negative reduction reduction potential potential becomes the becomes the oxidation…oxidation…

Zn Zn2+ + 2e- E = +0.76V

Zn + Cu2+ Zn2+ + Cu E0 = + 1.10 V

Line Line NotationNotation

Zn(Zn(ss) | Zn) | Zn2+2+((aqaq) || Cu) || Cu2+2+((aqaq) | Cu() | Cu(ss))

An abbreviated An abbreviated representation representation of an of an electrochemical electrochemical cellcell

Anode Anode solutionsolution

Anode Anode materialmaterial

Cathode Cathode solutionsolution

Cathode Cathode materialmaterial|| ||||||

Line notation is cool, just like AC

ZnZn((ss) ) ||ZnZn2+2+((aqaq) ) (1.0M) (1.0M) ||||HH++

((aqaq) ) (1.0M)(1.0M)||HH2(g)2(g) (1.00 atm) (1.00 atm) || PtPt((ss))

Calculating Calculating GG00 for a Cell for a Cell

).)()((Coulomb

Joules

emol

coulombsemolG 1014859620

GG00 = -nFE = -nFE00

nn = moles of electrons in balanced redox equation = moles of electrons in balanced redox equation

FF = Faraday constant = 96,485 coulombs/mol e = Faraday constant = 96,485 coulombs/mol e--

Zn + Cu2+ Zn2+ + Cu E0 = + 1.10 V

kJJoulesG 2122122670

The Nernst EquationThe Nernst Equation

)ln(0 QnF

RTEE

Standard potentials assume a concentration of Standard potentials assume a concentration of 1.0 M. The Nernst equation allows us to 1.0 M. The Nernst equation allows us to calculate potential when the two cells are not calculate potential when the two cells are not 1.0 M.1.0 M.

RR = 8.31 J/(mol = 8.31 J/(molK) K)

TT = Temperature in K = Temperature in K

nn = moles of electrons in balanced redox equation = moles of electrons in balanced redox equation

FF = Faraday constant = 96,485 coulombs/mol e = Faraday constant = 96,485 coulombs/mol e--

Nernst Equation SimplifiedNernst Equation Simplified

)log(0591.00 Qn

EE

At 25 At 25 C (298 K) the Nernst Equation C (298 K) the Nernst Equation is simplified this way:is simplified this way:

Equilibrium Constants and Cell Equilibrium Constants and Cell PotentialPotential

At equilibrium, forward and reverse reactions occur at equal rates, therefore:1.1. The battery is “dead”

2. The cell potential, E, is zero volts

)log(0591.0

0 0 Kn

Evolts

Modifying the Nernst Equation (at 25 C):

Zn + Cu2+ Zn2+ + Cu E0 = + 1.10 V

Calculating an Equilibrium Calculating an Equilibrium Constant from a Cell PotentialConstant from a Cell Potential

)log(2

0591.010.10 Kvolts

)log(0591.0

)2)(10.1(K

)log(2.37 K

372.37 1058.110 xK

ConcentratioConcentration Celln Cell

Step 1: Determine which side undergoes oxidation, and which side undergoes reduction.

Both sides have the same

components but at different

concentrations.

??????

ConcentratioConcentration Celln Cell

Both sides have the same

components but at different

concentrations.

The 1.0 M Zn2+ must decrease in concentration, and the 0.10 M Zn2+ must increase in concentrationZn2+ (1.0M) + 2e- Zn (reduction) Zn Zn2+ (0.10M) + 2e-

(oxidation)

??????

CathodeCathodeAnodeAnode

Zn2+ (1.0M) Zn2+

(0.10M)

Concentration CellConcentration Cell

)log(0591.00 Qn

EE

Step 2: Calculate cell potential using the Nernst Equation (assuming 25 C).

Both sides have the same

components but at different

concentrations.

??????

CathodeCathodeAnodeAnode

Zn2+ (1.0M) Zn2+

(0.10M)

ConcentratioConcentration Celln Cell

VoltsE 0.00 )0.1(

)10.0(Q2n

VoltsE 030.0)0.1

10.0log(

2

0591.00.0

Nernst CalculationsNernst CalculationsZn2+ (1.0M) Zn2+

(0.10M) )log(

0591.00 Qn

EE

Day 3 (dahditdadahditahh…Charge!)

Charging a BatteryCharging a BatteryWhen you charge a battery, you are forcing the electrons When you charge a battery, you are forcing the electrons backwards (from the + to the -). To do this, you will backwards (from the + to the -). To do this, you will need a higher voltage backwards than forwards. This is need a higher voltage backwards than forwards. This is why the ammeter in your car often goes slightly higher why the ammeter in your car often goes slightly higher while your battery is charging, and then returns to while your battery is charging, and then returns to normal.normal.

In your car, the battery charger is called an In your car, the battery charger is called an alternator. If you have a dead battery, it could be alternator. If you have a dead battery, it could be the battery needs to be replaced OR the the battery needs to be replaced OR the alternator is not charging the battery properly.alternator is not charging the battery properly.

Dry Cell BatteryDry Cell Battery

Anode (-)Anode (-)

Zn ---> ZnZn ---> Zn2+2+ + 2e- + 2e-

Cathode (+)Cathode (+)

2 NH2 NH44++ + 2e- ---> 2 NH + 2e- ---> 2 NH33 + H + H22

Alkaline BatteryAlkaline Battery

Nearly same Nearly same reactions as in reactions as in common dry cell, common dry cell, but under basic but under basic conditions.conditions.

Anode (-): Anode (-): Zn + 2 OHZn + 2 OH-- ---> ZnO + H ---> ZnO + H22O + 2e-O + 2e-

Cathode (+): Cathode (+): 2 MnO2 MnO22 + H + H22O + 2e- ---> MnO + 2e- ---> Mn22OO33 + 2 OH + 2 OH--

Mercury BatteryMercury Battery

Lead Storage BatteryLead Storage Battery

Ni-Cad BatteryNi-Cad BatteryAnode (-)Anode (-)

Cd + 2 OH- ---> Cd(OH)2 + 2e-

Cathode (+) Cathode (+)

NiO(OH) + H2O + e- ---> Ni(OH)2 + OH-

The positive electrode is made of Lithium cobalt oxide, or LiCoO2

The negative electrode is made of carbon.

HH22 as a Fuel as a Fuel

Cars can use electricity generated by H2/O2 fuel cells.H2 carried in tanks or generated from hydrocarbons

The Electrochemical Corrosion of Iron

Preventing Corrosion

• Coating to keep out air and water.– Galvanizing - Putting on a zinc coat

• Has a lower reduction potential, so it is more. easily oxidized.

• Alloying with metals that form oxide coats.• Cathodic Protection - Attaching large pieces of

an active metal like magnesium that get oxidized instead.

Cathodic Protection of an Underground Pipe

ElectrolytElectrolytic ic

ProcesseProcessess

A negative cell potential, (-E0)

A positive free energy change, (+G)

Electrolytic processes are NOT spontaneous. They have:

Electrolysis of Electrolysis of WaterWater

eHOOH 442 22 OHHeOH 4244 22

In acidic solutionIn acidic solution

Anode rxn:Anode rxn:

Cathode rxn:Cathode rxn:-1.23 V-1.23 V

-0.83 V-0.83 V

-2.06 V-2.06 V222 22 OHOH

Electroplating Electroplating of Silverof Silver

Anode reactionAnode reaction::

Ag Ag Ag Ag++ + e + e- -

Electroplating requirementsElectroplating requirements::1. Solution of the plating metal1. Solution of the plating metal

3. Cathode with the object to be plated3. Cathode with the object to be plated2. Anode made of the plating metal2. Anode made of the plating metal

4. Source of current4. Source of current

Cathode reactionCathode reaction::

AgAg++ + e + e-- Ag Ag

Step 1 – convert current and time to quantity of charge in coulombs

a. Amps x time = total charge transferred in coulombs

(Coulomb/sec) x sec = coulombsStep 2 – convert quantity of charge in coulombs to moles of electrons

coulombs /(96,485 coulombs/mol e-) = mol e-Step 3 – Convert moles of electrons to moles of substance

mol e- x (mole substance/mol e-) = mol substanceStep 4 – Convert moles of substance to grams of substance

mol substance x formula mass of substance = mass of substance

Calculating plating

1 mol Pb

Suppose that in starting a car on a cold morning a current of 125 amperes is drawn for 15.0 seconds from a cell of the type described above. How many grams of Pb would be consumed? (The atomic weight of Pb is 207.19.)

125 C 15 sec 1 mol e- 1 mol Pb

96 485 C 2 mol e-

207.19 g

1 sec

PbPb2+2+ + 2e + 2e- - Pb Pb

2.01 g Pb2.01 g Pb

Solving an Electroplating Solving an Electroplating ProblemProblem

Q: How many seconds will it take to Q: How many seconds will it take to plate out 5.0 grams of silver from a plate out 5.0 grams of silver from a solution of AgNOsolution of AgNO33 using a 20.0 Ampere using a 20.0 Ampere current?current?

5.0 g

AgAg++ + e + e- - Ag Ag

1 mol Ag

107.87 g

1 mol e-

1 mol Ag

96 485 C

1 mol e-

1 s

20.0 C

= 2.2 x 10= 2.2 x 1022 s s