Slide 1

Introductory ChemistryAS.030.102 Section 2 Spring 2015

Instructor: Dr. Thyagarajan 1First Class:January 26 2015

Class meets MWF 10:00 a.m. to 10:50 a.m. in Remsen 1

Office hours: M,F: 2:00 3:30 p.m. or by appointmentIn-class exams: February 23, March 30 and April 27Final Exam: May 11th - 9:00 a.m. to noon 2Calendar2Required textOxtoby, Gillis, and Campion, Principles of Modern Chemistry (7th edition)OWL (on-line web-based learning system)

OptionalStudent Solutions Manual3Syllabus34Three Units:Electrochemistry and KineticsChapters 17, 18 Quantum MechanicsChapters 4 6Organic, Inorganic, SpectroscopyChapters 7, 8, 20

Homework - Due wednesdays using OWL

Syllabus5Help Sessions:TA help sessions Learning Den Problem Solving PILOTGrading and exams: No scores will be dropped. You will be able to bring ONE sheet (one side filled) with equations/formulas/constants etc. Equations will NOT be provided.You WILL be given a periodic table.SyllabusTo Study for the TestsLook at the course objectives posted on the schedule.Know how to do problemsExamples done in classEnd-of-chapter homework problemsOptional tutorials and practice problems. Understand all vocabulary words / concepts.Try test questions from previous tests.

6Syllabus6Required math skillsAlgebra, including solving quadratic equations and solving two equations with two unknowns. Arithmetic, including exponents and logsMetric System Prefixes nano through kiloScientific NotationSmall amount of calculusBring to class each dayA calculatorYour periodic table7Syllabus78GradesHomework 20 %Mid-term Exams (3) 50 %Final Exam 30 % Clickers (class use) 10 points Ethics PolicyViolations: Homework, TestsPenalties: Class failure through expulsion

Syllabus9BlackboardSyllabus ScheduleDiscussion sectionLecture slides Course help information

Log in to My JHhttps://login.johnshopkins.edu

login: JHED ID Password: your JHU passwordSyllabusHow to Get Access to OWL

10Three ways to purchase: Card packaged with new text from bookstoreCard purchased separately from the bookstoreInstant access purchased online

Access codes are 25 characters long and look like this: kg609-qkj3y-2kd4k-h3kvq-hv5d4

Cards look like this:

How to Get an OWL Access Code

1112Questions?Chapter 17 - Electrochemistry13

Oxidation numbers and balancing equations

Electrochemical cells and cell parts

Energetics

1314

Oxidation Numbers (Review)

+1+3+2+4-3-2-10Oxidation Numbers (Review)Transition Metals

Oxidation Numbers (Review)17Determine the oxidation number of each atom

a) iron(III)chloride

b) nitrogen dioxide

c) sulfuric acid

d) potassium dichromate (K2Cr2O7)18Oxidation Reduction (Redox) ReactionsA reaction where the oxidation number of one or more atoms changes (transfer of electrons).

Not a redox reaction:

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

Redox reaction:2 H2 (g) + O2 (g) 2 H2O(l)00+1 -2+1 -1+1 -2 +1+1 -1+1 -219Redox ReactionsVERBS:OXIDATIONloss of electron(s) by a species; increase in oxidation number.REDUCTIONgain of electron(s); decrease in oxidation number.NOUNS:OXIDIZING AGENTelectron acceptor; species is reduced.REDUCING AGENTelectron donor; species is oxidized.Balance Redox ReactionsHalf-Reaction Method20

Step 1 Determine which atoms change oxidation number-2-2-2+5-2+2+2+2+6Note: All electrochemical reactions in this course are in water.Balance Redox ReactionsHalf-Reaction Method21

Step 2 Write out the half-reactions (not balanced, but with key compounds or ions present)Balance Redox ReactionsHalf-Reaction Method22Step 3 Balance charges by adding electrons to the right side for oxidation and to the left side for reduction. CuS(s) Cu2+(aq) + SO42-(aq) + 8 e-

NO3-(aq) + 3 e- NO(g)

-2+6+ 5+ 2Balance Redox ReactionsHalf-Reaction Method23Step 4 Balance atoms (mass) by adding H2O to balance oxygen and then H+ to balance hydrogen. 4 H2O(l) + CuS(s) Cu2+(aq) + SO42-(aq) + 8 H+(aq) + 8 e-

4 H+(aq) + NO3-(aq) + 3 e- NO(g) + 2 H2O(l)For acidic media (solutions): CuS(s) Cu2+(aq) + SO42-(aq) + 8 e- NO3-(aq) + 3 e- NO(g)Balance Redox ReactionsHalf-Reaction Method24Step 5 Multiply each half-reaction by an integerso the number of electrons in each is the same (so they cancel).4 H2O(l) + CuS(s) Cu2+(aq) + SO42-(aq) + 8 H+(aq) + 8 e-

4 H+(aq) + NO3-(aq) + 3 e- NO(g) + 2 H2O(l)For acidic media (solutions):x 3x 812 H2O(l) + 3CuS(s) 3Cu2+(aq) + 3SO42-(aq) + 24 H+(aq) + 24 e-

32 H+(aq) + 8 NO3-(aq) + 24 e- 8 NO(g) + 16 H2O(l)Balance Redox ReactionsHalf-Reaction Method25Step 6 Cancel and add the two half-reactions.For acidic media (solutions):12 H2O(l) + 3CuS(s) 3Cu2+(aq) + 3SO42-(aq) + 24 H+(aq) + 24 e-

32 H+(aq) + 8 NO3-(aq) + 24 e- 8 NO(g) + 16 H2O(l)483 CuS(s) + 8 NO3-(aq) + 8 H+(aq)

3 Cu2+(aq) + 3 SO42-(aq) + 8 NO(g) + 4 H2O(l)Balance Redox ReactionsHalf-Reaction Method26Solve for acidic media first.Step 7 Add OH- to both sides for every H+. H+ + OH- becomes H2O. Cancel water.For basic media (solutions):3 CuS + 8 NO3- + 8 H+ 3 Cu2+ + 3 SO42- + 8 NO + 4 H2O43CuS + 8NO3- + 8H2O 3Cu2+ + 3SO42- + 8NO + 4H2O + 8OH-3 CuS + 8 NO3- + 4 H2O 3 Cu2+ + 3 SO42- + 8 NO + 8 OH-27

Acid Solution:

VO2+ + Zn > VO2+ + Zn2+Examples for Practice28Electrochemistry

Oxidation numbers and balancing equations

Electrochemical cells and cell parts

Energetics

29Redox ReactionsClassifications:Direct No external circuit

Indirect external circuit

Galvanic / voltaic cell - Produces electrical energy

Electrolytic cell Requires electrical energyOxidizing and reducing agents in direct contact.Cu(s) + 2 Ag+(aq) > Cu2+(aq) + 2 Ag(s)Check Oxidation States andtheir changes

30

Direct Redox ReactionsOxidation: Zn(s) Zn2+(aq) + 2eReduction: Cu2+(aq) + 2e Cu(s) -------------------------------------------------------- Cu2+(aq) + Zn(s) Zn2+(aq) + Cu(s)

Electrons are transferred from Zn to Cu2+, but there is no useful electric current.With time, Cu plates out onto Zn metal strip, and Zn strip disappears. 31Direct Redox ReactionsTo obtain a useful current, we separate the oxidizing and reducing agents.- No direct contact. - Electron transfer occurs thru an external wire.

Indirect Redox ReactionsElectrochemical CellAn apparatus that allows a redox reaction to occur by transferring electrons through an external connector (circuit).

Galvanic / voltaic cell

Chemical change produces electric current

DGRxn < 0

Product favored reaction

Batteries are voltaic / galvanic cells33Electrochemical CellsElectrolytic cell

electric current used to cause chemical change in a non-spontaneous reaction

DGRxn > 0

Reactant favored reaction

Chrome plate car bumpers

Gold / silver plate jewelry, etc.Galvanic Cell PartsLeft-hand solution Convention Right-hand solutionZn(s) Zn2+(aq) + 2 e Cu2+(aq) + 2e Cu(s) Anode: Oxidation occurs at this electrodeCathode: Reduction occursat this electrodeElectrons flow in the wire from anode to cathode.Zn(s) Zn2+(aq) + 2 e Cu2+(aq) + 2e Cu(s) Cu2+(aq)+ Zn(s) Zn2+(aq)+ Cu(s)Anode:Cathode:

35Galvanic Cell Parts36

Galvanic Cell Parts37

Galvanic Cell Parts

Solutions [Cu(NO3)2 and AgNO3] are connected by a salt bridge, containing NaNO3. Porous plugs at ends of bridge prevent solutions from mixing but allows ions to pass through.Left:Cu(s) is oxidized - Cu2+ dissolves - pos. charge builds up - ions move

Right:Ag+ is reduced - Ag precipitates - neg. charge builds up - ions move 38Galvanic Cell PartsAmmeter: measures currentIntro Chemistry II 030.102 - Chapter 17Spring, 201038Anode on the leftCathode on the right.Electrons flow left to right.

Cu|Cu2+||Ag+|AgAnodeCathodeSalt Bridge39Writing Galvanic CellsCation in solutionat the anodeCation in solution at the cathodeIntro Chemistry II 030.102 - Chapter 17Spring, 201039Diagram the following galvanic cell, indicating the direction of flow of electrons in the external circuit and the motion of ions in the salt bridge:

Sn|Sn2+(aq)||Cl(aq)|Hg2Cl2(s)|Hg(l)

Write the half-reactions at each electrode and a balanced equation for the overall reaction in this cell. 40Intro Chemistry II 030.102 - Chapter 17Spring, 20104041Electrolytic Cell Parts

Anode, cathode, electrolyte solutionAll in one container, no salt bridge2NaCl(l) 2Na(l) + Cl2(g) 2 H2O(l) 2 H2(g) + O2(g) 42Electrochemistry

Oxidation numbers and balancing equations

Electrochemical cells and cell parts

Energetics

43EnergeticsMechanicsElectrostatics

Force

Potential Energy

Potential E

44EnergeticsIf the only change is the transfer of electrons, the change in energy of the reaction is the same as the change in energy of the electrons.

Energy = w = Q * V Q is charge V is voltage

Q = 1 CV = 1 volt Energy = 1 JQ = 1 electron (1.602 x 10-19 C) V = 1 volt Energy = 1 eV1 eV = 1.602 x 10-19 J

Electrochemical energetics are listed by voltage for 1 electron, not energy.Cu2+(aq)+ Zn(s) Zn2+(aq)+ Cu(s)45EnergeticsCu2+(aq)+ Zn(s) Zn2+(aq)+ Cu(s)Q is charge V is voltage

i = current (C/s)t = time (s)Q is the number of electrons that went through the circuit

If we measure the voltage (voltmeter), current (ammeter) and timethen we can calculate- number of electrons passed through the electrodes- amount of product produced (mass, moles, concentration)- amount of reactant consumed (mass, moles, concentration)- work performed

Electrons are driven from anode to cathode by an electromotive force or emf (change in potential).For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25 C and when [Zn2+] and [Cu2+] = 1.0 M.Zn and Zn2+,anodeCu and Cu2+,cathode

+1.10 V1.0 M1.0 M46Cu2+ + Zn Cu + Zn2++-voltmeterleadvoltmeterleadCell PotentialFor Zn/Cu cell, potential is +1.10 V at 25 C and when [Zn2+] and [Cu2+] = 1.0 M.

STANDARD CELL POTENTIAL, E o

A quantitative measure of the tendency of reactants to proceed to products when all are in their standard states, 1 atm (gases) and 1 M (solutions).

47Cell PotentialInstead of tabulating the potential for every possible electrochemical cell, half-reaction potentials are tabulated.

Balanced half-reactions can be added together to get overall, balanced equation.Zn(s) Zn2+(aq) + 2e-Cu2+(aq) + 2e- Cu(s)--------------------------------------------Cu2+(aq) + Zn(s) Zn2+(aq) + Cu(s)If we know Eo for each half-reaction, we could get Eo for net reaction.48Cell PotentialCant measure 1/2 reaction Eo directly. Therefore, measure it relative to a STANDARD HYDROGEN ELECTRODE, SHE. Also NHE2 H+ (aq, 1 M) + 2e H2 (g, 1 atm)

Eo = 0.0 V 49Cell PotentialZn/Zn2+ half-cell hooked to a SHE. Eo measured for the cell = + 0.76 V

Negative electrodeSupplier of electronsAcceptor of electronsPositive electrode2 H+ + 2e- H2ReductionCathodeZn Zn2+ + 2e- OxidationAnode50Net Reaction: Zn(s) + 2H+ Zn2+ + H2(g)Reduction of H+ by Zn

Zn Zn2+ (aq) + 2e- is - 0.76 V51Overall reaction is sum of two half-reactions:

Cathode: 2 H+ (aq) + 2 e- H2(g)

Anode:Zn(s) Zn2+(aq) + 2 e-Overall potential is: E o(cell) = E o(cathode) - E o(anode)+ 0.76 V = 0.00 V - E o(anode)Zn2+ (aq) + 2e- Zn is + 0.76 VStandard Reduction PotentialHalf-Reaction PotentialZn2+/Zn and 2H+/H2 Cell

Eo(cell) = + 0.34 VAcceptor of electronsSupplier of electronsCu2+ + 2e- CuReductionCathodeH2 2 H+ + 2e-OxidationAnodePositiveNegative52This Electrochemical Cell runs right-to-left !Cu2+/Cu and 2H+/H2 Cell

53Overall reaction is sum of two half-reactions:

Cathode:Cu2+ (aq) + 2 e- Cu(s)

Anode:H2(g) 2 H+(aq) + 2 e-

Overall potential is:E o(cell) = E o(cathode) - E o(anode)+ 0.34 V = E o(cathode) - 0.00 VCu2+ + 2e- Cu is + 0.34 VStandard Reduction PotentialHalf-Reaction PotentialCu2+/Cu and 2H+/H2 Cell Anode:Zn2+(aq) + 2e- Zn(s) Eo = - 0.76 VCathode:Cu2+(aq) + 2e- Cu(s)Eo = + 0.34 V-------------------------------------------------------------------------Cu2+(aq) + Zn(s) Zn2+(aq) + Cu(s)

Overall potential is:E o(cell) = E o(cathode) - E o(anode)= + 0.34 V - ( -0.76 V) = + 1.10 VCathode, positive, sink for electronsAnode, negative, source of electrons

+54Zn2+/Zn and Cu2+/Cu Cell

55Table of Standard Reduction PotentialsOrganize half-reactions by relative ability to act as oxidizing agentsStandard Reduction Potentials

Textbook (7th ed.) Appendix E

CRC Handbook Pages 8-20 thru 8-31

Potential Ladder for Reduction Half-ReactionsBest oxidizing agentsBest reducing agents56Any substance on the right will reduce any substance higher than it on the left.Zn can reduce H+ and Cu2+.H2 can reduce Cu2+ but not Zn2+Cu cannot reduce H+ or Zn2+.

57Table of Standard Reduction Potentials Cu2+ + 2e- > Cu +0.34+2 H + 2e- > H2 0.00Zn2+ + 2e- > Zn -0.76Northwest-southeast rule: product-favored reactions occur between reducing agent at southeast corner oxidizing agent at northwest cornerAny substance on the right will reduce any substance higher than it on the left.Ox. agentRed. agent58Standard Redox Potentials, Eo Cu2+ + 2e- > Cu +0.34+2 H + 2e- > H2 0.00Zn2+ + 2e- > Zn -0.76Ox. agentRed. agent59Standard Redox Potentials, Eo The REDUCING AGENT gets OXIDIZED at the ANODEZn Zn2+ + 2 e-

The OXIDIZING AGENT gets REDUCED at the CATHODECu2+ + 2 e- CuIn which direction do the following reactions go?Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s)

Fe3+ + e > Fe2+ E o = + 0.770 VSn2+ + 2 e > Sn(s) E o = 0.136 V60Using Standard Redox Potentials2 Fe2+(aq) + Sn2+(aq) 2 Fe3+(aq) + Sn(s)Cu2+ + 2e > Cu E o = + 0.340 VAg+ + e- > Ag E o = + 0.800 VGoes right as writtenGoes left, reverse of direction writtenWhat is E o(cell) for the overall reaction?Use Standard Reduction Potentials to predict:

- Direction of redox reactions (anode and cathode)

- Cell potentials

61Uses of E o Values Fe3+ + e > Fe2+ E o = + 0.770 V

Sn2+ + 2 e > Sn(s) E o = 0.136 V62Using Standard Redox Potentials2 Fe3+(aq) + Sn(s) 2 Fe2+(aq) + Sn2+(aq)What is E o(cell) for the overall reaction?

E o(cell) = E o(cathode) - E o(anode)

= 0.770 (-0.136) = + 0.906 V

ANODECATHODEDo NOT multiply by 2 or E o is an INTENSIVE property.

63Using Standard Redox PotentialsCd2+ + 2e > Cd E o = - 0.40 VFe2+ + 2e- > Fe E o = - 0.44 VWhat is the reducing agent?Which is the anode?Which direction do the electrons flow?What is the cell potentialFeFeFrom the Fe electrode to the Cd electrode. E o(cell) = E o(cathode) - E o(anode) = -0.40 (-0.44) = + 0.04 V

Assume I ion can reduce water. 2 H2O + 2e > H2 + 2 OH Cathode2 I > I2 + 2e Anode-------------------------------------------------2 I + 2 H2O > I2 + 2 OH + H2Negative E o means reaction occurs in opposite direction64Using Standard Redox Potentials2 H2O + 2 e- > H2 + 2 OH- E o = - 0.828 VI2 + 2 e- > 2 I- E o = + 0.535 V E o(cell) = E o(cathode) - E o(anode) = -0.828 (0.535) = - 1.363 VFe(s) | Fe2+(aq) || Al3+(aq) | Al(s)Fe2+ + 2e- > FeOrFe > Fe2+ + 2 e-Al > Al3+ + 3 e-orAl3+ + 3 e- > Al65Fe2+ + 2e- > Fe E o = - 0.44 VAl3+ + 3 e- > Al E o = - 1.66 V Al is the better reductant--It will reduce Fe2+

Fe2+ is the better oxidant--It will oxidize Al Fe(s) | Fe2+(aq) || Al3+(aq) | Al(s)

Fe2+ + 2e- > FeOrFe > Fe2+ + 2 e-Al > Al3+ + 3 e-orAl3+ + 3 e- > AlElectrons> or Fe E o = - 0.44 VAl3+ + 3 e- > Al E o = - 1.66 V Fe(s) | Fe2+(aq) || Al3+(aq) | Al(s)Fe2+ + 2e- > FeAl > Al3+ + 3 e-67Fe2+ + 2e- > Fe E o = - 0.44 VAl3+ + 3 e- > Al E o = - 1.66 V Al is the better reductant--It will reduce Fe2+

Fe2+ is the better oxidant--It will oxidize Al

Electrons flow FROM Al TO Fe2+

Fe2+ + 2e- > FeAl > Al3+ + 3 e-Electrons68Fe(s)1 M Fe(NO3)21 M Al(NO3)3Al(s)E o(cell) = + 1.22 V

CathodePositiveAnodeNegative+ 1.22 VWhich electrode gains mass?

Which electrode loses mass?Fe(s) | Fe2+(aq) || Al3+(aq) | Al(s)

Fe2+ + 2e- > FeAl > Al3+ + 3 e-Electrons69Fe(s)1 M Fe(NO3)21 M Al(NO3)3Al(s)Fe2+ + 2e- > Fe E o = - 0.44 VAl3+ + 3 e- > Al E o = - 1.66 V E o(cell) = E o(cathode) - E o(anode) = -0.44 (- 1.66) = + 1.22 V

CathodePositiveAnodeNegative+ 1.22 V