Lecture 9 Professor Hicks Inorganic Chemistry (CHE152) The - PDF document

Lecture 9 Professor Hicks Inorganic Chemistry (CHE152) • The following problems will NOT be on the final exam Chapter 17: 26, 36, 50, 54, 58, 62, 72 Add problem 20:49 to the list of HW problems required for the final exam review Oxidation-Reduction Reactions • oxidation is the loss of electrons • reduction is the gain of electrons • they must always happen together • referred to as redox reactions 1

review Oxidation-Reduction Reactions 4Na (s) + O 2 (g)  2Na 2 O (s) Na is oxidized O 2 is reduced Na is the reducing agent O 2 is the oxidizing agent How can we tell O has gained / Na lost electrons? Dissolve Na 2 O in water and the ions that form are 2Na + (aq) and O 2- (aq) review Oxidation-Reduction Reactions 2Ag + (aq) + Cu (s)  2Ag (s) + Cu 2+ (aq) • Cu donates electrons to Ag + • Cu is oxidized (loses electrons) • Ag + is reduced (gains electrons review More on Oxidation-Reduction Reactions  +  - 2C (s) + O 2 (g)  2CO (g) C O     CO does not break into ions when dissolved in water - it is a molecular not ionic compound this reaction is considered an oxidation-reduction reaction, based on: - large number of reactions product can be dissolved to yield O 2- ions - CO has a dipole moment 2

review Oxidation numbers in CO • O in CO is negative because it is more electronegative than C • if O is assigned a -2 charge as it forms in ionic compounds the C must be +2 • these are the Oxidation Numbers aka Oxidation States for O and C review Concept of oxidation numbers • keeps track of electrons • allows “charges” to be assigned  oxidation-reduction reactions identified review 3

review review review 4

review H 2 O, H + , OH - • redox reactions often involve H 2 O, H + , or OH - • powers in Q can be very large  strongly affect K eq • pH can have large effect PbO 2 (s) + 4H + (aq) + Sn (s)  Pb 2+ (aq) + 2H 2 O (l) + Sn 2+ (aq) Q = [Pb 2+ ][Sn 2+ ] [H + ] 4 Balancing redox reactions 5

Balance the previous reaction if it occurred in a basic solution 6

Balance the previous reaction if it occurred in a basic solution Balance the previous reaction if it occurred in a basic solution 7

Electricity = flow of electrons e- wire e- e- e- e- e- Voltage Joules • unit of the volt is Coulomb Alessandro Volta Which can be converted to Joules/electron x 1.6 x 10 -19 C 1.6 x 10 -19 J 1 Joules 1 volt = = 1 Coulomb 1 electron 1 electron • Coulomb (C) is a unit of electrical charge • 1 electron has a charge of 1.6 x 10 -19 C Voltage describes an amount of energy per electron Voltage is like the height of a building larger voltage = electrons at higher potential energy 5 volts e- highest energy lower energy increasing voltage heat (q) lower released energy charge on 1 electron = 5 volts x 1.6 x 10 -19 C = 8.0 x 10 -19 J heat (q) released 8

Every electron in an element has a different potential energy like the height of a different building Ag + (aq) + Cu (s)  Ag (s) + Cu + (aq) E o = +0.28 V e- Ag + (aq) + e-  Ag (s) E o = +0.80 V Cu (s)  Cu + (aq) + e- E o = -0.52 V 0.28 V heat (q) released or work done How much heat or work? 1.6  10 -19 C  0.28 V copper silver = 2.9  10 -20 J hydrogen used a reference V = 0 potential of electron is zero Galvanic cells = separate reduction and oxidation half reactions into different compartments (aka voltaic cells) wire e- Cu (s)  Cu + (aq) + e- Ag + (aq) + e-  Ag (s) electrons produced electrons consumed but salt bridge  (products) (reactants) charge balance negative charge positive charge would build up here would build up here – (spectator ion) NO 3 salt bridge reduction half cell oxidation half cell allows ions to flow Standard reduction potentials • non-metals towards top table • metals towards the bottom table • Reactants higher on table stronger oxidizing agents 9

Standard reduction potentials • reactions tabulated as reduction half reactions • voltage standard half reaction = E o Standard reduction potentials - voltage produced if half reaction (standard conditions) is attached to standard hydrogen electrode - Standard Hydrogen Electrode (SHE) 2H + (aq) + 2e-  H 2 (g) under standard conditions voltmeter voltage = E o aqueous species reaction standard 1.0 M [ ]‟s [H + ] = 1.0 M being hydrogen all gases P H2 = 1.0 atm tested electrode P gas = 1.0 atm salt bridge Standard hydrogen electrode • standard hydrogen electrode plays role that the standard state did in thermodynamics calculations of in  E o ,  H o ,  S o ,  G o standard hydrogen oxidation reduction electrode half reaction half reaction junction junction destination destination home station home station 10

Every electron in an element has a different potential energy like the height of a building e- 0.28 V e- copper silver hydrogen used a reference V = 0 potential of electron is zero How to use a table of reduction potentials to find out the voltage that will be produced for a reaction under standard conditions (E o ). • identify the half rxns involved • reverse the oxidation half reaction and the sign of its E o • add the two E o ‟s - positive voltage = spontaneous - negative voltage = non-spontaneous Determine the voltage produced under standard conditions by a cell based on the following unbalanced reaction. Ag + (aq) + Cu (s)  Ag (s) + Cu 2+ (aq) 1) find half reactions in the table of standard reduction potentials Ag + (aq) + e-  Ag (s) E o = +0.80 V Cu 2+ (aq) + 2e-  Cu (s) E o = +0.34 V 2) reverse the oxidation half reaction and the sign of its E o Ag + (aq) + e-  Ag (s) E o = +0.80 V Cu (s)  Cu 2+ (aq) + 2e- E o = -0.34 V reversed b/c Cu is oxidized in given rxn 4) add half reactions E o ‟s E o = 0.80 V + -0.34 V = 0.46 V 11

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1 Faraday of charge is the The Faraday charge on 1 mole of electrons e- e- e- e- e- e- 1 electron 2 electrons 3 electrons 1.6  10 -19 C 2  1.6  10 -19 C 3  1.6  10 -19 C = 3.2  10 -19 C = 4.8  10 -19 C e- e- e- e- e-e- e- e- e- e- e- e- e- e- e- e- e- e- e- e- e- e- e- e- e- e- e- e- e- e- e- e- e-e- e- e- e- e- e- e- e- e- e- e- e- e- e- 1 mole electrons 6.02  10 23  1.6  10 -19 C Michael Faraday = 96485 C Standard reduction potential vs free energy  G o = -nFE o both describe if a reaction is spontaneous • Why the negative sign? • positive E o „s  negative  G o „s spontaneous What is F? called the Faraday = the charge on 1 mole of electrons in Coulombs 6.02  10 23  1.6  10 -19 C = 96485 C • What/Why n? • n is usually moles – in electrochemistry it is the moles of electrons in the balanced reaction E o vs  G o n = moles of electrons F = Faradays constant 96485 C (charge on 1 mole electrons)  G o = -n F E o E o = standard reduction potential positive voltages  negative G’s  spontaneous!!! moles of e- ‟s Joules  Coulombs Joules  = mole rxn 1 mole e - Coulomb mole rxn 4 Au 3+ (aq) + 3Sn (s)  4 Au (s) + 3Sn 4+ (aq) E o = +1.35 V each Au 3+ requires 3 e- ’s voltage has dimensions of energy per electron n = 4  3 = 12 e-s if you double a reaction amount the voltage is not changed 8 Au 3+ (aq) + 6Sn (s)  8 Au (s) + 6Sn 4+ (aq) E o = +1.35 V each Au 3+ requires 3 e- ’s n = 8  3 = 24 e-s 13

Nernst Equation • Nernst Equation RT E = E o – nF ln(Q) Walther Nernst Walther Nernst describes how voltage changes with Q Nobel prize for 3rd Law of thermodynamics • n = number moles electrons • R gas constant • F charge of 1 mole of electrons Free energy vs cell voltage RT E = E o –  G =  G o + RTln(Q) nF ln(Q) • both describe how far a reaction is from equilibrium •  G negative / E positive = both spontaneous • doubling reaction amounts doubles  G • doubling reaction amounts does not change E! - nF factor makes E independent of # electrons in rxn (voltage is energy per electron) Voltage under non-standard conditions E = E o under standard conditions E = E o – RT RT RT E = E o – nF ln(Q) nF ln(Q) Q = 1  ln Q = 0 nF E o - 0 otherwise E = E o – RT RT RT this term adjusts voltage based E = E o – nF ln(Q) nF ln(Q) on concentrations/pressures nF 14

Reaction quotient and cell voltage • reaction moves forward if E is positive • at equilibrium voltage is zero Q  K eq Q  K eq E = + E = - G Q = K eq E = 0 volts 15

Electrolysis • if electrons forced against direction of spontaneous flow reaction will be driven backwards electrolytic cell • charges on cathode/anode reversed in electrolytic cells • anode is positive • cathode negative - + 16