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

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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

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

review

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Oxidation-Reduction Reactions

4Na (s) + O2 (g)  2Na2O (s)

Na is oxidized O2 is reduced Na is the reducing agent O2 is the oxidizing agent How can we tell O has gained / Na lost electrons? Dissolve Na2O in water and the ions that form are 2Na+ (aq) and O2- (aq)

review

Oxidation-Reduction Reactions

2Ag+ (aq) + Cu (s)  2Ag (s) + Cu2+ (aq)

• Cu donates electrons to Ag+
• Cu is oxidized (loses electrons)
• Ag+ is reduced (gains electrons

review

More on Oxidation-Reduction Reactions

2C (s) + O2 (g)  2CO (g) 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 O2- ions

• CO has a dipole moment

 

O

 

C

+ -

review

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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 
• xidation-reduction reactions identified

review review

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review review review

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review

H2O, H+, OH-

• redox reactions often involve H2O, H+, or

OH-

• powers in Q can be very large strongly

affect Keq

• pH can have large effect

PbO2 (s) + 4H+ (aq) + Sn (s)  Pb2+ (aq) + 2H2O (l) + Sn2+ (aq)

Q = [Pb2+][Sn2+] [H+]4

Balancing redox reactions

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Balance the previous reaction if it occurred in a basic solution

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

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Electricity = flow of electrons

wire e- e- e- e- e- e-

1.6 x 10-19 J 1 electron

Voltage

• unit of the volt is

1 Joules 1 Coulomb 1 volt =

• Coulomb (C) is a unit of electrical charge
• 1 electron has a charge of 1.6 x 10 -19 C

x 1.6 x 10-19 C 1 electron =

Voltage describes an amount of energy per electron

Alessandro Volta

Which can be converted to Joules/electron Joules Coulomb

highest energy lower energy lower energy heat (q) released

increasing voltage

Voltage is like the height of a building e-

5 volts

heat (q) released

= 5 volts x 1.6 x 10-19 C = 8.0 x 10 -19 J larger voltage = electrons at higher potential energy

charge on 1 electron

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Every electron in an element has a different potential energy like the height of a different building

e-

heat (q) released

• r work done

silver

hydrogen used a reference potential of electron is zero

V = 0

copper

Ag+ (aq) + e-  Ag (s) Eo = +0.80 V Cu (s) Cu+ (aq) + e- Eo = -0.52 V

Ag+ (aq) + Cu (s)  Ag (s) + Cu+ (aq) Eo = +0.28 V

0.28 V How much heat or work? 1.6 10-19 C  0.28 V = 2.9  10-20 J

Galvanic cells

Ag+ (aq) + e-  Ag (s) Cu (s) Cu+ (aq) electrons consumed (reactants) electrons produced (products)

e- wire

salt bridge allows ions to flow NO3

–

(spectator ion)

reduction half cell

• xidation half cell

e- + negative charge would build up here

but salt bridge  charge balance

= separate reduction and oxidation half reactions into different compartments positive charge would build up here

(aka voltaic cells)

Standard reduction potentials

• non-metals towards top table
• metals towards the bottom table
• Reactants higher on table stronger
• xidizing agents

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Standard reduction potentials

• reactions tabulated as reduction half

reactions

• voltage standard half reaction = Eo

Standard reduction potentials

• voltage produced if half reaction (standard

conditions) is attached to standard hydrogen electrode

• Standard Hydrogen Electrode (SHE)

2H+ (aq) + 2e-  H2 (g) under standard conditions

standard hydrogen electrode voltmeter

aqueous species 1.0 M [ ]‟s all gases Pgas = 1.0 atm

salt bridge voltage = Eo

[H+] = 1.0 M PH2= 1.0 atm

reaction being tested

Standard hydrogen electrode

• standard hydrogen electrode plays role

that the standard state did in thermodynamics calculations of in Eo, Ho, So, Go

home station home station junction junction destination destination

• xidation

half reaction reduction half reaction standard hydrogen electrode

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Every electron in an element has a different potential energy like the height of a building

e-

silver

hydrogen used a reference potential of electron is zero

V = 0

copper

0.28 V

e-

How to use a table of reduction potentials to find out the voltage that will be produced for a reaction under standard conditions (Eo).

• identify the half rxns involved
• reverse the oxidation half reaction and the

sign of its Eo

• add the two Eo‟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) + Cu2+ (aq) 1) find half reactions in the table of standard reduction potentials Ag+ (aq) + e-  Ag (s) Eo = +0.80 V Cu2+ (aq) + 2e-  Cu (s) Eo = +0.34 V Ag+ (aq) + e-  Ag (s) Eo = +0.80 V Cu (s) Cu2+ (aq) + 2e- Eo = -0.34 V 2) reverse the oxidation half reaction and the sign of its Eo Eo = 0.80 V + -0.34 V = 0.46 V 4) add half reactions Eo‟s reversed b/c Cu is oxidized in given rxn

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The Faraday

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-

e-

1 electron 1.6  10-19 C

e- e- e- e- e-

2 electrons 21.6  10-19 C = 3.2  10-19 C 3 electrons 3  1.6  10-19 C = 4.8  10-19 C 1 mole electrons 6.02  1023  1.6  10-19 C = 96485 C 1 Faraday of charge is the charge on 1 mole of electrons Michael Faraday

Standard reduction potential vs free energy

• Why the negative sign?
• positive Eo „s  negative Go „s spontaneous

What is F? called the Faraday = the charge on 1 mole

• f electrons in Coulombs

6.021023  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

Go = -nFEo

both describe if a reaction is spontaneous

Eo vs Go

Go = -n F Eo

4 Au3+ (aq) + 3Sn (s) 4 Au (s) + 3Sn4+ (aq) Eo = +1.35 V

voltage has dimensions of energy per electron

if you double a reaction amount the voltage is not changed 8 Au3+ (aq) + 6Sn (s) 8 Au (s) + 6Sn4+ (aq) Eo = +1.35 V

n = moles of electrons F = Faradays constant 96485 C (charge on 1 mole electrons) Eo = standard reduction potential

Joules mole rxn  Coulombs 1 mole e- Joules Coulomb 

positive voltages  negative G’s  spontaneous!!!

moles of e-‟s mole rxn =

each Au3+ requires 3 e- ’s n = 4  3 = 12 e-s each Au3+ requires 3 e- ’s n = 8  3 = 24 e-s

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Nernst Equation

• Nernst Equation

Walther Nernst Walther Nernst

RT nF ln(Q) E = Eo –

• n = number moles electrons
• R gas constant
• F charge of 1 mole of electrons

describes how voltage changes with Q

Nobel prize for 3rd Law

• f thermodynamics

Free energy vs cell voltage

RT nF ln(Q) E = Eo – G = Go + RTln(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 = Eo under standard conditions Q = 1 ln Q = 0

• therwise

RT nF ln(Q) E = Eo – RT nF RT nF ln(Q) E = Eo – RT nF ln(Q) E = Eo – RT nF RT nF ln(Q) E = Eo –

this term adjusts voltage based

• n concentrations/pressures

Eo -

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G

Reaction quotient and cell voltage

• reaction moves forward if E is positive
• at equilibrium voltage is zero

Q = Keq E = 0 volts Q  Keq E = - Q  Keq E = +

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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
• +

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• units coulombs/second
• can be converted to

electrons/second I 1 F  = rate electrons flowing

Coulombs second 1 mole e- 96486 Coulombs moles e- second

=

André-Marie Ampère

 2Ag+ (aq) + Cu (s)  2Ag (s) + Cu2+ (aq) 2Ag+ (aq) + 2e-  2Ag (s) Cu (s)  Cu2+ (aq) + 2 e- two half reactions reduction

• xidation

I F 1 moles Ag(s) 2 moles e-  = moles Ag  t

converts to moles product converts rate into amount

Electrolysis

• verall reaction

moles e- sec rate electrons flowing

mass compound

moles of formula units, molecules, atoms,

• r ions

number formula units, molecules, atoms,

• r ions

multiply molar mass divide molar mass multiply molar mass divide molar mass divide Avogadro's number multiply Avogadro's number divide Avogadro's number multiply Avogadro's number

n= PV RT n= PV RT PV RT

volume

It F

electrical current

I

n =

Molarity M

n=MV

(n = number of moles electrons) (n = number of moles gas) (n = number of moles dissolved substance)

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Electrolysis and reaction stoichiometry

2Ag+ (aq) + Cu (s)  2Ag (s) + Cu2+ (aq) spontaneous 2 electron reaction

How much copper will be deposited by electrolysis if a current of 1.1 amp flows for 30 seconds?

1.1 C s 1 mol e- 96485 C

30 s

    1 mole Cu 2 mole e- 63.55 g Cu 1 mole Cu Eo = 0.48 V = 0.0108 g Cu It F n = = moles of e-