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Chapter 20: Electrochemistry

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Chapter 20: Electrochemistry. Chemistry 1062: Principles of Chemistry II. Andy Aspaas, Instructor ... H3O becomes simply H when dealing with redox reactions ... – PowerPoint PPT presentation

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Title: Chapter 20: Electrochemistry


1
Chapter 20 Electrochemistry
Chemistry 1062 Principles of Chemistry II Andy
Aspaas, Instructor
2
Oxidation-Reduction reactions
  • Oxidation-reduction (redox) reaction transfer of
    electrons from one species to another
  • H3O becomes simply H when dealing with redox
    reactions to simplify balancing
  • (still the same species, just different notation)
  • Skeleton oxidation-reduction equation involves
    only the species being oxidized and reduced.
  • Write oxidation numbers above each species.
  • No spectator ions, no balancing
  • Half reaction shows only one oxidation OR one
    reduction
  • Most redox reactions are split into an oxidation
    half-reaction and a reduction half-reaction
  • LEO, GER

3
Balancing redox equations in acidic solutions
  • For each half reaction
  • Balance everything except H or O
  • Balance O by adding H2O to one side
  • Balance H by adding H to one side
  • Balance charge by adding e- to one side
  • Multiply each half reaction by a factor so that
    the electrons cancel when the two half reactions
    are added together (e- cannot appear in the final
    equation)
  • Add the reactions, cancel anything that appears
    on the left and right, and simplify the
    coefficients to the smallest integers

4
Practice balancing acidic redox reactions
  • Balance I2(s) NO3-(aq) ? IO3-(aq) NO2(g) in
    acidic solution
  • Half reactions
  • Cancel electrons
  • Add half-reactions
  • Simplify

5
Voltaic cells
  • A voltaic cell consists of two half-cells
  • Each half-cell contains a metal rod dipped in a
    solution containing that metal ion
  • Anode a species is being oxidized
  • Cathode a species is being reduced
  • Cell reaction redox reaction for entire voltaic
    cell
  • Zn(s) Cu2(aq) ? Zn2(aq) Cu(s)
  • Zn(s) ? Zn2(aq) oxidation half-reaction, anode
  • Cu2(aq) ? Cu(s) reduction half-reaction,
    cathode

6
Cell notation
  • Zn(s) Cu2(aq) ? Zn2(aq) Cu(s)
  • Cell notation Zn(s) Zn2(aq) Cu2(aq)
    Cu(s)
  • Anode Cathode
  • Write half reactions and cell reactions for the
    following cell
  • Tl(s) Tl(s) Sn2(aq) Sn(s)

7
emf, Standard Cell emf, Standard electrode
potential
  • Electromotive force, emf, Ecell electrical
    pressure across the conductors of an
    electrochemical cell
  • Unit Volt, V
  • Measure of the driving force of a cell reaction
  • Standard cell emf, Eocell solutes are 1 M,
    gases are 1 atm, temperature is 25 oC
  • Standard electrode potenital
  • By convention, the standard hydrogen electrode
    has an emf of 0 V
  • All reactions shown as reductions
  • Ecell Ecathode Eanode
  • Ecell is positive for spontaneous reactions as
    written

8
Practice calculating Ecell
  • Using standard potentials, calculate Ecell for
    Zn(s) Zn2(aq) Cu2(aq) Cu(s)
  • Standard cell potentials are an intensive
    property
  • Do not depend on quantity!
  • If you have to multiply a half-reaction to cancel
    electrons, do not multiply the Eo for that
    half-reaction

9
Free energy and K from Ecell
  • ?Go -nFEcell
  • n moles of electrons transferred
  • F Faradays constant, 96,500 C/mol e-
  • This gives an answer in J, since 1 J 1 CV
  • Convert to kJ since thats what ?Go is usually
    expressed in
  • Ecell (0.0592 / n) log K (Nernst equation)

10
Practice with ?Go and K
  • Calculate ?Go and K for the following cell
    Zn(s) Zn2(aq) Cu2(aq) Cu(s)
  • ?Go -nFEcell
  • Ecell (0.0592 / n) log K

11
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