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Electrochemistry

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Title: Electrochemistry


1
Electrochemistry
  • Chapter 20
  • Brown-LeMay

2
Review of Redox Reactions
  • Oxidation - refers to the loss of electrons by a
    molecule, atom or ion - LEO goes
  • Reduction - refers to the gain of electrons by an
    molecule, atom or ion GER
  • Chemical reactions in which the oxidation state
    of one or more substances changes are called
    oxidation-reduction reactions (or redox
    reactions)

3
Zn(s) 2H(aq)? Zn2(aq) H2(g)
  • Zn 0, Zn2 2 (LEO) reducing agent
  • H 1, H2 0 (GER) oxidizing agent
  • Thus, the oxidation number of both the Zn(s) and
    H(aq) change during the course of the reaction,
    and so, this must be a redox reaction
  • Review balancing redox-equations

4
Balancing Redox by Half Reactions
  • Half reactions are a convenient way of separating
    oxidation and reduction reactions.
  • Balance the titration of acidic solution of
    Na2C2O4 (colorless) with KMnO4(deep purple)
  • 1st Write the incomplete ½ reactions
  • MnO4(aq) ? Mn2(aq) is reduced (pale pink)
  • C2O4(aq) ? CO2(g) is oxidized

5
  • MnO4-(aq) ? Mn2(aq) 4H2O
  • C2O42-(aq) ? 2CO2(g)
  • Then bal H by adding H
  • 8H MnO4-(aq) ? Mn2(aq) 4H2O
  • C2O42-(aq) ? 2CO2(g)
  • finish by balancing the es
  • For the permanganate 7 left and 2 on the right
  • 5e- 8H MnO4-(aq) ? Mn2(aq) 4H2O
  • On the oxalate 2- on the right and o on the left
  • C2O42-(aq) ? 2CO2(g) 2e-

6
  • 2(5e- 8H MnO4-(aq) ? Mn2(aq) 4H2O)
  • 5(C2O42-(aq) ? 2CO2(g) 2e-)
  • 10e- 16H 2MnO4- (aq) ? 2Mn2(aq) 8H2O
  • 5C2O42-(aq) ? 10CO2(g) 10e-
  • 16H2 MnO4- 5C2O42- ? 2Mn2(aq)8H2O

  • 10CO2

7
Balancing Eq in Basic Solution
  • The same method is used but OH- is added to
    neutralize the H used.
  • The equation must again be simplified by
    canceling the terms on both sides of the equation.

8
Voltaic Cells
  • Spontaneous redox reactions may be use to perform
    electrical work
  • Voltaic or galvanic cells are devices that
    electron transfer occurs in an external circuit.

9
Zn(s) Cu2(aq) ? Zn2(aq) Cu(s)
  • Zn0 is spontaneously oxidized to Zn2
  • Cu2 is spontaneously reduced to Cu0
  • Oxidization half reaction at the anode
  • Zn(s) ? Zn2(aq) 2e-
  • Reduction half reaction at the cathode
  • Cu2(aq) 2e- ? Cu(s)

10
  • As oxidization occurs, Zn is converted to Zn2
    and 2e-.
  • The electrons flow toward the cathode, where they
    are used in the reduction reaction
  • We expect the Zn electrode to lose mass
  • Electrons flow from the anode to cathode so the
    anode is negative and the cathode is positive

11
  • Electrons cannot flow through the solution they
    have to be transported though an external wire.
  • Anions and cations move through a porous barrier
    or salt bridge
  • Cations move into the cathodic compartment to
    neutralize the excess negatively charged ions
    (cathode Cu2 2e- ? Cu, so the counter ion of
    Cu is in excess)

12
  • A build up of excess charge is avoided by
    movement of cations and anions through the salt
    bridge

13
  • The anions move into the anodic compartment to
    neutralize the excess Zn2 ions formed by the
    oxidation.
  • Molecular View
  • - Rules of voltaic cells
  • at the anode electrons are products
  • oxidization occurs
  • at the cathode electrons are reactants
  • reduction occurs

14
Cell EMF the driving force
  • Reactions are spontaneous because the cathode has
    a lower electrical potential energy than the
    anode
  • Potential difference difference in electrical
    potential measured in volts
  • One volt is the potential difference required to
    impart one joule (J) of energy to a charge of one
    coulomb (C)
  • 1V 1 J/C

15
Cell EMF the driving force
  • Electromotive force (emf) is the force required
    to push electrons though the external circuit
  • Cell potential Ecell is the emf of a cell
  • Ecell gt 0 for a spontaneous reaction
  • For 1 molar, 1 atm for gases at 250C(standard
    conditions), the standard emf
  • (standard cell potential) E0cell

16
Standard Reduction Potentials Cal. Cell
Potentials
  • Standard Reduction Potentials E0red are measured
    relative to a standard
  • The emf of a cell is
  • E0cell E0red(cathode) E0red(anode)
  • The standard hydrogen electrode is used as the
    standard (standard hydrogen electrode)
  • (SHE)
  • 2H(aq,1M)2e- ? H2(g,1atm) E0cell 0 V

17
The standard hydrogen electrode
  • The (SHE) is assigned a potential of zero
  • Consider the following half reaction
  • Zn(s) ? Zn2(aq) 2e-
  • We can measure E0cell relative to the SHE
  • In the cell the SHE is the cathode
  • It cons of a Pt electrode in a tube 1M Hsol
  • H2 is bubbled through the tube
  • E0cell E0red(cathode)-E0red(anode)
  • 0.76V 0V-E0red(anode)
  • Therefore E0red(anode) -0.76V

18
The standard hydrogen electrode
  • Standard electrode potentials are written as
    reduction reactions
  • Zn2(aq) (aq,1 M) 2e- ? Zn(s) E0 -0.76V
  • Since the reduction potential is negative in the
    presence of the SHE the reduction of Zn2 is
    non-spontaneous
  • However the oxidization of Zn2 is spontaneous
    with the SHE

19
Standard reduction potential
  • The standard reduction potential is an intensive
    property
  • Therefore, changing the stoichiometric
    coefficient does not affect E0red
  • 2Zn2(aq) 4e- ? 2Zn(s) E0red -0.76 V

20
  • E0red gt 0 are spontaneous relative to the SHE
  • E0red lt 0 are non- spontaneous relative to the
    SHE
  • The larger the difference between E0red values
    the larger the E0cell
  • The more positive the E0cell value the greater
    the driving force for reduction

21
Oxidizing and Reducing Agents
  • Consider the table of standard reduction
    potentials
  • We use the table to determine the relative
    strength of reducing and oxidizing agents
  • The more positive the E0red the stronger the
    oxidizing agent (written as the reactant)
  • The more negative the E0red the stronger the
    reducing agent (written as the product)

22
Oxidizing and Reducing Agents
  • We can use tables to predict if one reactant can
    spontaneously oxidize or reduce another
  • Example F2 can oxidize H2 or Li
  • Ni2 can oxidize Al(s)
  • Li can reduce F2

23
Spontaneity of Redox Reactions
  • E0cell E0red(red process) E0red(oxid process)
  • Consider the reaction
  • Ni(s) 2Ag(aq) ? Ni2(aq) 2Ag(s)
  • The standard cell potential is
  • E0cell E0red (Ag/Ag) E0red(Ni2/Ni)
  • E0cell (0.80 V) - (-0.28)
  • E0cell 1.08 V the value indicates the reaction
    is spontaneous

24
EMF and free energy change
  • Delta G -nFE
  • where delta G is the change in free energy
  • n the number of moles of electrons transferred
  • F Faradays constant
  • E emf of the cell

25
EMF and free energy change
  • F 96,500 C/mole- 96,500 J/(V)(mole-)
  • Since n and F are positive, if Delta G lt 0, then
    E gt 0 and the reaction will be spontaneous.
  • Effect of concentration on cell EMF
  • the cell is function until E0 at which
    point equilibrium has been reached and the cell
    is dead
  • The point at which E0 is determined by the
    concentrations of the species involved in the
    redox reaction

26
Walter Nernst (Nobel Prize 1920)
  • Nernst Equation
  • G G0 RTlnQ
  • -nFE -nFE0 RTlnQ
  • Solve the equation for
  • E give the Nernst Eq
  • E E0 - RT/nF lnQ
  • Or for base 10 log
  • E E0- 2.3RT/nF logQ

27
  • The nernst eq at 298K
  • E E0 0.0592/n log Q
  • Consider if you may
  • Zn(s) Cu2(aq) ? Zn2(aq) Cu(s)
  • If Cu2 5.0M and Zn2 0.05 M
  • Ecell 1.10 V 0.0592/2 log 0.05/5 1.16 V
  • Cell emf and chemical equilibrium
  • Log K nE0/0.0592
  • thus if we know cell emf, we can calc the
    equilibrium constant
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