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Electrochemistry

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Electrochemistry Marriage of redox and thermo Spontaneous electron-transfer reactions can result in spontaneous electric current if the reactants are separated ... – PowerPoint PPT presentation

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Title: Electrochemistry


1
Electrochemistry
  • Marriage of redox and thermo
  • Spontaneous electron-transfer reactions can
    result in spontaneous electric current if the
    reactants are separated by a wire
  • Voltaic (Galvanic) cells
  • We can use the spontaneity of the reaction to
    do electrical work

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  • We can push electrons through a cell in order
    to make a nonspontaneous redox reaction occur.
  • Electrolysis cell
  • Doing work to force chemical reaction to occur
    opposite of voltaic cell

3
Balancing Redox Equations
  • Deferred until later
  • For now just know that
  • A half reaction has electrons written as a
    reactant or a product
  • Oxidation half reaction A reactant gets
    oxidized (loses electrons) electrons appear as a
    product
  • Reduction half reaction A reactant gets reduced
    (gains electrons) electrons appear as a
    reactant
  • A balanced redox equation does not show electrons
    explicitly. e-s lost e-s gained (called
    n)

4
Voltaic Cells
  • Recall lab

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  • The spontaneous rxn occurs in the cell
  • e-s flow from to (get to go where they want
    to go)
  • Anode where ox occurs
  • Cathode where red occurs
  • Salt bridge prevents charge buildup (which would
    stop flow)

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You used graphite in place of Pt in lab for
Fe2/Fe3 and I2/I-cells. A cheaper inert
electrode.
Used when neither redox species in a half
reaction (or electrode) is a neutral metal.
Neither is a neutral metal
9
Standard Reduction Potentials (Ered)
  • Recall lab
  • Make a bunch of different cells, get different
    Ecell values (Eºcell if at standard state).
  • Clearly some reductions are more favorable than
    others
  • How do you know? Which direction did e-s
    flow?
  • By how much?
  • Rank them? (Must pick a zero as reference.)

10
Quick quiz
  • NOTE Every electrode compartment has one
    oxidizing agent and one reducing agent (this pair
    is called the redox couple)
  • If an electrode has Ni(s) and Ni2 ions in it,
    which species is the oxidizing agent and which
    the reducing agent (of the pair)?

Ni2 (b/c its more positive it has room for
an electron
Ox agent is ___
Ni (b/c its more negative it has an
electron to give)
Red agent is ___
11
Revisit Earlier CellLook at this as a
Competition for the electrons. Which oxidizing
agent wants them more?
Who is the (possible) oxidizing agent on the
left? _____
Who is the (possible) oxidizing agent on the
right? _____
Zn2
Cu2
Hint The two players are Zn and Zn2
Hint The two players are Cu and Cu2.
Who wins? (Which one got the electrons?) ____
Cu2
? Cu2 pulled harder
Cu2 2 e- ? Cu(s)
Sowhich of the half reactions shown at the right
is more favorable (greater tendency to happen)?
Zn2 2 e- ? Zn(s)
By how much?.....
12
Reducing Cu2 is more favorable than reducing
Zn2 by 1.10 V! (Measure it w/voltmeter!)
We define a standard reduction potential,
Ered, for every reduction half reaction such
that
Ecell Ered(cathode) - Ered(anode)
The more positive the E (Ecell, Ered, or Eox),
the more favorable the process
13
Reducing Cu2 is more favorable than reducing
Zn2 by 1.10 V! (Measure it w/voltmeter!)
Ecell Ered(cathode) - Ered(anode)
1.10 V Ered(Cu2/Cu) - Ered(Zn2/Zn)
NOTE
If Ered(Zn2/Zn) were 0 V, Ered(Cu2/Cu) would
be 1.10 V
If Ered(Zn2/Zn) were -1.0 V, Ered(Cu2/Cu)
would be 0.10 V
If Ered(Zn2/Zn) were 1.0 V, Ered(Cu2/Cu)
would be 2.10 V
?The zero is arbitrary, but must be chosen /
agreed upon!
14
This was chosen to be the zero of potential.
2 H 2 e- ? H2 (g) Ered 0.0 V
15
0.76 V Ered(SHE) - Ered(Zn2/Zn)
? 0.76 V 0 - Ered(Zn2/Zn)
? Ered(Zn2/Zn) - 0.76 V
Both as reductions
Ecell Ered Eox
OR Ecell Ecathode - Eanode
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Table 18.1 (continued)
19
Calculate E?cell (see board)
20
Lab interlude
  • See overhead / board
  • The lab manual initially asks you to pretend that
    the Ag/Ag reduction potential is 0.0 V just to
    show you the arbitraryness of this.
  • Then it tells you that in reality, Ag/Ag
    reduction potential is 0.80 V if the H/H2
    potential (SHE) is 0.0 V

21
Cu/Cu2 Fe3/Fe2
Cu/Cu2
0.45 V
-0.13 V
0.32 V
Cu ? Cu2 2e-
Fe3 e-? Fe2
Zn/Zn2 Ag/Ag
1.50 V
Zn/Zn2
Ag e-? Ag
0.0 V
1.50 V
Zn ? Zn2 2e-
Cu/Cu2 Zn/Zn2
1.05 V
Zn/Zn2
1.50 V
-0.45 V
Cu2 2e-? Cu
Zn ? Zn2 2e-
Circle the species that is the better oxidizing
agent
22
From Text Table
0.80 V
Ag e-? Ag
0.0 V
0.80 V
0.77 V
0.67 V
-0.13 V
Fe3 e-? Fe2
Fe3 e-? Fe2
-0.13 V
0.34 V
0.35 V
Cu2 2e-? Cu
-0.45 V
Ag e-? Ag
0.0 V
-0.76 V
-1.50 V
-0.70 V
Zn2 2e- ? Zn
Cu2 2e-? Cu
-0.45 V
Zn ? Zn2 2e-
1.50 V
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Table 18.1 (continued)
25
Excerpt from Voltaic Cell lab reading
26
Calculate E?cell (see board)
27
Calculate E?cell (see board)
28
Nernst Equation
  • See handout

29
Standard vs. Nonstandard Cell
30
Explain in detail an in a conceptual way why the
cell potential goes up when the NH3 is added. Is
the driving force for the cell rxn greater or
smaller after the NH3 is added?
31
Relationship between variables (at any
conditions)
32
Relationship between variables (at standard
state conditions)
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What mass of gold is plated in 25 minutes if the
current is 5.5 A?
Au3 (aq) 3 e- ? Au(s)
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