Please Sit with Your Group

1
Please Sit with Your Group

• Please be sure each member of your team has a
  copy of
• Electrochemical Determination of Equilibrium
  Constants Lecture Notes
• Electrochemical Equilibrium Problem Set
• Todays reporter is the person who began watching
  Super Bowl coverage the earliest.
• Next reading assignment
• Zumdahl Chapter 11

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If the voltmeter is connected for a long period
of time,

what will be the final concentrationsif the
volumes are the same?
3
If the voltmeter is connected for a long period
of time,

what will happen to the masses of the anode and
the cathode?
4

0.000 V
Cathode Electrode
Anode Electrode
Salt Bridge
Cu
Cu
0.55 M Cu2
0.55 M Cu2
The anode will lose mass
The cathode will gain mass
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Electrochemical Determination of Equilibrium
Constants

• Edward A. Mottel
• Integrated, First-Year Curriculum in Science,
  Engineering and Mathematics

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Electrochemical Determination of Equilibrium
ConstantsAgenda

• Equilibrium conditions
• Determination of equilibrium constants
• Solubility product
• Molar solubility

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The Observed Cell Potential is a Function of Two
Terms
This term remains constant throughout the
reaction, because the underlying reaction doesn't
change.
This term becomes more negative and its effect on
the observed cell potential (Ecell) becomes
greater as the reaction proceeds towards
equilibrium
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Would you like to buy a battery at equilibrium?
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At Equilibrium
observed cell potential is zero
reaction quotient equals the equilibrium constant
the cell is dead
Keq
0.00
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Procedure to Determine an Equilibrium Constant

• Divide the target reaction into an oxidation and
  a reduction half-cell.
• Determine the standard cell potential of the
  cell.
• Use the Nernst equation to determine the
  equilibrium constant.

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Determine the Solubility Product of Silver
Chloride
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Procedure

• Divide the target reaction into an oxidation and
  a reduction half-cell.
• Every reaction can be divided into at least one
  oxidation-reduction half-cell pair.
• Even reactions not normally considered redox
  equations can be described as a redox half-cell
  pair.

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Divide the Target Reaction into Reduction and
Oxidation Equations
reduction
oxidation
target
Find an equation that looks like the target
equation
Find the oxidation half-cell equation needed to
give the target equation
The oxidation and reduction reactions added
together give the target equation
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Determine the Standard Cell Potential of the
Target Equation
reduction
E½
0.222 V
oxidation
E½
- 0.800 V
target
Ecell
- 0.578 V
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At Equilibrium
Ecell 0.00V
Eºcell
Eºcell
Apply this to the silver chloride solubility
product.
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Solubility Product for AgCl
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Solubility Product for AgCl
Ksp 109.76 1.7 x 1010 M2
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What is Molar Solubility?

• Molar solubility is the maximum number of moles
  of solute which will dissolve to form a liter of
  solution.

What is the molar solubility of silver chloride
in pure water?
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Molar Solubility of Silver Chloride
-x
x
x
Ksp Ag Cl 1.7 x 1010 M2
Ksp x x 1.7 x 1010 M2
x 1.3 x 105 M
1.3 x 105 moles of AgCl will dissolve to form a
liter of solution.
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Determine the Equilibrium Constant for the
Reaction of Aluminum Metal and Copper(II) Ion

• Write the target equation for the reaction.
• Break the equation into a reduction and an
  oxidation half-cell.
• Determine the standard cell potential.
• Use the Nernst equation to solve for the
  equilibrium constant.

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Determine the Standard Cell Potential of the
Target Equation
reduction
E½0.340 V
oxidation
E½1.662 V
target
Ecell2.002 V
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At Equilibrium
Ecell 0.00 V
Al32
Eºcell
2.002
6
Cu23
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At Equilibrium
202.9
log Keq
Keq
10202.9
8 x 10202 M1
What exactly does 8 10202 mean?
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Example

• The solubility product of lead(II) chloride is
  1.7 105 M3.
• Using this equilibrium constant, determine a
  half-cell potential involving lead(II) chloride.
• Design an electrochemical cell to determine this
  value.

How are the half-cells and the cell
equation related to other terms?
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Relationship of the Cell Equations
reduction
Usually listed in a table of half-cell potentials
oxidation
target
Equals sum of oxidation and reduction half-cells.
Related to the equilibrium constant
Ksp Pb2 Cl2
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Use the Solubility Product Equation as the Target
Equation
target
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Break the Equation into a Reduction and an
Oxidation Half-cell
reduction
oxidation
target
The reduction potential isnt listed in the table.
Now what?
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Use the Equilibrium Constant to Determine the
Standard Cell Potential
(1.7 105 M3 )
2
Eºcell -0.141 V
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Work the Problem Backwards
reduction
- 0.267 V
E½
oxidation
0.126 V
E½
target
Ecell - 0.141 V
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Design an Electrochemical Cell to Determine this
Value
Can the potential of a non-spontaneous reaction
be measured?
No. It would need to be powered by the voltmeter.
Only spontaneous reactions can be measured.
What can you do?
Set up the reverse reaction, which is spontaneous.
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Reverse the Equations
reduction
E½ - 0.267 V
oxidation
E½ 0.126 V
target
Ecell - 0.141 V
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Reverse the Equations
oxidation
E½ 0.267 V
oxidation
E½ 0.126 V
target
Ecell - 0.141 V
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Reverse the Equations
oxidation
E½ 0.267 V
reduction
E½ -0.126 V
target
Ecell -0.141 V
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Reverse the Equations
oxidation
PbCl2(s) 2 e
Pb(s) 2 Cl(aq)
E½ 0.267 V
reduction
E½ - 0.126 V
Pb2(aq) 2 e
Pb(s)
target
Ecell 0.141 V
Pb2(aq) 2 Cl(aq)
PbCl2(s)
Write the shorthand notation for the cell.
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A Precipitation Cell is the Reverse of the
Solubility Product Cell
oxidation
PbCl2(s) 2 e
Pb(s) 2 Cl(aq)
E½ 0.267 V
reduction
E½ - 0.126 V
Pb2(aq) 2 e
Pb(s)
target
Ecell 0.141 V
Pb2(aq) 2 Cl(aq)
PbCl2(s)

• Pb(s) Cl(aq) (1.00 M) Pb2 (1.00 M) Pb(s)

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Pb(s) Cl(aq) (1.00 M) Pb2 (1.00 M) Pb(s)

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Consider the Following Electrochemical Cell
Involving an Inert Platinum Electrode
Cr(s) Cr2(0.400 M) Sn2(0.018 M),
Sn4(0.680 M) Pt(s)

• Identify the reaction occurring at each
  electrode.
• Determine the observed cell potential at 25 C.

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Possible Anode Reactions Cr(s) Cr2
Which reaction occurs at the anode?
What reactions might occur at the cathode?
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Possible Cathode Reactions Sn2, Sn4 Pt(s)
Which reaction occurs at the cathode?
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Analysis

• Chromium metal is oxidized to chromium(II) ion at
  the anode.
• Tin(IV) ion is reduced to tin(II) ion, but no
  precipitate is formed. Reduction occurs at the
  inert electrode.

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Overall Reaction
E½ 0.86 V
E½ 0.150 V
Ecell 1.01 V
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Cr(s) Cr2(0.400 M) Sn2(0.018
M), Sn4(0.680 M) Pt(s)
1.01
2
Ecell 1.07 V
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The End
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