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Electrochemical Equilibrium Problem Set

04/19/23

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

Cu(s) Cu2+(aq) + 2 e– Cu2+(aq) + 2 e– Cu(s)0.030 V

AnodeElectrode

CathodeElectrodeSalt Bridge

Cu

e–

1.00 M Cu2+

e–

Cu

0.10 M Cu2+

what will be the final concentrationsif the volumes are the same?

[Cu2+]o=0.10 M

[Cu2+]o=1.00 M

Cu CuCu Cu

0.020 V

Cu Cu

0.010 V

Cu Cu

0.000 V

0.55 M Cu2+

0.55 M Cu2+

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

Cu(s) Cu2+(aq) + 2 e– Cu2+(aq) + 2 e– Cu(s)0.030 V

AnodeElectrode

CathodeElectrodeSalt Bridge

Cu

e–e–

Cu

what will happen to the massesof the anode and the cathode?

0.020 V

Cu Cu

0.010 V

Cu Cu

0.000 V

Cu Cu

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AnodeElectrode

CathodeElectrodeSalt Bridge

0.000 VCu(s) Cu2+(aq) + 2 e– Cu2+(aq) + 2 e– Cu(s)

[Cu2+]o=0.10 M

[Cu2+]o=1.00 M

0.55 M Cu2+

Cu Cu

0.55 M Cu2+

The anode will lose mass.The cathode will gain mass.

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Electrochemical Determination of Equilibrium Constants

Reading assignment: Fine, Beall & Stuehr, Chapter 14.3-14.4

The Nernst equation can be used to determine equilibrium constants and related terms.

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Nernst Equation

Ecell = E °cell -RTnF · ln Q

Standard cellpotential

Concentrationterm

Under what conditions wouldthe observed cell be zero?

Would you like to buy a battery at equilibrium?

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At Equilibrium

observed cellpotential is zero

reaction quotientequals the equilibriumconstant

the cellis “dead”

standard cell potential equals the concentration term

Ecell = E °cell -RTnF · ln Q0.00 Keq

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Equilibrium Constant Determination

Determine the standard cell potential. Use the Nernst equation to determine the

equilibrium constant.

Solubility is an equilibrium process.

Determine the solubility product of silver chloride.

Procedure

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Determine the Numeric Value of the Solubility Product of Silver Chloride

Ksp =

Solubility product expression

Chemical equation that corresponds to this “target” process:

AgCl(s) Ag+(aq) + Cl–(aq)

[Ag+] [Cl–]

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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.

AgCl(s) Ag+(aq) + Cl–(aq)

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Divide the Target Reaction into Reduction and Oxidation Equations

targetAg+(aq) + Cl–(aq)AgCl(s)

Find an equation that “looks like” the target equation

reductionAgCl(s) + e– Ag(s) + Cl–(aq)

oxidation

Find the oxidation half-cell equation needed to complete the target equation

Ag+(aq) + e– Ag(s)

The oxidation and reduction reactions added together give the target equation

e– + Ag(s) + + Ag(s) + e–

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Determine the Standard Cell Potential of the Target Equation

AgCl(s) Ag+(aq) + Cl–(aq)

AgCl(s) + e– Ag(s) + Cl–(aq) E °½ =

Ag+(aq) + e– Ag(s)

+ 0.222 V

- 0.800 V

- 0.578 Vtarget

reduction

oxidation

E °½ =

E °cell =

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At Equilibrium

E cell = 0.00V

E °cell =0.0592

nproducts

reactants· log

E °cell =0.0592

n · log Keq

0.0592n

products

reactants· log= E °cell -

Apply this to the silver chloride solubility product.

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Solubility Product for AgCl

AgCl(s) Ag+(aq) + Cl–(aq)

Ecell = E ºcell -0.0592

nproducts

reactants· logE cell = -0.578 V -

0.05921 · log [Ag+][Cl–]

0.000 V = -0.578 V -0.0592

1 · log Ksp

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AgCl(s) Ag+(aq) + Cl–(aq)

log Ksp = - 0.5780.0592 = -9.76

Ksp = 10–9.76 = 1.7 x 10–10 M2

Solubility Product for AgCl

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Molar Solubility of Silver Chloride

AgCl(s) Ag+(aq) + Cl–(aq)-x +x +x

1.3 x 10–5 moles of AgCl will dissolve to form a liter of solution.

Ksp = [Ag+] [Cl–] = 1.7 x 10–10 M2

Ksp = [ x ] [ x ] = 1.7 x 10–10 M2

[ x ] = 1.3 x 10–5 M

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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

E °½ = +1.662 V

E °cell = +2.002 V

E °½ = +0.340 V

2 Al(s) + 3 Cu2+(aq) 2 Al3+(aq) + 3 Cu(s)target

reductionCu2+(aq) + 2 e– Cu(s)

oxidationAl3+(aq) + 3 e–Al(s)

3 ( )

2 ( )

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0.0592n

products

reactants· logE °cell =

At Equilibrium

[Al3+]2

[Cu2+]3 2.002

6

2.002 =0.0592

6 · log Keq

2 Al(s) + 3 Cu2+(aq) 2 Al3+(aq) + 3 Cu(s)

= E °cell -0.0592

nproducts

reactants· logEcell = 0.00 V

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2.002 =0.0592

6 · log Keq

log Keq =6 x 2.002

0.0592 = 202.9

Keq =

At Equilibrium2 Al(s) + 3 Cu2+(aq) 2 Al3+(aq) + 3 Cu(s)

What exactly does 8 × 10202 mean?

10202.9 = 8 x 10202 M–1

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Example

The solubility product of lead(II) chloride is 1.7 × 10–5 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?

reduction

oxidation

target

Usually listed in a tableof half-cell potentials

Related to theequilibrium constant

Equals sum ofoxidation and

reductionhalf-cells.

Relationship of the Cell Equations

Ksp = [Pb2+] [Cl–]2

PbCl2(s) Pb2+(aq) + 2 Cl–(aq)

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Use the Solubility Product Equation as the Target Equation

PbCl2(s) Pb2+(aq) + 2 Cl–(aq)target

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Break the Equation into a Reduction and an Oxidation Half-cell

reductionPbCl2(s) + 2 e– Pb(s) + 2 Cl–(aq)

oxidationPb2+(aq) + 2 e– Pb(s)

PbCl2(s) Pb2+(aq) + 2 Cl–(aq)target

The reduction potential isn’t listed in the table.Now what?

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Use the Equilibrium Constant to Determine the Standard Cell Potential

Ecell = 0.00 V = E °cell -0.0592

n · log Keq

E °cell =0.0592

n · log Keq(1.7 × 10–5 M3 )

E °cell = -0.141 V

2

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Work the Problem Backwards

PbCl2(s) + 2 e– Pb(s) + 2 Cl–(aq)

Pb2+(aq) + 2 e– Pb(s)

PbCl2(s) Pb2+(aq) + 2 Cl–(aq)

E °½ =

E °cell = - 0.141 V

E °½ =

reduction

oxidation

target

- 0.267 V

+0.126 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 poweredby 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

PbCl2(s) + 2 e– Pb(s) + 2 Cl–(aq)

Pb2+(aq) + 2 e– Pb(s)

PbCl2(s) Pb2+(aq) + 2 Cl–(aq)

E °½ = + 0.126 V

E °cell = - 0.141 V

E °½ = - 0.267 V

reduction

oxidation

target

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Reverse the Equations

Pb(s) + 2 Cl–(aq) PbCl2(s) + 2 e–

Pb2+(aq) + 2 e– Pb(s)

PbCl2(s) Pb2+(aq) + 2 Cl–(aq)

oxidation

oxidation

target

E °½ = + 0.267 V

E °½ = + 0.126 V

E °cell = - 0.141 V

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Reverse the Equations

Pb(s) + 2 Cl–(aq) PbCl2(s) + 2 e–

Pb2+(aq) + 2 e– Pb(s)

PbCl2(s) Pb2+(aq) + 2 Cl–(aq)

oxidation

reduction

target

E °½ = + 0.267 V

E °½ = - 0.126 V

E °cell = - 0.141 V

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Pb2+(aq) + 2 Cl–(aq)

Pb2+(aq) + 2 e–

Pb(s) + 2 Cl–(aq)

Reverse the Equations

PbCl2(s) + 2 e–

Pb(s)

PbCl2(s)

oxidation

reduction

target

Write the shorthand notation for the cell.

E °½ = + 0.267 V

E °½ = - 0.126 V

E °cell = + 0.141 V

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Pb2+(aq) + 2 Cl–(aq)

A Precipitation Cell is the Reverse of the Solubility Product Cell

PbCl2(s)

Pb2+(aq) + 2 e– Pb(s)

Pb(s) + 2 Cl–(aq) PbCl2(s) + 2 e–

Pb(s) | Cl–(aq) (1.00 M) | | Pb2+ (1.00 M) | Pb(s)

oxidation

reduction

target

E °½ = + 0.267 V

E °½ = - 0.126 V

E °cell = + 0.141 V

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Pb(s) | Cl–(aq) (1.00 M) | | Pb2+ (1.00 M) | Pb(s)

0.141 V

e– Salt Bridge

1 M NaCl

AnodeElectrode

Pb

1 M Pb(NO3)2

CathodeElectrodee–

Pb

Pb(s) + 2 Cl–(aq) PbCl2(s) + 2 e– Pb2+(aq) + 2 e– Pb(s)

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Pb(s) | Cl–(aq) (1.00 M) | | Pb2+ (1.00 M) | Pb(s)

0.141 V

e– Salt Bridge

1 M NaCl

AnodeElectrode

Pb

1 M Pb(NO3)2

CathodeElectrodee–

Pb

Pb(s) + 2 Cl–(aq) PbCl2(s) + 2 e– Pb2+(aq) + 2 e– 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+

Cr(s) Cr2+(aq) + 2 e– E°½= +0.86 VCr(s) Cr2+(aq) + 2 e– E °½ = +0.86 V

Cr(s) Cr3+(aq) + 3 e– E °½ = +0.74 V

Cr2+(aq) Cr3+(aq) + e– E °½ = +0.41 V

Which reaction occurs at the anode?

What reactions might occur at the cathode?

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Possible Cathode Reactions Sn2+, Sn4+ | Pt(s)

Sn2+(aq) + 2 e– Sn(s) E °½ = -0.136 V

Sn4+(aq) + 2 e– Sn2+(aq) E °½ = +0.150 VSn4+(aq) + 2 e– Sn2+(aq) E °½ = +0.150 V

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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E °½ = +0.150 V

E °½ = +0.86 V

E °cell = +1.01 V

oxidation

reduction

net reaction

Overall Reaction

Cr(s) + Sn4+(aq) Sn2+(aq) + Cr2+(aq)

Sn4+(aq) + 2 e– Sn2+(aq)

Cr(s) Cr2+(aq) + 2 e–

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Cr(s) | Cr2+(0.400 M) | | Sn2+(0.018 M), Sn4+(0.680 M) | Pt(s)

E cell = E °cell -0.0592

nproducts

reactants· log

[Sn2+] [Cr2+]

[Sn4+] 2

E cell = 1.07 V

[0.018] [0.400]

[0.680] 1.01

Cr(s) + Sn4+(aq) Sn2+(aq) + Cr2+(aq)

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The End

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