Electrochemistry - York · PDF fileReview: Chapter 5 questions 21-26. Chapter 21 questions:...

York University CHEM 1001 3.0 Electrochemistry - 1

Electrochemistry

Reading: from Petrucci, Harwood and Herring (8th edition):

Required for Part 1: Sections 21-1 through 21-4.

Examples for Part 1: 21-1 through 21-10.

Problem Set for Part 1:

Review: Chapter 5 questions 21-26.

Chapter 21 questions: 15-17, 32, 34, 43, 53

Additional problems from Chapter 21:


Applications of Electrochemistry

C Spontaneous chemical reactions can be used to producean electric current and do work. (batteries, fuel-cells)

C An electric current can be used to force non-spontaneouschemical reactions to occur. (electrolysis)

C Reactions can be made to occur in a specific place.(electroplating, electropolishing)

C The voltage produced by a reaction can be used as ananalytical tool. (pH electrodes)

C The current produced by a reaction can be used as ananalytical tool.


)G and non-PV Work

For a reversible process at constant T and P:

)H = qP + wE (definition of enthalpy)

wE = non-PV work done on the system

qP = T)S (reversible, constant T)

Let wmax = maximum non-PV work done by the system

wmax = -wE (reversible)

So wmax = -)H + T)S = -)G

Conclusion: The maximum non-PV work that can beobtained from a process is equal to -)G.


Work from Chemical Reactions

Spontaneous chemical reactions can be used to do work.

How?

One possibility:

C burn fuel to release heat

C boil water

C use the expanding steam to do work

Disadvantages:

C inefficient (only part of heat can be turned into work)

C can not readily carry out the reverse process


Work from Redox Reactions

Another method: Use redox reactions.

Cu(s) + 2Ag+ W Cu2+ + 2Ag(s) )rG° = -88.43 kJ mol-1

The Cu(s) is oxidized (gives up electrons).

Half-reaction: Cu(s) 6 Cu2+ + 2e-

The Ag+(aq) is reduced (receives electrons).

Half-reaction: Ag+ + e- 6 2Ag(s)

This reaction is spontaneous. If we can transfer the electronsthrough an external circuit, we can use it do electrical work.


Electrochemical Cells

C Cu is oxidized atone electrode:Cu(s) 6 Cu2+ + 2e-

C Ag+ is reduced atthe other electrode:e- + Ag+ 6 Ag(s)

C Electrons travelthrough the wire.

C Ions travel throughthe salt bridge.


Atomic View of an Electrochemical Cell


Electrochemical Cells - Terminology

C The anode is the electrode at which oxidation occurs.

C The cathode is the electrode at which reduction occurs.

C The cell potential is the voltage difference when nocurrent flows between the electrodes.

Cell potential is also called cell voltage or EMF(electromotive force).

C Cells in which spontaneous reactions produce a currentare called voltaic cells or galvanic cells.

C In electrolytic cells electricity is used to force a non-spontaneous reaction to occur.


Cell Diagrams C Anode (oxidation) is placed on left side of diagram.

C Cathode (reduction) is placed on right side of diagram.

C Single vertical line, | , indicates a boundary betweendifferent phases (i.e., solution | solid).

C Double vertical line, || , indicates a boundary (salt bridge)between different half-cell compartments.

Example: Cu(s) * Cu2+ 2 Ag+ * Ag(s)

At the anode: Cu(s) 6 Cu2+ + 2e-

At the cathode: Ag+ + e- 6 Ag(s)

Overall: Cu(s) + 2Ag+ 6 Cu2+ + 2Ag(s)


Cell Diagrams - examples

Diagram the cell in which the following overall reactionoccurs:

Pb(s) + 2AgCl(s) W PbCl2(s) + 2Ag(s)

Answer: Pb(s)*Cl-(aq)*PbCl2(s))2AgCl(s)*Ag(s)*Cl-(aq)*

Write the half-cell reactions for the following cell:

Ag(s)*Ag+(aq)2Cl-(aq)*AgCl(s)*Ag(s)

Answer:

Anode (oxidation): Ag(s) 6 Ag+(aq) + e-

Cathode (reduction): AgCl(s) + e- 6 Ag(s) + Cl-(aq)


Balancing Redox Reactions - ReviewExample: SO3

2- + MnO4- W SO4

2- + Mn2+ (unbalanced)

S from +4 to +6 (oxidized). Mn from +7 to +2 (reduced).

(1) Write balanced half-reactions for oxidation andreduction.

Oxidation half-reaction

Skeleton reaction: SO32- W SO4

2- + 2e-

Balanced: SO32- + H2O(l) W SO4

2- + 2H+ + 2e-

Reduction half-reaction

Skeleton reaction: MnO4- + 5e- W Mn2+

Balanced: MnO4- + 8H+ + 5e- W Mn2+ + 4H2O(l)


Balancing Redox Reactions - continued

(2) Adjust coefficients so the two half-reactions have thesame numbers of electrons.

5SO32- + 5H2O(l) W 5SO4

2- + 10H+ + 10e-

2MnO4- + 16H+ + 10e- W 2Mn2+ + 8H2O(l)

(3) Add the two half-reactions.

5SO32- + 2MnO4

- + 6H+ W 5SO42- + 2Mn2+ + 3H2O(l)

10 electrons are transferred (important for later).

(4) Check that the reaction is balanced for both atoms andcharge.


Redox Reactions in Basic Solution

5SO32- + 2MnO4

- + 6H+ W 5SO42- + 2Mn2+ + 3H2O(l)

Since this balanced reaction involves H+, it is appropriatefor acidic solution

For the reaction in basic solution, add

6 × {H2O(l) W H+ + OH-}

This gives:

5SO32- + 2MnO4

- + 3H2O(l) W 5SO42- + 2Mn2+ + 6OH-

This method can also be used to get half-reactions in basicsolution.


Current and Charge C Current is the amount of charge transferred per unit time.

C The amount of charge transferred is determined by thestoichiometry of the cell reaction.

C The charge on one mole of electrons is 96,485 coulombs.This is called the Faraday constant, F.

F = 96,485 C mol-1 = 96,485 J V-1 mol-1

Example: Cu(s) + 2Ag+ 6 Cu2+ + 2Ag(s)

1.93×105 C are transferred per mole of Cu oxidized.

C The current produced by a cell is determined by the reactionkinetics and the resistance of the circuit.


Electrical WorkThe work done in an electrical circuit is

welec = charge × (potential difference)

Units: coulombs × volts = joules

The maximum possible non-PV work equals -)G. Let

n / moles of electrons transferred per mole of reaction

nF = total charge (coulombs) transferred

Ecell / maximum possible cell potential (requires zerocurrent)

Then: )rG = -nFEcell


Electrical Work - continued

)rG = -nFEcell

There is a fundamental connection between )rG and cellpotential. Because of this:

C Cell potentials depend on concentrations.

C Electrochemical cells can be used to measureconcentrations. (pH electrodes, for example)

C Electrochemical cells can be used to measure )rG (andto determine )rG°).

C Tabulated thermodynamic data can be used to determinecell potentials.


Cell Potential - ExampleA cell is constructed in which the half-cell reactions are:

Anode: H2(g) 6 2H+(aq) + 2e-

Cathode: Cl2(g) + 2e- 6 2Cl-(aq)

When PH2 = PCl2 = 1.000 atm and [H+] = [Cl-] = 0.0100 M,the cell potential is found to be 1.4813 volts. Find )rGunder these conditions for

H2(g) + Cl2(g) W 2H+(aq) + 2Cl-(aq)

Solution: )rG = -nFEcell and n = 2 for the overall reaction.

)rG = -2(96,485 J V-1 mol-1)(1.4813 V) = -285.85 kJ mol-1


Cell Potentials and Spontaneity

)rG = -nFEcell

C If )rG < 0, the reaction is spontaneous as written.

C If )rG > 0, the reaction is non-spontaneous as written(reverse reaction is spontaneous).

If the reaction proceeds as written, then n > 0.

Therefore:

C If Ecell > 0, the reaction is spontaneous as written.

C If Ecell < 0, the reaction is non-spontaneous as written.


Standard Cell Potentials

C Definition: The standard cell potential, E°cell, is the cellpotential that would obtain if all reactants and products werein their standard states.

C Therefore:

)rG° = -nFE°cell

C Standard states may be hypothetical. The standard cellpotential is used in calculations of actual cell potentials.

C The standard cell potential is the sum of standard potentialsfor the individual half-cells.


Standard Cell Potential - exampleThe cell with the overall reaction

H2(g) + Cl2(g) W 2H+(aq) + 2Cl-(aq)

has a standard cell potential of 1.3604 V. Determine )fG°for Cl-(aq).

Solution: )rG° = -nFE°cell

)rG° = -2(96,485 C mol-1)(1.3604 V) = -262.52 kJ mol-1

For H2(g), Cl2(g), and H+(aq); )fG° = 0.

Y )rG° = 2)fG°(Cl-(aq))

Y )fG°(Cl-(aq)) = -131.26 kJ mol-1


Standard Hydrogen Electrode

C To create a voltage, we need two half-cells. So wecan't measure individual half-cell potentials.

C Convention: The standard hydrogen electrode isassigned a half-cell potential of zero.

2H+(aq) + 2e- 6 H2(g) E° = 0 volts

Standard states (a=1): [H+] . 1 M, PH2 = 1 bar . 1 atm

C The half-cell potential will differ from zero if H+ and/orH2 are not in their standard states.


Standard Hydrogen Electrode - continued

C H2(g) at one barbubbled over aplatinum electrode.

C Pt acts a catalyst forthe reaction.

C 2H+ + 2e- 6 H2(g)

C Used as a basis forcalculations. Notreally very practical.


Standard Electrode PotentialsThe standard electrode potential for a half-cell is thepotential when all species are in their standard states.

C refers to reduction at the electrode (these days)

C measured relative to a standard hydrogen electrode asthe anode

Example: Cell for measuring E° for Cu2+/Cu.

Anode: H2(g) 6 2H+ + 2e- (oxidation)

Cathode: Cu2+(1 M) + 2e- 6 Cu(s) (reduction)

Cell Diagram: Pt*H2(1 bar)*H+(1 M)2Cu2+(1 M)*Cu(s)

Cell potential is 0.340 V. So E° = 0.340 V for Cu2+/Cu.


Standard Electrode Potentials - continued

Reduction Half-Reaction E° (volts)

F2(g) + 2e- 6 2F-(aq) 2.866

O2(g) + 4H+(aq) + 4e- 6 2H2O(l) 1.229

2H+(aq) + 2e- 6 H2(g) 0.000

Zn2+(aq) + 2e- 6 Zn(s) -0.763

Li+(aq) + e- 6 Li(s) -3.040

F2 is easiest to reduce (largest E°). F- is hardest to oxidize.

Li+ is hardest to reduce. Li is easiest to oxidize.

F2 is best oxidizing agent; Li is best reducing agent.


Using Standard Electrode Potentials

A standard cell potential, E°cell, may be calculated from thestandard electrode potentials for the cathode, E°cathode, andanode, E°anode:

E°cell = E°cathode - E°anode

C The anode potential is subtracted since the potential is forreduction and the anode reaction is oxidation.

C Standard electrode potentials are listed in tables.

C Standard electrode potentials do not depend on how areaction is written since they are related to )rG° per moleof electrons. ()rG° = -nFE°cell)


Standard Electrode Potentials - exampleFind the standard cell potential for the reaction:

Zn(s) + Cl2(g) W Zn2+(aq) + 2Cl-(aq)

Solution: Write half-cell reactions and find E° values.

Oxidation: Zn(s) 6 Zn2+(aq) + 2e-

Reduction: Cl2(g) + 2e- 6 2Cl-(aq)

From Table 21.1:

Zn2+(aq) + 2e- 6 Zn(s) E° = -0.763 V

Cl2(g) + 2e- 6 2Cl-(aq) E° = 1.358 V

E°cell = E°cathode - E°anode = 1.358 - (-0.763) = 2.121 V


Cell Potential and Equilibrium Constant

We have derived the following two equations:

)rG° = -nFE°cell and )rG° = -RTln Keq

Combining these gives

E°cell = lnKeq

Uses of this equation:

C Calculating Keq from standard half-cell potentials (seeexample 21-7 in text).

C Relating E°cell for different reactions.


Cell Potential and Equilibrium - Example

At 298.15 K, the standard reduction potential for O2(g) inacidic solution is 1.229 V:

O2(g) + 4H+(aq) + 4e- 6 2H2O(l) E1° = 1.229 V

Find the standard reduction potential for O2(g) in basicsolution:

O2(g) + 2H2O(l) + 4e- 6 4OH-(aq) E2° = ?

Solution: The second reaction is equal to the first plus

4H2O(l) W 4H+(aq) + 4OH-(aq) Keq = KW4

So K2 = K1KW4, E2° = (RT/nF)ln(K1KW

4), n = 4

Y E2° = E1° + (RT/F)lnKW = 1.229 + 0.02569 ln(1.0×10-14)

E2° = 0.401 V


Dissolving Metals with Acids

Many metals are dissolved by acids with the evolutionof H2(g).

Oxidation: M(s) 6 Mn+(aq) + ne- E° = E°M

Reduction: 2H+(aq) + 2e- 6 H2(g) E° = 0

Overall: M(s) + nH+(aq) 6 Mn+(aq) + (n/2)H2(g) E° = -E°M

Conclusions:

C The more negative the standard reduction potential of themetal ion, the easier the metal is to dissolve.

C Lowering the pH promotes the dissolution of metals.


Dissolving Metals with Acids - exampleDetermine the concentrations of each of the followingmetals that will dissolve at pH = 7.00 and pH = 0.00.

Cu2+(aq) + 2e- 6 Cu(s) E° = 0.340 V

Pb2+(aq) + 2e- 6 Pb(s) E° = -0.125 V

Zn2+(aq) + 2e- 6 Zn(s) E° = -0.763 V

Solution: M(s) + 2H+(aq) 6 M2+(aq) + H2(g)

n = 2 E°cell = -E° T = 298.15 K PH2 . 1 bar

E°cell = (RT/nF)lnKeq Y Keq = exp(-(77.85 V-1)E°)

Y [M2+] . [H+]2 Keq = [H+]2 exp(-(77.85 V-1)E°)


Dissolving Metals with Acids - continued

[M2+] . [H+]2 exp(-(77.85 V-1)E°)

pH = 7 pH = 0 E° (V)

[Cu2+]eq = 3.2×10-26 M 3.2×10-12 M 0.340

[Pb2+]eq = 1.7×10-10 M 1.7×104 M -0.125

[Zn2+]eq = 6.3×1011 M 6.3×1025 M -0.763

C Metals with E° o 0 are difficult to dissolve even instrong acids.

C Metals with E° - 0 will dissolve in strong acids.

C Metals with E° n 0 will dissolve in water.


Enhancing Dissolution of Metals C Concentrated HNO3 will dissolve Cu(s):

Cu2+(aq) + 2e- 6 Cu(s) E° = 0.340 V

NO3- + 4H+ + 4e- 6 NO(g) + H2O(l) E° = 0.956 V

For the overall reaction, E° = (0.956 - 0.340) V = 0.616 V.This is very favorable.

C Gold can be dissolved using aqua regia (1 part HNO3 to3 parts HCl):

Au3+(aq) + 3e- 6 Au(s) E° = 1.52 V

Au3+(aq) + 4Cl-(aq) W [AuCl4]-(aq)


The Nernst EquationWe have shown that

)rG = -nFEcell

But )rG depends on the concentrations of reactants andproducts:

)rG = )rG° + RTlnQ

Therefore, Ecell also depends on concentrations. Combiningthese gives nFEcell = nFE°cell - RTlnQ. So

This is known as the Nernst Equation.


The Nernst Equation - continued

Here: R = 8.314 J mol-1 K-1

F = 96,485 C mol-1 = 96,485 J V-1 mol-1

lnQ = ln(10)×logQ = 2.303 logQ

So Ecell = E°cell - (1.984×10-4 V K-1)(T/n)logQ

If T = 298.15 K (25 °C), then

Ecell = E°cell - (1/n)(0.05916 V)logQ


Using the Nernst EquationFind Ecell at 298 K for the cell

Pt*Fe2+(0.10 M),Fe3+(0.20 M)2Ag+(1.0 M)*Ag(s)

Solution: First find E°cell, then use the Nernst Equation.

Anode: Fe2+ 6 Fe3+ + e- E° = 0.771 V

Cathode: Ag+ + e- 6 Ag(s) E° = 0.800 V

Cell: Fe2+ + Ag+ 6 Fe3+ + Ag(s)

E°cell = E°cathode - E°anode = 0.029 V n = 1

Q = [Fe3+] / [Fe2+][Ag+] = (0.20) / (0.10)(1.0) = 2.0

Ecell = E°cell - (1/n)(0.05916 V)logQ = 0.011 V


Concentration Cells

We can make a cellwith the samereaction occuring atboth electrodes.

Y E°cell = 0

The cell voltage isdue to the differencein concentration.


Determining Ksp

With saturated AgI(aq) at the anode and [Ag+] = 0.100 M atthe cathode, Ecell = 0.417 V. Use this to find Ksp.

Anode: Ag(s) 6 Ag+(aq, sat. AgI) + e-

Cathode: Ag+(aq, 0.1M) + e- 6 Ag(s)

Cell: Ag+(aq, 0.1M) 6 Ag+(aq, sat. AgI)

E°cell = 0. Q = [Ag+]sat,KI / [0.1 M]. n = 1.

The Nernst equation becomes (at 298.15 K):

0.417 V = Ecell = - (0.05916 V) log([Ag+]sat,KI / [0.1 M])

Y [Ag+]sat,KI = [0.1 M] 10-7.049 = 8.94×10-9 M

Y Ksp = [I-][Ag+] = (8.94×10-9)2 = 7.99×10-17


Electrochemistry Basics - Summary

C Electrochemical cells permit us to couple electricalwork to chemical reactions.

C Cell potential and )rG are directly related:

)rG = -nFEcell.

C Standard reduction potentials are tabulated and maybe used to compute E°cell and )rG°.

C The effect of concentration on cell potential is givenby the Nernst equation:

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Electrochemistry - York · PDF fileReview: Chapter 5 questions 21-26. Chapter 21 questions:...

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