1 Zn + Cu 2 + Zn 2 + + Cu. 2 Applied Electrochemistry Voltaic (or galvanic) cells: spontaneous redox reaction electricity (or electrical work)redox

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Zn + Cu2+ Zn2+ + CuZn + Cu2+ Zn2+ + Cu

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Applied ElectrochemistryApplied Electrochemistry

Voltaic (or galvanic) cells: spontaneous Voltaic (or galvanic) cells: spontaneous • redox reaction redox reaction electricity (or electrical work) electricity (or electrical work)

Electrolytic cells: non-spontaneousElectrolytic cells: non-spontaneous• electricity electricity redox redox

To produce electricity, we must direct the To produce electricity, we must direct the electron flow through an external circuit. electron flow through an external circuit. We cannot have direct redox.We cannot have direct redox.

Daniell cellDaniell cell Zn – Cu rxn:Zn – Cu rxn:

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Lemon BatteryLemon Battery

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17.1 Galvanic Cells17.1 Galvanic Cells

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Galvanic CellsGalvanic Cells

To produce electricity, we need:To produce electricity, we need: Isolated half-reactions, using half-cellsIsolated half-reactions, using half-cells Conductive solids (electrodes) connected Conductive solids (electrodes) connected

by external circuitsby external circuits• May consist of a reactant/product or be May consist of a reactant/product or be

an inert substance such as platinum or an inert substance such as platinum or graphitegraphite

• Anode: oxidation half-reactionAnode: oxidation half-reaction• Cathode: reduction half-reactionCathode: reduction half-reaction

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• Externally, the anode has the negative Externally, the anode has the negative charge; internally, it has a positive charge; internally, it has a positive chargecharge

• Anions flow towards the anode; Anions flow towards the anode; cations move away from it and cations move away from it and towards the cathode.towards the cathode.

• Two half-cells must be connected to Two half-cells must be connected to pass ionspass ions• Salt bridge or porous glassSalt bridge or porous glass

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Molecular View of Electrode Processes

Molecular View of Electrode Processes

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17.2 Cell Potentials17.2 Cell Potentials

If the half-reactions are carried out If the half-reactions are carried out separately (but coupled), we find they separately (but coupled), we find they generate an electrical current generate an electrical current characterized by a voltagecharacterized by a voltage

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Cell PotentialsCell Potentials

The voltage produced by a voltaic cell is called the The voltage produced by a voltaic cell is called the cell potential, Ecell potential, Eoo

cellcell (also the reaction potential, (also the reaction potential,

EEoorxnrxn, when the half-reactions are not separated), when the half-reactions are not separated)

Under standard conditions, the voltage is also Under standard conditions, the voltage is also called the standard electromotive force (emf), Ecalled the standard electromotive force (emf), Eoo..

Unit = Volt (V)Unit = Volt (V) Volt = 1 Joule of energy / coulomb of charge Volt = 1 Joule of energy / coulomb of charge

transferred = 1 J/Ctransferred = 1 J/C• J = JouleJ = Joule C = coulombC = coulomb

• 1 mol e- = 96,500 coulombs1 mol e- = 96,500 coulombs

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Reference for EReference for Eoo is the standard hydrogen is the standard hydrogen electrode, using the reaction:electrode, using the reaction:

2H2H++(aq) + 2e(aq) + 2e-- H H22(g) E(g) Eoo = 0 V = 0 J/C = 0 V = 0 J/C

1 M1 M 1 atm 1 atm (std (std conditions)conditions)

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Then get other half-reaction potentials Then get other half-reaction potentials from measured Efrom measured Eoo

cellcell values. values. Zn + 2HZn + 2H++ Zn Zn22++ + H + H22 EEoo

cellcell = 0.76 = 0.76 VV

Zn Zn Zn Zn22++ + 2e + 2e-- EEoooxox = 0.76 V = 0.76 V

2H2H++ + 2e + 2e-- H H22 EEooredred = 0.00 V = 0.00 V

How do these reactionsHow do these reactionsrelate?relate?

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Cell potential values can thus be determined Cell potential values can thus be determined relative to the standard hydrogen electrode, with relative to the standard hydrogen electrode, with EEoo = 0 V = 0 V

EEooredred = cell potential for the reduction half-rxn = cell potential for the reduction half-rxn

EEoooxox = cell potential for the oxidation half-rxn = cell potential for the oxidation half-rxn

Thus, EThus, Eoocellcell = E = Eoo

redred + E + Eoooxox

We can measure EWe can measure Eoocellcell, using the standard , using the standard

reference of 0 V, we can measure Ereference of 0 V, we can measure Eoooxox and E and Eoo

redred for half reactions paired with the Hfor half reactions paired with the H22 half half reaction.reaction.

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Determining Cell Potentials Values

Determining Cell Potentials Values

If we reverse a half-reaction, what happens to If we reverse a half-reaction, what happens to the sign of Ethe sign of Eoo..


ZnZn22++ + 2e + 2e-- Zn Zn EEooredred = ? = ?

Now consider data for ENow consider data for Eoocellcell = 0.63 V for the = 0.63 V for the

following reaction.following reaction.

Zn + PbZn + Pb22++ Zn Zn22++ + Pb + Pb What is the EWhat is the Eoo

redred of of

PbPb22++ + 2e + 2e-- Pb ? Pb ?

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Cell Potential ValuesCell Potential Values


PbPb22++ + 2e + 2e-- Pb Pb EEooredred = ? = ?

Zn + PbZn + Pb22++ Zn Zn22++ + Pb + Pb EEoocellcell = 0.63 V = 0.63 V

EEoocellcell = E = Eoo

redred + E + Eoooxox

0.63 V = E0.63 V = Eooredred + 0.76 V + 0.76 V

EEooredred = 0.63 V - 0.76 V = -0.13 V = 0.63 V - 0.76 V = -0.13 V

Values determined in this way are listed in Table Values determined in this way are listed in Table 17.1 and Appendix 5.517.1 and Appendix 5.5

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Reduction PotentialsReduction Potentials

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Relative Strengths of Oxidizing and Reducing Agents

Relative Strengths of Oxidizing and Reducing Agents

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Cell PotentialCell Potential

Reduction or oxidation values can also be Reduction or oxidation values can also be measured from Emeasured from Eoo

cellcell with other known half- with other known half-

cells.cells. AgCl + eAgCl + e-- Ag + Cl Ag + Cl-- EEredred

oo = 0.22 V = 0.22 V HgHg22ClCl22 + 2e + 2e-- 2Hg + 2Cl 2Hg + 2Cl-- EEredred

oo = 0.2802 V = 0.2802 V Half-cells such as these are used as Half-cells such as these are used as

reference electrodes. The Ag/AgCl reference electrodes. The Ag/AgCl electrode, along with a glass electrode, is electrode, along with a glass electrode, is used in a pH meter.used in a pH meter.

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17.2 Describing Galvanic Cells17.2 Describing Galvanic Cells

What will happen if we place a piece of Zn What will happen if we place a piece of Zn and a piece of Cu in a solution that and a piece of Cu in a solution that contains a mixture of Zncontains a mixture of Zn22++ and Cu and Cu22++??

Two possibilities:Two possibilities:• Zn + CuZn + Cu22++ Zn Zn22++ + Cu + Cu• Cu + ZnCu + Zn22++ Cu Cu22++ + Zn + Zn

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Spontaneous RedoxSpontaneous Redox

Two possible reduction half-reactions:Two possible reduction half-reactions:• ZnZn22++ + 2e + 2e-- Zn Zn EEoo

redred = -0.76 V = -0.76 V• CuCu22++ + 2e + 2e-- Cu Cu EEoo

redred = 0.34 V = 0.34 V Two possible oxidation half-reactions:Two possible oxidation half-reactions:

• Zn Zn Zn Zn22++ + 2e + 2e-- E Eoooxox = 0.76 V = 0.76 V

• Cu Cu Cu Cu22++ + 2e + 2e-- E Eoooxox = -0.34 V = -0.34 V

One way to combine them:One way to combine them: ZnZn22++ + 2e + 2e-- Zn Zn EEoo

redred = -0.76 V = -0.76 V Cu Cu Cu Cu22++ + 2e + 2e-- EEoo

oxox = -0.34 V = -0.34 V———————— ———————————————— ———————— Cu + ZnCu + Zn22++ Cu Cu22++ + Zn + Zn EEoo

rxnrxn = -1.10 V = -1.10 V

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Spontaneous RedoxSpontaneous Redox

Other way to combine them:Other way to combine them: CuCu22++ + 2e + 2e-- Cu Cu EEoo

redred = 0.34 V = 0.34 V Zn Zn Zn Zn22++ + 2e + 2e-- EEoo

oxox = 0.76 V = 0.76 V———————— ——————————————— ——————— Zn + CuZn + Cu22++ Zn Zn22++ + Cu + Cu EEoo

rxnrxn = 1.10 V = 1.10 V Which combination is observed to be Which combination is observed to be

spontaneous.spontaneous.

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Stability in Aqueous SystemsStability in Aqueous Systems

Reaction with WaterReaction with Water Reduce hydronium ion to release hydrogen Reduce hydronium ion to release hydrogen

gas:gas: 2H2H++(aq) + 2e(aq) + 2e-- HH22(g) (g) E°E°redred = 0.000 V = 0.000 V Any substance with EAny substance with Eoo

oxox > 0 will reduce H > 0 will reduce H++ to H to H22

Examples are V, VExamples are V, V22++, Cr, Cr, Cr, Cr22++, Mn, Mn The ions will react, but tend to react only very The ions will react, but tend to react only very

slowly. There seems to be a kinetic factor that slowly. There seems to be a kinetic factor that results in a fast reaction only if Eresults in a fast reaction only if Eoo

rxnrxn > 0.4-0.5 V > 0.4-0.5 V (called an overvoltage).(called an overvoltage).

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Reaction with WaterReaction with Water Oxidize water to release oxygen gas:Oxidize water to release oxygen gas: 2H2H22O(l) O(l) OO22(g) + 4H(g) + 4H++(aq) + 4e(aq) + 4e-- E° E°oxox = -1.23 V = -1.23 V Any EAny Eoo

redred > 1.23 V will result in production of O > 1.23 V will result in production of O22. . Generally need EGenerally need Eoo

rxnrxn > 0.4-0.5 V for fast reaction. > 0.4-0.5 V for fast reaction. Examples are CrExamples are Cr22OO77

22-- (very slow), (very slow),

MnOMnO4422-- (disproportionates faster), (disproportionates faster),

MnMn33++ (disproportionates faster) (disproportionates faster)

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Oxidation by OOxidation by O22 in Air in Air OO22(g) + 4H(g) + 4H++(aq) + 4e(aq) + 4e-- 2H2H22O(l) E°O(l) E°redred = 1.23 V = 1.23 V Any EAny Eoo

oxox > -1.23 V will result in oxidation by air. Many > -1.23 V will result in oxidation by air. Many substances fall into this category (Esubstances fall into this category (Eoo

rxnrxn > 0.4-0.5 V for fast > 0.4-0.5 V for fast reaction).reaction).

• VV CrCr MnMn• VV22++ CrCr22++ not Mnnot Mn22++

• VV33++ not Crnot Cr33++ MnMn33++

• VOVO22++ (very slow) (very slow) not MnOnot MnO22

• MnOMnO4422--

(disproportionates faster)(disproportionates faster)

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Eocell and SpontaneityEocell and Spontaneity

We have seen three criteria for spontaneity:We have seen three criteria for spontaneity:• EEoo > 0 > 0• GGoo < 0 < 0 1 V = 1 J/C, so 1 J = 1 C x 1 V1 V = 1 J/C, so 1 J = 1 C x 1 V

These criteria are related:These criteria are related:• K >> 1K >> 1• GGoo = - RT lnK = - RT lnK• GGoo = - n = - n F F EEoo, , G = - nG = - nFFEEwhere n = number of ewhere n = number of e-- transferred and transferred andFF = Faraday constant (charge on 1 mole e = Faraday constant (charge on 1 mole e--))1 1 FF = 96,500 coul/mol e = 96,500 coul/mol e-- = 96,500 J/V mol e = 96,500 J/V mol e--

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ThermodynamicsThermodynamics

These relationships work for half-These relationships work for half-reactions or complete redox reactions.reactions or complete redox reactions.

Zn + CuZn + Cu22++ Zn Zn22++ + Cu + Cu EEoo = 1.10 = 1.10 n = n =

22 GGoo = -2mol e = -2mol e-- x 96500 J/V mol e x 96500 J/V mol e-- x 1.10 V x 1.10 V GGoo = -212,300 J = -212.3 kJ = -212,300 J = -212.3 kJ GGoo depends on the number of moles, but depends on the number of moles, but

EEoo does not does not

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Voltage and MolesVoltage and Moles

Note that different size alkaline cells all Note that different size alkaline cells all deliver the same voltage, in spite of deliver the same voltage, in spite of different number of moles of reactants.different number of moles of reactants.

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We can add EWe can add Eoooxox to E to Eoo

redred to give E to give Eoocellcell or E or Eoo

rxnrxn in the in the same way that we can add half-reactions to give an same way that we can add half-reactions to give an overall reaction.overall reaction.

Fe Fe Fe Fe22++ + 2e + 2e-- EEoooxox = +0.44 V = +0.44 V

GGoooxox = - 2 x 96500 x 0.44 = -84900 J = - 2 x 96500 x 0.44 = -84900 J

ClCl22 + 2e + 2e-- 2Cl 2Cl-- EEooredred = 1.36 V = 1.36 V

GGooredred = - 2 x 96500 x 1.36 = -262500 J = - 2 x 96500 x 1.36 = -262500 J

Fe + ClFe + Cl22 Fe Fe22++ + 2Cl + 2Cl-- EEoorxnrxn = 1.80 V = 1.80 V

GGoo = - 2 x 96500 x 1.80 = -347400 J = - 2 x 96500 x 1.80 = -347400 J GGoo

rxnrxn = = GGoooxox + + GGoo

redred = -84900 + -262500 = -347400 J= -84900 + -262500 = -347400 J

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From Chapter 16, we know that From Chapter 16, we know that GGoo values are additive when we add reactions.values are additive when we add reactions.

EEoos are additive when we add half-s are additive when we add half-reactions to give a complete reaction reactions to give a complete reaction because the value of n is the same for the because the value of n is the same for the half-reactions and the complete reaction. half-reactions and the complete reaction.

EEoos are not additive when adding two half-s are not additive when adding two half-reactions to give a third half-reaction reactions to give a third half-reaction because the value of n is not constant.because the value of n is not constant.

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We can add We can add GGoo under all circumstances: under all circumstances:

GGoo33 = = GGoo

11 + + GGoo22

-n-n33FFEEoo33 = -n = -n11FFEEoo

11 - n - n22FFEEoo22

nn33EEoo33 = n = n11EEoo

11 + n + n22EEoo22

EEoo33 = (n = (n11EEoo

11 + n + n22EEoo22)/n)/n33

V V V V22++ + 2e + 2e-- EEoo11 = 1.20 V = 1.20 V

VV22++ VV33++ + e + e-- EEoo22 = 0.26 V = 0.26 V

V V VV33++ + 3e + 3e-- EEoo33 < 1.20 + 0.26 < 1.20 + 0.26

EEoo33 = (2 x 1.20 + 1 x 0.26)/3 = 0.887 V = (2 x 1.20 + 1 x 0.26)/3 = 0.887 V

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We can calculate KWe can calculate Keqeq from E from Eoo::

GGoo = - n = - nFFEEoo = - RT ln K = - RT ln K EEoo = (RT/n = (RT/nFF) ln K = 2.303 (RT/n) ln K = 2.303 (RT/nFF) log K) log K At 25At 25ooC, 2.303 RT/C, 2.303 RT/FF = 0.05916 = 0.05916 EEoo = (0.05916/n) log K at 25 = (0.05916/n) log K at 25ooCC Thus, we can measure EThus, we can measure Eoo for a redox for a redox

reaction and then calculate the reaction and then calculate the equilibrium constant for that reaction.equilibrium constant for that reaction.

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17.4 Effect of Concentration on Cell EMF

17.4 Effect of Concentration on Cell EMF

So far, we have been using standard state So far, we have been using standard state conditions, but we don’t always have 1 M conditions, but we don’t always have 1 M solutions. We can correct Esolutions. We can correct Eoo to E by using to E by using the Nernst equation.the Nernst equation.

G = G = GGoo + RT ln Q + RT ln Q But, But, G = - nG = - nFFE and E and GGoo = -n = -nFFEEoo, so, so - n- nFFE = -nE = -nFFEEoo + 2.303 RT log Q + 2.303 RT log Q E = EE = Eoo - (2.303 RT/n - (2.303 RT/nFF) log Q) log Q At 25At 25ooC, E = EC, E = Eoo - (0.05916/n) log Q - (0.05916/n) log Q

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Nernst EquationNernst Equation

At 25At 25ooC, E = EC, E = Eoo - (0.05916/n) log Q - (0.05916/n) log Q E = EE = Eoo if Q = 1 if Q = 1 When the system reaches equilibrium, Q = K, When the system reaches equilibrium, Q = K,

and E = 0, because Eand E = 0, because Eoo = (0.05916/n) log K, = (0.05916/n) log K, and the cell has “run down”.and the cell has “run down”.

Consider the Zn/CuConsider the Zn/Cu22++ reaction if more reaction if more CuCu22++ is added to the cell. The voltage is added to the cell. The voltage becomes greater than 1.10 V.becomes greater than 1.10 V.

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What is E of the Zn/CuWhat is E of the Zn/Cu22++ reaction if [Cu reaction if [Cu22++] ] = 0.010 M and [Zn= 0.010 M and [Zn22++] =1.99 M? Note that ] =1.99 M? Note that this corresponds to starting with standard this corresponds to starting with standard conditions and changing to 99% conditions and changing to 99% completion of reaction. Ecompletion of reaction. Eoo = 1.10 V (with = 1.10 V (with [Cu[Cu22++] = [Zn] = [Zn22++] = 1.00 M)] = 1.00 M)

Zn + CuZn + Cu22++ Zn Zn22++ + Cu + Cu For this reaction, n = 2.For this reaction, n = 2. Q = [ZnQ = [Zn22++]/[Cu]/[Cu22++]]

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E = EE = Eoo - (0.05916/n) log Q - (0.05916/n) log Q E = 1.10 V - (0.05916/2) log (1.99/0.010)E = 1.10 V - (0.05916/2) log (1.99/0.010) E = 1.10 V - (0.05916/2) log 199E = 1.10 V - (0.05916/2) log 199 E = 1.10 - 0.068 = 1.03 VE = 1.10 - 0.068 = 1.03 V For 99.9% reaction (1.999 M ZnFor 99.9% reaction (1.999 M Zn22++, 0.001 , 0.001

M CuM Cu22++), E = 1.10 - 0.098 = 1.00 V), E = 1.10 - 0.098 = 1.00 V For 99.99% reaction, 1.9999 M ZnFor 99.99% reaction, 1.9999 M Zn22++, ,

0.0001 M Cu0.0001 M Cu22++), E = 1.10 - 0.127 = 0.97 V), E = 1.10 - 0.127 = 0.97 V

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Concentration CellsConcentration Cells

We can generate a voltage with a cell that We can generate a voltage with a cell that contains the same materials in the cathode contains the same materials in the cathode and anode compartments, but at different and anode compartments, but at different concentrations.concentrations.

Documents

1 Zn + Cu 2 + Zn 2 + + Cu. 2 Applied Electrochemistry Voltaic (or galvanic) cells: spontaneous redox reaction electricity (or electrical work)redox