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Ch 11 El t h i tCh 11 Electrochemistry

Copyright©2000 by Houghton Mifflin Company. All rights reserved.

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Electrochemistry is best defined as the study of theElectrochemistry is best defined as the study of the interchange of chemical & electrical energy. It is primarily concerned with two processes that involve oxidation-reduction reactions : (1) the generation of( ) g fan electric current from a chemical reaction, & (2) th f t t d h i l hthe use of a current to produce chemical change.


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11.1 Galvanic cells

Oxidation-reduction (redox) reaction : involves a transfer of electrons from the reducing agent totransfer of electrons from the reducing agent to the oxidizing agent.

Oxidation : loss of electronsReduction : gain of electrons


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Half-Reactions

The overall reaction is split into two half-reactions,one involving oxidation & one reduction.g

8H+(aq) + MnO4−(aq) + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)

Reduction: 8H+ + MnO4− + 5e− → Mn2+ + 4H2OOxidation: 5Fe2+ → 5Fe3+ + 5e−


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Figure 11.1: Schematic of a method to separate the oxidizing and reducing agents of a redoxthe oxidizing and reducing agents of a redox

reaction.

Copyright©2000 by Houghton Mifflin Company. All rights reserved. Copyright © Houghton Mifflin Company. All rights reserved. 17b–5

Figure 11.2: Galvanic cells can contain a lt b id i ( ) di k tisalt bridge in (a) or a porous-disk connection

as in (b).


Galvanic Cell

A device in which chemical energy is changed to electrical energy. A galvanic cell uses a spontane-gy g pous redox reaction to produce a current that can be used to do work.be used to do work.

Anode and Cathode

Oxidation occurs at the anode.Reduction occurs at the cathode.


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Figure 11.3: An electrochemical process involves electron transfer at the interfaceinvolves electron transfer at the interface between the electron and the solution.


Cell potentialCell potential

Cell potential or electromotive force (emf 電動勢) :Cell potential or electromotive force (emf, 電動勢) :The “pull” or driving force on the electrons is called the cell potential (Εcell).


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Figure 11.4: Digital voltmeters draw only a li ibl d inegligible current and are convenient to use.

VoltmeterVoltmeterPotentiometerDigital voltmeterDigital voltmeter


11.2 Standard Reduction PotentialsThe cell potential can be obtained by summing the half-cell potentials.

2H+(aq) + Zn(s) → Zn2+(aq) + H2(g)Oxidation : Zn → Zn2+ + 2e-, [Zn2+] = 1MR d ti 2H+ + 2 - → H [H+] 1MReduction : 2H+ + 2e- → H2, [H+] = 1M

[2H+ + 2e- → H2 (standard hydrogen electrode / Pt)][2H + 2e → H2 (standard hydrogen electrode / Pt)]


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Although we can measure the total potential of this cell (0 76 V) there is no way to measure thecell (0.76 V), there is no way to measure the potentials of the individual electrodes. p


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[H+] = 1 M & PH = 1 atm as having a potential[H ] 1 M & PH2 1 atm as having a potential exactly 0 V, then the reaction

Zn → Zn2+ + 2e- will have a potential of 0.76 V.

E°cell = E°H+→H2 + E°Zn→Zn2+ (° : standard state)

0.76 V 0 V 0.76 V


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Figure 11.5: (a) Galvanic cell involving the reactions Zn → Zn2+ + 2e-

(at the anode) and 2H+ + 2e- → H2 (at the cathode) has a potential of 0 6 (b) h d d h d l d h i d0.76 V. (b) The standard hydrogen electrode where H2(g) at 1 atm is passed

over a platinum electrode in contact with 1 M H+ ions.


Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)A d Z Z 2+ 2Anode : Zn → Zn2+ + 2e-

Cathode : Cu2+ + 2e- → CuCathode : Cu + 2e → CuE°cell = E°Zn→Zn2+ + E°Cu2+→Cu1.10 V = 0.76 V + 0.34 V

2H+ + 2e- → H2 (E°H+→H2 = 0 has been universally accepted the value of 0)accepted the value of 0)


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Figure 11.6: Galvanic cell involving the half reactions Zn → Zn2+ + 2e- (anode) and Cu2+( )

+ 2e- → Cu (cathode).


E° values corresponding to these half-reactions are called standard reduction potentials Combiningcalled standard reduction potentials. Combining two half-reactions to obtain a balanced oxidation-

d ti ti ft i treduction reactions often requires two manipulations :

1. One of the reduction half-reactions must be reversed (a redox reaction contains both oxidation & reduction reactions.) The net potential of the ) p fcell will be the difference between the two.


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Since the reduction process occurs at the cathode & th id ti t th d& the oxidation process occurs at the anode, we can write

E°cell = E°(cathode) - E°(anode)

“change the sign & add”


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2. Since the # of electrons lost must equal the #qgained, the half-reactions must be multiplied byintegers as necessary to achieve electron balanceintegers as necessary to achieve electron balance.However, the value of E° is not changed when

h lf i i l i li d b i (E°a half-reaction is multiplied by an integer (E°is an intensive property).


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Consider a galvanic cell based on the reaction

Fe3+(aq) + Cu(s) → Cu2+(aq) + Fe2+(aq)

Fe3+(aq) + e- → Fe2+(aq) E° = 0.77 VCu2+(aq) + 2e- → Cu E° = 0.34 V(aq)

E°cell = E°(cathode) - E°(anode)cell ( ) ( )= 0.77 – 0.34 = 0.43 V


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Fe2+ + 2e- → Fe E° = -0.44 VMnO4- + 5e- + 8H+ → Mn2+ + 4H2O E° = 1 51 VMnO4 + 5e + 8H → Mn + 4H2O E 1.51 V

Fe → Fe2+ + 2e- anode reactionMnO4- + 5e- + 8H+ → Mn2+ + 4H2O cathode reaction

E°cell = E°(cathode) - E°(anode)= 1.51 – (-0.44) = 1.95 V( )

A cell will always run spontaneously in the directiony p ythat produces a positive cell potential.


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Fe → Fe2+ + 2e- (x -5)MnO4- + 5e- + 8H+ → Mn2+ + 4H2O (x 2)

2MnO4-(aq) + 5Fe(s) + 16H+(aq) →5F 2+ 2M 2+ 8H O5Fe2+(aq) + 2Mn2+(aq) + 8H2O(l)


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Figure 11.7: Schematic of a galvanic cell based on specific half-reactions.


Summary

1. The cell potential (always positive for a galvaniccell) & the balanced reactioncell) & the balanced reaction.

2. The direction of electron flow, obtained by , yinspecting the half-reactions & using the directionsthat give a positive E°that give a positive E cell.


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3 Designation of the anode & cathode3. Designation of the anode & cathode

4 The nature of each electrode & the ions present4. The nature of each electrode & the ions present in each compartment (a chemically inert conductor is required if none of the substance participating in the half reaction is a conducting solid - Pt)the half reaction is a conducting solid Pt).


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Ex 11 1 Describe completely the galvanic cell on theEx. 11.1 Describe completely the galvanic cell on the following half-reactions under standard conditions :

Ag+ + e- → Ag E° = 0.80 V (1)F 3+ + F 2+ E° 0 77 V (2)Fe3+ + e- → Fe2+ E° = 0.77 V (2)


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Since a positive E° is required, reaction (2) must be run in reverse :

A + A E°( h d ) 0 80 V (1)Ag+ + e- → Ag E°(cathode) = 0.80 V (1)Fe2+ → Fe3+ + e- -E°(anode) = -0.77 V (2)

Cell reaction : Ag+(aq) + Fe2+(aq) →Fe3+(aq) + Ag(s) E° = 0.03 V


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Figure 11.8: Schematic diagram for the galvanic cell based on the half-reactions Ag+ + e-→ Ag; g g

Fe2+ → Fe3+ + e-


11.3 Cell potential, electrical work, & free energy

The goal : to explore the relationship between thermodynamics & electrochemistrythermodynamics & electrochemistry.

The driving force (emf) is defined in terms of potential difference (in volts) between two points in the circuit.

emf = potential difference (V) = work (J) / charge (C)= work (J) / charge (C)


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E = -W/q (w : work, q : charge) ⇒ -W = qEFrom this equation, we can see the max. work in acell is obtained at the max cell potential :cell is obtained at the max. cell potential :

-W = qE or W = -qEWmax qEmax or Wmax qEmaxIn any real, spontaneous process some energy is always wasted - the actual work realized is alwaysless than the calculated maxless than the calculated max.


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The above principle is a consequence of the factThe above principle is a consequence of the factthe entropy of the universe must increase in any spontaneous process - second law of thermo-dynamics (a max work could be only realized fordynamics (a max work could be only realized forthe hypothetical reversible process).


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In Ch10 (P. 141):

Thus q which is pathway dependent variesThus qP, which is pathway-dependent, varies between ΔH (when wuseful = 0) & TΔS (wuseful =

)wusefulmax).

TΔS represents the minimum heat flow that mustTΔS represents the minimum heat flow that must accompany the process under consideration. That is, TΔS h i i h bTΔS represents the minimum energy that must be “wasted” through heat flow as the process occurs.


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To evaluate the efficiency of a real process basedTo evaluate the efficiency of a real process basedon the cell reaction.S t i l i ll h t ti lSuppose a certain galvanic cell has a max potentialof 2.50 V. In a particular experiment 1.33 moles ofl h h hi ll lelectrons pass through this cell at an average actual

potential of 2.10 V. The actual work done is

W = -qEq = nF = 1.33 mol e- x 96,485 C/mol e-

W = -qE = -(1 33 mole- x 96 485 C/mole-) x (2 10 J/C)W = -qE = -(1.33 mole x 96,485 C/mole ) x (2.10 J/C)= -2.69 x 105 J


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W = qE = (1 33 mol e- x 96 485 C/mol e-)Wmax = -qEmax = - (1.33 mol e x 96,485 C/mol e )x (2.50 J/C)

= -3.21 x 105 J

/ 100% ( 2 69 105 / 3 21 105 ) 100%W/Wmax x 100% = (-2.69x105 J/-3.21x105 J)x100%= 83.8 %83.8 %


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To relate the potential of a galvanic cell to free energy:W ΔG ( t t t T & P th h i fWmax = ΔG (at constant T & P, the change in freeenergy equals the max useful work)

For a galvanic cellWmax = -qEmax = ΔG (q = nF)ΔG = -qE = -nFE (ignore max)ΔG = -qEmax = -nFEmax (ignore max)

For standard state : ΔG° = -nFE°


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In Ch10 (P. 134):

Q i i lQuantitatively

The change in free energy is important quantita-The change in free energy is important quantita-tively because it can tell us how much work can be done through a given process The maximumbe done through a given process. The maximum possible useful work obtained from a process at

t t t & i l t th hconstant temp & pressure is equal to the change in free energy :

wusefulmax = ΔG


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The equation states the max cell potential is directlyl d h f diff b hrelated to the free energy difference between the

reactants & products in the cell.

It provides an experimental means of obtaining ΔGf i d l fi l i llfor a reaction, and also confirms a galvanic cell runs in the direction that a positive Ecell value corresponding to a negative ΔG value, which is the condition for spontaneity.condition for spontaneity.


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Ex 11.2 Calculate ΔG° for the reactionCu2+( ) + Fe( ) → Cu( ) + Fe2+( )Cu (aq) + Fe(s) → Cu(s) + Fe (aq)Is this reaction spontaneous ?

Cu2+ + 2e- → Cu E°(cathode) = 0.34 VFe → Fe2+ + 2e- E°(anode) = 0.44 V

Cu2+ + Fe → Fe2+ + Cu E° = 0 78 VCu2 + Fe → Fe2 + Cu E cell = 0.78 V

ΔG° = -nFE° = -(2mole-)(96,485 C/mole-)(0.78 J/C)( )( , )( )= -1.5 x 105 J (spontaneous)


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Ex 11 3 Predict whether 1M HNO will dissolveEx 11.3 Predict whether 1M HNO3 will dissolvegold metal to form a 1 M Au3+ solution.

NO3- + 4H+ + 3e- → NO + 2H2OE°( th d ) 0 96 VE°(cathode) = 0.96 V

Au → Au3+ + 3e- -E°(anode) = -1 50 VAu → Au + 3e E (anode) 1.50 V


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The sum of these half-reactions gives the requiredg qreaction :

3Au(s) + NO3-(aq) + 4H+(aq) →Au3+(aq) + NO(g) + 2H2O(l)E° = 0 96 V 1 50 V = 0 54 VE cell = 0.96 V - 1.50 V = -0.54 V

Since E° is negative, the process will not occur underg pstandard conditions.


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11.4 Dependence of the cell potential on conc.

Cu(s) + 2Ce4+(aq) → Cu2+(aq) + 2Ce3+(aq)All conc. at 1.0 M & cell potential of 1.36 V. What will the cell potential be if [Ce4+] is greaterWhat will the cell potential be if [Ce4+] is greaterthan 1.0 M ?

Le Châtelier’s principle : The cell potential willincrease.


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Ex. 11.4 For the reaction 2 32Al(s) + 3Mn2+(aq) → 2Al3+(aq) + 3Mn(s) E°cell = 0.48 V

P di t h th E ill b l ll thPredict whether Ecell will be larger or smaller than E°cell for the following cases.cell

a. [Al3+] = 2.0 M, [Mn2+] = 1.0 M (< 0.48 V)b. [Al3+] = 1.0 M, [Mn2+] = 3.0 M (> 0.48 V)


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The Nernst equationThe dependence of the cell potential on concThe dependence of the cell potential on concresults directly from the dependence of free energy on conc.

ΔG = ΔG° + RTln(Q)ΔG = ΔG + RTln(Q)Since ΔG = -nFE & ΔG° = -nFE°-nFE = -nFE° + RTln(Q)

E = E° - (RT/nF)ln(Q)


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E = E° - (0.0591/n)log(Q)

Using this relationship we can calculate the potentialf ll ( d d ) i hi h ll fof a cell (non-standard state) in which some or all of

the components are not in their standard states.the components are not in their standard states.


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I.e., E°cell is 0.48 V for the galvanic cell based onthe reactionthe reaction

2Al(s) + 3Mn2+(aq) → 2Al3+(aq) + 3Mn(s)(s) (aq) (aq) (s)Consider a cell in which[Mn2+] = 0.50 M & [Al3+] = 1.50 M

E E° (0 0591/ )l (Q)Ecell = E°cell – (0.0591/n)log(Q)

(E° = 0.48 V & Q = 1.502/0.503 = 18)(E 0.48 V & Q 1.50 /0.50 18)


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2Al → 2Al3+ + 6e- & 3Mn2+ + 6e- → 3Mn

We know that n=6

E = 0 48 (0 0591/6)log(18) = 0 47 VEcell = 0.48 - (0.0591/6)log(18) = 0.47 V

Memo :Memo :

1. 0.47 V < 0.48 V (Le Châtelier’s principle)( p p )2. The potential calculated from the Nernst eq. isthe max potential before any current flow occursthe max potential before any current flow occurs.


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As the cell discharges & current flows from anodeto cathode the conc changes; as a result Eto cathode, the conc changes; as a result, Ecellchanges. In fact, the cell will spontaneously dis-h til it h ilib i t hi h i tcharge until it reaches equilibrium, at which point

Q = K & E = 0 (no driving force at all)Q K & Ecell 0 (no driving force at all)A “dead” battery is one in which the cell reaction yhas reached equilibrium; there is no longer any chemical driving force to push e- through the wirechemical driving force to push e through the wire.


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I th d t ilib i th t iIn other words, at equilibrium the components in the two cell compartments have the same free p fenergy; that is, ΔG = 0 for the cell reaction at the equilibrium concentration.


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Ex. 11.5 Describe the cell based on the following h lf tihalf-reactions :

VO2+ + 2H+ + e- → VO2+ + H2O E° = 1 00 VVO2 + 2H + e → VO + H2O E 1.00 VZn2+ + 2e- → Zn E° = -0.76 V

Where T = 25 °C [VO2+] = 2 0 M [VO2+] = 1 0 x 10-2 M[VO2 ] 2.0 M [VO ] 1.0 x 10 M[H+] = 0.50 M [Zn2+] = 1.0 x 10-1 M


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VO2+ + 2H+ + e- → VO2+ + H2O x 2 E° = 1.00 VZn2+ + 2e- → Zn x -1 E° = -0.76 V

Cell reaction :Cell reaction :

2VO + + 4H+ + Zn →2VO2 (aq) + 4H (aq) + Zn(s) →2VO2+(aq) + 2H2O(l) + Zn2+(aq) E°cell = 1.76 V ( q) ( ) ( q)

E = E° - (0.0591/n)log(Q)


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E = E° - (0.0591/n)log(Q)= 1 76 - (0 0591/2)log([Zn2+][VO2+]2/[VO +]2[H+]4)= 1.76 - (0.0591/2)log([Zn ][VO ] /[VO2 ] [H ] )= 1.76 - (0.0591/2)log(4 x 10-5)= 1.76 + 0.13 = 1.89 V


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Figure 11.9: Schematic diagram of the cell described in Sample Exercise 11 7described in Sample Exercise 11.7.


Ion-selective electrodes

Because the cell potential is sensitive to the f th t t & d t i l d i thconc of the reactants & products involved in the

cell reaction, measured potentials can be used to determine the conc of an ion. A pH meter is a familiar example of an instrument that measures pconc from an observed potential.


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Th H t h th i tThe pH meter has three main components : a standard electrode of known potential, a specialglass electrode that changes potential depending on the conc of H+ & a potentiometer thaton the conc of H , & a potentiometer that measures the potential between two electrodes.


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Th l l t d t i f l tiThe glass electrode contains a reference solution of dilute hydrochloric acid in contact with a thin glass membrane. The electrical potential of the glass electrode depends on the difference in [H+]glass electrode depends on the difference in [H ](conc difference). Thus, the electrical potential varies with the pH of the solution.


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Figure 11.10: A glass electrode


Source: American Color

Glass electrodes can be made sensitive to ionsGlass electrodes can be made sensitive to ions, i.e. Na+, K+, or NH4+ by changing the composition of the glass. Other ions can be detected if an appropriate crystalline solids replaces the glassappropriate crystalline solids replaces the glass membrane (i.e., LaF3, Ag2S…).


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Calculations of equilibrium constants for redoxreactions

F ll t ilib i E 0 & Q KFor a cell at equilibrium, Ecell = 0 & Q = K

E = E° (0 0591/n)log(Q)E = E° - (0.0591/n)log(Q)0 = E° - (0.0591/n)log(K)g

log(K) = (nE°/0.0591) at 25°C


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Ex. 11.6 For the oxidation-reduction reaction

S O 2 C 2+ C 3+ S O 2S4O62-(aq) + Cr2+(aq) → Cr3+(aq) + S2O32-(aq)The appropriate half-reactions areThe appropriate half reactions are

S4O62- + 2e- → 2S2O32- E° = 0.17 V (1) Cr3+ + e- → Cr2+ E° = -0.50 V (2)

Balance the redox reaction & calculate E° & K (at 25 °C)( )


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(1) - 2 x (2)(1) 2 x (2)

S4O62-(aq) + 2 Cr2+(aq) → 2Cr3+(aq) + 2 S2O32-4 6 (aq) (aq) (aq) 2 3(aq)E° = 0 67 VE 0.67 V

log(K) = nE°/0.0591 = 2(0.67)/0.0591 = 22.6g( ) ( )K = 4 x 1022


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Concentration cells (濃度/差電池)Concentration cells (濃度/差電池)

B ll t ti l d dBecause cell potentials depend on conc, we canconstruct galvanic cells where both compartments co s uc g v c ce s w e e bo co p e scontain the same components but at different conc.


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I e in figure 11 11 both compartments containI.e., in figure 11.11 both compartments contain aqueous AgNO3, but with different molarities.

Ag+ + e- → Ag E° = 0.80 V

If the cell had 1 M Ag+ in both compartments,

E° 0 80 0 80 0E°cell = 0.80 – 0.80 = 0


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To calculate the potential at 25°C (Ag+ : 1M / 0.1M)E = E° - (0 0591/n)log(Q) (n =1 & E° = 0) E° refersE E - (0.0591/n)log(Q) (n 1 & E 0). E refersto standard conditions, like figure 11.11, except that[Ag+] = 1 M in both compartments. Such a cell hasa potential of zero. That is, E° = 0.a potential of zero. That is, E 0.

Ag+ (1M) → Ag+ (0.1M)g g

E = E° - (0.0591/n)log(Q) ( / )l ( / )= 0 - (0.0591/1)log(0.10/1.0) = 0.0591 V


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A cell in which both compartments have the sameA cell in which both compartments have the samecomponents, but at different conc, is called aconcentration cell. The difference in conc is the

l f (d i i f ) h d llonly factor (driving force) that produces a cellpotential, & the voltages are typically small.potential, & the voltages are typically small.


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Figure 11.11: A concentration cell that contains a silver electrode and aqueous silver q

nitrate in both compartments.


Ex. 11.7 Determine the direction of electron flow, d i h d & h d & l l hdesignate the anode & cathode, & calculate the potential at 25°C for the cell represented in Fig. 11.12.

(1) A d (l f ) h d ( i h )(1) Anode (left) - cathode (right)

(2) E = E° - (0 0591/n)log(Q)(2) E E - (0.0591/n)log(Q) = 0 - (0.0591/2)log(0.01/0.10) = 0.0296 V


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Figure 11.12: Concentration cell containing iron electrodes and different concentrationsiron electrodes and different concentrations

of Fe2+ ion in the two compartments.


Ex. 11.8 A silver concentration cell similar to the h i Fi 11 11 i 25°C i h 1 0one shown in Fig. 11.11 is set up at 25°C with 1.0

M AgNO3 in the left compartment & 1.0 M NaCl along with excess AgCl(s) in the right compartment. The measured cell potential is 0.58 V. Calculate theThe measured cell potential is 0.58 V. Calculate the Ksp value for AgCl at 25 °C.


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In this case at 25°C

E = 0.58 V = 0 - (0.0591/1)log([Ag+]/1.0)

[Ag+] represents the equilibrium conc of Ag+ in the[Ag ] represents the equilibrium conc of Ag in the compartment containing 1.0 M NaCl & AgCl(s).

log[Ag+] = - 0.58/0.0591 = -9.80 [Ag+] = 1 6 x 10-10 M[Ag ] 1.6 x 10 M

Ksp = [Ag+][Cl-] = (1.6 x 10-10)(1.0) = 1.6 x 10-10


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11.5 Battery

A battery is a galvanic cell or, more commonly, a group of galvanic cells connected in series, where the potential of the individual cells add towhere the potential of the individual cells add to give the total battery potential, i.e., lead storageb tt d ll b tt i & f l ll (燃燒電池)battery, dry cell batteries, & fuel cells (燃燒電池).


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A fuel cell is a galvanic cell in which the reactantsare continuously supplied.

CH + 2O CO + 2H O +CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) + energy

In a fuel cell designed to use this reaction theIn a fuel cell designed to use this reaction, the energy is used to produce an electric current; theelectrons flow from the reducing agent (CH4) tothe oxidizing agent (O2) through a conductor. g g 2 g


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US space program (shuttle missions)US space program (shuttle missions)

2H2(g) + O2(g) → 2H2O(l)2(g) 2(g) 2 (l)Anode reaction : 2H2 + 4 OH- → 4H2O + 4e-

Cathode reaction : 4e- + O2 + H2O → 4OH-


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f h bl i d i h hBecause of the problems associated with the storageof H2 (in addition to high cost), scientists are alsoof H2 (in addition to high cost), scientists are alsoworking to develop fuel cells that use gasoline ormethanol to generate the H2 needed for the fuel cell.


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Figure 11 16:Figure 11.16: Schematic of th H O f lthe H2-O2 fuel cell


11.6 CorrosionCorrosion can be viewed as the process of returningCorrosion can be viewed as the process of returningmetals to its natural state (ores).

Corrosion of ironIn the anodic regions : Fe → Fe2+ + 2e-In the cathodic regions : O2 + 2H2O + 4e- → 4OH-In the cathodic regions : O2 + 2H2O + 4e → 4OH4Fe2+(aq) + O2(g) + (4+2n)H2O(l) →

2Fe2O3.nH2O(s) + 8H+(aq)


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Figure 11.17: Electrochemical corrosion of iron


Prevention of corrosionTh i f t ti i th li tiThe primary means of protection is the application of a coating, most commonly paint or metal plating, to protect the metal from oxygen & moisture.

C tiCoating :

Fe → Fe2+ + 2e- -E° = 0 44 VFe → Fe + 2e E 0.44 VZn → Zn2+ + 2e- -E° = 0.76V


79

The reaction with the most positive standard potential has the greatest thermodynamic tendency to occur. Thus zinc acts as a “sacrificial” coating f gon steel.

All i i l d i S i lAlloying is also used to prevent corrosion. Stainless steel contains chromium & nickel, both of which form oxide coatings that change steel’s reduction potential to one characteristic of the noble metals.potential to one characteristic of the noble metals.


80

Cathodic protection is a method most often used to protect steel in buried fuel tanks & pipelines. An active metal, such as magnesium (a better reducing , g ( gagent than iron), is connected by a wire to the pipeline or tank to be protected As oxidationpipeline or tank to be protected. As oxidation occurs, the Mg anode dissolves, & so it must be

l d i di llreplaced periodically.


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Figure 11.18: Cathodic protection of an underground pipe


11.7 Electrolysisy

A galvanic cell produces current when an oxidation-reduction reaction proceeds spontane-ously A similar apparatus, an electrolytic cell,ously. A similar apparatus, an electrolytic cell, uses electrical energy to produce chemical hchange.


83

The process of electrolysis involves forcing a current through a cell to produce a chemical g pchange for which the cell potential is negative; that is electrical work causes a nonthat is, electrical work causes a non-spontaneous chemical reaction to occur.


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Figure 11 19: (a) A standard galvanic cellFigure 11.19: (a) A standard galvanic cell and (b) A standard electrolytic cell


Stoichiometry of electrolytic processes – how much chemical change occurs with the flow of a given current for a specified time ?p

Suppose we wish to determine the mass of copperthat is plated out when a current of 10.0 A is passedfor 30.0 min through a solution containing Cu2+ (Q g g (Q= I x t).


86

In this case each Cu2+ ion requires two electrons to b t f t lbecome an atom of copper metal : Cu2+(aq) + 2e- → Cu(s)Cu (aq) 2e → Cu(s)This reduction process occurs at the cathode of the l l i llelectrolytic cell.

current & time → quantity of charge (in coulombs)current & time → quantity of charge (in coulombs)→moles of electrons → moles of copper (analyte)→ grams of copper (analyte)→ grams of copper (analyte)


87

Electrolysis of water

Anode : 2H2O → O2 + 4H+ +4e- -E° = -1.23 VCathode : 4H2O + 4e- → 2H2 + 4OH- E° = -0 83 VCathode : 4H2O + 4e → 2H2 + 4OH E 0.83 V

Net reaction : 2H2O → 2H2 + O2 E° = -2.06 V2 2 2([H+] = [OH-] = 1 M)

In pure water, where [H+] = [OH-] = 10-7 M, the potential for the overall process is 1 23 Vpotential for the overall process is -1.23 V.


88

Electrolysis of mixtures of ions

Suppose a solution in an electrolytic cell contains the ions Cu2+ Ag+ & Zn2+the ions Cu2+, Ag+, & Zn2+.

Ag+ + e- → Ag E° = 0.80 VAg e → Ag E 0.80 VCu2+ + 2e- → Cu E° = 0.34 VZn2+ + 2e- → Zn E = 0 76 VZn2+ + 2e → Zn E = -0.76 V


89

Remember that the more positive the E° value, the more the reaction has a tendency to proceed in the direction indicated. The order of oxidizing ability :direction indicated. The order of oxidizing ability :

Ag+ > Cu2+ > Zn2+

This means that silver will plate out first as the i l i i d f ll d b &potential is increased, followed by copper, &

finally zinc.


90

In the electrolysis of an aqueous solution of sodiumchloride, we should be able to use E° values to predict which products are expected.p p p

2Cl- → Cl2 + 2e- -E° = -1.36 V2H2O → O2 + 4H+ + 4e- -E° = -1.23 V

I f t th Cl i i th fi t t b idi dIn fact, the Cl- ion is the first to be oxidized.


91

A much higher potential than expected is required g p p qto oxidize water. The voltage required in excess of the expected value (called the overvoltage) is muchthe expected value (called the overvoltage) is much greater for the production of O2 than for Cl2, whichexplains why chlorine is produced at the lower voltagevoltage.


92

B i ll th h i d b diffi ltiBasically, the phenomenon is caused by difficulties in transferring electrons from the species in the f g f psolution to the atoms on the electrode across the electrode-solution interface.


93

ch 11 el t h i tch 11 electrochemistry - chem.ccu.edu.tbct/general chem (pdf)/ch 11.pdf ·...

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