Chapter 13 &14

Electrochemistry and

Electrode Potentials

Oxidation: a loss of electrons to an oxidizing agent

Reduction: a gain of electrons from a reducing agent

16A Principles Reduction-oxidation reaction

(redox reaction)Ox1 + Red2 Red1 + Ox2

An oxidizing substance:Ma+ + ne- M(a-n)+

An reducing substance: Ma+ M(a-n)+ + ne-

16B Electrochemical Cells(1) Galvanic (Voltaic) cell:

a chemical reaction spontaneously occurs to produce electrical energy.

Ex: lead storage battery

(2) Electrolytic cell:electrical energy is used to force a nonsponta

neous chemical reaction to occur.Ex: electrolysis of water

the anode: oxidation occurs

the cathode: reduction occurs

Salt bridge: allows charge transfer through the solutions but prevents mixing of the solutions.

Fig 16-2

AgeAg:cathode the

e2CuCu:anode thereaction-half

CuAg2Cu2Ag

-

2

2

Cu: reducing agent

Ag+: oxidizing agent

Electrode potential: the tendency of the ions to give off or take on electrons.

Normal Hydrogen Electrode (NHE)or Standard Hydrogen Electrode (SHE)

2H+ + 2e- = H2 or H+ + e- = 1/2H2

Eo ( 標準電位 ) = 0.000 V

Table 16.1( 於 1953, the 17th IUPAC meeting 決定半反應以「還原反應」來表示 )

Potential are dependent on [con]. & temp.

Standard reduction potential: activity=1

The more positive the electrode potential, the greater the tendency of the oxidized form to be reduced.

The more negative the electrode, the greater the tendency of the reduced form to be oxidized.

Related to free energy

-nFεΔG

:condition standard

nFεΔG

ΔG w& nFq & qε w

q

w-ε

(C) charge

(J)work (V) emf

max

The Nernst Equation

lnK RTnFEΔG

EEE

FeCuFeCu

ΔG calculate

0rxn

0ox

0red

0rxn

(aq)2

(s)(s)(aq)2

Ex:Predict whether 1M HNO3 will dissolve

gold metal to form 1M Au3+?

0.54Vεεε

-1.50Vε 3eAuAu

0.96Vε

O2HNO3e4HNO

0ox

0re

0cell

0ox

3

0re

23

Cell representation

anodesolutioncathode

ionfor titratpoint end sharp 0.3~0.2E

0.617V0.1540.771

E E E :Ex

E-EE-EE-EE

Pt|)(CFe ),(CFe||)(CSn ),(CSn|Pt

Pt|)(CCe ),(CCe||)(CFe ),(CFe|Pt :Ex

cell

Sn,SnFe,Fecell

-anodecathodeleftrightcell

42

33

24

12

43

34

23

12

2423

The Nernst Equation & [C] effect

Activities should be used in the Nernst equation. We will use concentrations here because titrations de

al with large potential changes, and the errors are small by doing so.

Table 16-1, 每個離子之活性（ activity ）皆為 1 ，叫「標準還原電位」

Dependence of the cell potential on [C]

a

b

[Ox][Red]

lognF

2.3026RTEE

bRedneaOx

E is the reduction potential at the specified concentrations n: the number of electrons involved in the half-reactionR: gas constant (8.3143 V coul deg-1mol-1)T: absolute temperatureF: Faraday constant (96,487 coul eq-1)

at 25°C 2.3026RT/F=0.05916

ex:[C] & Ecell

standard conditions: [C]=1M what if [C]≠1M?

a)[Al3+]=2.0M, [Mn2+]=1.0M Ecell<0.48Vb)[Al3+]=1.0M, [Mn2+]=3.0M Ecell>0.48V

0.48VE

3Mn2Al3Mn2Al 0cell

(s)(aq)3

(aq)2

(s)

0E mequilibriuat

][I][Fe

][I][Felog

2

0.059EE

][I

][Ilog

2

0.059E

][Fe

][Felog

2

0.059E

0EEE

I2Fe3I2Fe :17.6Ex

cell

3233

22

I,IFe,Fe

3

3

I,I

23

22

Fe,Fe

I,IFe,Fecell

32-3

323

3

23

323

After 兩個半反應 reached eq., the cell voltage necessarily becomes zero and the reaction is complete.

Ex: One beaker contains a solution of 0.020 M KMnO4, 0.005 M M

nSO4, and 0.500 M H2SO4; and a second beaker contains 0.150

M FeSO4 and 0.0015 M Fe2 (SO4)3. The 2 beakers are connecte

d by a salt bridge and Pt electrodes are placed one in each. The electrodes are connected via a wire with a voltmeter in between.

What would be the potential of each half-cell (a) before reaction and (b) after reaction?

What would be the measured cell voltage (c) at the start of the reaction and (d) after the reaction reaches eq.?

Assume H2SO4 to be completely ionized and equal volumes in e

ach beaker.

5Fe+2 + MnO4- + 8H+ = 5Fe+3 + Mn+2 + 4H2O

 Pt/Fe+2(0.15 M), Fe+3(0.003 M)//MnO4-(0.02 M), Mn+2(0.005 M), H+(1.00 M)/Pt

(a) EFe = EoFe(III)/Fe(II) – (0.059/1) log [Fe+2]/[Fe+3]

= 0.771 – 0.059 log (0.150)/(0.0015 × 2) = 0.671 V

EMn = EoMnO4-/Mn+2 – (0.059/5)log [Mn+2]/[MnO4

-][H+]8

  = 1.51 – 0.059/5 log (0.005)/(0.02)(1.00) 8 = 1.52 V 

(b) At eq., EFe = EMn, 可以含鐵之半反應來看，先找出平衡時兩個鐵離子的濃度，得EFe = 0.771 – 0.059 log (0.05)/(0.103) = 0.790 V 

(c) Ecell = EMn - EFe = 1.52 – 0.671 = 0.849 V

(d) At eq., EFe = EMn, 所以 Ecell = 0 V

16C-7 Limitation to use E0

The sources of differences: For Fe3+ + e- Fe2+

(1) Use [conc] vs ax (activities) Fe(II)/ Fe(III) = 0.4/0.18 at = 0.1M

(2) Other equilibria:complexes Fe(III) with Cl-, SO4

-2 are more stable than those of Fe(II).

16C-7 Formal PotentialEx: Ce4+ + e- Ce3+ E°=1.6V

with H+A- E°≠1.61V則酸之陰離子會與之起某種程度的螯合，而影響他們的活性 

再是標準還原電位 Eo 改叫形式電位 Eo`

Formal potential: (E°’)The standard potential of a redox couple with the oxidized and r

educed forms at 1M concentrations and with the solution conditions specified.

Ex: Ce4+/Ce3+ in 1M HCl E°’=1.28V

Dependence of potential on pH

0.059pHE'E

]AsO[H

]AsO[Hlog

2

0.059-0.059pHE

]AsO[H

]AsO[Hlog

2

0.059-]0.059log[HE

][H ]AsO[H

]AsO[Hlog

2

0.059-EE

OHAsOH2e2HAsOH :Ex

43

33

43

33

243

33

233-

43

Many redox reactions involved protons, and their potentials are influenced greatly by pH.

Dependence of potential on complexation:

][FeCl

][Fe0.059log-0.70E

constant[HCl] If

4ClFeeFeCl

FeCl iscomplex theIf

0.70V'E HCl 1MIn

0.771VE /FeFe :Ex

-4

2

2-4

-4

23

Complexing one ion reduces its effective concentration, which changes the potential.

In effect, we’ve stabilized the Fe+3 by complexing it, make it more difficult to reduce.

Ex: Systems involving ppt Calculate Ksp for AgCl at 25℃ & E =0.58V

Ag Ag

1MAgNO 3(aq)

1M NaCl(s)

& AgCl(s)

Sol:

1.0Msp

0

ClAgK

1.0

Aglog

1

0.05920

0.58Vε

(why?) 0ε

Ex 17-4: Calculate the cell potential for Ag AgCl(sat’d), HCl(0.0200 M) H∣ ∣ 2(0.800 atm), Pt

Sol: 2H+ + 2e- H2(g) E0H+/H2

= 0.000 V

AgCl(s) + e- Ag(s) E0AgCl/Ag = 0.222 V

32

Hright

)0200.0(

800.0log

2

0592.0

]H[log

2

0592.0000.0E

2

P

= -0.0977 V

Eleft = 0.222 – 0.0592 log[Cl-] = 0.222 – 0.0592 log 0.0200

= 0.3226 V

Ecell = Eright – Eleft = -0.0977 – 0.3226 = -0.420 V

2H+ + 2Ag(s) H2(g) + 2AgCl(s)

17B Calculating Redox Equilibrium Constants

Ex 17-6: Calculate the equilibrium constant for the reaction

2Fe3+ + 3I- 2Fe2+ + I3-

Sol: 2Fe3+ + 2e- 2Fe2+ E0 = 0.771 V

I3- + 2e- 3I- E0 = 0.536 V

23

22

/FeFe0

/FeFe ]Fe[

]Fe[log

2

0592.0EE 23

23

][I

]I[log

2

0592.0EE

3

3

/II0

/II 33

]I[

]I[log

]Fe[

]Fe[log

0.0592

)E-2(E

]I[

]I[log

2

0592.0E

]Fe[

]Fe[log

2

0592.0E

EE

3

3

23

22/II

0/FeFe

0

3

3

/II0

23

22

/FeFe0

/II/FeFe

--3

23

323

323

3233

22

]I[]Fe[

]I[][Felog

0592.0

)EE(2

]I[]Fe[

]I[][Felog

/II0

/FeFe0

3233

223

23

7eq

/II0

/FeFe0

eq

107.894.7loganti

94.70592.0

)536.0771.0(2

0592.0

)EE(2log 3

23

K

K

Example 17-7

Calculate the equilibrium constant for the reaction

2MnO4- + 3Mn2+ + 2H2O 5MnO2(s) + 4H+

Sol: 2MnO4- + 8H+ +6e- 2MnO2(s) + 4H2O E0 = +1.695 V

3MnO2(s) + 12H+ + 6e- 3Mn2+ + 6H2O E0 = +1.23 V

EMnO4-/MnO2

= EMnO2/Mn2+

12

32

824 ]H[

]Mn[log

6

0592.023.1

]H[]MnO[

1log

6

0592.0695.1

83224

12

]H[]Mn[]MnO[

]H[log

0592.0

)23.1695.1(6

eq3224

4

log]Mn[][MnO

]H[log1.47 K

47eq 1011.47loganti K

32

12

824 ]Mn[

]H[log

]H[]MnO[

1log

0592.0

)23.1695.1(6

17C Constructing Redox Titration Curves

Example 17-8

Obtain an expression for the equivalence-point potential in the titration of 0.0500 M U4+ with 0.1000 M Ce4+. Assume

that both solutions are 1.0 M in H2SO4.

U4+ + 2Ce4+ + 2H2O UO22+ + 2Ce3+ + 4H+

Sol: UO22+ + 4H+ + 2e- → U4+ + 2H2O E0 = 0.334 V

Ce4+ + e- Ce3+ E0' = 1.44 V422

4

/UUO0

eq]][HUO[

]U[log

2

0592.0EE 42

2

]Ce[

]Ce[log

1

0592.0EE

4

3

/CeCe0

eq34

422

4

/UUO0

eq]H][UO[

]U[log0592.0E2E2 42

2

4422

34

/CeCe0

/UUO0

eq]H][Ce][UO[

]Ce][[Ulog0592.0EE2E3 3442

2

2

]Ce[]U[

44

2

]Ce[]UO[

32

2

443

34/CeCe

0/UUO

0

eq ]H][[Ce]Ce[2

]Ce][Ce[2log

3

0592.0

3

EE2E

34422

4

/CeCe0

/UUO0

]H[

1log

3

0592.0

3

EE2 34422

Electrode Potential versus SHE in Titrations with 0.100 M Ce4+

Potential, V vs. SHE

Reagent Volume, mL

50.00 mL of 0.0500 M Fe2+

50.00 mL of 0.02500 M U4+

5.00 0.64 0.316

15.00 0.69 0.339

20.00 0.72 0.352

24.00 0.76 0.375

24.90 0.82 0.405

25.00 1.06 ←Equivalence →

Point

0.703

25.10 1.30 1.30

26.00 1.36 1.36

30.00 1.40 1.40

Table 17-1

Note: H2SO4 concentration is such that [H+] = 1.0 M throughout.

17D Oxidation/Reduction Indicators Self-indication:

If the titrant is highly colored, this color may be used to detect the end point.

Ex : MnO4- Mn2+

purple faint pink

Starch indicator:This indicator is used for titrations invol

ving iodineStarch + I2 dark-blue color complex

Redox Indicators:These are highly colored dyes that are weak r

educing or oxidizing agents that can be oxidized or reduced

Oxind + ne- Redind

ind

ind0indind Ox

Redlog

n

0.059EE

10

1

]In[

]In[

ox

red 10

]In[

]In[

ox

red

A potential equal to 2×(0.059/n)V is required for a sharp color change

n = 1 0.12Vn = 2 0.060V

The redox indicator reaction must be rapid and reversible.

Table 17.2Ex:

(1) Ferroin: [tris(1,10-phenanthroline)ion(II) sulfate]

for titrations with cerium(IV)(2) Starch/Iodine soln.

18B Reducing AgentsThiosulfate: stable to air oxidation Iron(II): E0 = 0.771V for titration of cerium(IV), chromium(V

I), vanadium(V) indicator:

ferroin or diphenylamine sulfonate.

18C Oxidizing AgentsPotassium permanganate (KMnO4)

E0=1.51In neutral solution: MnO4

-MnO2

In acid solution: MnO4-Mn2+

Autocatalytic decomposition: Standardization: Na2C2O4

5H2C2O4+2MnO4-+6H+

10CO2+2Mn2++8H2O

Cerium (IV): Ce4+ / H2SO4: E0 = 1.44V; Ce4+ / HClO4: E0 = 1.70V

1) the rate of oxidation of chloride ion is slow2) is stable in H2SO4

3) (NH4)2Ce(NO3)6 can be obtained as a primary standard.

indicator: Ferroin

Potassium dichromate: K2Cr2O7

a slightly weaker oxidizing agent than KMnO4 primary standard

Cr2O72- Cr3+

E0 = 1.33~1.00V in 1M HCl