Chapter 6 Redox - National Taiwan Universitysfcheng/inchem94/Chapter 6 Redox.pdf · Electrochemical...

Chapter 6 Slide 1 of 63

Chapter 6

Oxidation and Reduction

(Redox)


Oxidation-Reduction:The Transfer of Electrons

Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)

Cu metal

AgNO3solution


Half-Reactions

• In any oxidation-reduction reaction, there are two half-reactions.

Cu(s) → Cu2+(aq) + 2e-

Ag+(aq) + e- → Ag(s)

Oxidation: losing electrons

Reduction: gaining electrons

Overall reactions

Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)

half-reactions

Chapter 6 Slide 4 of 63electrochemical cell

Voltaic cell =

= Battery


A Zinc-Copper Voltaic Cell

Cathode

Cu2+(aq) + 2e- → Cu(s)

Anode

Zn(s) → Zn2+(aq) + 2e-


Strong oxidizing agent

Strong reducing agent

Electrochemical series


Standard Hydrogen Electrode

Activity of H3O+ = 1Or, [H3O+ ] ˜ 1 M

Standard Reduction Potential2H+(1M) + 2e- → H2 (g, 1 atm)

Eo = 0 V, at 298K


Measuring the Standard Potentialof the Zn2+ / Zn Electrode

Eocell = Eo (cathode) - Eo (anode)

= Eo (H2) - Eo (Zn)

0.763 V = 0 - Eo (Zn)

Eo (Zn) = -0.763 Vfor Zn2+ (1M) + 2e- → Zn (s)

cathode anode


Activity Series of Some Metals


Criteria for SpontaneousChange in Redox Reactions

• If Ecell > 0, forward reaction is spontaneous.• If Ecell < 0, forward reaction is non-

spontaneous, and the reverse reaction is spontaneous.

• If Ecell = 0, the system is at equilibrium.


Equilibrium Constantsfor Redox Reactions

• Eocell : the standard cell potential

• R : the gas constant (8.3145 J mol-1 K-1)• T : Kelvin temperature• n : the number of moles of electrons involved in the

reaction• F : Faraday constant (96,500 Coulomb)

∆Go = -RT lnKeq = - n F Eocell

Eocell = (RTlnKeq)/nF = (RT/nF)lnKeq


Summarizing the Important Relationships


- n F Ecell = -n F Eocell + RT lnQ

Ecell = Eocell - (RTlnQ)/nF

Ecell = Eocell - (RT/nF)lnQ Nernest equation

For an electrochemical cell

Zn(s) + Cu 2+ (aq) → Zn2+(aq) + Cu

E cell = Eocell - (0.0257/2) ln ([Zn 2+ ]/[Cu 2+ ])

At 298K, Ecell = Eocell - (0.0257/n)lnQ

or, Ecell = Eocell - (0.0592/n)logQ

For a reaction αA +βB → γC + δD [ ] [ ][ ] [ ]βα

δγ

BADC

Q =

[ ] [ ][ ] [ ]βα

δγ

BADC

RTGQRTGG lnln 00 +∆=+∆=∆


Nernest eq. also applied to Half reactionsE = Eo - (0.0257/n)lnQ

Anode

Zn(s) → Zn2+(aq) + 2e-

E (Zn) = Eo(Zn) - (0.0257/ 2)ln [Zn 2+ ]

Hydrogen electrode

2 H+(aq) + 2e- → H2(g)

E (H+, H2) = Eo(H+, H2) - (0.0592/ 2)log (PH2/[H+]2)

For PH2=1,

E (H+, H2) = 0 -0.0592 log(1/[H+]) = -0.0592 pH


(i) OverpotentialReaction with potentialfor a one-electron transfer is readily to proceed.

V6.0>∆E

(ii) One-electron transfer reactionOuter-sphere electron transferInner-sphere electron transfer

Usually the more favorable the potential, the faster the reaction

V44.0)44.0(0),(),(

metal? Feby reduced besolution neutralin Can 2

2 +=−−=− ++

+

FeFeEHHE

H

Kinetic factors for the redox in solutions

Not seen!


Outer-sphere electron transfer


Inner-sphere electron transfer


•Non-complementary redox reactions are often slow.

Non-complementary redox reactions : the change in oxidation number of the oxidizing and reducing agents are unequal.

)(3

)(2

)()(3 22 aqaqaqaq TlFeTlFe ++++ +→+

)(3

)(2

)(2

)(3

)(2

)(2

)()(3

)2(

)1(

aqaqaqaq

aqaqaqaq

TlFeTlFe

TlFeTlFe

Mechanism

++++

++++

+→+

+→+ E0= -1.4 Vslow reaction

•Formation or consumption of diatomic molecules e.g. O2, N2, H2 are often slow.

E0= 0.771-1.25 = -0.48 V


(iii) atom transfer reaction

)()(*

2)(*

)(2 aqaqaqaq ClONOOClNO −−−− +→+

−−−− +−→⋅⋅⋅−→−⋅⋅⋅ ClONOClONOClONO *2

2*2

2*2

atom central ofnumber oxidation 1

∞rate

−−−− <<< ClOClOClOClO 234

−−− << 24

244 HPOSOClO

atom central theof size ∞rate−−− << 333 IOBrOClO


Metallurgy:Some General Considerations

• An ore is a solid deposit containing a sufficiently high percentage of a mineral to make extraction of a metal economically feasible.

• Native ores are free metals and include gold and silver.• Oxides or Silicates include iron, manganese, aluminum,

and tin.• Sulfides include copper, nickel, zinc, lead, and mercury.• Carbonates include sodium, potassium, and calcium.• Chlorides (often in aqueous solution) include sodium,

potassium, magnesium, and calcium.


Stone Age

Bronze Age earliest ~3500 BCBronze- the ancient name for a broad range of alloys of copper, usually with tin as the main additive.

Iron Age earliest ~1800-1200 BC

Because Fe2O3 is not as easily reduced as Cu2O, Iron Age was much later than Bronze Age.

r4Cu2O(s) + C(s) 4Cu (l)+ CO2 (g)


Electrothermal reduction

Al2O3 is dissolved in a molten cryolite, Na3AlF6. AlF3 is also present to reduce the melting point of the cryolite.

Hall-Héroult process (1886) electrolysis2 Al2O3 (in cryolite) + 3 C(s) 4 Al (l) + 3 CO2 (g)

950 ~ 980°C


Electrolytic process

Cathode: Mg 2+ + 2e-→Mganode: 2Cl-→ Cl2 (gas) + 2e-

Dow Process


Pidgeon process - invented in early 1940's by Dr. Lloyd Montgomery Pidgeon

Si(s) + MgO(s) ? SiO2(s) + Mg(g) (high temperature, distillation boiling zone)

Mg(g) ? Mg(liq, s) (low temperature, distillation condensing zone)

The usual metallurgic carbon as the deoxidising reducing agent instead of silicon cannot be used because CO2 is a gas too.

This would be impractically slow at low temperatures.

Silicothermic reactions

Carbothermic reaction


Extractive Metallurgy

• Metallurgy is the general study of metals.• Extractive metallurgy focuses on the activities required to

obtain a pure metal from one of its ores. ores

concentration & physically separatingwaste rock

roasting

reductionslag

refininglow purity metal

high purity metal

metal oxide


Extractive Metallurgy(Continued)

• Slag is a lower-melting, glassy product. Slag formation plays a crucial role in the metallurgy of iron.

• Refining is the process of removing impurities from a metal by any of a variety of chemical orphysical means. Several metals are refined by electrolysis.


Concentration of an Oreby Flotation

Froth containing ore


Pyrometallurgy & Hydrometallurgy

• Pyrometallurgy - uses high temperature to transform metals and their ores. Often pyrometallurgical processes are autogenous, and so the energy required to heat the minerals comes from the exothermic reaction of the minerals in the process and no further energy is required.

• Hydrometallurgy - metallurgical methods that involve processing aqueous solutions of metallic compounds. This process involves leaching the metal ions with water, acids, bases, or salt solutions, followed by purification and/or concentration which removes impurities, and finally byprecipitation and reduction to the desired metal.


Hydrometallurgy

reduction


PyrometallurgyMxO + C gM + CO

rG = rG (C, CO) -rG (M, MxO) < 0

⇒rG (C, CO) < rG (M, MxO)

for spontaneous rxn.


erature with tempnegligibly change ? and SH∆


< 0 for spontaneous reactions


Ellingham diagram

Elements extracted by pyrometallurgy are Fe, Co, Ni, Cu, Zn.

SiO2(l) + C (s) Si (l) + CO (g)>15000C


Elements extracted by oxidation

V 1.358- 1.358-0 ==∆E

V 1.229- 1.229-0 ==∆E

Should be more easily to proceed.However, it has high overpotential.


A Diaphragm Chlor-Alkali Cell


•2X-(aq) + Cl2(g) g X2 + 2Cl-(aq) rE >0for X= Br and I

•F2(g) was prepared by electrolysis of a mixture of KF + HF


Claus process for production of S from H2S

One of the very few metals obtained by oxidation

Purify Au from the low-grade ores


Redox stability in water

Reduction of H+ to form H2

Oxidation to form O2

)(2)(2)(2 22 aqOHgHelOH −− +→+V 0.826-V)pH 059.0(−=E

V)pH 059.0(V 23.1),( 22 −=OHOE

V)pH 059.0(),( 2 −=+ HHE

Reduction of H+ to form H2),( 2HHEE +<

),( 22 OHOEE > Oxidation to form O2


[ ][ ]?222

Kw

pH)V 059.0( )(22

322

2

12

=+→+

=+↔

−=→+

−−

−+−+

−+

E OH(g)HeOH

OHHOHHOH

EgHeH

KwF

FEKw

FFE

lnRT 2-pH)V 059.0(2

G2G2- GlnRT- G

pH)V 059.0(22- G

2133

2

11

=

∆+∆==∆=∆

==∆

V 0.826pH)V 059.0( log 0.059pH)V 059.0(

lnRT

pH)V 059.0(3

−−=+−=

+−=

Kw

KwF

E


Oxidation by water

Reductionn by water

M= s-block metals except Be, first row d-series Group 4-7

Other metals


Pourbaix diagram-thermodynamically stable species as a function of pH and potential (i)

(ii)

(iii)(i)

(ii)

(iii)

(iv)

(iv)


V)pH 177.0( ]V)log[Fe 059.0(

V)pH 059.03( ]V)log[Fe 059.0(

][H][Fe

V)log 059.0(

20

20

3

20

−−=

×−−=

−=

+

+

+

+

E

E

EE

For rxn (iii)

For rxn (iv)

V)pH 059.0(

][H1

V)log 059.0(

0

0

−=

−= +

E

EE

Slope of the profile in Pourbaix diagram


MnO2 is important in well aerated water (near the air-water boundary).


Oxidation by atmospheric oxygen

pH = 7, E= (0.82 V) - (0.77 V) = +0.05 V

pH = 0, E= (1.23 V) - (0.77 V) = +0.45 Vstill slow due to overpotential

OHOMnHeMnO 2322 222 +→++ +− V 146.00 =E2232 42 MnOOOMn →+

pH = 7, E= (0.82 V) - (0.146 V) = +0.674 V

pH = 0, E= (1.23 V) - (0.146 V) = +1.08 V ready to occur


Diagrams presenting potential data

Latimer diagrams

pH= 0

pH= 14

0AE

0BE


V 37.0

)(OH2)(ClO)(OH2)(ClO0

324

+=

+→++ −−−−

E

aqaqleaq

pH= 0

pH= 14


Nonadjacent species

and

V 89.02

36.1142.010 =×+×

=E


Oxoanions are stronger oxidizing agents in acidic than in basic solution.


In biochemical state, pH ~ 7

reactions in the ions H of

t coefficien tricstoichiome the:+

+HνpH

059.0

][H1

lnF][H

1ln

F

H0

H00

H

nE

nRT

EnRT

EE

+

++

+=

+=+= ++

ν

νν

Ox + ne- + νH+g Red + ν/2 H2O

0 0,pH EE == Θ

nEE

nE

nEE

+

++

+=

+=+==

Θ⊕

Θ⊕

H

HH0

414.0

059.07

059.07 7,pH

ν

νν


Disproportionation

(i)

(ii)

0)L()R( if sspontaneou 000 >−=∆ EEE

)L()R( 00 EE >⇒


Frost diagrams- a plot of NE0 for the X(N)/X(0) against oxidation number

00 F ENG −=∆

F

00 G

NE∆

−=


0AE

0BE


Standard reduction potential of N’g N” species


Tend to disproportionation

Chapter 6 Redox - National Taiwan Universitysfcheng/inchem94/Chapter 6 Redox.pdf · Electrochemical...

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