Post on 10-Mar-2018
Chapter 6 Slide 1 of 63
Chapter 6
Oxidation and Reduction
(Redox)
Chapter 6 Slide 2 of 63
Oxidation-Reduction:The Transfer of Electrons
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
Cu metal
AgNO3solution
Chapter 6 Slide 3 of 63
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
Chapter 6 Slide 5 of 63
A Zinc-Copper Voltaic Cell
Cathode
Cu2+(aq) + 2e- → Cu(s)
Anode
Zn(s) → Zn2+(aq) + 2e-
Chapter 6 Slide 6 of 63
Strong oxidizing agent
Strong reducing agent
Electrochemical series
Chapter 6 Slide 7 of 63
Standard Hydrogen Electrode
Activity of H3O+ = 1Or, [H3O+ ] ˜ 1 M
Standard Reduction Potential2H+(1M) + 2e- → H2 (g, 1 atm)
Eo = 0 V, at 298K
Chapter 6 Slide 8 of 63
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
Chapter 6 Slide 9 of 63
Activity Series of Some Metals
Chapter 6 Slide 10 of 63
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.
Chapter 6 Slide 11 of 63
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
Chapter 6 Slide 12 of 63
Summarizing the Important Relationships
Chapter 6 Slide 13 of 63
Chapter 6 Slide 14 of 63
- 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 +∆=+∆=∆
Chapter 6 Slide 15 of 63
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
Chapter 6 Slide 16 of 63
(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!
Chapter 6 Slide 17 of 63
Outer-sphere electron transfer
Chapter 6 Slide 18 of 63
Inner-sphere electron transfer
Chapter 6 Slide 19 of 63
•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
Chapter 6 Slide 20 of 63
(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
Chapter 6 Slide 21 of 63
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.
Chapter 6 Slide 22 of 63
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)
Chapter 6 Slide 23 of 63
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
Chapter 6 Slide 24 of 63
Electrolytic process
Cathode: Mg 2+ + 2e-→Mganode: 2Cl-→ Cl2 (gas) + 2e-
Dow Process
Chapter 6 Slide 25 of 63
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
Chapter 6 Slide 26 of 63
Chapter 6 Slide 27 of 63
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
Chapter 6 Slide 28 of 63
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.
Chapter 6 Slide 29 of 63
Concentration of an Oreby Flotation
Froth containing ore
Chapter 6 Slide 30 of 63
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.
Chapter 6 Slide 31 of 63
Hydrometallurgy
reduction
Chapter 6 Slide 32 of 63
PyrometallurgyMxO + C gM + CO
rG = rG (C, CO) -rG (M, MxO) < 0
⇒rG (C, CO) < rG (M, MxO)
for spontaneous rxn.
Chapter 6 Slide 33 of 63
erature with tempnegligibly change ? and SH∆
Chapter 6 Slide 34 of 63
< 0 for spontaneous reactions
Chapter 6 Slide 35 of 63
Ellingham diagram
Elements extracted by pyrometallurgy are Fe, Co, Ni, Cu, Zn.
SiO2(l) + C (s) Si (l) + CO (g)>15000C
Chapter 6 Slide 36 of 63
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.
Chapter 6 Slide 37 of 63
A Diaphragm Chlor-Alkali Cell
Chapter 6 Slide 38 of 63
•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
Chapter 6 Slide 39 of 63
Claus process for production of S from H2S
One of the very few metals obtained by oxidation
Purify Au from the low-grade ores
Chapter 6 Slide 40 of 63
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
Chapter 6 Slide 41 of 63
[ ][ ]?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
Chapter 6 Slide 42 of 63
Oxidation by water
Reductionn by water
M= s-block metals except Be, first row d-series Group 4-7
Other metals
Chapter 6 Slide 43 of 63
Chapter 6 Slide 44 of 63
Chapter 6 Slide 45 of 63
Pourbaix diagram-thermodynamically stable species as a function of pH and potential (i)
(ii)
(iii)(i)
(ii)
(iii)
(iv)
(iv)
Chapter 6 Slide 46 of 63
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
Chapter 6 Slide 47 of 63
MnO2 is important in well aerated water (near the air-water boundary).
Chapter 6 Slide 48 of 63
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
Chapter 6 Slide 49 of 63
Diagrams presenting potential data
Latimer diagrams
pH= 0
pH= 14
0AE
0BE
Chapter 6 Slide 50 of 63
V 37.0
)(OH2)(ClO)(OH2)(ClO0
324
+=
+→++ −−−−
E
aqaqleaq
pH= 0
pH= 14
Chapter 6 Slide 51 of 63
Nonadjacent species
and
V 89.02
36.1142.010 =×+×
=E
Chapter 6 Slide 52 of 63
Oxoanions are stronger oxidizing agents in acidic than in basic solution.
Chapter 6 Slide 53 of 63
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
Chapter 6 Slide 54 of 63
Disproportionation
(i)
(ii)
0)L()R( if sspontaneou 000 >−=∆ EEE
)L()R( 00 EE >⇒
Chapter 6 Slide 55 of 63
Frost diagrams- a plot of NE0 for the X(N)/X(0) against oxidation number
00 F ENG −=∆
F
00 G
NE∆
−=
Chapter 6 Slide 56 of 63
0AE
0BE
Chapter 6 Slide 57 of 63
Chapter 6 Slide 58 of 63
Standard reduction potential of N’g N” species
Chapter 6 Slide 59 of 63
Chapter 6 Slide 60 of 63
Chapter 6 Slide 61 of 63
Chapter 6 Slide 62 of 63
Tend to disproportionation
Chapter 6 Slide 63 of 63