Chapter 6
Oxidation and Reduction
(Redox)
Chapter 6
Slide 1 of 63
Oxidation-Reduction:
The Transfer of Electrons
Cu metal
AgNO3
solution
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
Chapter 6
Slide 2 of 63
Half-Reactions
Oxidation: losing electrons
Cu(s) → Cu2+(aq) + 2e-
Reduction: gaining electrons
half-reactions
Ag+(aq) + e- → Ag(s)
Overall reactions
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
• In any oxidation-reduction reaction, there are two
half-reactions.
Chapter 6
Slide 3 of 63
Voltaic cell =
= Battery
electrochemical cell
Chapter 6
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A Zinc-Copper Voltaic Cell
Anode
Cathode
Zn(s) → Zn2+(aq) + 2e-
Cu2+(aq) + 2e- → Cu(s)
Chapter 6
Slide 5 of 63
Electrochemical series
Strong oxidizing agent
Strong reducing agent
Chapter 6
Slide 6 of 63
Standard Hydrogen Electrode
Standard Reduction Potential
2H+(1M) + 2e- → H2 (g, 1 atm)
E o = 0 V, at 298K
Activity of H3O+ = 1
Or, [H3O+ ] ˜ 1 M
Chapter 6
Slide 7 of 63
Measuring the Standard Potential
of the Zn2+ / Zn Electrode
Eocell = Eo (cathode) - Eo (anode)
= Eo (H2) - Eo (Zn)
cathode
anode
0.763 V = 0 - Eo (Zn)
Eo (Zn) = -0.763 V
for Zn2+ (1M) + 2e- → Zn (s)
Chapter 6
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Activity Series of Some Metals
Chapter 6
Slide 9 of 63
Criteria for Spontaneous
Change in Redox Reactions
• If Ecell > 0, forward reaction is spontaneous.
• If Ecell < 0, forward reaction is nonspontaneous, and the reverse reaction is
spontaneous.
• If Ecell = 0, the system is at equilibrium.
Chapter 6
Slide 10 of 63
Equilibrium Constants
for Redox Reactions
∆Go = -RT lnKeq = - n F Eocell
Eocell = (RTlnKeq)/nF = (RT/nF)lnKeq
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)
•
•
•
•
Chapter 6
Slide 11 of 63
Summarizing the Important
Relationships
Chapter 6
Slide 12 of 63
Chapter 6
Slide 13 of 63
γ
δ
[
C
]
[
D
]
For a reaction αA +βB → γC + δD
Q= α β
[A] [B ]
γ
δ
[
C ] [D ]
0
0
∆G = ∆G + RT ln Q = ∆G + RT ln α β
[ A] [B ]
- n F Ecell = -n F Eocell + RT lnQ
Ecell = Eocell - (RTlnQ)/nF
Ecell = Eocell - (RT/nF)lnQ
Nernest equation
At 298K, Ecell = Eocell - (0.0257/n)lnQ
or, Ecell = Eocell - (0.0592/n)logQ
For an electrochemical cell
Zn(s) + Cu 2+ (aq) → Zn2+(aq) + Cu
E cell = Eo cell - (0.0257/2) ln ([Zn 2+ ]/[Cu 2+ ])
Chapter 6
Slide 14 of 63
Nernest eq. also applied to Half reactions
E = Eo - (0.0257/n)lnQ
Anode
Zn(s) → Zn2+(aq) + 2eE (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 15 of 63
Kinetic factors for the redox in solutions
Usually the more favorable the potential, the faster
the reaction
(i) Overpotential
Reaction with potential ∆E > 0.6V
for a one-electron transfer is readily to proceed.
Can H + in neutral solution be reduced by Fe metal?
E ( H + , H 2 ) − E ( Fe 2+ , Fe) = 0 − (−0.44) = +0.44V
Not seen!
(ii) One-electron transfer reaction
Outer-sphere electron transfer
Inner-sphere electron transfer
Chapter 6
Slide 16 of 63
Outer-sphere electron transfer
Chapter 6
Slide 17 of 63
Inner-sphere electron transfer
Chapter 6
Slide 18 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.
2 Fe 3+ ( aq ) + Tl + ( aq) → 2 Fe 2 + ( aq) + Tl 3+ ( aq)
E0= 0.771-1.25 = -0.48 V
Mechanism
E0= -1.4 V
(1) Fe 3+ ( aq) + Tl + ( aq) → Fe 2+ (aq ) + Tl 2+ (aq )
slow reaction
3+
2+
2+
3+
( 2) Fe ( aq) + Tl ( aq) → Fe ( aq) + Tl (aq )
•Formation or consumption of diatomic molecules e.g. O2, N2,
H2 are often slow.
Chapter 6
Slide 19 of 63
(iii) atom transfer reaction
NO2
−
+ * OCl − ( aq) → NO2 O − ( aq) + Cl − (aq )
*
( aq )
O2 N ⋅ ⋅ ⋅
*
O − Cl 
2−
→ O2 N − O ⋅ ⋅ ⋅ Cl  → O2 N −* O − + Cl −
*
2−
1
rate ∞
oxidation number of central atom
−
−
−
−
2−
< HPO4
ClO4 < ClO3 < ClO2 < ClO −
ClO4 < SO4
2−
rate ∞ size of the central atom
−
−
ClO3 < BrO3 < IO3
−
Chapter 6
Slide 20 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 21 of 63
Stone Age
Bronze Age earliest ~3500 BC
Bronze- the ancient name for a broad range of alloys of copper,
usually with tin as the main additive.
r
4Cu2O(s) + C(s)
4Cu (l)+ CO2 (g)
Iron Age earliest ~1800-1200 BC
Because Fe2O3 is not as easily reduced as Cu2O, Iron Age
was much later than Bronze Age.
Chapter 6
Slide 22 of 63
Electrothermal reduction
Hall-Héroult process (1886) electrolysis
2 Al2O3 (in cryolite) + 3 C(s)
4 Al (l) + 3 CO2 (g)
950 ~ 980°
C
Al2O3 is dissolved in a molten cryolite, Na3 AlF6. AlF3 is also
present to reduce the melting point of the cryolite.
Chapter 6
Slide 23 of 63
Electrolytic process
Cathode: Mg 2+ + 2e- → Mg
anode: 2Cl- → Cl2 (gas) + 2eDow Process
Chapter 6
Slide 24 of 63
Pidgeon process - invented in early 1940's by Dr. Lloyd
Montgomery Pidgeon
Silicothermic reactions
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.
Carbothermic reaction
This would be impractically slow at low temperatures.
Chapter 6
Slide 25 of 63
Chapter 6
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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.
waste rock
ores
concentration & physically separating
roasting
metal oxide
reduction
slag
low purity metal
refining
high purity metal
Chapter 6
Slide 27 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 or
physical means. Several metals are refined by
electrolysis.
Chapter 6
Slide 28 of 63
Concentration of an Ore
by Flotation
Froth
containing ore
Chapter 6
Slide 29 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 by
precipitation and reduction to the desired metal.
Chapter 6
Slide 30 of 63
Hydrometallurgy
reduction
Chapter 6
Slide 31 of 63
Pyrometallurgy
MxO + C g M + CO
rG = rG (C, CO) - rG (M, MxO) < 0
⇒ rG (C, CO) < rG (M, MxO)
for spontaneous rxn.
Chapter 6
Slide 32 of 63
∆H and ? S change
negligibly with temp erature
Chapter 6
Slide 33 of 63
< 0 for spontaneous reactions
Chapter 6
Slide 34 of 63
Ellingham diagram
Elements extracted by
pyrometallurgy are Fe,
Co, Ni, Cu, Zn.
>15000C
SiO2(l) + C (s)
Si (l) + CO (g)
Chapter 6
Slide 35 of 63
Elements extracted by oxidation
∆E = 0 - 1.358 = - 1.358 V
∆E = 0 - 1.229 = - 1.229 V
Should be more easily to proceed.
However, it has high overpotential.
Chapter 6
Slide 36 of 63
A Diaphragm Chlor-Alkali Cell
Chapter 6
Slide 37 of 63
•F2(g) was prepared by electrolysis of a mixture of KF + HF
•2X-(aq) + Cl2(g) g X2 + 2Cl-(aq)
Chapter 6
rE >0
for X= Br and I
Slide 38 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 39 of 63
Redox stability in water
Reduction of H+ to form H2
E ( H + , H 2 ) = −(0.059 V)pH
2 H 2O(l ) + 2e − → H 2 ( g ) + 2OH − ( aq )
E = −(0.059 V)pH - 0.826 V
Oxidation to form O2
E (O2 , H 2O) = 1.23 V − (0.059 V)pH
E < E (H + , H 2 )
Reduction of H+ to form H2
E > E (O2 , H 2O)
Oxidation to form O2
Chapter 6
Slide 40 of 63
2H + + 2e− → H 2 ( g )
E1 = −(0.059 V)pH
H 2O ↔ H + + OH −
Kw = H + OH −
[ ][
2H 2O + 2e − → H 2(g) + 2OH −
]
E3 = ?
∆G1 = -2FE1 = 2 F (0.059 V)pH
∆G 2 = - RT ln Kw
∆G 3 = - 2 FE3 = ∆G1 + 2∆G 2
= 2F (0.059 V) pH - 2 RT ln Kw
RT
E3 = −(0.059 V)pH +
ln Kw
F
= −(0.059 V)pH + 0.059 log Kw
= −(0.059 V )pH − 0.826 V
Chapter 6
Slide 41 of 63
Oxidation by water
M= s-block metals except Be, first row d-series Group 4-7
Other metals
Reductionn by water
Chapter 6
Slide 42 of 63
Chapter 6
Slide 43 of 63
Chapter 6
Slide 44 of 63
Pourbaix diagramthermodynamically stable species as a
function of pH and potential
(i)
(ii)
(i)
(iii)
(ii)
(iv)
(iii)
(iv)
Chapter 6
Slide 45 of 63
For rxn (iii)
2+
[Fe
]
0
E = E − (0.059 V)log + 3
[H ]
= E 0 − (0.059 V)log[Fe 2+ ] − (3 × 0.059 V)pH
= E 0 − (0.059 V)log[Fe 2+ ] − ( 0.177 V)pH
For rxn (iv)
1
E = E − (0.059 V)log +
[H ]
0
Slope of the profile in
Pourbaix diagram
= E 0 − (0.059 V)pH
Chapter 6
Slide 46 of 63
MnO2 is important in well
aerated water (near the airwater boundary).
Chapter 6
Slide 47 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 V
still slow due to overpotential
2Mn2O3 + O2 → 4MnO2
2 MnO2 + 2e − + 2 H + → Mn 2O3 + H 2 O
E 0 = 0.146 V
pH = 7, E= (0.82 V) - (0.146 V) = +0.674 V
pH = 0, E= (1.23 V) - (0.146 V) = +1.08 V
Chapter 6
ready to occur
Slide 48 of 63
Diagrams presenting potential data
Latimer diagrams
EA0
pH= 0
EB 0
pH= 14
Chapter 6
Slide 49 of 63
pH= 0
pH= 14
−
−
ClO4 (aq) + 2e − + H 2 O(l ) → ClO3 (aq) + 2OH − (aq)
E 0 = +0.37 V
Chapter 6
Slide 50 of 63
Nonadjacent species
and
1× 0.42 + 1×1.36
E =
= 0.89 V
2
0
Chapter 6
Slide 51 of 63
Oxoanions are stronger oxidizing agents in acidic than in basic
solution.
Chapter 6
Slide 52 of 63
In biochemical state, pH ~ 7
Ox + ne- + νH+ g Red + ν/2 H2O
RT
1
RT
1
0
E=E +
ln
= E +ν H +
ln +
ν +
+
nF [H ] H
nF [H ]
0.059
0
ν H : the stoichiome tric coefficient
= E +ν H +
pH
n
+
0
+
of H ions in the reactions
pH = 0, E Θ = E 0
0.059
0.059
Θ
pH = 7, E = E + 7ν H +
= E + 7ν H +
n
n
ν H+
⊕
Θ
E = E + 0.414
n
⊕
0
Chapter 6
Slide 53 of 63
Disproportionation
(i)
(ii)
spontaneous if ∆E 0 = E 0 (R ) − E 0 (L) > 0
⇒ E ( R ) > E ( L)
0
Chapter 6
0
Slide 54 of 63
Frost diagrams- a plot of NE0 for the X(N)/X(0) against
oxidation number
∆G 0 = − NFE 0
0
∆
G
NE 0 = −
F
Chapter 6
Slide 55 of 63
EA0
EB 0
Chapter 6
Slide 56 of 63
Chapter 6
Slide 57 of 63
Standard reduction potential
of N’g N” species
Chapter 6
Slide 58 of 63
Chapter 6
Slide 59 of 63
Chapter 6
Slide 60 of 63
Chapter 6
Slide 61 of 63
Tend to disproportionation
Chapter 6
Slide 62 of 63
Chapter 6
Slide 63 of 63