Redox Chemistry
M.Sc. (Tech.) Otto Mustonen
Outline
Solution Redox Chemistry
Redox potentials & thermodynamics
Latimer, Frost and Pourbaix diagrams
Redox Chemistry in Solids
Ellingham diagrams
Oxygen non-stoichiometry
Mixed valence
Charge ordering
Revisiting Redox Reactions:
Pair/group exercise
Form groups of 2-3 with the people sitting next to you. Discuss the
following concepts based on your previous studies in chemistry.
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•
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•
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What is reduction? What is oxidation?
What is oxidation state/number?
What are half (cell) reactions?
What is (standard) reduction potential E°?
What is the Gibbs free energy change ΔG?
Half reactions and reduction potential
• Redox reactions can be divided into two half-reactions
• Reaction is spontaneous if sum of redox potentials of the half
reactions is positive
2Ag+(aq) +2e- => 2Ag(s)
E° = +0.8 V
Cu(s) => Cu2+(aq) + 2eE° = -0.34 V
______________________________________________
2Ag+ + Cu(s) => 2Ag(s) + Cu2+(aq)
E° = 0.46 V
Standard reduction potential & thermodynamics
• Standard reduction potential is a thermodynamic quantity
• It is linked to Gibbs free energy change by:
ΔG = -nFE°
where
n is the number of electrons transferred
F is the Faraday constant (96485 C/mol)
Combining reduction potentials
• Standard reduction potentials cannot be added together when
combining half-reactions of the same type (red/ox)
• However, Gibbs free energy changes can be added up
• ΔG = -nFE°
ΔG = -1 * 0.77 FV
Fe3+(aq) + e- => Fe2+(aq)
E° = +0.77 V
ΔG = -2 * -0.44 FV
Fe2+(aq) + 2e- => Fe(s)
E° = -0.44 V
Fe3+(aq) +3e- => Fe(s)
E° = ?
ΔG = -0.77 FV + 0.88 FV
E° = - (0.11 FV) / 3F
E° ~ -0.04 V
Latimer diagram
• An illustrative representation of the standard reduction
potentials (E°) between different oxidation states of an element
• The highest oxidation state is on the left, the lowest on the right
• The more positive E°, the more readily the species on the left is
reduced to the species on the right
+6
FeO42-
+2.20 V
+3
Fe3+
+0.77 V
+2
Fe2+
-0.04 V
-0.44 V
0
Fe
Disproportionation
0
O2
+0.68 V
-1
H2O2
+1.78 V
-2
H2O
+1.23 V
• A species might both reduce and oxidize at the same time
H2O2(aq) + 2H+(aq) + 2e- => 2H2O(l)
E° = +1.78 V
H2O2(aq) => O2(g) + 2H+(aq) + 2eE° = -0.68 V
_______________________________
2H2O2(aq) => 2H2O(l) + O2(aq)
E° = +1.10 V
Frost diagrams
• Latimer diagrams get complicated when many oxidation states
are included
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•
•
•
•
ΔG = -nFE°
Frost diagrams plot ΔG (in units of F) against oxidation state
Allow us to obtain information visually without calculating
ΔG of metal/pure element is set to 0
What we can learn:
relative stability of different oxidation states, reducing/oxidizing
power, disproportionation
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Interpret the diagram in a group of 2-3
• More stable species are at the
bottom
• Species on a convex curve tend to
disproportionate
• Species on a concave curve do not
disproportionate
• Species high on the left will be
highly oxidizing
• Species high on the right will be
highly reducing
Frost diagrams are pH dependent
Pourbaix diagrams
• Pourbaix diagrams shows the most thermodynamically stable
species as a function of potential and pH in aqueous solution
• Commonly utilized in applications related to environment,
corrosion, etc.
• A vertical line describes an acid-base equilibrium
• Other lines describe a redox equilibrium
• Redox equilibrium does not involve H+/OH• Diagonal line: redox equilibrium involves H+/OH-
• Broken lines indicate the water stability region
Pourbaix diagram of manganese
Lecture exercise: Frost diagrams of bromine
Latimer diagrams for bromine in acidic and basic solutions are:
+1.82 V
+1.49 V
+1.59 V
+1.07 V
BrO4- -----------® BrO3- ----------® HBrO -----------® Br2 -----------® Br+0.99 V
+0.54 V
+0.45 V
+1.07 V
BrO4- -----------® BrO3- ----------® BrO- -----------® Br2 -----------® Br-
Construct the corresponding Frost diagrams, and answer:
a) Which species tend to disproportionate
b) Calculate E° for the reduction of BrO3-(aq) to Br2(l)
c) Why is the last reduction potential the same in both conditions?
Hints
ΔG = -nFE°
Frost diagram plots ΔG as a function of oxidation state
ΔG can be given in units of F
ΔG for elements (= oxidation state 0) is zero
Standard potentials cannot be added together, but ΔG ’s can
Ellingham diagrams
• Ellingham diagrams are a plot of ΔG° of oxidation of metals into
metal oxides as a function of temperate (ΔG° vs T plot)
• They are used in metallurgy in order to reduce oxides to metals
• ΔG° = ΔH° – TΔS°
• ΔG = ΔG° - RTln(pO2)
• ΔG° = RTln(Kp)
Kp = pO2 at equilibrium
• Below ΔG = 0 oxide is stable, above 0 it will be reduced to metal
• Metal oxides become less stable as T increases (ΔS° term)
• Oxides lower in the diagram are more stable than those above
Ellingham diagrams
Carbothermic reduction
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Reducing oxides to metals just by heating is hard!
Use a reducing agent: carbon!
At high temperatures carbon will burn into CO
CO itself is a reducing agent, as it burn into CO2
Defects in solids
Crystalline solids consist of repeating arrangements of atoms
These arrangements are not perfect, they have defects
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Oxygen non-stoichiometry
• Materials with oxygen nonstoichiometry have non-integer
amounts of oxygen per formula unit
• The degree of oxygen nonstoichiometry is described by δ
Origin of nonstoichiometry is in defects:
• Interstitial oxygen atoms: e.g. La2CuO4+δ
• Cation vacancies: e.g. La1-xMn1-xO3
• Oxygen vacancies: YBa2Cu3O7-δ
• Interstitial cations: e.g. Zn1+xO
Reduction of oxides
• May occur in in a stepwise fashion
• Or it can occur gradually, i.e. changes in oxygen nonstoichiometry
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Nonstoichiometry affects properties
• An example: effect of oxygen
content on the superconducting
transition temperature in
YBa2Cu3O7-δ
Atmosphere
Atmosphere used changes the
oxygen content
• a: In O2
• b: In N2
Engineering of the oxygen content
possible by controlling T and pO2!
Mixed valency
• In the previous example, tuning the amount of oxygen
nonstoichiometry induced mixed valency in Cu
• Thus Cu became a mix of Cu2+ and Cu3+
Definition of mixed valency
• the formal oxidation state for (at least) one of the elements is
fractional, or
• (at least) one of the elements is at two different oxidation states
Class I Mixed valency
Examples:
• GaCl2 (GaI & GaIII)
• KCr3O8 (CrIII & CrVI)
• mixed-valence atoms have clearly different environments (=
coordination spheres)
• large energy required for electron transfer between the mixedvalent atoms
• no interaction between the species of different oxidation states
• no special properties
Class II Mixed valency
Examples:
• Eu3S4 (EuII & EuIII)
• NaxWO3 (x < 0.3) (WV & WVI)
• Ag2O2 (AgI & AgIII)
• The different oxidation states are associated with different
environments, but the two sites are sufficiently similar that
electron transfer requires only a small energy
• semiconductive
• optical absorption (intervalence absorption)
Class III Mixed valency
Examples:
• NaxWO3 (0.3 < x < 0.9) (W6-x)
• Ag2F (Ag0.5)
• all atoms of the mixed-valence element are in an identical
environment and at an identical fractional oxidation state
• electrons delocalized between the mixed-valence atoms
• ==> metallic conductivity
Charge ordering
• Class III mixed valent compounds can have a phase transition, in
which the mixed valent cation orders into separate sites
• Mx+0.5 => Mx + Mx+1
• The compound changes from Class III to Class I material
Verwey transition of magnetite Fe3O4
• Magnetite has two sites for iron: one contains only Fe2+ and one
contains a 1:1 mix of Fe2+ and Fe3+ (Fe2.5)
• At 120 K the second site orders into separate Fe2+ and Fe3+
sites
• A significant drop in electrical conductivity is observed
Ordered structure finally explained
Mark S. Senn, Jon P. Wring, J. Paul Attfield,
Nature 481, 173–176 (2012)
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