REDOX REACTIONS
Inorganic Chemistry
Final Version
160:371
Fall 2010
M.Greenblatt
The Periodic Table
Redox Reactions
2Mg (s) + O2 → 2MgO
Oxidation
loss of electrons Mg → Mg2+ + 2eaccompanied by an increase in oxidation number
Reduction
gain of electrons O2 + 4e- → 2O2accompanied by a decrease in oxidation number
oxidizing agent
substance that cause oxidation by being reduced
reducing agent
substance that cause reduction by being oxidized
O2 oxidizing agen
O2 +4e- → 2O2-
Reducing
C reducing agent
C → C4+ + 4e-
strong
weak
weak
strong
-acid (or base)
_Sum:______________________________________________________________
14 H+ + 3Cu2O +2NO3- +6e- ⇔ 6Cu2+ + 2NO +7H2O + 6e-
Pu4+ → Pu5+ + ePu 4+ +e- → Pu3+
____________
2Pu4+ → Pu3+ + Pu5+
cell, electrical work
Half cells and galvanic cells In a galvanic
is done by the system
Measure Ecell (V) under std conditions:
∆G° = -zFE°cell
z = # mol e transferred/mol reaction
F = 96,485 C/mol (Faraday’s const)
E° = standard cell potential (V)
∆G° (J/mol)
Oxidation at the anode
Standard State Conventions
for gases: 1 atm
for solid and liquids: pure solid or liquid at 1 atm pressure
for solutions: activity of 1 molar
Temperature: 298 K
Reduction at the cathode
Spontaneous reaction:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
J-unit of energy, energy expended, when 1 C flows for
1 s in a resistance of 1 Ohm;
C-coulomb, charge of 1 ampere/sec
V = IR
V
V
V
VI
V
VIt
IVIt Vq
V/I
Thermodynamic Aspects of Electrochemistry
∆G° = -zFE°cell
∆G° = - RTln K
ln K = zFE°cell / RT
R = 8.314 JK-1mol-1
∆G° - perfect predictor of spontaneity
Electrochemical cell is thermodynamically favorable
- a reaction is spontaneous when ∆Grxo is negative
- negative value corresponds to K > 1
- E°cell > 1
Do ex. 8.1 and self-study exercises, pg 214
- equilibrium is not usually attained
- thus processes with K < 1 can occur
- if the products are swept away
+ Self-study
Relationship between equilibrium constant, K and standard cell potential, E0
ln K = zFE°cell / RT
Strong oxidizing agents
Strong reducing agents
MnO4- is reduced, Zn is oxidized, net:
5Zn (s) +2MnO4- (aq) + 16 H+ → 5Zn2+ +2Mn2+(aq) + 8H2O
Ecell = 1.51+0.76 = +2.27 V
Cannot measure E or E° for a half cell. We need to connect two half cells
How were the values in Table above obtained?
Hydrogen Electrode
Standard Hydrogen Half-Cell
consists of a platinum electrode
covered with a fine powder of
platinum in 1M H3O+ (aq) at
25 °C around which H2 (g) at P=1
atm, is bubbled. The potential
of this std. H+/H2 half cell is
defined as zero volts.
2 H+ (aq, 1M) + 2 ebar)
↔
reversible reaction
E°cell ≡ 0
H2 (g, 1
Galvanic Cell-for E°cell (Zn2+/Zn)
In this cell, H2(g) evolves
Zn(s) → Zn2+ (aq, 1M) + 2e2 H+ (aq, 1M) + 2 e- ↔ H2 (g, 1 bar)
To find correct sign of E°cell for half
cell:
net: 2H+ + Zn ↔ H2 + Zn2+
E°cell measured = 0.76 V
E°cell = [E° (reduction) - E°
(oxidation)]
= 0 – (0.76) V
E°cell ≡ -0.76
Cell Potentials
cell potential (Ecell): the potential difference, in volts,
between the electrodes of an electrochemical cell
standard cell potential (E°cell):the potential difference, in
volts, between the electrodes of an electrochemical cell
when the concentrations of all solutes is 1 molar, all the
partial pressures of any gases are 1 atm, and the
temperature at 25oC
cell diagram the shorthand representation of an
electrochemical cell showing the two half-cells
connected by a salt bridge or porous barrier, such as:
Zn(s)/ZnSO4(aq)//CuSO4(aq)/Cu(s)
anode
cathode
Cu2+ + 2e- → Cu(s)
Zn(s) → Zn2+ + 2e-
Note: Standard reduction cell potential of O2 in acid and base
note*
*
Section 5.3: Trends in std. potential
Inorganic Chemistry Chapter 1: Figure 5.2
© 2009 W.H. Freeman
Inorganic Chemistry Chapter 1: Table 5.1
© 2009 W.H. Freeman
Standard Cell Potentials
Direction of Oxidation-Reduction Reactions
E°cell positive value indicates a spontaneous reaction under stand. cond
But careful! ∆G° < 1 is true test of spontaneity
Consider reaction between Fe and Cl2 (aq)
E° (V)
-0.44
Fe2+ (aq) + 2e- → Fe (s)
Fe3+ (aq) + 3e- → Fe (s)
-0.04
Cl2 (aq) + 2e- → 2Cl- (aq)
+1.36
___________________________________________________
1. Fe (s) + Cl2 (aq) → Fe2+ (aq) + 2Cl- (aq)
1.80 V
2. 2Fe (s) + 3Cl2 (aq) →2 Fe3+ (aq) + 6Cl- (aq) 1.40 V
1. ∆G° = -zFE° = -2 x 96,485 x 1.80 = -347 kJ/molBased on E0cell
2. ∆G° = -zFE° = -6 x 96,485x1.40 = -810
might expect
reaction 1 to be favored over #2
But per mole of Fe, ∆G0 is -173 kJ/mol
kJ/mol
For Eq. 1 and -405 kJ/mol for Eq. 2
+ neEcel= E0cell
(aq)
E0cell= -0.763 V
Ecell=
Nernst Equation
Ecell = Ecell° - (RT/nF) ln Q
Note: Q is not K
Effect of pH on the oxidizing power of MnO4MnO4- (aq) + 8H+ (aq) +5e- → Mn2+ (aq) +4H2O
E° = 1.51 V
RT/F = 2.303(8.3145 J K-1 mol-1)(298.15 K)/9.6485 x104 C mol-1 = 0.05916 V
Ecell = 1.51- (0.0592/5)log [Mn2+]/[MnO4-][H+]8
Oxidizing power of MnO4- is lower in dilute acid than in conc. acid
(E is more positive as [H+] increases pH is lower)
Reason why MnO4- will not oxidize Cl- in neutral solution,
but liberate Cl2 in conc. HCl
Cl2(g)+ 2e- → 2Cl-
E° = +1.36 V
Inorganic Chemistry Chapter 1: Box 5.1
Example 5.5
© 2009 W.H. Freeman
H2O acting as a oxidizing agent when it is reduced to H2
H2O(l) + e- → H2(g)+ OH-(aq)
or
e.g., of complex formation:
2H+ (aq, 1M) + 2 e- ↔ H2 (g, 1 bar)
Zn2+ (aq) + 2 e- → Zn(s)
Show what is the reduction potential of 2H+/H2 at
H2 (1 bar) in neutral water and at pH=14
only applies to hydrated Zn2+
E° = -0.76 V
In basic solution (pH=14; [OH-]=1):
E (2H+/H2) = E0(2H+/ H2) –(RT/2F)ln1/[H+]2
[Zn(OH)4]2- (aq) + 2e- → Zn(s) + 4OH- (aq)
E = 0 - (0.0592/2)log(1/[H+]2 = (0.0592)log[H+]=-(0.0592 V) pH
E° [OH-]=1 = -1.20V
Reduction of H2O @ any pH:
E = -(0.0592 V) pH
at pH = 7
E (H+,H2) =(0.059)7= -0.41 V
at pH = 14
E[OH-]=1 = 0.0592 x14 = -0.83 V
Mn+/M system more positive relative to 2H+/H2 (under given pH) should reduce H2O
however, some systems that should reduce H O do not, because they form a hydroxide coating, or
H2Oacting as a reducing agent
Consider the reduction O2 to H2O or oxidation of H2O to O2 at pH = 0 ([H+] = 1) by species in
the cell:
O2 (g) + 4H+ (aq) + 4e- → 2H2O (l) E° = +1.23 V
Nernst Eq. shows that (work it out!): E = E0 – [RT/4F]ln{1/[H+]4 = 1.23 V – (0.0592)pH
at pH= 7 & PO2 =1atm: Ecell= +0.82 V
at pH= 14 & PO2 =1at : Ecell =+ 0.40 V
Thus O2 in aqueous solution, thermodynamically should oxidize any system that:
at pH= 0 in H2O, with reduction potential (Ered) less positive than +1.23 V
or at pH =7 with Ered that is less positive than +0.82 V
or at pH =14 with Ered that is less positive than +0.40 V
conversely,
system with E more positive (>) than +1.23 V should oxidize H2O to O2 at pH=0 etc….
Bear in mind that the reduction Potentials in Table apply only to std. conditions for half cell,
and for any other condition the Nernst Eq. must be applied to determine the value of
Ecell before use
Inorganic Chemistry Chapter 1: Figure 5.3
Stability field of H2O
as a function of E/N
and pH
H2O as oxidizing
agent; reduced by
MnO4-/Mn2+; Ce 4+/Ce3+
Strong reducing
agents, group I & II
M, E < 0 will reduce
H2O to H2
© 2009 W.H. Freeman
Disproportionation Reactions
oxidation
2Cu+(aq) ⇔ Cu2+(aq) + Cu(s)
reduction
Cu+ is unstable in H2O to disproportionation; can be stabilized by precipitation of CuCl
Ksp = 10-7
Disproportionation reactions
Self-study 3, p.225 using data from Appendix 11, show that H2O2 is
unstable and disproportionates to O2 and H2O
2H2O2 → O2 + 2H2O
O22- → O2 + 2eO22- + 2e- → 2H2O
Given that:
O2 (g) + 2H2O (l) + 4e- → 4[OH-] (aq)
O2 (aq) + 2H2O(l) + 2e- → H2O2 aq) + 2[OH-](aq)
E° = +0.40 V
E° = -0.15 V
Show that ∆G° is spontaneous for the disproportionation of
H2O2 into O2 and H2O
Latimer diagrams of Mn reduction half cell reactions in acid and base
E°cell = +0.90 for the reaction:
Instead of
writing out all
possible detailed
half cell reactions
as in Table of
Reduction Potentials
Oxidation state of Mn
MnO4- (aq) + H+ + e- → HMnO4- (aq)
show that HMnO4- disproportionates to MnO4- and MnO2
7+
6+
4+
3+
2+
0
Diagram is pH dependent; for every pH
MnO2(s) + 2H2O (aq) + 2e- → Mn(OH)2 (s) + 2OHnew diagram is needed
Note: MnO4- → MnO2 E°(acid) = +1.69 V
MnO4- → MnO2 E°(base) = +0.59 V
E° = -0.04 V
HMnO4- disproportionates
2MnO4- (aq) + H+ + e- → HMnO4- (aq))
E0 =+0.90 V
HMnO4- (aq) + 3H+ (aq) + 2e- → MnO2(s) +2H2O
E0 =+2.10 V
_____________________________________________________
3[HMnO4]- (aq) + H+ (aq) → 2[MnO4]- (aq) + MnO2 (s) + 2H2O
Ecell0 = +1.20 V
∆G0 = -2x96,485x1.20 = -231kJ/mol
In general, when E(R) > E (L) in the Latimer diagram of a species, it is
Uunstable to disproportionation
Similarly, Mn3+ disproportionates in acid to Mn2+ and MnO2
Deriving E°cell for non-adjacent couples of the Latimer
diagram
∆G° = -nFE°
The overall ∆G° for two or more successive steps
∆G° = ∆G1° + ∆G2° +……. ∆Gf°
∆G = -Σ niFEi
= -(n1FE1 + n2FE2 + n3FE3 + * *)
-ΣniFEi = -(n1FE1 + n2FE2 + n3FE3
ΣniFEi = F(n1E1 + n2E2 + n3E3
+ ……….)
+…………..)
ΣniEi = (n1E1 + n2E2 + n3E3 + ………….
(n1E1 + n2E2 + n3E3 + * * *)
Et = ------------------------------------Σni
(n1E1o + n2E2o + n3E3o + * * *)
Eto = --------------------------------------nt
Example: Calculate Eo for ClO4- → Cl2
(n1E1o + n2E2o + n3E3o + * * *)
Eto = --------------------------------------nt
Latimer diagram in acid
Oxid.
State 7+
Cl
5+
4+
3+
1+
0
1-
Eo = [2(1.19) + 1(1.15) + 1(1.28) + 2(1.65) + 1(1.61)]V/7 = 1.23 V
Frost-Ebsworth (F-B) diagram - relation to potenial diagram
Graphical representation of -∆G°/F =zE° vs N (the oxidation state of M(N)) for the formation
of M(N) from M(0) For the reaction Mn+ + ne- → M where N is n+
•Fig. a and b are for pH = 0 in aqueous solution; for any other
•pH a different F-B applies of course (E0 is different)
•All points on the F-B refer to stability with respect of -∆G°/F =0;
•Any two points may be connected (not just neighboring poins)
5.21 V
4.31 V
MnO4- + 4H+ +3e- →MnO2 + 2H2O
Mn2+ (aq) + 2e-→ Mn (s) E° = -1.19 V
Mn3+ (aq) + e- → Mn2+ (aq) E° = +1.54 V
∆G°/F = -2 x (-1.19) /96,485 =+2.38 V; - ∆G°/F= -2.38
∆G°/F = -1 x (+1.54) /96,485 =-1.54 V;
Relative to Mn(0): - ∆G°/F= 1.54 – 2.38 = -0.84 V
•Relative thermodynamic stability is indicated –nothing about
kinetic stability
5.21 V
highest point, least stable
Strongest oxidizing
agent thermodynamically
most unstable species
MnO4Moving down, less stable
From gradient of line between two
points, E° for couple can be gotten:
Mn2+ (aq) + 2e-→ Mn (s)
E° = gradient of line/# e- = -2.38/2
=-1.19 V
+ gradient for reduction means E°
is +; negative gradient, E0 Most stable oxidation state, lowest point – Mn2+; most stable species. why/?
Inorganic Chemistry Chapter 1: Figure 5.5
© 2009 W.H. Freeman
2[MnO4- + H+ + e- → HMnO4- ] E° = +0.90 V
HMnO4- + 2e- + 3H+→ MnO2 + 2H2O E° = +2.10 V
2
Any state represented by a “convex” point
is unstable to disproportionation
Inorganic Chemistry Chapter 1: Figure 5.9
Comproportionation: Mn(s) + 2Mn 3+ → 3Mn2+
Ag2+(aq) + Ag(s) → 2 Ag+(aq)
E0 = +1.18; K = 1020 at 298 K© 2009 W.H. Freeman
What does this diagram tell us?
Powerful oxidizing
agent, reduced to
Cr3+
Most stable oxidation state of Cr?
Strongest oxidizing agent?
Any species unstable to disproportionation?
E° of the Cr2O72-/Cr3+ couple is? +
E° of Cr3+/Cr2+ and Cr2+/Cr is? -
Cr2+ is reducing agent
oxidized to Cr3+
What do we know from this diagram?
Inorganic Chemistry Chapter 1: Figure 5.6
© 2009 W.H. Freeman
Inorganic Chemistry Chapter 1: Figure 5.7
© 2009 W.H. Freeman
Disproportionation reactions
Self-study 3, p.225 using data from Appendix 11, show that H2O2 is
unstable and disproportionates to O2 and H2O
2H2O2 → O2 + 2H2O
O22- → O2 + 2eO22- + 2e- → 2H2O
Given that:
O2 (g) + 2H2O (l) + 4e- → 4[OH-] (aq)
O2 (aq) + 2H2O(l) + 2e- → H2O2 aq) + 2[OH-](aq)
E° = +0.40 V
E° = -0.15 V
Show that ∆G° is spontaneous for the disproportionation of
H2O2 into O2 and H2O
Inorganic Chemistry Chapter 1: Figure 5.11
Pourbaix diagrams
E & pH map of
stability of species in
H2O
© 2009 W.H. Freeman
Application of redox reactions to extraction of elements from
their ores
Ellingham Diagram
Earth’s is oxidizing
environment ores:
Fe2O3
Fe3O4
SnO2
PbS
SnO2 + C → Sn + CO
PbS + O2 → PbO + SO2
PbO + C → Pb + CO
xM(s) +1/2O2 (g) → MxO(s) ∆G°(M, MxO)
C(s) + ½O2(g)  CO(g) ∆Go(C,CO)
For metal oxide to be reduced to metal
∆Go(C,CO) must be more negative than
∆Go(M, MxO)
Ellingham Diagram for the
Reduction of Metal Oxides,
Pyrometallurgy
1.as T increeases
stability of MxOy
Increases
2. CO is more stable
as T increases
3. C can reduce any
oxide to its element
when ∆Go(C,CO)
is more negative
than ∆Go(M, MxO)
For Al2O3 to be reduced
need > 2000 °C
See Box 6.1
for production
of Fe (steel)

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