Transcript Document

Oxidation-Reduction
Biology
Industry
Environment
Biology
Biology
0
Industry
Extraction of elements Synthesis of different compounds
Environment
Redox reactions - transfer of electrons between species.
All the redox reactions have two parts:
Oxidation
Reduction
• The Loss of Electrons is Oxidation.
• An element that loses electrons is said to be oxidized.
• The species in which that element is present in a
reaction is called the reducing agent.
• The Gain of Electrons is Reduction.
• An element that gains electrons is said to be reduced.
• The species in which that element is present in a
reaction is called the oxidizing agent.
Mg
Mg2+
Cu2+
Cu
Balancing Redox Equations
Fe2+ + MnO4- + H+
1.
2.
3.
4.
5.
6.
7.
8.
9.
Mn2+ + Fe3+ + H2O
Assign oxidation numbers to each atom.
Determine the elements that get oxidized and reduced.
Split the equation into half-reactions.
Balance all atoms in each half-reaction, except H and O.
Balance O atoms using H2O.
Balance H atoms using H+.
Balance charge using electrons.
Sum together the two half-reactions, so that: e- lost = egained
If the solution is basic, add a number of OH- ions to each
side of the equation equal to the number of H+ ions shown
in the overall equation. Note that
H+ + OH-  H2O
Example
Fe2+ + MnO4- + H+
MnO4-
Mn2+
(+7)
(+2)
Fe2+
Fe3+
MnO4- + 8H+ + 5e
5Fe2+
Mn2+ + Fe3+ + H2O
Reduction half reaction
Oxidation half reaction
Mn2+ + 4H2O
5Fe3+ +5e
5Fe2+ + MnO4- + 8H+
Mn2+ + 5Fe3+ + 4H2O
Nernst Equation
aOx1 +bRed2
Q=
[Red1]a’ [Ox2]b’
[Ox1]a [Red2]b
a’Red1 + b’Ox2
E = E0 - RT ln Q
nF
E0 = Standard Potential
R = Gas constant 8.314 J/K.mol
F- Faraday constant = 94485 J/V.mol
n- number of electrons
G0 = - n F  E0
Note: if G0 < 0, then  E0 must be >0
A reaction is favorable if  E0 > 0
(a) 2H+ (aq) + 2e
(b) Zn2+ (aq) + 2e
(a-b) 2H+ (aq) + Zn(s)
E0 (H+, H2) = 0
H2(g)
Zn(s)
E0 (Zn2+, Zn) = -0.76 V
Zn2+(aq) + H2(g)
Reaction is favorable
E0 = +0.76 V
Hydrogen Electrode
• consists of a platinum
electrode covered with
a fine powder of
platinum
around
which H2(g) is bubbled.
Its potential is defined
as zero volts.
Hydrogen Half-Cell
H2(g) = 2 H+(aq) + 2 ereversible reaction
Galvanic Cell
-
Diagrammatic presentation of potential data
Latimer Diagram
Frost Diagram
Latimer Diagram
* Written with the most oxidized species on the left, and the most
reduced species on the right.
* Oxidation number decrease from left to right and the E0 values
are written above the line joining the species involved in the
couple.
A+5
w
B+3
x
C+1
y
D0
z
E-2
0.44
What happens when Fe(s) react with H+?
Iron
+2 and +3
2+
Fe
+ 2e
Fe2+
Fe3+ + 3e
Fe
+0.036
-2 x F x -0.44 = 0.88 V
-0.440-0.440
Fe
Fe3+ + e
-0.771 +0.77
-1 x F x +0.771 = -0.771 V
+ 0.109 F
= -3 x F x –0.036
Fe3+
Fe
+0.44
Fe2+
-0.036
Fe3+
+0.77 Fe2+ -0.44
G = -nFE
Fe
-0.036
Fe3+ +0.77 Fe2+ -0.44
0.36
[Fe(CN)6]3-
[Fe(CN)6]4-
Fe
-1.16
Oxidation of Fe(0) to Fe(II) is considerably more favorable in the
cyanide/acid mixture than in aqueous acid.
(1) Concentration
(2) Temperature
(3) Other reagents which are not inert
Oxidation of elemental copper
Latimer diagram for chlorine in acidic solution
ClO4
+7
- +1.2
ClO3
- +1.18
+5
HClO2
+1.65
HClO
+1
+3
- +1.2
ClO4
+1.63
Cl2
0
+1.36
Cl-1
ClO3-
Can you balance the equation?
HClO
+1.63
Cl2
balance the equation
2 HClO(aq) + 2 H+(aq) + 2 e-
Cl2(g) + 2 H2O(l) E0 = +1.63 V
How to extract E0 for nonadjacent oxidation state?
E0 =?
ClO4
+7
- +1.2
ClO3
- +1.18
+5
HClO2
+1.65
+3
HClO
+1.63
+1
+1
Identify the two reodx couples
Cl2
+1.36
Cl-1
-1
00
1
2
Find out the oxidation state of chlorine
Write the balanced equation for the first couple
HClO(aq) + H+(aq) + e
½ Cl2(g) + H2O(l)
+1.63 V
Write the balanced equation for the second couple
G = G’ + G’’
’FE’ - ’’FE’’
-FE
=

½ Cl2(g) + e
Cl (l)
+1.36 V
E = ’E’+ ’’E’’
’+ ’’
E = 1.5 V
Latimer diagram for chlorine in basic solution
ClO4
- +1.2
ClO3
+5
+7
ClO4
- +1.18
-
+0.37
ClO3
HClO2
+1.65
HClO
+1
+3
-
+0.3
ClO2
-
+1.63
+0.68
ClO-
Cl2
+1.36
Cl-1
0
+0.42
Cl2
+1.36
Cl-
+0.89
+0.89
ClO-
+0.42
Cl2
Balance the equation…
2ClO- (aq) + 2H2O(l) + 2eFind out the E0
+0.89
Cl2(g) + 4OH-(aq)
E0 = 0.42 V
Disproportionation
Element is simultaneously oxidized and reduced.
2 M+(aq)
E0
ClO4
+0.37
ClO3
-
E0’
2M+(aq)
M(s)
-
M(s) + M2+(aq)
+0.3
ClO2
-
+0.68
ClO-
M2+(aq)
+0.42
Cl2
+1.36
Cl-
‘the potential on the left of a species is less positive than that on
the right- the species can oxidize and reduce itself, a process known
as disproportionation’.
+0.42
ClO-
Cl2
+1.36
Cl-
ClO4
-
+0.37
+0.42
ClO-
ClO3
Cl2
-
+0.3
+1.36
ClO2
+0.68
ClO-
+0.42
Cl2
+1.36
Cl-
Cl-
Cl2(g) + 2 e2Cl-(aq)
2ClO- (aq) 2H2O(l) +2e-
Cl2 + 2OH-
-
+1.36
Cl2(g) + 4OH-(aq)
ClO- + Cl- + H2O
E = E0 (Cl2/Cl-) –E0 (ClO-/Cl2) = 1.36 - +0.42 = 0.94
Reaction is spontaneous
+0.42
Latimer diagram for Oxygen
1.23 V
Disproportionation
the potential on the left of a species is less positive than that on the
right- the species can oxidize and reduce itself, a process known
as disproportionation.
Is it spontaneous?
H2O2(aq) + 2H+ (aq) +2e-
2H2O(aq)
+1.76 V
O2(g) + 2H+(aq) +2 e-
H2O2(aq)
+0.7 V
H2O2(aq)
O2 (g) + H2O(l)
Yes the reaction is spontaneous
+0.7 V
Another example…
2 Cu+(aq)
Cu2+(aq) + Cu(s)
Cu+(aq) + e-
Cu(s)
E0 = + 0.52 V
Cu2+(aq) + e-
Cu+(aq)
E0 = =0.16 V
Cu(I) undergo disproportionation in aqueous solution
Comproportionation reaction
Reverse of disproportionation
Ag2+(aq) + Ag(s)
2Ag+(aq)
E0 = + 1.18 V
…we will study this in detail under Frost diagram
Frost Diagram
Arthur A. Frost
Graphically illustration of the stability of different oxidation
states relative to its elemental form (ie, relative to oxidation
state= 0)
XN + Ne-
NE0 = -G0/F
X0
Look at the Latimer diagram of nitrogen in acidic solution
a
b
c
f
g
d
e
h
c
a
e
b
d
g
G = G’ + G’’
-nFE = -n’FE’ - n’’FE’’
E = n’E’+ n’’E’’
n’+n’’
N2
f
h
a NO3- + 6H+ + 5e-
½ N2 + 3H2O
E0 = 1.25V
b ½ N2O4 + 4H+ + 4e-
½ N2 + 2H2O
E0 = 1.36V
c HNO2 + 3H+ + 3e-
½ N2 + 2H2O
E0 = 1.45V
d NO + 2H+ + 2e-
½ N2 + H2O
E0 = 1.68V
e ½ N2O + H+ + e-
½ N2 + ½ H2O
E0 = 1.77V
f ½ N2 + 2H+ + H2O + e-
NH3OH+
E0 = -1.87V
g ½ N2 + 5/2 H+ + 2e-
½ N2H5+
E0 = -0.23V
h ½ N2 + 4H+ + 3e-
NH4+
E0 = 0.27V
Oxidation state: species
NE0, N
N(V): NO3-
(5 x 1.25, 5)
N(IV): N2O4
(4 x 1.36, 4)
N(III): HNO2
(3 x 1.35, 3)
N(II): NO
(2 x 1.68, 2)
N(I): N2O
(1 x 1.77, 1)
N(-I): NH3OH+
[-1 x (-1.87), -1]
N(-II): N2H5+
[-2 x (-0.23), -2]
N(-III): NH4+
(-3 x 0.27, -3)
Frost Diagram – N2
What do we really get from the Frost diagram?
the lowest lying species corresponds to
the most stable oxidation state of the
element
Slope of the line joining any two points is equal to the std
potential of the couple.
N’E0’
N’
N”E0’’
N’’
0’-N”E0”
N’E
Slope = E0=
N’-N”
E0 of a redox couple
HNO2/NO
3, 4.4
2, 3.4
0’-N”E0”
N’E
Slope = E0=
N’-N”
1V
Oxidizing agent? Reducing agent?
The oxidizing agent - couple with more positive slope - more
positive E
The reducing agent - couple with less positive slope
If the line has –ive slope- higher lying species – reducing agent
If the line has +ive slope – higher lying species – oxidizing agent
Identifying strong or weak agent?
NO – Strong oxidant than HNO3
Disproportionation
Element is simultaneously oxidized and reduced.
2 M+(aq)
E0
M(s)
M(s) + M2+(aq)
2M+(aq)
E0’
M2+(aq)
‘the potential on the left of a species is less positive than that on
the right- the species can oxidize and reduce itself, a process known
as disproportionation’.
Disproportionation
What Frost diagram tells about this reaction?
A species in a Frost diagram is unstable with respect to disproportionation
if its point lies above the line connecting two adjacent species.
Disproportionation…. another example
Comproportionation reaction
Comproportionation is spontaneous if the intermediate species
lies below the straight line joining the two reactant species.
Favorable?
NE0
Disproportionation
Comproportionation
In acidic solution…
Mn and MnO2
Mn2+
Rate of the reaction hindered
insolubility?
In basic solution…
MnO2 and Mn(OH)2
Mn2O3
From the Frost diagram for Mn….
* Thermodynamic stability is found at the bottom of the diagram.
Mn (II) is the most stable species.
* A species located on a convex curve can undergo disproportionation
example: MnO43MnO2 and MnO42- (in basic solution)
•Any species located on the upper right side of the diagram will be
a strong oxidizing agent. MnO4- - strong oxidizing agent.
•Any species located on the upper left side of the diagram will be
a reducing agent. Mn - moderate reducing agent.
* Although it is thermodynamically favorable for permanganate
ion to be reduced to Mn(II) ion, the reaction is slow except in the
presence of a catalyst. Thus, solutions of permanganate can be
stored and used in the laboratory.
* Changes in pH may change the relative stabilities of the species.
The potential of any process involving the hydrogen ion will
change with pH because the concentration of this species is
changing.
* Under basic conditions aqueous Mn2+ does not exist. Instead
Insoluble Mn(OH)2 forms.
*All metals are good reducing agents
*Exception: Cu
*Reducing strength: goes down
smoothly from Ca to Ni
*Ni- mild reducing agent
*Early transition elements: +3 state
Latter +2 state
*Fe and Mn – many oxidation states
*High oxidation state:
Strong oxidizing agents