Electron Transfer Reactions. Electron transfer reactions occur by one of two fundamental mechanisms...
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![Page 1: Electron Transfer Reactions. Electron transfer reactions occur by one of two fundamental mechanisms In an inner sphere mechanism, there is a common bridging.](https://reader036.fdocuments.in/reader036/viewer/2022062407/56649e4e5503460f94b4491d/html5/thumbnails/1.jpg)
Electron Transfer Reactions
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Electron transfer reactions occur by one of two fundamental mechanisms
In an inner sphere mechanism, there is a common bridging ligand, and the electron is transferred from the reductant to the oxidant through the bridging ligand
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In an outer sphere mechanism, there is an encounter between the reductant and the oxidant. The electron is transferred from one to the other whilst there is no change in the coordination sphere of either.
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Common bridging ligands
Cl
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Common bridging ligands
Oxide and hydroxide. But water is a very poor bridging ligand
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Common bridging ligands
Ligands which have more than one donor atom (called ambident nucleophiles)
S C N_
Other examples
N
O O
_S
S
OO
O
2-
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Common bridging ligands
Ligands which have more than one donor atom separated by a delocalised electron system
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How do we distinguish an inner sphere from an outer sphere mechanism?
Henry Taube’s classic experiment
[CoIII(NH3)5Cl]2+
[CrII(H2O)6]2+
Inert: d6 Co(III)
Labile: d4 Cr(II)
Cl- is a bridging ligand; neither H2O nor NH3 are
Observation: products (in acidic medium) are
[CoII(H2O)6]2+ + [CrIII(H2O)5Cl]2+
and this allowed him to deduce the mechanism
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The reaction could have occurred through an inner sphere pathway:
[CoIII(NH3)5Cl]2+ + [CrII(H2O)6]2+ [CoIII(NH3)5ClCrII(H2O)5]4+
electron transfer
[CoII(NH3)5ClCrIII(H2O)5]4+
break apart
Co(II) is labileCr(III) is inert
[CoII(NH3)5(H2O)]2+ + [CrIII(H2O)5Cl]2+
hydrolysis, in acid
[CoII(H2O)6]2+ + 5NH4+
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Or it could have gone through an outer sphere pathway:
[CoIII(NH3)5Cl]2+ + [CrII(H2O)6]2+
electron transfer
Co(II) is labileCr(III) is inert
hydrolysis, in acid
[CoII(H2O)6]2+ + 5NH4+ + Cl
[CoII(NH3)5Cl]2+ + [CrIII(H2O)6]2+
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[CoII(H2O)6]2+ [CrIII(H2O)6]2+
Observation: products (in acidic medium) are
[CoII(H2O)6]2+ + [CrIII(H2O)5Cl]2+
So there would have had to be a subsequent anation of [CrIII(H2O)6]2+ by Cl
Rate of [CrIII(H2O)6]2+ by Cl : k = 2.9 10-8 M-1 s-1
Rate of electron transfer: k = 6 10+5 M-1 s-1
Rate of electron transfer is 13 orders to magnitude faster than rate of anation
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Rate of electron transfer is 13 orders to magnitude faster than rate of anation
The reaction could not have proceeded through an outer sphere mechanism
Taube’s postulate:
A reaction will have proceeded through an inner sphere mechanism if one of the products is substitution inert and it retains the bridging ligand
i.e., the rate of the electron transfer reaction is much faster than the rate of formation of the product by subsequent anation
Corollary:
If the rate of electron transfer is much faster than the rate of ligand substitution on either metal ion, the reaction must proceed through an outer sphere mechanism
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ExampleV2+ is inert (d3 ion, high LFSE – like Cr3+)Ru3+ is inert (2nd transition series)
Now
[RuIII(NH3)5Br]2+ + [VII(H2O)6]2+
e transfer, k = 5.1 103 M-1 s-1
[RuII(NH3)5Br]+ + [VIII(H2O)6]3+
But[VII(H2O)6] + Br [VII(H2O)5Br]+ + H2O
k = 5.0 101 s-1
Hence cannot form the inner sphere complex fast enough – the anation reaction is too slow. The reaction must have been outer sphere.
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Barriers to electron transfer
Donor Acceptor
e-
ΨD ΨA
Rate of an electronic transition HDA>2 where
DADA A DˆH H d
(Atkins, 8th ed., Chapter 9; 9th ed., Chapter 8) Hamiltonian operator that describes the coupling of the two wavefunctions
• Distance
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If the coupling is relatively weak,2 o 2
DA DArH H e
edge-to-edge distance
between D and A
Parameter that measures the
sensitivity of the coupling to
distance
Electron coupling when A and D are in direct
contact (r = 0)
and it turns out that the rate constant for electron transfer between D and A is
2 1/ 2o 3DA /
ET
2
4
r
G RTH e
k eh RT
(Atkins, 8th ed., p. 897; not in 9th ed.)
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2 1/ 2o 3DA /
ET
2
4
r
G RTH e
k eh RT
n
D
A
Will be a constant if D and A are the same
ETln( ) constantk r
ln kET
r
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Nature often uses large, conjugated macrocycles to do electron transfer
Examples:
Porphyrins Chlorophylls
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Effectively increases radius of D and A, cutting down separation, and hence increasing rate of e- transfer
effective distance
distance between metals
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Q-cytochrome c oxidoreductase – Complex III or the bc1 complex
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2 1/ 2o 3DA /
ET
2
4
r
G RTH e
k eh RT
/ET N E
G RTk e
which is often written in simplified form as
nuclear frequency
factor
electronic factor
0 ≤ κ ≤ 1
• For fast electron transfer, maximise κE
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1/ 23
E 4
RT
• For fast electron transfer, maximise κE
minimise the reorganisation energy, λ, of inner and outer sphere
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use appropriate electronic configurations
When an e- is transferred from the D to the A molecule, it cannot change its spin.
In many cases this is not a problem:
[Co(phen)3]3+ + [Co(bipy)3]2+ → [Co(phen)3]2+ + [Co(bipy)3]3+
low spin low spin low spin low spin
eg
t2g
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But in some cases - especially if there is a change of spin state - this is a barrier to electron transfer
[Co(NH3)4Cl2]3+ + [Co(OH2)6]2+ → [Co(NH3)4Cl2]2+ + [Co(OH2)6]3+
low spin high spin high spin low spinS = 0 S = 3/2 S = 3/2 S = 0
eg
t2g
This cannot be a single step
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[Co(NH3)4Cl2]3+ + [Co(OH2)6]2+ → {[Co(NH3)4Cl2]3+}* + [Co(OH2)6]2+
low spin high spin excited state1 high spin
eg
t2g
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{[Co(NH3)4Cl2]3+}* + [Co(OH2)6]2+→ {[Co(NH3)4Cl2]3+}* + [Co(OH2)6]2+
excited state1 high spin excited state2 high spin
eg
t2g
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{[Co(NH3)4Cl2]3+}* + [Co(OH2)6]2+→ [Co(NH3)4Cl2]2+ + [Co(OH2)6]3+
excited state2 high spin high spin excited state3
eg
t2g
[Co(OH2)6]3+
low spin
and finally
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Faster electron transfer occurs if an electron is removed from and added to a non-bonding orbital (less reorganisational energy λ)
Recall that in complexes with σ only ligands the t2g orbitals are non-bonding and eg orbitals are antibonding
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Compare self-exchange rate constants:
[Cr(OH2)6]3+/2+ t2g3/t2g
3eg1 1 × 10-5 M-1s-1
[Fe(OH2)6]3+/2+ t2g3eg
2/t2g4eg
2 1.1 × 10-5 M-1s-1
[Ru(OH2)6]3+/2+ t2g5/t2g
6 20 × 10-5 M-1s-1
electrons going in and out of the t2g
orbitals makes for fast
electron transfer
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The Inner Sphere Mechanism
The rate-determining step could be
• the formation of the bridged complex (i.e., the precursor complex)
• the electron transfer step (most commonly rate determining)
• the break-up of the successor complex
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We can often rationalise which step will be rate-determining
[RuIII(NH3)5Cl]2+ + [CrII(H2O)6]2+ [RuIII(NH5)5ClCrII(H2O)5]4+ + H2OK
e transfer
[RuII(NH5)5ClCrIII(H2O)5]4+ + H2O
k1
[RuII(NH3)5(H2O)]2+ + [CrIII(H2O)5Cl ]2+
Both Ru(II) and Cr(III) are inert – so we would expect the breakup of the successor complex to be rate-limiting
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Changes in mechanism are often accompanied by substantial changes in rate
The following reactions must be outer sphere reactions (why?)
oxidant reductant k
[CoIII(NH3)5(H2O)]3+ [RuII(NH3)6]2+ 3.0 M-1 s-1
[CoIII(NH3)5(OH)]2+ [RuII(NH3)6]2+ 0.04 M-1 s-1
and the hydroxo complex reacts some 100 times slower than the aqua complex
oxidant reductant k
[CoIII(NH3)5(H2O)]3+ [CrII(H2O)6]2+ 0.1 M-1 s-1
[CoIII(NH3)5(OH)]2+ [CrII(H2O)6]2+ 1.5 106 M-1 s-1
The hydoxo complex in this case must be reacting through an inner sphere mechanism
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For the first row of the d block:
If electron transfer is rate-determining, then the rate depends markedly on
1. the identity of the metal ion
These are the same factors that controlled rate in the outer sphere mechanism (see later)
2. the nature of the bridging ligands
The ability of the ligand to act as an electron conductor
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2. The ability of the ligand to act as an electron conductor
N [RuII(NH3)4(H2O)]
O
O[(H3N)5CoIII]
N [RuII(NH3)4(H2O)]
O
O
[(H3N)5CoIII]
para
meta
k = 100 M-1 s-1
k = 1.6 10-3 M-1 s-1
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If formation of the precursor complex is rate determining, then the rate is usually not very sensitive to the nature of the bridging ligand
This is because the ligand substitution reactions of the first row d metals are usually dissociative
hence does not depend strongly on the nature of the entering ligand
[L5MoxX] + [L5MredY]
[L5Mox] + X + [L5MredY]
[L5MoxYMredL5]
[L5MredYMoxL5]
etc
rate limiting
Example: V2+(aq) is oxidised to V3+
(aq) by a long series of Co3+ oxidants with different bridging ligands
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Inner sphere mechanism always suspected if good bridging ligands are available:
Cl- Br- I- N3- CN-
N
N N
N
N
NMe2
pyrazine 4,4'-bipyridyl N,N-dimethylaminopyridine
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The Outer Sphere Mechanism
Rudolph Marcus
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Energy changes during electron transfer – the Frank-Condon Principle
Electron transfer is fast compared to nuclear motion Hence the nuclei are essential frozen in space during the electron transfer step
Now consider the following situation:
[FeII(H2O)6]2+ + [*FeIII(H2O)6]3+ [FeIII(H2O)6]3+ + [*FeII(H2O)6]2+
Fe(II)-O = 2.02-2.07 Å Fe(III)-O = 2.00 Å
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So suppose
Fe(II)OH2 + *Fe(III)OH2
e
Fe(III)OH2 + *Fe(II)OH2
bond too long forFe(III)
bond too short forFe(II)
spontaneous (exothermic)
Fe(III)OH2 + *Fe(II)OH2
getting energy from nothing – which would be a violation of the First Law
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e transfer
What actually happens:
Fe(II)OH2 + *Fe(III)OH2
shrinks stretches ENDOTHERMIC
Fe(III)OH2 + *Fe(II)OH2
rearrangement EXOTHERMIC
Fe(III)OH2 + *Fe(II)OH2
G‡
Frank-Condon Energy
Fe(II)OH2 + *Fe(III)OH2
bonds now about the same length
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A given system can (in principle) be represented by a wavefunction,
Represents hypersurface of the reactants, reac
represents changes to all structural parameters (bond lengths, angles, torsions, etc) during the reaction
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20 0( )E k x x E
Parabolic function because bond stretching and angle bending terms can be approximated by Hooke’s law behaviour:
x0
E0
For example, [Fe3+(H2O)6] with a short Fe–O bond, xo
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products
For example, [Fe2+(H2O)6] with a long Fe–O bond
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Electron transfer from the reactants to products can occur when the reactant deforms along the reaction coordinate until it structurally resembles the product (at )
For example, Fe3+––O must stretch
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λ is the reorganisation energy, the energy that would be expended to reorganise the reactant form to the product form if no electron transfer took place
λ
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For an exothermic (exergonic) reaction:
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‡
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‡
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and rearranging:
Show, using similar reasoning, that for an endothermic (endergonic) reaction
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So in general
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2o
2o
o 2
2
2 o o2
14
4
( )
41
( 2 )4
GG
G
G
G G
So when ΔGo << λ 2 o
o
o
1( 2 )
4
( 2 )41
( 2 )4
G G
G G
G G
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0
10
20
30
40
50
60
70
-40 -30 -20 -10 0 10 20 30 40
Circles o1
( 2 )4 G G
Diamonds:
ΔGo /kJ mol-1
λ = 200 kJ mol-1
ΔG
‡ /k
J m
ol-1
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-300
-200
-100
0
100
200
300
400
500
600
700
-600 -400 -200 0 200 400 600
Circles o1
( 2 )4 G G
Diamonds:
ΔGo /kJ mol-1
λ = 200 kJ mol-1
ΔG
‡ /k
J m
ol-1
Simplified eqt only good in this region, i.e., when |ΔGo| << λ
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G‡ Go
But Go = nFEo
G‡ Eo
G‡ Eo
The height of the activation energy barrier to electron transfer becomes smaller and the reaction becomes faster as the redox potential of the couple increases
o1( 2 )
4G G
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Reaction becomes activationless when ΔGo = -λ
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As ΔGo becomes very large (ΔGo < -λ), ΔG‡ increases again. Hence kET goes through a maximum when ΔGo = -λ
/2
/2
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0
100
200
300
400
500
600
700
-600 -400 -200 0 200 400 600
ΔGo /kJ mol-1
λ = 200 kJ mol-1
ΔG
‡ /k
J m
ol-1
ΔG‡ = 0 when Go = -λ (here, -200 kJ mol-1)...
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Rel
ativ
e ra
te
Go+ve -ve
Inverted Marcus region
...so RATE goes through a maximum
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Inverted Marcus region demonstrated experimentally by Harry Gray (Caltech) 1990 for an iridium complex
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• For fast electron transfer, minimise the reorganisation energy, λ, of inner and outer sphere
Reorganisation of solvent often major component
Hexaaqua ions: λ > 100 kJ mol-1
Redox centres in proteins (buried, shielded from solvent):λ ~ 25 kJ mol-1
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Reorganisation of solvent often major component
Hexaaqua ions: λ > 100 kJ mol-1
Redox centres in proteins (buried, shielded from solvent):λ ~ 25 kJ mol-1
Fe porphyrin cytochrome c
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Bulky, hydrophobic chelating ligands shield the metal from solvent,lowering λ
N N
bipyridyl (bipy)
[Ru(OH2)6]3+/2+ kET = 20 M-1s-1
[Ru(bipy)3]3+/2+ 4 × 108 M-1s-1
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The speed of an electron transfer reaction depends on
• the reactivity of the complex
This can be measured by the self-exchange rate
[FeII(H2O)6]2+ + [*FeIII(H2O)6]3+ [FeIII(H2O)6]3+ + [*FeII(H2O)6]2+
• Eo
because G‡ Eo
• other factors such as the collision geometry and the collision frequency (the pre-exponential factor in the Arrhenius equation)
The Marcus cross-relationship
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Suppose, as a special case, we have a redox reaction where the electron donor, D, and the acceptor A, reversibly form an encounter complex; where the rate determining step is the electron transfer step; and where break up of the successor complex is fast
DA
ET + -
+ - + -
D + A DA
DA D A
D A D + A
K
k
fast
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DA
ET + -
+ - + -
D + A DA
DA D A
D A D + A
K
k
fast
ET obs 0 0
d[ ] d[ ] = [DA] = [D] [A]
d d
P Pk k
t t
DA t t 0 0
DA 0 0
[DA] [DA] [D] [A] [D] [A]
[DA] = [D] [A]
K
K
Assume [D]t ~ [D]0
[A]t ~ [A]0
ET DA 0 0
d[ ] = [D] [A]
d
Pk K
t
obs ET DA= k k K
DA /obs DA N E= G RTk K e
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2
2
2 22
2
= 1 + 4
4
1 2
4
4 2 4
o
o
o o
o o
GG
G
G G
G G
If ΔGo « λ then ΔGo2/4λ is negligibly small
=4 2
oGG
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For the reactionA + A → A+ + A-
ΔGo = 0, so
AAAA =
4G
AA DD/ 4 / 4AA AA DD DD
JJ N(JJ) E(JJ)
= =
=
RT RTk Z e k Z e
where Z
DD AADA
+ =
2
and similarly DDDD =
4G
Hence
We will assume that λDA is the arithmetic mean of λAA and λDD
DD AADA = + +
2 8 8
oDAG
G
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DA DD AA/ 2 /8 /8obs DA DA =
oG RT RT RTk K Z e e e
DA /DA DA DAln( ) = - / or =
oG RToK G RT K e
Since
oDA /o
DA DA DA
1/2 1/21/2 DD AA
obs DA DA DA 1/2 1/2DD AA
1/2 1/2 1/2DA DADA DD AA1/2 1/2
DD AA
2 2DA DA
obs DA DD AADD AA
ln( ) = / or =
=
=
=
G RTK G RT K e
k kk K Z K
Z Z
K ZK k k
Z Z
K Zk K k k f where f
Z Z
Marcus Cross Relationship
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obs DA DD AA
2 2DA DA
DD AA
=
1
k K k k f
where
K Zf
Z Z
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Example
Predict kobs for
[CoIII(bipy)3]3+ + [CoII(terpy)2]2+ [CoII(bipy)3]2+ + [CoIII(terpy)2]3+
N N
N
N
N
bipy
terpy
Data (0 oC)
Self-exchange rate constants[CoII/III(bipy)3]2+/3+kAA = 9.0 M-1s-1
[CoII/III(terpy)2]2+/3+ kDD = 48 M-1s-1
Redox potentials:bipy complex 0.34 Vterpy complex 0.31 V
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cell 0.34 V - 0.31 V = 0.03 VE
12lnG RT K nFE
DA
-1 -1
-1 -1
exp
1 96485 C mol 0.03 J Cexp
8.315 J K mol × 273 K
3.58
nFEK
RT
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12 AA DD DA
-1 -1
9 48 3.58 1
39 M s
k k k K f
Experimental value: 62 M-1 s-1
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Example 2
For [CoIII(NH3)5Cl]2+ + [EuII(H2O)8]2+ [CoII(NH3)5Cl]+ + [EuIII(H2O)8]3+
[CoII(H2O)6]2+
apply Marcus equation and find k12 = 2.7 M-1s-1
The experimental value is 390 M-1s-1
The Marcus theory was developed for an outer sphere reaction. The experimental value is very different to the theoretical value. This suggests that this reaction did not occur through an outer-sphere mechanism, but rather through an inner sphere mechanism with Cl as bridging ligand