Engineering Chemistry S Datta

236 ENGINEERING CHEMISTRY
MnO
–
4
+ 4H + + 3e MnO 2
+ 2H 2
O, ∈° = 1.70 V
MnO
–
4
+ 8H + + 5e Mn 2+ + 4H 2
O, ∈° = 1.51 V
The whole information can be summarised in the compact manner in a collinear diagram
as follows.
0.56 V 2.26 V 0.95 V 1.52 V –1.18V
MnO
–
4
⎯⎯→ MnO4
2–
⎯⎯→ MnO2 ⎯⎯→ Mn
3+
⎯⎯→ Mn
2+
⎯⎯→ Mn
In this diagram, the oxidation states of Mn decrease gradually from left to right. The
species H + , H 2
O etc. are not included in the diagram. The couple such as MnO – 4
/MnO 2
or
MnO – 4
/Mn 2+ or MnO 2
/Mn 2+ etc. are called skip couples and the corresponding potentials are
called skip-step potentials.
A Latimer diagram helps us to derive the standard potentials of such non-adjacent
couples. We know that ∆G° values corresponding to different steps can be added but the ∈°
values cannot be so added. Again ∆G° is related to ∈° as follows, ∆G° = – nF∈°. The overall ∆G°
of two successive steps will be the sum of the individual ∈° values to ∆G° values by multiplying
by the relevant – nF factor, adding them together and converting the sum back to ∈° for the
non-adjacent couple dividing by – nF for the overall electron transfer. Since the nF factor
cancels in this way, we can write
n ∈° → + n ∈°
∈° 1→
3 =
n + n
1 1 2 2 2→3
1 2
Besides calculating the ∈° values of non-adjacent couples, the Latimer diagram is also
helpful in predicting whether a particular species will undergo disproportionation or
comproportionation.
A species will disproportionate into its neighbour species if the potential on the left is
lower than that of the right species. On the other hand, a species will undergo
comproportionation if the potential on the left is higher than that of the right species in the
diagram. For example, we know that H 2
O 2
disproportionates into H 2
O and O 2
under acidic
conditions.
0.70 volt
1.76 volt
O ⎯⎯⎯→ H O ⎯⎯⎯→ H O−2
2
2 2
2
The feasibility of disproportionation of H 2
O 2
can be predicted by considering the
change in the ∈° values for the following reactions:
Overall reaction:
2H 2
O 2
⎯⎯→ 2H 2
O + O 2
(a) Oxidation: H 2
O 2
⎯⎯→ O 2
+ 2H + + 2e ∈ ° Ox = – .70 V
∴ ∆G° 1
= – nF∈° = – 2F × (– .70) = 1.4 F
(b) Reduction: H 2
O 2
+ 2H + + 2e ⎯⎯→ 2H 2
O ∈° Red
= 1.76 V
∴ ∆G° 2
= – nF∈° = – 2F × 1.76 = – 3.52 F
Adding (a) and (b),
2H 2
O 2
⎯⎯→ 2H 2
O + O 2
∆G° = – 2.12 F
Since the process is accompanied by decrease in Gibbs free energy, therefore disproportionation
will be spontaneous. On the other hand, for the change Ag 2+ ⎯⎯→
1.98 V Ag + ⎯⎯→
0.80 V
Ag
it can be shown by similar calculation that disproportionation of Ag + will not take place because
it will be accompanied by increase in ∆G° value. Hence, in this case the comproportionation
reaction Ag 2+ + Ag = 2Ag + will take place.
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