Engineering Chemistry S Datta

ELECTROCHEMICAL CELLS 261
Ans. We have
Adding the above two equations,
∆G° (i)
= – nFE° = – 2F (– 0.44) = 0.88 F
∆G° (ii)
= – nFE° = – F(0.77) = – 0.77 F
∆G° (iii)
= ∆G° (i)
+ ∆G° (ii)
= 0.88 + (– 0.77) F = + 0.11 F
So ∆G° (iii)
is the value for the reaction,
Fe +3 + 3e ⎯⎯→ Fe
∈° = ∆ G ° . F
= + 011
= – 0.04 V
−nF
− 3 F
The Latimer diagram of the above system is Fig. 10.19.
Fe +3 0.77
Fe +2 –0.44
Fe
–0.04
Fig. 10.19 Latimer diagram for Fe species.
It is conventional to put highly oxidised form of the element on the left and the elements
with lower oxidation state are put successively in the Latimer diagram.
Q. 26. In alkaline solution (pH = 14) the Latimer diagram for chlorine is
– + 0.37 – + 0.30 – + 0.68 – + 0.42 + 1.36 –
CO l 4 ClO3 ClO2 ClO Cl2
Cl
Calculate from the above diagram ∈° value for the reaction,
ClO – ⎯⎯→ Cl – .
Ans. ClO – + e 1
⎯⎯→
2 Cl 2 , ∈° = + 0.42 V; ∆G° I
= – (+ 0.42) F = – 0.42 F
1
2 Cl 2 ⎯⎯→
+ e
Cl – , ∈° = + 1.36 V; ∆G° II
= – (+ 1.36) F = – 1.36 F
Adding total ∆G for the reaction = – 1.78 F
From the relation, ∆G° = – nFE°
∈° (ClO – /Cl – ) = ∆G
I + ∆G
− nF
II
. F
= − 178
= 0.89 V
− 2 F
Q. 27. Use the following data to draw a Frost diagram.
∈°(Mn +2 , Mn) = – 1.18 V,
∈°(Mn +3 , Mn +2 ) = 1.51 V.
Ans. ∈º(Mn +2 , Mn) = – 1.18 V
∈°(MnO 2
, Mn +3 ) = 0.95 V
3∈°(Mn +3 , Mn) = – 0.85 V
4∈°(MnO 2
, Mn) = 0.10 V
?
2∈° = – 2.36 V