JEE Main | 2014ElectrochemistryMedium
Question
Given below are the half-cell reactions
Mn2+ + 2e- → Mn ; Eo = - 1.18 V
(Mn3+ + e- → Mn2+) ; Eo = + 1.51 V
The Eo for 3 Mn2+ → Mn + 2Mn3+ will be
Mn2+ + 2e- → Mn ; Eo = - 1.18 V
(Mn3+ + e- → Mn2+) ; Eo = + 1.51 V
The Eo for 3 Mn2+ → Mn + 2Mn3+ will be
Options
A.- 2.69 V; the reaction will not occur
B.- 2.69 V; the reaction will occur
C.- 0.33 V; the reaction will not occur
D.- 0.33 V; the reaction will occur
Solution
(1) Mn2+ + 2e → Mn ; Eo = - 1.18V ;
ᐃG1o = - 2F (- 1.18) = 2.36 F
(2) Mn3+ + e → Mn2+ ; Eo = + 1.51 V;
ᐃG2o = - F(1.51) = - 1.51F
(1) - 2 × (2)
3Mn2+ → Mn + 2Mn3+ ;
ᐃG3o = ᐃG1o - 2ᐃG2o
= [2.36 - 2(-1.51)] F
= (2.36 + 3.02) F
= 5.38 F
But ᐃG3o = - 2FEo
⇒ 5.38F = - 2FEo
⇒ Eo = - 2.69 V
As Eo value is negative reaction is non spontaneous.
ᐃG1o = - 2F (- 1.18) = 2.36 F
(2) Mn3+ + e → Mn2+ ; Eo = + 1.51 V;
ᐃG2o = - F(1.51) = - 1.51F
(1) - 2 × (2)
3Mn2+ → Mn + 2Mn3+ ;
ᐃG3o = ᐃG1o - 2ᐃG2o
= [2.36 - 2(-1.51)] F
= (2.36 + 3.02) F
= 5.38 F
But ᐃG3o = - 2FEo
⇒ 5.38F = - 2FEo
⇒ Eo = - 2.69 V
As Eo value is negative reaction is non spontaneous.
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