Given
1- Cu2+ + 2e- ----- Cu, E(standard) = 0.337V
2- Cu2+ + e- ------- Cu+ , E(standard) = 0.153V
Electrode potential , E(standard) for the reaction , Cu+ + e- ------ Cu, will be
1- 0.38V 2- 0.52V
3- 0.90V 4-0.30V
For reaction 1.
Cu2+ + 2e- →Cu Eº1 = 0.337 V
Number of electron transfer (n) = 2
∆Gº for this reaction is :nFEº1
∆Gº1 = -2FEº1
For reaction 2.
Cu2+ + e- →Cu+ Eº2 = 0.153 V
Number of electron transfer (n) = 1
∆Gº for this reaction is :nFEº2
∆Gº2 = -1FEº2
For reaction 3
Cu+ + e- →Cu Eº3 = ?
Number of electron transfer (n) = 1
∆Gº for this reaction is :nFEº3
∆Gº3 = -1FEº3
Now ∆Gº3 = ∆Gº1- ∆Gº2
-1FEº3 = -2FEº1-(-1FEº2)
- Eº3 = -2Eº1+Eº2
Eº3 = 2Eº1 - Eº2
= 2 x 0.337 V - 0.153 V
= 0.521 V
Correct option is 2.
Cu2+ + 2e- →Cu Eº1 = 0.337 V
Number of electron transfer (n) = 2
∆Gº for this reaction is :nFEº1
∆Gº1 = -2FEº1
For reaction 2.
Cu2+ + e- →Cu+ Eº2 = 0.153 V
Number of electron transfer (n) = 1
∆Gº for this reaction is :nFEº2
∆Gº2 = -1FEº2
For reaction 3
Cu+ + e- →Cu Eº3 = ?
Number of electron transfer (n) = 1
∆Gº for this reaction is :nFEº3
∆Gº3 = -1FEº3
Now ∆Gº3 = ∆Gº1- ∆Gº2
-1FEº3 = -2FEº1-(-1FEº2)
- Eº3 = -2Eº1+Eº2
Eº3 = 2Eº1 - Eº2
= 2 x 0.337 V - 0.153 V
= 0.521 V
Correct option is 2.