NEET | 2017ElectrochemistryEasy

Question

In the electrochemical cell :-
Zn|ZnSO4(0.01M)| |CuSO4(1.0 M)|Cu, the emf of this Daniel cell is E1. When the concentration of ZnSO4 is changed to 1.0M and that of CuSO4 changed to 0.01M, the emf changes to E2. From the followings, which one is the relationship between E1 and E2? (Given, RTF= 0.059)

Options

A.

E1 < E2

B.

E1 > E2

C.

E2 = 0 ≠E1

D.

E1 = E2

Solution

For cell
Zn|ZnSO4(0.01M)||CuSO4(1M)|Cu
Cell reaction →Zn + Cu+2 → Zn+2 + Cu
E1 = Eo - 0.0592logZn+2Cu+2
E1 = Eo - 0.0592log0.011
=Eo-0.0592log1100    ......(1)
For cell
Zn|ZnSO4(1M)||CuSO4(0.01M)|Cu
E2 = Eo– - 0.0592log10.01
=Eo-0.0592log 100 ....(2)     E1 >E2
Option (2)

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