NEET2017ChemistryElectrochemistryActual
In the electrochemical cell: Zn ZnSO 4 0 .01 M CuSO 4 1 .0   M Cu , the emf of this Daniel cell is E 1 . When the concentration ZnSO 4 is changed to 1 .0 M and that of CuSO 4 changed to 0 .01 M , the emf changes to E 2 . From the following, which one is the relationship between E 1 and E 2 ? (Given, RT F = 0 .059 )
Options
- AE 1 = E 2
- BE 1 < E 2
- CE 1 > E 2
- DE 2 = 0   ≠   E 1
Correct answer
C. E 1 > E 2
Step-by-step solution
Zn ZnSO 4 0 .01   M CuSO 4 1 .0 M Cu E = E cell o − 2 .303 RT 2 F log [ Zn + 2 ] [ Cu + 2 ] ∴   E 1 = E cell o - 2 .303 RT 2 × F × log ⁡ 0 .01 1 When concentrations are changed ∴   E 2 = E cell o - 2 .303 RT 2 F × log ⁡ 1 0 .01 i.e., E 1 > E 2