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,…

Question

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 )

Aicharya · Accepted Answer

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

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…

Solution