NEET | 2017ElectrochemistryHard

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)

Create a free account to view solution

View Solution Free
Topic: Electrochemistry·Practice all Electrochemistry questions

More Electrochemistry Questions

The molar conductance of a strong electrolyte at infinite dilution...A gas Cl2 at 1 atm is bubbled through a solution containing a mixture of 1 M Br-1 and 1 M F-1 at 25oC. If the reduction ...Which of the following facts is not true ?...A constant current flowed for 2 h through a potassium iodide solution, oxidizing the iodide ion to iodine. At the end of...When the electric current is passed through a cell having an electrolyte, the positive ions move towards cathode and neg...