wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

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

A
E1<E2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
E1>E2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
E2=E1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
E1=E2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B E1>E2
Ecell=E0cell0.059n log[Zn2+][Cu2+]
E1=E00.0592 log0.010
E1=E0.0592 (2)=E+0.059
E2=E0.0592 log10.01=E0.059
Hence, E1>E2

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Nernst Equation
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon