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

A galvanic cell is constructed by coupling of an iron and a nickel electrode. If at 298 K
ENi2+(aq)/Ni(s)=0.24 V
EFe2+(aq)/Fe(s)=0.44 V
then calculate the EMF of the cell

A
+0.68 V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.68 V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
+0.2 V
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
0.2 V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C +0.2 V
Given that :
ENi2+(aq)/Ni(s)=0.24 V
EFe2+(aq)/Fe(s)=0.44 V
The standard reduction potentials of both are given. Comparing both,
ENi2+(aq)/Ni(s)>EFe2+(aq)/Fe(s)
So, nickel will reduce at cathode and iron will oxidise at anode

Ecell=SRP of substance reducedSRP of substance oxidisedEcell=0.24(0.44)=0.2 V

flag
Suggest Corrections
thumbs-up
3
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon