CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Consider the following relations for emf of an electrochemical cell :


(i) EMF of cell = (Oxidation potential of anode) - (Reduction potential of cathode)
(ii) EMF of cell = (Oxidation potential of anode) + (Reduction potential of cathode)
(iii) EMF of cell = (Reduction potential of anode) + (Reduction potential of cathode)
(iv) EMF of cell = (Oxidation potential of anode) - (Oxidation potential of cathode)
Which of the above relation(s) is correct?

A
(iii) and (i)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(i) and (ii)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(iii) and (iv)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(ii) and (iv)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D (ii) and (iv)
The EMF of a cell is given by the following expressions:

EMF of a cell = Reduction potential of cathode - Reduction potential of anode

= Reduction potential of cathode + Oxidation potential of anode

= Oxidation potential of anode - Oxidation potential of cathode

Hence, the relations (ii) and (iv) are correct.

Hence, the correct option is D.

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