Q. Consider the following relations for emf of an electrochemical cell: (a) emf of cell = (Oxidation potential of anode) − (Reduction potential of cathode) (b) emf of cell = (Oxidation potential of anode) + (Reduction potential of cathode) (c) emf of cell = (Reduction potential of anode) + (Reduction potential of cathode) (d) emf of cell = (Oxidation potential of anode) − (Oxidation potential of cathode) Which of the above relations are correct?
A) (c) and (a)
B) (a) and (b)
C) (c) and (d)
D) (b) and (d)
Explanation
[Solution from source] emf = E(red, cathode) − E(red, anode). Since oxidation potential = −reduction potential, this equals (oxidation potential of anode) + (reduction potential of cathode), and also (oxidation potential of anode) − (oxidation potential of cathode). The relation subtracting the reduction potential of the cathode from the anode's oxidation potential gets the sign wrong. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.