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 = (Reductional potential of anode) + (Reduction potential of cathode)
(iv) EMF of cell = (Oxidation potential of anode) - (Oxidation potential of cathode)
Which of the above relations are correct?
[1 mark]

• A. (iii) and (i)
• B. (i) and (ii)
• C. (iii) and (iv)
• D. (ii) and (iv)

Answer 1

The correct option is D (ii) and (iv)

The difference in potentials of two half cells is known as the electromotive force (EMF) of the cell
EMF of cell = Reduction potential of cathode - Reduction potential of anode
= Reduciton potential of cathode + Oxidation potential of anode
= Oxidation potential of anode - Oxidation potential of cathode.