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 relations are correct?
- (iii) and (i)
- (i) and (ii)
- (iii) and (iv)
- (ii) and (iv)
Answer
(4) (ii) and (iv)
Solution
As per NCERT, Ecell = Ecathode - Eanode (both reduction potentials).
∵ Oxidation potential of an electrode = − (its reduction potential)
⇒ -Eanode = oxidation potential of anode
∴ EMF = (Oxidation potential of anode) + (Reduction potential of cathode) ⇒ (ii) correct
Also Ecathode = − (oxidation potential of cathode)
∴ EMF = (Oxidation potential of anode) − (Oxidation potential of cathode) ⇒ (iv) correct
(i) and (iii) do not follow from this relation.
Answer: (4)