Question

Consider the following relations for emf of an electrochemical cell:
$\text{(i) EMF of cell = (Oxidation potential of anode) – (Reduction potential of cathode)}\text{(ii) EMF of cell = (Oxidation potential of anode) + (Reduction potential of cathode)}\text{(iii)) EMF of cell = (Reductional potential of anode) + (Reduction potential of cathode)}\left(iv\right)\text{) EMF of cell = (Oxidation potential of anode) – (Oxidation potential of cathode)}$

• A. $\left(iii\right)\text{and}\left(i\right)$
• B. $\left(i\right)\text{and}\left(ii\right)$
• C. $\left(iii\right)\text{and}\left(i\right)$
• D. $\left(ii\right)\text{and}\left(iv\right)$

Answer 1

The correct option is D $\left(ii\right)\text{and}\left(iv\right)$

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.