NTA Abhyas NEET2020ChemistryElectrochemistryPractice
Consider the following relations for emf of an electrochemical cell ( i ) EMF of cell = (oxidation potential of anode) - (Reduction potential of cathode) ( i i ) EMF of cell = (oxidation potential of anode) + (Reduction potential of cathode) ( i i i ) EMF of cell = (reduction potential of anode) + (oxidation potential of cathode) ( i v ) EMF of cell = (oxidation potential of anode) - (oxidation potential of cathode)
Options
- A(iii) and (i)
- B(i) and (ii)
- C(iii) and (iv)
- D(ii) and (iv)
Correct answer
D. (ii) and (iv)
Step-by-step solution
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.