NEET2010ChemistryElectrochemistryActual
Consider the following relations for emf of a 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
Options
- A(C) and (A)
- B(A) and (B)
- C(C) and (D)
- D(B) and (D)
Correct answer
D. (B) and (D)
Step-by-step solution
aligned & E _ cell = cathode D - (red) E _ anode ^ & or E_ cell = cred) E _ cathode ^ + (oxide) E_ Anode ^ & or E _ cell = (oxide) E anode E - E _ cathode ^ (oxide) aligned