Electrochemistry — Easy JEE chemistry MCQ
Given E°(Zn²⁺/Zn) = -0.76 V and E°(Cu²⁺/Cu) = +0.34 V, the EMF of a Daniell cell is:
- A. 0.42 V
- B. 1.10 V
- C. -1.10 V
- D. -0.42 V
Solution
E°cell = E°cathode - E°anode = E°(Cu²⁺/Cu) - E°(Zn²⁺/Zn) = 0.34 - (-0.76) = 1.10 V.