. Given that
E°(Zn2+/Zn) = – 0.763V and E° (Cd2+/Cd) = – 0.403V, the emf of the following cell:
Zn/Zn2+ (a = 0.04) ||Cd2+ (a = 0.2) / Cd is given by,
A. E = –0.36 + [0.059/2] [log(0.004/0.2)]
B. E = + 0.36 + [0.059/2] [log (0.004/0.2)]
C. E = – 0.36 + [0.059/2] [log(0.2/0.004)]
D. E = + 0.36 + [0.059/2][log (0.2/0.004)]
Best Answer
Ans: B
Sol: Sine E°Cd2+/Cd > E°Zn2+/Zn , therefore Zn electrode acts as anode and Cd electrode as cathode.
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