Redox Reactions And Electrochemistry Ques 176
Question: The oxidation potentials of following half-cell reactions are given $ Zn\to Z{n^{2+}}+2{e^{-}};E^{o}=0.76V $ , $ Fe\to F{e^{2+}}+2{e^{-}};E^{o}=0.44V $ what will be the emf of cell, whose cell-reaction is $ F{e^{2+}}(aq)+Zn\to Z{n^{2+}}(aq)+Fe $ [MP PMT 2003]
Options:
A) - 1.20 V
B) + 0.32 V
C) - 0.32 V
D) + 1.20 V
Show Answer
Answer:
Correct Answer: B
Solution:
$ F{e^{+2}}+Zn\to Z{n^{2+}}+Fe $ $ EMF={E_{cathode}}-{E_{anode}} $ $ =0.44-(0.76)=+0.32,V $ .
