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 $ .



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