question_answer

The oxidation potentials of following half-cell reactions are given \[Zn\to Z{{n}^{2+}}+2{{e}^{-}};{{E}^{o}}=0.76\,\,V\],\[Fe\to F{{e}^{2+}}+2{{e}^{-}};{{E}^{o}}=0.44\,\,V\] 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]

A)                 ? 1.20 V               

B)                 + 0.32 V

C)                 ? 0.32 V               

D)                 + 1.20 V

Correct Answer: B

Solution :

           \[F{{e}^{+2}}+Zn\to Z{{n}^{2+}}+Fe\]                 \[EMF={{E}_{\text{cathode}}}-{{E}_{\text{anode}}}\]\[=0.44-(0.76)=+0.32\,V\].