Categories > electrochemistry > Electrode potential > The oxidation potentials of following half-cell reactions are given $Zn\to Z{{n}^{2+}}+2{{e}^{-}};\,{{E}^{0}}=0.76\,V$ $Fe\to F{{e}^{2+}}+2{{e}^{-}};\,{{E}^{0}}=0.44\,V$ what will be the emf of cell, whose cell reaction is $F{{e}^{2+}}(aq)+Zn\to Z{{n}^{2+}}(aq)+Fe$

electrochemistry

Question: The oxidation potentials of following half-cell reactions are given $Zn\to Z{{n}^{2+}}+2{{e}^{-}};\,{{E}^{0}}=0.76\,V$ $Fe\to F{{e}^{2+}}+2{{e}^{-}};\,{{E}^{0}}=0.44\,V$ what will be the emf of cell, whose cell reaction is $F{{e}^{2+}}(aq)+Zn\to Z{{n}^{2+}}(aq)+Fe$


1) – 1.20 V
2) + 0.32 V
3) – 0.32 V
4) + 1.20 V
Solution: Explanation: No Explanation
Electrode potential Ecell Nernst equation and ECS

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