The oxidation potentials of following half-cell reactions are given Zn arrow Zn 2++2 e- ; E°=0.76 V , Fe arrow Fe 2++ 2 e- ; E°=0.44 V what will be the emf of cell, whose cell reaction is Fe 2+( aq .)+ Zn arrow Zn 2+( aq .)+ Fe

Question

The oxidation potentials of following half-cell reactions are given Zn arrow Zn 2++2 e- ; E°=0.76 V , Fe arrow Fe 2++ 2 e- ; E°=0.44 V what will be the emf of cell, whose cell reaction is Fe 2+( aq .)+ Zn arrow Zn 2+( aq .)+ Fe (A) -1.20 V (B) -0.32 V (C) +0.32 V (D) +1.20 V

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/c7a275d636407cb581bd7e4d14e266d1-.png" /> EMF =E text cathode -E text anode =0.44-(0.76)=+0.32 V

Q. The oxidation potentials of following half-cell reactions are given $Z n \to Z n^{2 +} + 2 e^{-}; E^{\circ} = 0.76 V, F e \to F e^{2 +} +$ $2 e^{-}; E^{\circ} = 0.44 V$ what will be the emf of cell, whose cell reaction is $F e^{2 +} (a q .) + Z n \to Z n^{2 +} (a q .) + F e$

-1.20 V

-0.32 V

+0.32 V

+1.20 V

$EMF = E_{cathode} - E_{anode} = 0.44 - (0.76) = + 0.32 V$

Q. The oxidation potentials of following half-cell reactions are given Zn→Zn2++2e−;E∘=0.76V,Fe→Fe2++ 2e−;E∘=0.44V what will be the emf of cell, whose cell reaction is Fe2+(aq.)+Zn→Zn2+(aq.)+Fe