MCQ
The standard reduction potential for the half reactions are as $Zn = Z{n^{2 + }} + 2{e^ - };\,\,{E^o} = + 0.76\,V$ $Fe = F{e^{2 + }} + 2{e^ - };\,\,{E^o} = + 0.41\,V$ The $EMF$ for cell reaction $F{e^{2 + }} + Zn\, \to \,Z{n^{2 + }} + Fe$ is ............ $\mathrm{V}$
  • A
    $ - 0.35$
  • $ + 0.35$
  • C
    $ + 1.17$
  • D
    $ - 1.17$

Answer

Correct option: B.
$ + 0.35$
b
(b)In this reaction $F{e^{2 + }} + \mathop {Zn}\limits^0 \to \mathop {Zn}\limits^{2 + } + \mathop {Fe}\limits^0 $

$EMF = {E_{{\rm{cathode}}}} - {E_{{\rm{anode}}}}$$ = - 0.41 - ( - 0.76)$

$EMF = +\,0.35$V.

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