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The standard reduction potentials,$E^{\large\circ}$,for the half reactions are $Zn^{2+}+2e^-=Zn,E^{\large\circ}=-0.76V$ and $Fe^{2+}+2e^-=Fe,E^{\large\circ}=-0.41V$.The emf of the cell involving the reaction $Fe^{2+}+zn\rightarrow Zn^{2+}+Fe$ is

$\begin{array}{1 1}-0.35V\\1.17V\\0.35V\\-1.17V\end{array} $

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Answer : 0.35V
In the given reaction Zn is oxidised to $Zn^{2+}$ and $Fe^{2+}$ is reduced to Fe.Hence,
$E_{cell}=E^{\large\circ}_{Fe^{2+}|Fe}-E^{\large\circ}_{Zn^{2+}|Zn}=-0.41V-(-0.76V)=0.35V$
answered Jul 18, 2014 by sreemathi.v
 

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