Question

Given the standard half-cell potentials $\left(E^{\circ}\right)$ of the following as $Zn \longrightarrow Zn ^{2+}+2 e^{-} ; E^{\circ}=+0.76 \,V$ $Fe \longrightarrow Fe ^{2+}+2 e^{-} ; \quad E^{\circ}=0.41 \,V$ Then the standard e.m.f. of the cell with the reaction $Fe ^{2+}+ Zn \longrightarrow Zn ^{2+}+ Fe$ is

• A. -0.35 V
• B. +0.35 V
• C. +1.17 V
• D. -1.17 V

Answer 1

Answer: (B)

+0.35 V

Answer 2

Given,

$Zn \longrightarrow Zn ^{2+}+2 e^{-} ; E^{\circ}=+0.76\, V\,\,\,\,\,\,\,\dots(i)$
$Fe \longrightarrow Fe ^{2+}+2 e^{-} ; E^{\circ}=+0.41\, V \,\,\,\,\,\,\,\dots(ii)$

On reversing the above equation (i) and (ii), we get

$Zn ^{2+}+2 e^{-} \longrightarrow Zn ; E^{\circ}=-0.76 \,V$
$Fe ^{2+}+2 e^{-} \longrightarrow Fe ; E^{\circ}=-0.41 \,V$

[where, $E^{\circ}=$ standard reduction potential ]

To find, the standard emf of the cell with the reaction.

So, $E_{\text {cell }}^{\circ} =E^{\circ}_{ Fe^{2+} / Fe} -E_{ Zn ^{2+} / Zn}$
$=-0.41 \,V +0.76 V =+0.35 \,V$