Question

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

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

Answer 1

Answer: (B)

Given,

$Zn⟶Z{n}^{2+}+2e^{-};E^{\circ }=+0.76V\dots (i)$

$Fe⟶F{e}^{2+}+2e^{-};E^{\circ }=+0.41V\dots (ii)$

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

$Z{n}^{2+}+2e^{-}⟶Zn;E^{\circ }=-0.76V$

$F{e}^{2+}+2e^{-}⟶Fe;E^{\circ }=-0.41V$

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

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



So, $E_{\text{cell }}^{\circ }=E_{F{e}^{2+}/Fe}^{\circ }-E_{Z{n}^{2+}/Zn}$

$=-0.41V+0.76V=+0.35V$