Electrochemistry - Result Question 33
####35. Consider the half-cell reduction reaction :
$Mn^{2+}+2 e^{-} \to Mn, E^{\circ}=-1.18 V$
$Mn^{2+} \to Mn^{3+}+e^{-}, E^{\circ}=-1.51 V$
The $E^{\circ}$ for the reaction $3 Mn^{2+} \to Mn^{0}+2 Mn^{3+}$, and possibility of the forward reaction are, respectively
[NEET Kar. 2013]
(a) $-2.69 V$ and no
(b) $-4.18 V$ and yes
(c) $+0.33 V$ and yes
(d) $+2.69 V$ and no
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Solution:
(a) $\Delta E^{\circ}=E _{\text{red }}^{\circ}+E _{\text{oxd }}^{\circ}=-1.81-1.51=-2.69$ Since $\Delta E^{\circ}$ is negative
$\therefore \Delta G=-n F E^{\circ}, \Delta G$ will have positive value so, forward reaction is not possible.
The value of electrode potential of a half-cell reaction does not depend on stoichiometric coefficient. e.g.
$ \begin{aligned} & Mn^{2+} \longrightarrow Mn^{3+}+e^{-} ; E^{\circ}=-1.51 V \\ & 2 Mn^{2+} \longrightarrow 2 Mn^{3+}+2 e^{-} ; E^{\circ}=-1.51 V \end{aligned} $
