Electrochemistry Ques 28
28. 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 the possibility of the forward reaction are, respectively
[NEET Kar. 2013]
(a) $-2.69 V$ and no
(b) $-4.18, \text{V}$ and yes
(c) $+0.33,V$ and yes
(d) $+2.69 V$ and no
Show Answer
Solution:
(a) $\Delta E^{\circ}=E _{\text{red }}^{\circ}+E _{\text{oxd }}^{\circ}=-1.81+(-1.51)=-3.32$ Since $\Delta E^{\circ}$ is negative
$\therefore \Delta G=-n F E^{\circ}, \Delta G$ will have negative value so, forward reaction is not possible.
The value of electrode potential of a half-cell reaction does depend on stoichiometric coefficient. e.g.
$ \begin{alignedat} & 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} $
BETA
