Question

Given below are the half-cell reactions
$Mn ^{2+}+2 e ^{-} \rightarrow Mn ; E ^{\circ}=-1.18 \,eV$
$2\left( Mn ^{3+}+ e ^{-} \rightarrow Mn ^{2+}\right) ; E ^{\circ}=+1.51 \,eV$
The $ E ^{\circ} $ for $3 \,Mn ^{2+} \rightarrow Mn +2 Mn ^{3+}$ will be

• A. $-2.69\, V$; the reaction will not occur
• B. $- 2.69\, V$; the reaction will occur
• C. $-2.69 \,V$; the reaction will occur
• D. $- 0.33\, V$; the reaction will occur

Answer 1

Answer: (A)

Standard electrode potential of reaction $\left[E^{\circ}\right]$ can be calculated as
$E_{\text {cell }}^{\circ}=E_{R}-E_{P}$
where, $E_{R}=$ SRP of reactant,$E_{P}=$ SRP of product
If $E_{\text {cell }}^{\circ}=+v e$, then reaction is spontaneous otherwise non-spontaneous.
$Mn ^{3+} \xrightarrow{E_{1}^{0}=1.51 V } Mn ^{2+}$
$Mn ^{2+} \xrightarrow{E_{2}^{0}=-1.18 V } Mn$
$\therefore$ For $Mn ^{2+}$ disproportionation,
$E^{\circ}=-1.51\, V -1.18 \,V =-2.69\, V < 0$
Thus, all reaction will not occur.