$Mn^{2+} + 2e^- \rightarrow Mn,\, $$E^o = - 1.18\, V$
$Mn^{2+} \rightarrow Mn^{3+} + e^-,$ $ E^o = - 1.51 \,V$
The $E^o$ for the reaction $3Mn^{2+} \rightarrow Mn^o + 2Mn^{3+},$
and possibility of the forward reaction are respectively
$2\left(\mathrm{Mn}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}\right), E^{\circ}=+1.51 \mathrm{V}$.... $(ii)$
Subtracting Eq. $(ii)$ from Eq. $(i)$, we get
${3 \mathrm{Mn}^{2+} \longrightarrow \mathrm{Mn}+2 \mathrm{Mn}^{3+}}$
${E^{\circ}=-1.18-(+1.51)=-2.69 \mathrm{V}}$
since, the value of $E^{\circ}$ is -ve, therefore the reaction is non-spontaneous
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$(i)$ $P{b^{2 + }} > P{b^{4 + }},T{l^ + } < T{l^{3 + }}$ $(ii)$ $Bi^{3+} < Sb^{3+} , Sn^{2+} < sn^{4+}$
$(iii)$ $P{b^{2 + }} > P{b^{4 + }},B{i^{3 + }} > B{i^{5 + }}$ $(iv)$ $T{l^{3 + }} < I{n^{3 + }},S{n^{2 + }} > S{n^{4 + }}$
$(v)$ $S{n^{2 + }} < P{b^{2 + }},S{n^{4 + }} > P{b^{4 + }}$ $(vi)$ $S{n^{2 + }} < P{b^{2 + }},S{n^{4 + }} < P{b^{4 + }}$
Reactant $ X$ Product $Y$