$Mn^{2+} +2e- \rightarrow Mn;\, E^o = -1.18\,V$
$2(Mn^{3+} +e^- \rightarrow Mn^{2+} )\,;\,E^o=+1.51\,V$
The $E^o$ for $3Mn^{2+} \rightarrow Mn+ 2Mn^{3+}$ will be :
(b) $\quad M n^{3+}+e \rightarrow M n^{2+} ; \quad E^{\circ}=-1.51 V ; \ldots(i i)$
Now multiplying equation $(i i)$ by two and subtracting from equation $(i)$
$3 M n^{2+} \rightarrow M n^{+}+2 M n^{3+}$
$E^{\circ}=E_{O_{X}}+E_{\mathrm{Red} .}=-1.18+(-1.51)=-2.69 \mathrm{V}$
$[-\text { ve value of } E M F(\text { i.e., } \Delta G=+v e)$$ \text { shows that the reaction is non-spontaneous }]$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
(1)${H_2} \to {H^ \bullet } + {H^ \bullet }$
(2)$B{r_2} \to B{r^ \bullet } + B{r^ \bullet }$
(3)$B{r^ \bullet } + HBr \to {H^ \bullet } + B{r_2}$
(4)${H^ \bullet } + B{r_2} \to HBr + B{r^ \bullet }$
(5)$B{r^ \bullet } + B{r^ \bullet } \to B{r_2}$