Chemistry - Redox Reactions Question with Solution | TestHub

$\mathrm{M}{\mathrm{n}}^{2+}+2{\mathrm{H}}_2\mathrm{O}\to \mathrm{M}\mathrm{n}{\mathrm{O}}_2+4{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}$
Since 1 mol of$\mathrm{M}\mathrm{n}\mathrm{S}{\mathrm{O}}_4$, when converted to$\mathrm{M}\mathrm{n}{\mathrm{O}}_2$, loses 2 mol electron, therefore, its equivalent weight would be one-half of its molecular weight. It is not so in A, C and D.
$\mathrm{A}.$$2\mathrm{M}{\mathrm{n}}^{2+}+3{\mathrm{H}}_2\mathrm{O}\to \mathrm{M}{\mathrm{n}}_2{\mathrm{O}}_3+6{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}$
$\mathrm{C}.$$\mathrm{M}{\mathrm{n}}^{2+}+4{\mathrm{H}}_2\mathrm{O}\to \mathrm{M}\mathrm{n}{\mathrm{O}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}$
$\mathrm{D}.$$\mathrm{M}{\mathrm{n}}^{2+}+4{\mathrm{H}}_2\mathrm{O}\to \mathrm{M}\mathrm{n}{\mathrm{O}}_4^{2-}+8{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}$