Question

The equivalent weight of MnSO$_4$ is half of its molecular weight, when it converts to

• A. $Mn_2O_3$
• B. $MnO_2$
• C. $MnO_4^-$
• D. $MnO_4^{2-}$

Answer 1

Answer: (B)

$MnO_2$

Answer 2

Equivalent weight in redox system is defined as :
\hspace20mm E =\frac{Molar \, mass}{n-factor}
Here n-factor is the net change in oxidation number per formula
unit of oxidising or reducing agent. In the present case, n-factor
is 2 because equivalent weight is half of molecular weight. Also,
$n-factor \, MnSO_4 \rightarrow \, \frac{1}{2}Mn_2O_3 \, \, 1(+2 \rightarrow \, +3)$
$\, \, \, \, \, \, MnSO_4 \rightarrow \, \, \, MnO_2 \, \, \, 2(+2 \rightarrow \, \, \, +4)$
$\, \, \, \, \, \, MnSO_4 \rightarrow \, \, \, MnO_4^- \, \, \, 5(+2 \rightarrow \, \, \, +7)$
$\, \, \, \, \, \, MnSO_4 \rightarrow \, \, \, MnO_4^2- \, \, \, 4(+2 \rightarrow \, \, \, +6)$
Therefore , MnS$O_4$ convertsa to $MnO_2$