Question:
The equivalent weight of MnSO$_4$ is half of its molecular
weight, when it converts to
The equivalent weight of MnSO$_4$ is half of its molecular
weight, when it converts to
Updated On: Jun 23, 2023
- $Mn_2O_3$
- $MnO_2$
- $MnO_4^-$
- $MnO_4^{2-}$
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The Correct Option is B
Solution and Explanation
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 $MnO_2$
$\hspace20mm E =\frac{Molar \, mass}{n-factor}$
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 $MnO_2$
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