Question

Eqivalent Weight of Mn3+ in the following reaction is (Mn=55) Mn3+ gives Mn2+ +Mno2

Answer 1

equivalent weight of a compound in a redox reaction =molecular wt/no of ... eq wt of Mn=at wt/change in oxdn number=55/2=27.5

Answer 2

Answer : The equivalent weight of $Mn^{3+}$ is 110 g.eq

Explanation:

Equivalent weight : It is the molar mass of a substance divided by the number of equivalents in the substance.

The given redox reaction is :

$Mn^{3+}\rightarrow Mn^{2+}+MnO_2$

The half oxidation-reduction reactions are:

Oxidation reaction : $Mn^{3+}+2H_2O\rightarrow MnO_2+4H^++1e^-$

Reduction reaction : $Mn^{3+}+1e^-\rightarrow Mn^{2+}$

Given:

Molar mass of $Mn$ = 55 g/mole

$\text{Equivalent weight of oxidizing agent}=\frac{\text{Molar mass of }Mn}{\text{Number of electrons gained by 1 molecule}}=\frac{55}{1}=55g.eq$

and,

$\text{Equivalent weight of reducing agent}=\frac{\text{Molar mass of }Mn}{\text{Number of electrons lost by 1 molecule}}=\frac{55}{1}=55g.eq$

So,

$\text{Equivalent weight of }Mn^{3+}=55+55=110g.eq$

Therefore, the equivalent weight of $Mn^{3+}$ is 110 g.eq