Question

Alkaline solution of KMnO4 reacts as follows:
$2KMnO_{4}+2KOH \longrightarrow 2K_{2}MnO_{4}+H_{2}O+[O]$
Calculate the equivalent weight of $KMnO_{4}$ in basic medium.

Answer 1

Mass of one equivalent is known as gram equivalent or equivalent weight that is the of a given substance associate with or disturb a fixed quantity of another substance.

The balanced reaction given in the question is

$2KMnO_{ 4 }+2KOH\longrightarrow 2K_{ 2 }MnO_{ 4 }+H_{ 2 }O+O$

If the number is authorized to an atom of the substance then the number is known as oxidation number.  The number of the oxidation could be “positive, negative or zero” which points out the gained and lost of electrons.  

Oxidation number of Mn in $KMnO_4$

(+1) + (x) + 4(-2) = 0  

1 + x - 8 = 0  

x = 7  

Oxidation number of Mn in $K_2 MnO_4$  

2(+1) + x + 4(-2) = 0  

2 + x - 8 = 0  

x - 6 = 0  

x = 6  

Change in oxidation number of $KMnO_4$ and $K_2 MnO_4$ = 1

The formula to find the gram equivalent =  $\quad \frac { Molecular\quad weight }{ Change\quad in\quad oxidation\quad number }$

Molecular weight of $KMnO_4$  is 158 g/mole.

Equivalent weight of $KMnO_4$ = $\quad \frac { Molecular\quad weight\quad of\quad KMnO_ 4 }{ Change\quad in\quad oxidation\quad number }$ = $\frac { 158\quad g/mole }{ 1\quad eq/mole }$ = 158g/eq.

Therefore, the gram equivalent of $KMnO_4$ is 158g/eq.