Question

What will be the equivalent weight of KMnO4 in acidic, basic and neutral medium?

Answer 1

in acidic 158/5=31.6  in basic 158/3=52.66

Answer 2

Answer:

in acidic medium:  $MnO_4^-+8H^++5e^-\rightarrow Mn^{2+++4H_2O$

in neutral medium:  $MnO_4^-+4H^++3e^-\rightarrow MnO_2+2H_2O$

in basic medium: $MnO_4^-=e^-\rightarrow MnO_4^{2-}$

Equivalent weight=$\frac{\text {Molecular weight}}{n}$

For acidic medium, n=5 Thus equivalent weight=$\frac{158}{5}=31.6$

For neutral medium , n=3, Thus equivalent weight=$\frac{158}{3}=52.6$

For basic medium, n=1, Thus equivalent weight=$\frac{158}{1}=158$