Question

Equivalent weight of $Mn{O}_4{}^{-}$ in acidic, basic, neutral medium is in the ratio is:

Answer 1

In acidic medium :

${MnO}_4^{-}+{8H}^{+}+5e^{-}={Mn}^{2+}+4H_{2}O$

$\therefore$ Equivalent weight $=\frac{M_{{MnO}_4^{-}}}{5}=\frac{158}{5}=31.6g{mol}^{-1}$

In neutral or feebly alkaline medium,

${MnO}_4^{-}+2H_{2}O+3e^{-}\rightleftharpoons {MnO}_2+4{OH}^{(-)}$

$\therefore$ Equivalent weight $=\frac{158}{3}=52.68gm{mol}^{-1}$

In strongly alkaline medium :

${MnO}_4^{-}+e^{-}={MnO}_4^{2-}$

$\therefore$ Equivalent weight $=\frac{158}{1}=158gm/mol$

$\therefore$ Equivalent weight of ${MnO}_4=$ in acidic, basic and neutral medium is $5:1:3$