Question

Equivalent mass of $KMn{O}_4$ in acidic, basic and neutral are in the ratio of:

• A. $3:5:15$
• B. $5:3:1$
• C. $5:1:3$
• D. $3:15:5$

Answer 1

The correct option is D $3:15:5$

In acidic medium: ${MnO}_4^{-}+{8H}^{+}+5e^{-}\to {Mn}^{+2}+4H_{2}O$

In basic medium: ${MnO}_4^{-}+e^{-}\to {MnO}_4^{2-}$

In neutral medium: ${MnO}_4^{-}+{4H}^{+}+3e^{-}\to Mn{O}_2+2H_{2}O$

Equivalent mass is molecular mass divided by number of electrons gained.

In acidic medium, equivalent weight $=$$\frac{Molecularweight}{5}$

In basic medium, equivalent weight $=$$\frac{Molecularweight}{1}$

In neutal medium, equivalent weight $=$$\frac{Molecularweight}{3}$

Ration of equivalent mass $=$$\frac{Molecularweight}{5}$:$\frac{Molecularweight}{1}$:$\frac{Molecularweight}{3}$

Ratio of equivalent mass $=3:15:5$