Question

An equal volume of a reducing agent is titrated separately with $1 \,M \,KMnO KM _{4}$ in acid, neutral and alkaline medium. The volumes of $KMnO _{4}$ required are $20\, mL$ in acid, $33.3 \,mL$ in neutral and $100 \,mL$ in alkaline media. Find out the oxidation state of manganese in each reduction product. Give the balanced equations for all the three half reaction. Find out the volume of $1 \,M$ $K _{2} Cr _{2} O _{7}$ consumed, if the same volume of the reducing agent is titrated in acid medium.

Answer 1

Let the $n$-factor of $KMnO _{4}$ in acid, neutral and alkaline media are $N_{1}, N_{2}$ and $N_{3}$ respectively. Also, same volumes of reducing agent is used everytime, same number of equivalents of $KMnO _{4}$ would be required every time.
$\Rightarrow 20\, N_{1}=\frac{100}{3} N_{2}=100 \,N_{3} \Rightarrow N_{1}=\frac{5}{3} N_{2}=5 N_{3}$
Also, $n$-factors are all integer and greater than or equal to one but less than six, $N_{3}$ must be $1$ .
$\Rightarrow N_{1}=5, N_{2}=3$
$\therefore$ In acid medium $ MnO _{4}^{-} \longrightarrow Mn ^{2+}$
In neutral medium $ MnO _{4}^{-} \longrightarrow Mn ^{4+}$
In alkaline medium $MnO _{4}^{-} \longrightarrow Mn ^{6+}$
$\Rightarrow$ meq of $K _{2} Cr _{2} O _{7}$ required $=100$
$\Rightarrow 100=1 \times 6 \times V(n$-factor $=6)$
$\Rightarrow V=100 / 6=16.67\, mL$