Question

An equal volume of a reducing agent is titrated separately with $1MKMnOK{M}_4$ in acid, neutral and alkaline medium. The volumes of $KMn{O}_4$ required are $20mL$ in acid, $33.3mL$ in neutral and $100mL$ 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 $1M$ $K_{2}C{r}_2{O}_7$ consumed, if the same volume of the reducing agent is titrated in acid medium.

Answer 1

Let the $n$-factor of $KMn{O}_4$ 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 $KMn{O}_4$ would be required every time.
$\Rightarrow 20N_1=\frac{100}{3}{N}_2=100N_3\Rightarrow N_1=\frac{5}{3}{N}_2=5N_3$
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 $Mn{O}_4^{-}⟶M{n}^{2+}$
In neutral medium $Mn{O}_4^{-}⟶M{n}^{4+}$
In alkaline medium $Mn{O}_4^{-}⟶M{n}^{6+}$
$\Rightarrow$ meq of $K_{2}C{r}_2{O}_7$ required $=100$
$\Rightarrow 100=1\times 6\times V(n$-factor $=6)$
$\Rightarrow V=100/6=16.67mL$