Some Basic Concepts of Chemistry 2 Question 45
45. An equal volume of a reducing agent is titrated separately with $1 \mathrm{M} \mathrm{KMnO}{4}$ in acid, neutral and alkaline medium. The volumes of $\mathrm{KMnO}{4}$ required are $20 \mathrm{~mL}$ in acid, $33.3 \mathrm{~mL}$ in neutral and $100 \mathrm{~mL}$ 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 $1 \mathrm{M} \mathrm{K}{2} \mathrm{Cr}{2} \mathrm{O}_{7}$ consumed, if the same volume of the reducing agent is titrated in acid medium.
(1989, 5M)
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Solution:
- Let the $n$-factor of $\mathrm{KMnO}{4}$ 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 $\mathrm{KMnO}_{4}$ 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 .
$$ \begin{aligned} & \Rightarrow \quad N_{1}=5, N_{2}=3 \ & \therefore \text { In acid medium } \quad \mathrm{MnO}{4}^{-} \longrightarrow \mathrm{Mn}^{2+} \ & \text { In neutral medium } \quad \mathrm{MnO}{4}^{-} \longrightarrow \mathrm{Mn}^{4+} \ & \text { In alkaline medium } \quad \mathrm{MnO}{4}^{-} \longrightarrow \mathrm{Mn}^{6+} \ & \Rightarrow \text { meq of } \mathrm{K}{2} \mathrm{Cr}{2} \mathrm{O}{7} \text { required }=100 \ & \Rightarrow \quad 100=1 \times 6 \times V(n \text {-factor }=6) \ & \Rightarrow \quad V=100 / 6=16.67 \mathrm{~mL} \end{aligned} $$
