Some Basic Concepts of Chemistry - Result Question 40
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 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 $1 \mathrm{M} \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ consumed, if the same volume of the reducing agent is titrated in acid medium.
(1989,5 M)
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Answer:
Correct Answer: 45. (16.67 mL)
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, $\boldsymbol{n}$-factors are all integer and greater than or equal to one but less than six, $N_3$ must be 1 .
$\begin{aligned} & \Rightarrow \quad N_1=5, N_2=3 \\ & \therefore \text { In acid medium } \quad \mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{2+} \\ & \text { In neutral medium } \mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{4+} \\ & \text { In alkaline medium } \mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{6+} \\ & \Rightarrow \text { meq of } \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \text { required }=100 \\ & \Rightarrow \quad 100=1 \times 6 \times V(n \text {-factor }=6) \\ & \Rightarrow \quad V=100 / 6=16.67 \mathrm{~mL} \\ \end{aligned}$
BETA
