Equivalent weights of $\mathrm{KMnO}_4$ and $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ in acidic medium are…

Equivalent weights of $\mathrm{KMnO}_4$ and $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ in acidic medium are respectively. (Molecular weight of $\mathrm{KMnO}_4=M_A$ and Molecular weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=M_B$ ).
  1. $\frac{M_A}{3} \cdot \frac{M_B}{6}$
  2. $\frac{M_A}{6} \cdot \frac{M_B}{5}$
  3. $\frac{M_A}{3} \cdot \frac{M_B}{5}$
  4. $\frac{M_A}{5} \cdot \frac{M_B}{6}$

Solution

Given that, Molecular weight of $\mathrm{KMnO}_4=M_A$ $\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}$ $\therefore$ Equivalent weight of $\mathrm{KMnO}_4$ $=\frac{\text { Molecular weight }}{5}=\frac{M_A}{5}$ Similarly, molecular weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=M_B$ $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+14 \mathrm{H}^{+}+6 e^{-} \longrightarrow 2 \mathrm{~K}^{+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}$ $\therefore$ Equivalent weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ $=\frac{\text { Molecular weight }}{6}=\frac{M_B}{6}$

Asked in: AP EAMCET 2022 (06 Jul Shift 2)

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