TS EAMCET · Chemistry · Redox Reactions
What is the equivalent weight of \(\mathrm{KMnO}_4\) in acidic medium? (Molecular weight of \(\mathrm{KMnO}_4=158 \mathrm{~g}\) )
- A \(158 \mathrm{~g}\)
- B \(52.7 \mathrm{~g}\)
- C \(31.6 \mathrm{~g}\)
- D \(39.5 \mathrm{~g}\)
Answer & Solution
Correct Answer
(C) \(31.6 \mathrm{~g}\)
Step-by-step Solution
Detailed explanation
\(\because\) In acidic medium, \(\mathrm{KMnO}_4\), i.e. \(\mathrm{MnO}_4^{-}\) changes to \(\mathrm{Mn}^{2+}\) as follows : \(\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}\) Hence, equivalent weight \(=\frac{158}{5}\)…
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