Search any question & find its solution
Question:
Answered & Verified by Expert
$\mathrm{KMnO}_4$ acts as an oxidising agent in acidic medium. The number of moles of $\mathrm{KMnO}_4$ that will be required to react with one mole of oxalate ions (to form $\mathrm{CO}_2$ ) in acidic solution is
Options:
Solution:
2507 Upvotes
Verified Answer
The correct answer is:
$\frac{2}{5}$
$$
\begin{aligned}
\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e^{-} & \left.\longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}\right] \times 2 \\
\mathrm{C}_2 \mathrm{O}_4^{2-} \longrightarrow & \left.\mathrm{CO}_2+2 e^{-}\right] \times 5
\end{aligned}
$$
$$
\begin{aligned}
& \hline 2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_4^{2-}+16 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{2+}+\mathrm{10CO}_2+8 \mathrm{H}_2 \mathrm{O} \\
& 5 \text { moles of } \mathrm{C}_2 \mathrm{O}_4^{2-} \text { need } 2 \text { moles of } \mathrm{KMnO}_4 . \\
& 1 \text { mole of } \mathrm{C}_2 \mathrm{O}_4^{2-} \text { would need }=\frac{2}{5} \text { mole of } \mathrm{KMnO}_4 .
\end{aligned}
$$
\begin{aligned}
\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e^{-} & \left.\longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}\right] \times 2 \\
\mathrm{C}_2 \mathrm{O}_4^{2-} \longrightarrow & \left.\mathrm{CO}_2+2 e^{-}\right] \times 5
\end{aligned}
$$
$$
\begin{aligned}
& \hline 2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_4^{2-}+16 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{2+}+\mathrm{10CO}_2+8 \mathrm{H}_2 \mathrm{O} \\
& 5 \text { moles of } \mathrm{C}_2 \mathrm{O}_4^{2-} \text { need } 2 \text { moles of } \mathrm{KMnO}_4 . \\
& 1 \text { mole of } \mathrm{C}_2 \mathrm{O}_4^{2-} \text { would need }=\frac{2}{5} \text { mole of } \mathrm{KMnO}_4 .
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.