Question

KMnO$_4$ acts as an oxidising agent in acidic medium. The number of moles of KMnO$_4$ that will be required to react with one mole of oxalate ions (to form CO$_2$) in acidic solution is:

• A. $\dfrac{2}{5}$
• B. $\dfrac{5}{2}$
• C. 5
• D. $\dfrac{5}{4}$

Hint:

Always balance redox reactions using the ion-electron method in acidic or basic medium. Identify oxidation and reduction half-reactions separately.

Answer 1

The Correct Option is A

Step 1: Write the balanced redox reaction in acidic medium.
In acidic medium, KMnO$_4$ acts as a strong oxidizing agent and gets reduced as:
\[ \text{MnO}_4^- \rightarrow \text{Mn}^{2+} \] This involves a gain of 5 electrons. On the other hand, oxalate ion (C$_2$O$_4^{2-}$) gets oxidized to carbon dioxide: \[ \text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 \] This involves a loss of 2 electrons. Step 2: Balance the electrons transferred.
To equalize the electron exchange, we use LCM of 5 and 2 = 10 electrons. \[ \text{5 MnO}_4^- + 16H^+ + 10e^- \rightarrow 5Mn^{2+} + 8H_2O \] \[ \text{2 C}_2\text{O}_4^{2-} \rightarrow 4CO_2 + 4e^- \] To balance the electrons: \[ \text{5 MnO}_4^- \text{ reacts with } \text{2 C}_2\text{O}_4^{2-} \] Step 3: Calculate moles required.
From the balanced reaction: \[ 2 \text{MnO}_4^- + 5 \text{C}_2\text{O}_4^{2-} + 16 H^+ \rightarrow 2 Mn^{2+} + 10 CO_2 + 8 H_2O \] This means: \[ \text{2 mol KMnO}_4 \text{ reacts with 5 mol oxalate} \] So, for 1 mol of oxalate, KMnO$_4$ needed is: \[ \frac{2}{5} \text{ mol} \] Thus, the required number of moles of KMnO$_4$ = $\dfrac{2}{5}$ mol.