Question

The oxidation of ferrous oxalate by \(KMnO_4\) in acidic medium is given by the following equation, \(3MnO_4^- + 5FeC_2O_4 + 24H^+ \to 3Mn^{2+} + 5Fe^{3+} + 10CO_2 + 12H_2O\). What is the volume of \(0.01 \text{ mol dm}^{-3}\) \(KMnO_4\) required to oxidise \(15 \text{ cm}^3\) of an acidified solution of \(0.01 \text{ mol dm}^{-3}\) ferrous oxalate?

• A. \(18 \text{ cm}^3\)
• B. \(9 \text{ cm}^3\)
• C. \(15 \text{ cm}^3\)
• D. \(18 \text{ cm}^3\)

Hint:

In acidic medium, \(KMnO_4\) (\(Mn^{+7} \to Mn^{+2}\)) has an n-factor of 5. For \(FeC_2O_4\), \(Fe^{2+} \to Fe^{3+}\) (1 electron) and \(C_2O_4^{2-} \to 2CO_2\) (2 electrons), giving a total n-factor of 3. Use \(N_1V_1 = N_2V_2\) where \(N = M \times \text{n-factor}\).
\((0.01 \times 5) \times V_1 = (0.01 \times 3) \times 15 \implies V_1 = (3 \times 15) / 5 = 9\).

Answer 1

The Correct Option is B