The metal that cannot be obtained on reduction of its oxide by aluminium is — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Introduction to Aluminothermic Reduction In aluminothermic (thermite) reduction, metallic aluminium ($\ce{Al}$) reduces metal oxides at high temperatures. The reaction is feasible only when the target metal's oxide is less stable than $\ce{Al2O3}$, i.e., the target metal is less reactive than aluminium. Step 2 - Reactivity Comparison The decreasing order of reactivity: $$\text{Reactivity Order: } \ce{K} > \ce{Al} > \ce{Mn} > \ce{Cr} > \ce{Fe}$$ Potassium ($\ce{K}$) is above aluminium in the reactivity series. Step 3 - Evaluation of Options * **Option (A) is correct:** $\ce{K}$ is more reactive than $\ce{Al}$. The hypothetical reduction: $$\ce{3 K2O(s) + 2 Al(s) -> Al2O3(s) + 6 K(g)}$$ is thermodynamically non-spontaneous ($\Delta G^\circ > 0$). K cannot be obtained by aluminothermic reduction. * **Option (B) is incorrect:** Mn is less reactive than Al. $\ce{MnO2}$ is readily reduced by Al: $$\ce{3 MnO2(s) + 4 Al(s) -> 2 Al2O3(s) + 3 Mn(l)}$$ * **Option (C) is incorrect:** Cr is less reactive than Al. $\ce{Cr2O3}$ is easily reduced: $$\ce{Cr2O3(s) + 2 Al(s) -> Al2O3(s) + 2 Cr(l)}$$ * **Option (D) is incorrect:** Fe is much less reactive than Al. The thermite reaction: $$\ce{Fe2O3(s) + 2 Al(s) -> Al2O3(s) + 2 Fe(l)}$$ is highly exothermic and widely used in welding. $$\text{Correct Answer: } \boxed{\text{A}}$$