Using same quantity of current, which among Na, Mg and Al is deposited more (by mass) during electro — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Write the Cathodic Reduction Reactions and Determine the Valency Factors ($n$-factors) During the electrolysis of molten salts of sodium ($\ce{Na}$), magnesium ($\ce{Mg}$), and aluminium ($\ce{Al}$), their metal cations migrate to the cathode where they undergo reduction to deposit as solid metals. The corresponding reduction half-reactions are: 1. **For Sodium ($\ce{Na}$):** $$\ce{Na^+ + e^- -> Na(s)}$$ The valency factor ($n$-factor, $n_{\ce{Na}}$) is: $$n_{\ce{Na}} = 1$$ 2. **For Magnesium ($\ce{Mg}$):** $$\ce{Mg^{2+} + 2e^- -> Mg(s)}$$ The valency factor ($n$-factor, $n_{\ce{Mg}}$) is: $$n_{\ce{Mg}} = 2$$ 3. **For Aluminium ($\ce{Al}$):** $$\ce{Al^{3+} + 3e^- -> Al(s)}$$ The valency factor ($n$-factor, $n_{\ce{Al}}$) is: $$n_{\ce{Al}} = 3$$ Step 2 - Calculate the Chemical Equivalent Weights ($E$) of the Metals The chemical equivalent weight ($E$) of any element is defined as its atomic mass ($M$) divided by its valency factor ($n$): $$E = \frac{M}{n}$$ Using the standard atomic masses ($M_{\ce{Na}} \approx 23\text{ g/mol}$, $M_{\ce{Mg}} \approx 24\text{ g/mol}$, and $M_{\ce{Al}} \approx 27\text{ g/mol}$): * **Equivalent weight of Sodium ($E_{\ce{Na}}$):** $$E_{\ce{Na}} = \frac{23\text{ g/mol}}{1} = 23\text{ g/eq}$$ * **Equivalent weight of Magnesium ($E_{\ce{Mg}}$):** $$E_{\ce{Mg}} = \frac{24\text{ g/mol}}{2} = 12\text{ g/eq}$$ * **Equivalent weight of Aluminium ($E_{\ce{Al}}$):** $$E_{\ce{Al}} = \frac{27\text{ g/mol}}{3} = 9\text{ g/eq}$$ Step 3 - Apply Faraday's Laws of Electrolysis According to Faraday's Laws of Electrolysis, the mass ($W$) of a substance deposited or liberated at an electrode is directly proportional to its chemical equivalent weight ($E$) when the same quantity of electricity ($Q$) is passed: $$W \propto E$$ Since the question specifies that the **same quantity of current** is passed for the same duration, the total electric charge ($Q$) passed through each molten electrolyte is identical. Therefore, the ratio of the masses deposited is directly equal to the ratio of their chemical equivalent weights: $$W_{\ce{Na}} : W_{\ce{Mg}} : W_{\ce{Al}} = E_{\ce{Na}} : E_{\ce{Mg}} : E_{\ce{Al}}$$ $$W_{\ce{Na}} : W_{\ce{Mg}} : W_{\ce{Al}} = 23 : 12 : 9$$ Comparing the equivalent weights: $$E_{\ce{Na}}\ (23\text{ g/eq}) > E_{\ce{Mg}}\ (12\text{ g/eq}) > E_{\ce{Al}}\ (9\text{ g/eq})$$ Thus, the mass of sodium ($\ce{Na}$) deposited will be the greatest: $$W_{\ce{Na}} > W_{\ce{Mg}} > W_{\ce{Al}}$$ Step 4 - Evaluate and Explain the Options * **Option (A) is correct:** Since sodium has the highest chemical equivalent weight ($23\text{ g/eq}$), the mass of sodium deposited is the greatest when the same quantity of electricity is passed. * **Option (B) is incorrect:** Magnesium has a lower equivalent weight ($12\text{ g/eq}$) than sodium, so a smaller mass of magnesium ($12\text{ g}$ of Mg per Faraday) will be deposited compared to sodium ($23\text{ g}$ of Na per Faraday). * **Option (C) is incorrect:** Aluminium has the lowest equivalent weight ($9\text{ g/eq}$), meaning it will deposit the least mass ($9\text{ g}$ of Al per Faraday). * **Option (D) is incorrect:** The masses deposited are different because they depend directly on the equivalent weights of the respective elements, which are unequal. $$\text{Correct Option: } \boxed{\text{A}}$$