Three faradays of electricity is passed through molten , aqueous solutions of and molten . The amoun — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand Faraday's First Law of Electrolysis According to Faraday's laws of electrolysis, the number of moles of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. The relation between the charge passed in Faradays ($Q$) and the moles of substance deposited ($n$) is: \begin{equation*} n = \frac{Q}{z} \end{equation*} where $Q = 3\text{ F}$ is the charge passed, and $z$ is the valency factor (number of electrons required to reduce one metal ion). Step 2 - Calculate Moles of Aluminum (\ce{Al}) Deposited In molten \ce{Al2O3}, aluminum is present as \ce{Al^3+} ions. The reduction reaction at the cathode is: \begin{equation*} \ce{Al^3+ + 3e^- -> Al} \end{equation*} The valency factor is $z_{\ce{Al}} = 3$. Substituting into the formula: \begin{equation*} n_{\ce{Al}} = \frac{3\text{ F}}{3\text{ F/mol}} = 1\text{ mol} \end{equation*} Step 3 - Calculate Moles of Copper (\ce{Cu}) Deposited In aqueous \ce{CuSO4} solution, copper is present as \ce{Cu^2+} ions. The reduction reaction at the cathode is: \begin{equation*} \ce{Cu^2+ + 2e^- -> Cu} \end{equation*} The valency factor is $z_{\ce{Cu}} = 2$. Substituting into the formula: \begin{equation*} n_{\ce{Cu}} = \frac{3\text{ F}}{2\text{ F/mol}} = 1.5\text{ mol} \end{equation*} Step 4 - Calculate Moles of Sodium (\ce{Na}) Deposited In molten \ce{NaCl}, sodium is present as \ce{Na+} ions. The reduction reaction at the cathode is: \begin{equation*} \ce{Na^+ + e^- -> Na} \end{equation*} The valency factor is $z_{\ce{Na}} = 1$. Substituting into the formula: \begin{equation*} n_{\ce{Na}} = \frac{3\text{ F}}{1\text{ F/mol}} = 3\text{ mol} \end{equation*} Step 5 - Determine the Molar Ratio The molar ratio of the deposited metals (\ce{Al} : \ce{Cu} : \ce{Na}) is: \begin{equation*} \text{Molar Ratio} = n_{\ce{Al}} : n_{\ce{Cu}} : n_{\ce{Na}} \end{equation*} \begin{equation*} \text{Molar Ratio} = 1 : 1.5 : 3 \end{equation*} This corresponds to Option (C). Therefore, the molar ratio of the deposited metals is: \begin{equation*} \boxed{1 : 1.5 : 3} \end{equation*}