From an electrolyte, one mole of electron will deposit at cathode — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand Faraday's First Law of Electrolysis According to Faraday's First Law of Electrolysis, the mass ($W$) of a substance deposited or liberated at any electrode during electrolysis is directly proportional to the quantity of chemical charge ($Q$) passed through the electrolyte: $$W = Z \cdot Q$$ Where: * $Q = I \cdot t$ (current in amperes multiplied by time in seconds). * $Z$ is the electrochemical equivalent of the substance, defined as: $$Z = \frac{E}{F}$$ * $E$ is the equivalent weight of the substance, and $F$ is Faraday's constant ($F \approx 96,500\text{ C/mol}$). Since $1\text{ mole of electrons}$ carries exactly $1\text{ Faraday}$ ($1\text{ F}$) of electrical charge, the passage of $1\text{ mole of electrons}$ will deposit exactly **one equivalent weight** of any substance at the cathode: $$\text{Mass deposited by } 1\text{ mole of } e^- = E = \frac{\text{Molar Mass}}{\text{Valency factor } (v)}$$ Step 2 - Analyze Option (A): Deposit of Copper (\ce{Cu}) In a standard aqueous electrolyte (such as copper sulfate, $\ce{CuSO4}$), copper exists as the divalent cupric cation ($\ce{Cu^{2+}}$). The reduction half-reaction at the cathode is: $$\ce{Cu^{2+}(aq) + 2e^- -> Cu(s)}$$ * Valency factor ($v$) for $\ce{Cu^{2+}}$ = $2$ * Molar mass of copper = $63.5\text{ g/mol}$ * Equivalent weight ($E$) of copper: $$E_{\ce{Cu}} = \frac{63.5\text{ g/mol}}{2} = 31.75\text{ g/eq}$$ Thus, $1\text{ mole of electrons}$ deposits $31.75\text{ g}$ of copper. *(Note: If the electrolyte were specified as a monovalent copper(I) salt like $\ce{CuCl}$, then $1\text{ mole of electrons}$ would deposit $63.5\text{ g}$ of copper. However, under standard conditions in general chemistry, copper electrolytes refer to standard divalent $\ce{Cu^{2+}}$ solutions. Hence, $63.5\text{ g}$ of copper requires $2\text{ moles of electrons}$, making this option incorrect in this context).* Step 3 - Analyze Option (B): Deposit of Magnesium (\ce{Mg}) Magnesium always exists as a divalent cation ($\ce{Mg^{2+}}$) in its electrolytes (such as molten $\ce{MgCl2}$). The reduction half-reaction at the cathode is: $$\ce{Mg^{2+}(l) + 2e^- -> Mg(s)}$$ * Valency factor ($v$) for $\ce{Mg^{2+}}$ = $2$ * Molar mass of magnesium = $24\text{ g/mol}$ * Equivalent weight ($E$) of magnesium: $$E_{\ce{Mg}} = \frac{24\text{ g/mol}}{2} = 12\text{ g/eq}$$ Thus, $1\text{ mole of electrons}$ deposits $12\text{ g}$ of magnesium. Depositing $24\text{ g}$ of magnesium would require $2\text{ moles of electrons}$. Therefore, Option (B) is incorrect. Step 4 - Analyze Option (C): Deposit of Sodium (\ce{Na}) Sodium always exists as a monovalent cation ($\ce{Na^+}$) in its electrolytes (such as molten $\ce{NaCl}$). The reduction half-reaction at the cathode is: $$\ce{Na^+(l) + e^- -> Na(s)}$$ * Valency factor ($v$) for $\ce{Na^+}$ = $1$ * Molar mass of sodium = $23\text{ g/mol}$ * Equivalent weight ($E$) of sodium: $$E_{\ce{Na}} = \frac{23\text{ g/mol}}{1} = 23\text{ g/eq}$$ Thus, $1\text{ mole of electrons}$ deposits $23\text{ g}$ of sodium. Depositing $11.5\text{ g}$ of sodium would require only $0.5\text{ moles of electrons}$. Therefore, Option (C) is incorrect. Step 5 - Analyze Option (D): Deposit of Aluminium (\ce{Al}) Aluminium always exists as a trivalent cation ($\ce{Al^{3+}}$) in its electrolytes (such as molten alumina, $\ce{Al2O3}$ in the Hall-Héroult process). The reduction half-reaction at the cathode is: $$\ce{Al^{3+}(l) + 3e^- -> Al(s)}$$ * Valency factor ($v$) for $\ce{Al^{3+}}$ = $3$ * Molar mass of aluminium = $27\text{ g/mol}$ * Equivalent weight ($E$) of aluminium: $$E_{\ce{Al}} = \frac{27\text{ g/mol}}{3} = 9.0\text{ g/eq}$$ Thus, $1\text{ mole of electrons}$ deposits exactly $9.0\text{ g}$ of aluminium. This perfectly matches the statement in Option (D), making it correct. $$\text{Correct Option: } \boxed{D}$$