Sodium amalgam is prepared by electrolysis of aqueous using 10 g mercury as cathode. How many Farada — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Composition of Sodium Amalgam and Set up Mass Relationship Sodium amalgam is an alloy formed by dissolving sodium ($\ce{Na}$) in liquid mercury ($\ce{Hg}$). We are given the following parameters: * Mass of the mercury cathode ($m_{\ce{Hg}}$) = $10\text{ g}$ * Mass percentage of sodium in the prepared amalgam = $18.7\%$ Let $x$ be the mass (in grams) of sodium metal deposited at the cathode and dissolved into the mercury. The total mass of the resulting sodium amalgam ($m_{\text{amalgam}}$) is the sum of the masses of sodium and mercury: $$m_{\text{amalgam}} = x + m_{\ce{Hg}} = (x + 10)\text{ g}$$ Step 2 - Calculate the Mass of Sodium Deposited ($x$) The mass fraction of sodium in the amalgam can be expressed mathematically as: $$\text{Weight \% of Na} = \frac{\text{Mass of Na}}{\text{Total Mass of Amalgam}} \times 100\%$$ Substituting the given values into the equation: $$18.7\% = \frac{x}{x + 10} \times 100\%$$ $$\frac{x}{x + 10} = 0.187$$ Now, solve for $x$ algebraically: $$x = 0.187(x + 10)$$ $$x = 0.187x + 1.87$$ $$x - 0.187x = 1.87$$ $$0.813x = 1.87$$ $$x = \frac{1.87}{0.813}\text{ g}$$ $$x \approx 2.30\text{ g}$$ Thus, the mass of sodium deposited at the cathode is approximately $2.30\text{ g}$. Step 3 - Calculate the Moles of Sodium and Theoretical Charge Required The number of moles of sodium ($n_{\ce{Na}}$) deposited is calculated by dividing its mass by its atomic weight ($M_{\ce{Na}} = 23\text{ g/mol}$): $$n_{\ce{Na}} = \frac{x}{M_{\ce{Na}}}$$ $$n_{\ce{Na}} = \frac{2.30\text{ g}}{23\text{ g/mol}} = 0.1\text{ mol}$$ The cathodic reduction half-reaction for the deposition of sodium ions ($\ce{Na^+}$) onto the mercury cathode is: $$\ce{Na^+(aq) + e^- -> Na(amalgam)}$$ According to the stoichiometry of this reaction, $1\text{ mole}$ of electrons is required to deposit $1\text{ mole}$ of sodium. Since $1\text{ mole}$ of electrons carries a charge of exactly $1\text{ Faraday}$ ($1\text{ F}$), the theoretical quantity of electricity ($Q_{\text{theoretical}}$) required to deposit $0.1\text{ mole}$ of sodium is: $$Q_{\text{theoretical}} = n_{\ce{Na}} \times 1\text{ F/mol}$$ $$Q_{\text{theoretical}} = 0.1\text{ mol} \times 1\text{ F/mol} = 0.1\text{ F}$$ Step 4 - Calculate the Actual Quantity of Electricity Required Considering Current Efficiency Current efficiency ($\eta$) represents the fraction of the total passed electric current that actually performs the desired chemical reduction. It is defined as: $$\eta = \frac{Q_{\text{theoretical}}}{Q_{\text{actual}}}$$ We are given: * Current efficiency ($\eta$) = $50\% = 0.50$ By rearranging the formula to solve for the actual quantity of electricity ($Q_{\text{actual}}$): $$Q_{\text{actual}} = \frac{Q_{\text{theoretical}}}{\eta}$$ $$Q_{\text{actual}} = \frac{0.1\text{ F}}{0.50}$$ $$Q_{\text{actual}} = \boxed{0.2\text{ F}}$$ Step 5 - Evaluate and Explain the Options * **Option (A) is incorrect:** This value ($0.1\text{ F}$) represents the theoretical charge required if the system had $100\%$ current efficiency, which ignores the specified electrical losses. * **Option (B) is correct:** As calculated, when accounting for the $50\%$ current efficiency, the actual charge required to obtain the desired amalgam is exactly $0.2\text{ F}$. * **Option (C) is incorrect:** This value ($0.05\text{ F}$) is obtained if the theoretical charge is incorrectly multiplied by the efficiency ($0.1\text{ F} \times 0.50$) rather than being divided by it. * **Option (D) is incorrect:** This represents a computational error in setting up the mass fraction or applying the efficiency factor. $$\text{Correct Option: } \boxed{\text{B}}$$