[Single-digit Integer] An alloy of lead (valency = 2)-thallium (valency = 1) containing 70% Pb and 3 — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Calculate the Mass of Each Component in the Alloy We are given that the total mass of the alloy to be electroplated onto the cathode is: $$m_{\text{alloy}} = 5.0\text{ g}$$ The alloy contains $70\%$ Lead ($\ce{Pb}$) and $30\%$ Thallium ($\ce{Tl}$) by weight. We calculate the mass of each metal deposited: * $$\text{Mass of Lead } (m_{\ce{Pb}}) = 5.0\text{ g} \times \frac{70}{100} = 3.5\text{ g}$$ * $$\text{Mass of Thallium } (m_{\ce{Tl}}) = 5.0\text{ g} \times \frac{30}{100} = 1.5\text{ g}$$ Step 2 - Calculate the Moles of Lead and Thallium Deposited Using the given atomic masses of Lead ($\text{Pb} = 208\text{ g mol}^{-1}$) and Thallium ($\text{Tl} = 204\text{ g mol}^{-1}$), we calculate the moles of each metal: * $$\text{Moles of Lead } (n_{\ce{Pb}}) = \frac{m_{\ce{Pb}}}{\text{Molar mass of Pb}}$$ $$n_{\ce{Pb}} = \frac{3.5\text{ g}}{208\text{ g mol}^{-1}} \approx 0.01683\text{ mol}$$ * $$\text{Moles of Thallium } (n_{\ce{Tl}}) = \frac{m_{\ce{Tl}}}{\text{Molar mass of Tl}}$$ $$n_{\ce{Tl}} = \frac{1.5\text{ g}}{204\text{ g mol}^{-1}} \approx 0.00735\text{ mol}$$ Step 3 - Write the Reduction Half-Reactions and Determine the Total Moles of Electrons Required The reduction half-reactions taking place at the cathode are: * **For Lead ($\text{valency} = 2$):** $$\ce{Pb^2+(aq) + 2e^- -> Pb(s)}$$ From the stoichiometry, $1\text{ mole}$ of $\ce{Pb}$ requires $2\text{ moles}$ of electrons. $$\text{Moles of electrons for Pb } (n_{e^-,\ce{Pb}}) = 2 \times n_{\ce{Pb}}$$ $$n_{e^-,\ce{Pb}} = 2 \times 0.01683\text{ mol} = 0.03366\text{ mol}$$ * **For Thallium ($\text{valency} = 1$):** $$\ce{Tl^+(aq) + e^- -> Tl(s)}$$ From the stoichiometry, $1\text{ mole}$ of $\ce{Tl}$ requires $1\text{ mole}$ of electrons. $$\text{Moles of electrons for Tl } (n_{e^-,\ce{Tl}}) = 1 \times n_{\ce{Tl}}$$ $$n_{e^-,\ce{Tl}} = 1 \times 0.00735\text{ mol} = 0.00735\text{ mol}$$ Summing these up, the total moles of electrons ($n_{e^-}$) required to electroplate both metals is: $$n_{e^-} = n_{e^-,\ce{Pb}} + n_{e^-,\ce{Tl}}$$ $$n_{e^-} = 0.03366\text{ mol} + 0.00735\text{ mol} = 0.04101\text{ mol}$$ Step 4 - Calculate the Total Quantity of Electrical Charge ($Q$) Required Using Faraday's constant ($F \approx 96,500\text{ C mol}^{-1}$), we calculate the total electrical charge ($Q$) in coulombs: $$Q = n_{e^-} \times F$$ $$Q = 0.04101\text{ mol} \times 96,500\text{ C mol}^{-1} = 3957.5\text{ C}$$ Step 5 - Calculate the Electroplating Time in Hours The relationship between charge, current ($I$), and time ($t$) in seconds is: $$Q = I \times t \implies t = \frac{Q}{I}$$ Substituting the given current of $I = 1.10\text{ A}$: $$t = \frac{3957.5\text{ C}}{1.10\text{ A}} \approx 3597.7\text{ s}$$ Converting the time from seconds to hours: $$t_{\text{hours}} = \frac{t}{3600}$$ $$t_{\text{hours}} = \frac{3597.7\text{ s}}{3600\text{ s hour}^{-1}} \approx 0.999\text{ hours} \approx \boxed{1}\text{ hour}$$ As the question specifies a single-digit integer format: $$\boxed{1}$$