This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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This problem involves Faraday's Laws of Electrolysis, which relate the amount of substance deposited during electrolysis to the quantity of electricity passed through the solution. Since the same quantity of electricity is used for both copper and lead deposition, we can relate the moles of electrons transferred. Given: Mass of copper deposited (m_Cu) = 2.70 g Molar mass of copper (M_Cu) = 63.5 g/mol Molar mass of lead (M_Pb) = 207 g/mol Step 1: Write the half-reactions for the deposition of copper and lead. For copper (II) salt, copper ions are Cu^2+: Cu^2+(aq) + 2e^- Cu(s) This reaction shows that 1 mole of copper metal is deposited by 2 moles of electrons. For lead (II) salt, lead ions are Pb^2+: Pb^2+(aq) + 2e^- Pb(s) This reaction shows that 1 mole of lead metal is deposited by 2 moles of electrons. Step 2: Calculate the moles of copper deposited. We use the given mass of copper and its molar mass: n_Cu = m_CuM_Cu n_Cu = 2.70 g63.5 g/mol n_Cu ≈ 0.04252 mol Step 3: Calculate the moles of electrons required to deposit the copper. From the copper half-reaction, 1 mole of Cu requires 2 moles of electrons. n_e^- = n_Cu × 2 n_e^- = 0.04252 mol × 2 n_e^- ≈ 0.08504 mol This is the quantity of electricity (in moles of electrons) that was passed through the solution. Step 4: Calculate the moles of lead that will be deposited by the same quantity of electricity. Since the same quantity of electricity (i.e., the same moles of electrons) is passed through the lead (II) solution, we use the moles of electrons calculated in Step 3. From the lead half-reaction, 1 mole of Pb requires 2 moles of electrons. n_Pb = n_e^-2 n_Pb = 0.08504 mol2 n_Pb ≈ 0.04252 mol Step 5: Calculate the mass of lead deposited. We use the moles of lead and its molar mass: m_Pb = n_Pb × M_Pb m_Pb = 0.04252 mol × 207 g/mol m_Pb ≈ 8.79164 g Rounding to three significant figures (consistent with 2.70 g): m_Pb ≈ 8.79 g The mass of lead that will be deposited is 8.79 g.