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.
ScanSolve AI Answer
Here's how to solve the problem using the formula (m)/(mm) = (It)/(nF): The problem states that the same quantity of electricity is used for both copper and lead. This means that the term (It)/(F) is constant for both metals. From the formula (m)/(mm) = (It)/(nF), we can rearrange it to (m · n)/(mm) = (It)/(F). Since (It)/(F) is constant, we can write: m_Cu · n_Cumm_Cu = m_Pb · n_Pbmm_Pb Step 1: Identify the known values for copper and lead. For Copper (Cu): Mass of copper (m_Cu) = 2.70 g Molar mass of copper (mm_Cu) = 63.5 g/mol Number of electrons transferred for copper (II) (n_Cu): The half-reaction is Cu^2+(aq) + 2e^- Cu(s), so n_Cu = 2. For Lead (Pb): Molar mass of lead (mm_Pb) = 207 g/mol Number of electrons transferred for lead (II) (n_Pb): The half-reaction is Pb^2+(aq) + 2e^- Pb(s), so n_Pb = 2. Mass of lead (m_Pb) = ? (This is what we need to calculate) Step 2: Substitute the known values into the equation. 2.70 g × 263.5 g/mol = m_Pb × 2207 g/mol Step 3: Solve for m_Pb. Since n_Cu = n_Pb = 2, we can cancel out the '2' from both sides of the equation: (2.70)/(63.5) = m_Pb207 Now, isolate m_Pb: m_Pb = (2.70)/(63.5) × 207 m_Pb = 0.042519685 × 207 m_Pb = 8.7965748 g Step 4: Round the answer to an appropriate number of significant figures (3 significant figures, based on 2.70 g). m_Pb ≈ 8.80 g The mass of lead that will be deposited is 8.80 g. Last free one today — make it count tomorrow, or type /upgrade for unlimited.