[Four-digit Integer] An alloy weighing 2.70 mg of Pb-Ag was dissolved in desired amount of and volum — Electrochemistry Chemistry Question

💡 Solution & Explanation

\textbf{Setup:} Pb-Ag alloy (2.70 mg) dissolved in HNO\textsubscript{3}, diluted to 250 mL. Silver electrode in this solution. Cell: Pt, H\textsubscript{2}(1 bar) $|$ H\textsuperscript{+} $\|$ Ag\textsuperscript{+}(aq) $|$ Ag. \textbf{Standard reduction potentials given:} $E^\circ(\text{Ag}^+/\text{Ag}) = 0.80$ V; $E^\circ(\text{H}^+/\text{H}_2) = 0.00$ V; $2.303RT/F = 0.06$ V. \textbf{Step 1: Apply Nernst equation.} For the silver half-cell ($n = 1$): \[ E_{\text{cell}} = E^\circ_{\text{cell}} - 0.06 \log\left(\frac{1}{[\text{Ag}^+]}\right) = 0.80 + 0.06 \log[\text{Ag}^+] \] The measured $E_{\text{cell}}$ allows solving for $[\text{Ag}^+]$. \textbf{Step 2: Calculate moles of Ag in solution.} \[ n(\text{Ag}^+) = [\text{Ag}^+] \times V = [\text{Ag}^+] \times 0.250\ \text{L} \] \textbf{Step 3: Calculate mass of Ag in alloy.} \[ m(\text{Ag}) = n(\text{Ag}^+) \times 108\ \text{g/mol} \] \textbf{Step 4: Find Pb content.} \[ m(\text{Pb}) = 2.70\ \text{mg} - m(\text{Ag}) \] \[ \%\ \text{Pb} = \frac{m(\text{Pb})}{2.70} \times 100 \] \textit{Note: The specific measured $E_{\text{cell}}$ value required to complete the numerical calculation was not recorded in the available question data. Apply the Nernst equation with the measured EMF to obtain the final answer.}