An alloy weighing 2.70 mg of Pb–Ag was dissolved in desired amount of and volume was made 250 ml. A — Electrochemistry Chemistry Question
Question
An alloy weighing 2.70 mg of Pb–Ag was dissolved in desired amount of $HNO_3$ and volume was made 250 ml. A silver electrode was dipped in solution and $E_{cell}$ of the cell: $Pt, H_2(1 \text{ bar}) \| H^+(1 \text{ M}) \|\| Ag^+ \| Ag$ was 0.50 V at 298 K. The percentage of lead in alloy is (Given: $E^\circ_{Ag^+\|Ag} = 0.80 \text{ V}, \text{Ag} = 108, 2.303 RT/F = 0.06$)
💡 Solution & Explanation
For the given cell, the anode is SHE (E=0 V) and cathode is $Ag^+\|Ag$. $E_{cell} = E^\circ_{Ag^+\|Ag} - 0.06 \log(1/[Ag^+])$. $0.50 = 0.80 + 0.06 \log[Ag^+] \Rightarrow \log[Ag^+] = -0.30 / 0.06 = -5 \Rightarrow [Ag^+] = 10^{-5}$ M. The volume is 250 mL = 0.25 L, so moles of Ag = $10^{-5} \times 0.25 = 2.5 \times 10^{-6}$ mol. Mass of Ag = $2.5 \times 10^{-6} \times 108 = 270 \times 10^{-6}$ g = 0.27 mg. The total alloy mass is 2.70 mg. Mass of Pb = $2.70 - 0.27 = 2.43$ mg. Percentage of Pb = $(2.43 / 2.70) \times 100 = 90\%$. Therefore, correct answer is 0090.