250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 M AgNO3 and 0.1 M AuCl. The solution…

Strategy: Preferential deposition occurs for the species with the higher reduction potential ($E^\circ_{\text{red}}$). Calculate the total charge passed and compare it with the charge required for the full deposition of the preferred metal.

Step 1: Determine the order of deposition

• $E^\circ_{\ce{Au^+/Au}} = 1.69\text{ V}$ (Higher, deposits first)
• $E^\circ_{\ce{Ag^+/Ag}} = 0.80\text{ V}$ (Lower)

Step 2: Calculate total charge passed ($Q$) $I = 1\text{ A}$, $t = 15\text{ min} = 900\text{ s}$. $Q = I \times t = 1 \times 900 = 900\text{ C}$

Step 3: Charge required for all Gold ($Q_{\ce{Au}}$) Moles of Gold in solution $= 0.1\text{ M} \times 0.250\text{ L} = 0.025\text{ moles}$. For $\ce{Au^+ + e^- \rightarrow Au}$, $n=1$. $Q_{\ce{Au}} = n \times F = 0.025 \times 96500 = 2412.5\text{ C}$

Conclusion: Sin\ce $Q\ (900\text{ C}) < Q_{\ce{Au}}\ (2412.5\text{ C})$, only some Gold will be deposited. Silver will not begin to deposit until most of the Gold is removed from solution.

$\boxed{\text{Answer: (a) only gold}}$