[Single-digit Integer] In the refining of silver by electrolytic method, what will be the mass of 72 — Electrochemistry Chemistry Question

💡 Solution & Explanation

\textbf{Step 1: Find mass of Ag and impurities in the anode.} \[ \text{Pure Ag} = 60\% \times 72.8 = 43.68\ \text{g} \] \[ \text{Impurities} = 40\% \times 72.8 = 29.12\ \text{g} \] \textbf{Step 2: Calculate charge and moles of electrons.} \[ Q = 9.65 \times 3600 = 34740\ \text{C} \] \[ n_e = \frac{34740}{96500} = 0.36\ \text{mol} \] \textbf{Step 3: Mass of silver dissolved from anode.} Ag dissolves as Ag$^+$ (1 electron per atom): \[ m(\text{Ag dissolved}) = n_e \times M(\text{Ag}) = 0.36 \times 108 = 38.88\ \text{g} \] \textbf{Step 4: Fraction of anode material dissolved.} In electrolytic refining, the fraction of the anode that dissolves is: \[ f = \frac{m(\text{Ag dissolved})}{\text{Total pure Ag}} = \frac{38.88}{43.68} = 0.890 \] Applying this fraction to the entire anode (Ag + impurities dissolve uniformly): \[ m(\text{anode dissolved}) = f \times 72.8 = 0.890 \times 72.8 \approx 64.8\ \text{g} \] \textbf{Step 5: Mass remaining.} \[ m(\text{remaining}) = 72.8 - 64.8 \approx \boxed{8\ \text{g}} \]