[Four-digit Integer] An amount of 19 g of molten is electrolysed for some time. Inert electrodes are — Electrochemistry Chemistry Question

💡 Solution & Explanation

\textbf{Step 1: Initial moles of SnCl\textsubscript{2}.} $M(\text{SnCl}_2) = 119 + 2 \times 35.5 = 190\ \text{g mol}^{-1}$ \[ n(\text{SnCl}_2) = \frac{19}{190} = 0.10\ \text{mol} \] \textbf{Step 2: Cathode — Sn deposited.} \[ n(\text{Sn}) = \frac{1.19}{119} = 0.01\ \text{mol} \] $\text{Sn}^{2+} + 2e^- \rightarrow \text{Sn}$ \quad $\Rightarrow$ $n_e = 2 \times 0.01 = 0.02\ \text{mol}$ \textbf{Step 3: Anode — Sn²⁺ oxidised to Sn⁴⁺.} The same 0.02 mol electrons are released at the anode: \[ \text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2e^- \quad \Rightarrow \quad n(\text{Sn}^{4+}) = 0.01\ \text{mol} \] These Sn$^{4+}$ ions combine with Cl$^-$ in solution to form SnCl$_4$: \[ n(\text{SnCl}_4) = 0.01\ \text{mol} \] \textbf{Step 4: Remaining SnCl\textsubscript{2}.} Total Sn$^{2+}$ consumed = 0.01 (cathode) + 0.01 (anode) = 0.02 mol \[ n(\text{SnCl}_2\ \text{remaining}) = 0.10 - 0.02 = 0.08\ \text{mol} \] \textbf{Step 5: Calculate masses and ratio.} $M(\text{SnCl}_4) = 119 + 4 \times 35.5 = 261\ \text{g mol}^{-1}$ \[ m(\text{SnCl}_2) = 0.08 \times 190 = 15.2\ \text{g} \] \[ m(\text{SnCl}_4) = 0.01 \times 261 = 2.61\ \text{g} \] \[ \text{Ratio} = \frac{15.2}{2.61} = \frac{1520}{261} \] Given the ratio is $261 : x$, we have $x = \boxed{1520}$.