Question

260.75 g lump of pure $SnCl_2$ is electrolysed for 16 minutes 5 sec using Pt electrodes. 3.57 g of Sn is deposited at cathode. If no substance is lost during electrolysis the mass ratio of $SnCl_2$ and $SnCl_4$ is

$\frac{m_{SnCl_2}}{m_{SnCl_4}} =$ ...... (Atomic mass of Sn = 119 g/mol) (nearest integer)

Answer 1

32

Answer 2

The problem describes the electrolysis of a lump of pure $SnCl_2$ where $Sn$ is deposited at the cathode. A crucial piece of information is "no substance is lost during electrolysis". This implies that if $Sn^{2+}$ is reduced to $Sn$ at the cathode, then at the anode, $Sn^{2+}$ must be oxidized to $Sn^{4+}$ (forming $SnCl_4$), rather than $Cl^-$ being oxidized to $Cl_2$ gas (which would be lost).

1. Determine the balanced chemical reaction:

   At the cathode (reduction): $Sn^{2+} + 2e^- \to Sn(s)$

   At the anode (oxidation): $Sn^{2+} \to Sn^{4+} + 2e^-$

   Combining these half-reactions gives the overall reaction:

   $2Sn^{2+} \to Sn(s) + Sn^{4+}$

   In terms of compounds:

   $2SnCl_2 \to Sn(s) + SnCl_4$

2. Calculate the molar masses:

   Given atomic mass of $Sn = 119$ g/mol. Atomic mass of $Cl = 35.5$ g/mol (standard value).

   Molar mass of $SnCl_2 = 119 + 2 \times 35.5 = 119 + 71 = 190$ g/mol.

   Molar mass of $SnCl_4 = 119 + 4 \times 35.5 = 119 + 142 = 261$ g/mol.

3. Calculate moles of $Sn$ deposited:

   Mass of $Sn$ deposited = $3.57$ g.

   Moles of $Sn = \frac{\text{Mass of Sn}}{\text{Molar mass of Sn}} = \frac{3.57 \text{ g}}{119 \text{ g/mol}} = 0.03$ mol.

4. Calculate the mass of $SnCl_2$ reacted and $SnCl_4$ produced using stoichiometry:

   From the reaction $2SnCl_2 \to Sn(s) + SnCl_4$:

   • For every 1 mole of $Sn$ produced, 2 moles of $SnCl_2$ are reacted.

     Moles of $SnCl_2$ reacted = $2 \times \text{Moles of Sn} = 2 \times 0.03 \text{ mol} = 0.06$ mol.

     Mass of $SnCl_2$ reacted = Moles of $SnCl_2$ reacted $\times$ Molar mass of $SnCl_2 = 0.06 \text{ mol} \times 190 \text{ g/mol} = 11.4$ g.

   • For every 1 mole of $Sn$ produced, 1 mole of $SnCl_4$ is produced.

     Moles of $SnCl_4$ produced = $1 \times \text{Moles of Sn} = 0.03$ mol.

     Mass of $SnCl_4$ produced = Moles of $SnCl_4$ produced $\times$ Molar mass of $SnCl_4 = 0.03 \text{ mol} \times 261 \text{ g/mol} = 7.83$ g.

5. Calculate the mass of $SnCl_2$ remaining:

   Initial mass of $SnCl_2 = 260.75$ g.

   Mass of $SnCl_2$ remaining = Initial mass of $SnCl_2 -$ Mass of $SnCl_2$ reacted

   Mass of $SnCl_2$ remaining = $260.75 \text{ g} - 11.4 \text{ g} = 249.35$ g.

6. Calculate the mass ratio of $SnCl_2$ and $SnCl_4$:

   $\frac{m_{SnCl_2}}{m_{SnCl_4}} = \frac{\text{Mass of } SnCl_2 \text{ remaining}}{\text{Mass of } SnCl_4 \text{ produced}} = \frac{249.35 \text{ g}}{7.83 \text{ g}} \approx 31.845$.

7. Round to the nearest integer:

   The nearest integer to $31.845$ is $32$.

The time given (16 minutes 5 sec) is extra information not required for this calculation, as the amount of product ($Sn$ deposited) is directly provided.

The final answer is $\boxed{32}$

Explanation of the solution:

The electrolysis of $SnCl_2$ with no substance loss implies the reaction $2SnCl_2 \to Sn + SnCl_4$. Calculate moles of $Sn$ deposited. Use stoichiometry to find the mass of $SnCl_2$ reacted and $SnCl_4$ produced. Subtract reacted $SnCl_2$ from the initial mass to get remaining $SnCl_2$. Finally, calculate the ratio of remaining $SnCl_2$ to produced $SnCl_4$ and round to the nearest integer.