If 0.50L of a 0.60M `SnSo_4` solution is electrolysed for a period of 30.0min using a current of 4.60 A. If inert electrods are used, what is the final concentration of `Sn^(2+)` remaining in the solution?[at.mass of Sn=119]

Options

A. 0.342M
B. 0.544M
C. 0.389M
D. 0.514M

Answer

To find the final concentration of `Sn^(2+)` remaining in the solution after electrolysis, we can follow these steps:

### Step 1: Calculate the initial moles of `Sn^(2+)`
Given:
- Volume of solution = 0.50 L
- Concentration of `SnSO4` = 0.60 M

Using the formula:
\[ \text{Number of moles} = \text{Concentration} \times \text{Volume} \]
\[ \text{Number of moles of } Sn^{2+} = 0.60 \, \text{mol/L} \times 0.50 \, \text{L} = 0.30 \, \text{moles} \]

### Step 2: Calculate the total charge passed during electrolysis
Given:
- Current (I) = 4.60 A
- Time (t) = 30.0 min = 30.0 × 60 s = 1800 s

Using the formula:
\[ \text{Charge (Q)} = I \times t \]
\[ Q = 4.60 \, \text{A} \times 1800 \, \text{s} = 8280 \, \text{C} \]

### Step 3: Calculate the number of gram equivalents of `Sn`
Using Faraday's law:
\[ \text{Number of gram equivalents} = \frac{Q}{F} \]
Where \( F \) (Faraday's constant) = 96500 C/mol.

\[ \text{Number of gram equivalents} = \frac{8280 \, \text{C}}{96500 \, \text{C/mol}} = 0.0858 \, \text{equivalents} \]

### Step 4: Calculate the number of moles of `Sn` deposited
The n-factor for `Sn^(2+)` is 2 (since it gains 2 electrons), so:
\[ \text{Number of moles} = \frac{\text{Number of gram equivalents}}{n} \]
\[ \text{Number of moles of } Sn = \frac{0.0858}{2} = 0.0429 \, \text{moles} \]

### Step 5: Calculate the remaining moles of `Sn^(2+)`
\[ \text{Remaining moles of } Sn^{2+} = \text{Initial moles} - \text{Moles deposited} \]
\[ \text{Remaining moles of } Sn^{2+} = 0.30 - 0.0429 = 0.2571 \, \text{moles} \]

### Step 6: Calculate the final concentration of `Sn^(2+)`
Using the formula:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume}} \]
\[ \text{Final concentration of } Sn^{2+} = \frac{0.2571 \, \text{moles}}{0.50 \, \text{L}} = 0.5142 \, \text{M} \]

### Final Answer:
The final concentration of `Sn^(2+)` remaining in the solution is approximately **0.514 M**.

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