Electrolysis of an acetate solution produces ethane according to the Kolbe reaction :
`2CH_(3)COO^(-) rarr C_(2)H_(6)+2CO_(2)+2e^(-)`
What volume of ethane is produced at `27^(@)C` and 740 mm Hg, If a current of 0.5 ampere were passed through the solution for 7 hours and the electrode reaction is `82%` efficient?

Options

A. 2.70 L
B. 5. 4L
C. 1.35 L
D. 0.65 L

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the total charge passed during electrolysis.
The total charge (Q) can be calculated using the formula:
\[ Q = I \times t \]
Where:
- \( I \) = current in amperes (0.5 A)
- \( t \) = time in seconds

Convert time from hours to seconds:
\[ t = 7 \text{ hours} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} = 25200 \text{ seconds} \]

Now, calculate the total charge:
\[ Q = 0.5 \, \text{A} \times 25200 \, \text{s} = 12600 \, \text{C} \]

### Step 2: Calculate the moles of electrons transferred.
Using Faraday's constant (\( F = 96500 \, \text{C/mol} \)):
\[ \text{Moles of electrons} (n) = \frac{Q}{F} \]
\[ n = \frac{12600 \, \text{C}}{96500 \, \text{C/mol}} \approx 0.1305 \, \text{mol} \]

### Step 3: Adjust for efficiency.
The efficiency of the electrode reaction is given as 82%. Thus, the effective moles of electrons used in the reaction will be:
\[ n_{\text{effective}} = 0.1305 \, \text{mol} \times 0.82 \approx 0.107 \, \text{mol} \]

### Step 4: Relate moles of acetate to moles of ethane.
From the Kolbe reaction:
\[ 2 \, \text{CH}_3\text{COO}^- \rightarrow \text{C}_2\text{H}_6 + 2 \, \text{CO}_2 + 2 \, e^- \]
This shows that 2 moles of acetate produce 1 mole of ethane. Therefore, the moles of ethane produced can be calculated as:
\[ \text{Moles of ethane} = \frac{n_{\text{effective}}}{2} = \frac{0.107}{2} \approx 0.0535 \, \text{mol} \]

### Step 5: Calculate the volume of ethane produced.
Using the ideal gas law:
\[ PV = nRT \]
Where:
- \( P \) = pressure in atm
- \( V \) = volume in liters
- \( n \) = moles of ethane (0.0535 mol)
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin (27°C = 300 K)

Convert pressure from mm Hg to atm:
\[ P = \frac{740 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} \approx 0.973 \, \text{atm} \]

Now, rearranging the ideal gas law to find volume:
\[ V = \frac{nRT}{P} \]
Substituting the values:
\[ V = \frac{0.0535 \, \text{mol} \times 0.0821 \, \text{L·atm/(K·mol)} \times 300 \, \text{K}}{0.973 \, \text{atm}} \]

Calculating the volume:
\[ V \approx \frac{1.318 \, \text{L·atm}}{0.973 \, \text{atm}} \approx 1.35 \, \text{L} \]

### Final Answer:
The volume of ethane produced is approximately **1.35 liters**.

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