To calculate the volume of gas produced at the anode in an electrolytic cell, we can use Faraday's second law of electrolysis. The fundamental steps for this calculation are as follows.
Given:
- Current (I) = 1.25 A
- Time (t) = 40 minutes = 40 × 60 seconds = 2400 seconds
- Faraday's constant (F) = 96500 C
- Molar volume of gas at STP = 22.4 L (or 22.4 dm³)
Step 1: Calculate the total charge (Q)
The total charge passed through the cell can be calculated using the formula:
\[ Q = I \times t \]
Substituting the values:
\[ Q = 1.25 , \text{A} \times 2400 , \text{s} = 3000 , \text{C} \]
Step 2: Calculate the moles of electrons transferred
To find the number of moles of electrons, we use Faraday's constant:
\[ \text{Moles of electrons} = \frac{Q}{F} \]
Substituting in the values:
\[ \text{Moles of electrons} = \frac{3000 , \text{C}}{96500 , \text{C/mol}} \approx 0.0311 , \text{mol} \]
Step 3: Determine the reaction at the anode
In the case of electrolyzing dilute \( \text{H}_2\text{SO}_4 \), the gas produced at the anode is typically oxygen gas (\( \text{O}_2 \)). The half-reaction for the oxidation of water to oxygen is:
\[ 2 , \text{H}_2\text{O} \rightarrow 4 , \text{H}^+ + \text{O}_2 + 4e^- \]
From this reaction, we can see that 4 moles of electrons produce 1 mole of oxygen gas. Hence, the moles of oxygen gas produced is:
\[ \text{Moles of } \text{O}_2 = \frac{\text{Moles of electrons}}{4} = \frac{0.0311}{4} \approx 0.007775 , \text{mol} \]
Step 4: Calculate the volume of gas at STP
Now we can calculate the volume of oxygen gas produced using the molar volume at STP:
\[ \text{Volume of gas} = \text{Moles of } \text{O}_2 \times \text{Molar Volume} \]
Substituting in the values:
\[ \text{Volume of gas} = 0.007775 , \text{mol} \times 22.4 , \text{L/mol} \approx 0.174 , \text{L} \]
Final Answer
Therefore, the volume of gas (oxygen) produced at the anode at STP is approximately 0.174 L or 174 mL.
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